url
stringlengths 17
172
| text
stringlengths 44
1.14M
| metadata
stringlengths 820
832
|
|---|---|---|
http://mathoverflow.net/questions/10267/distribution-of-non-solvable-group-orders
|
## distribution of non-solvable group orders
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
let $M$ be the set of natural numbers such that there is a group of this order, which is not solvable. what is the minimal distance $D$ of two numbers in $M$?
the examples $660$ and $672$ show $D \leq 12$. the famous theorem of feit-thompson implies $D>1$.
-
## 2 Answers
By the Euclidean algorithm, the answer is the gcd of all orders of all non-abelian finite simple groups. I believe that this is 4 (looking at the groups listed in Wikipedia, one can see that it is at most 4 since once can get down to 12 on the tables of low order groups, and the Suzuki groups have order not divisible by 3). My recollection is that a finite simple group actually cannot have cyclic 2-Sylow, and thus must have order divisible by 4.
-
That's right -- it's an elementary exercise that a simple group cannot have order twice an odd number: a permutation representation argument shows that a group of order 2 mod 4 has a subgroup of index 2. – Pete L. Clark Dec 31 2009 at 14:09
how do you arrive at simple groups? – Martin Brandenburg Dec 31 2009 at 14:43
A group is nonsolvable iff it has at least one nonabelian simple group as a composition factor. Moreover, if G is a finite nonabelian simple group of order a, then for all positive integers x, ax is the order of a nonsolvable group: G x Z_x. – Pete L. Clark Dec 31 2009 at 14:47
1
Pete is also implicitly using the odd order theorem, to know that the nonabelian simple factor cannot be odd. – David Speyer Dec 31 2009 at 16:22
2
To avoid confusion: when Ben writes "the Suzuki group", he means a member of the family of Suzuki groups of Lie type, not the Suzuki sporadic group Suz. – Bjorn Poonen Dec 31 2009 at 19:48
show 1 more comment
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The encyclopedia of integer sequences gives the following criteria for a number being a non-solvable number:
A positive integer n is a non-solvable number if and only if it is a multiple of any of the following numbers:
a) $2^p (2^{2p}-1)$, p any prime.
b) $3^p (3^{2p}-1)/2$, p odd prime.
c) $p(p^2-1)/2$, p prime greater than 3 and congruent to 2 or 3 mod 5
d) $2^4 3^3 13$
e) $2^{2p}(2^{2p}+1)(2^p-1)$, p odd prime.
It's not hard to check that all these orders are divisible by 4, so there will never be two non-solvable numbers differing by less than 4.
In fact, they're all divisible by 12 except those generated by (e), which are all divisible by 20.
So, for example, all numbers of the form 29120n are nonsolvable, since $29120 = 2^6 (2^6+1) (2^3-1)$. And all numbers of the form 25308n are nonsolvable, since $25308 = 37(37^2-1)/2$. We have the prime factorizations $25308 = 2^2 3^2 19^1 37^1$ and $29120 = 2^6 5^1 7^1 13^1$.
So we just need to find multiples of 29120 and 25308 which differ by 4. From the Euclidean algorithm, $29120 \cdot 2483 = 72304960$ and $25308 \cdot 2857 = 72304956$.
I haven't searched exhaustively, so it's possible that there's a smaller pair of non-solvable numbers that differ by 4; I chose 25308 and 29120 by just looking at the prime factorizations of the numbers generated by (a) through (e) until I found two that had gcd 4.
-
again, I wish I could accept two answers. thank you both :-) – Martin Brandenburg Dec 31 2009 at 19:10
Ben's answer is better, in the sense that he actually knows group theory. I couldn't have done this without reading his answer. – Michael Lugo Dec 31 2009 at 19:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9328679442405701, "perplexity_flag": "head"}
|
http://en.wikipedia.org/wiki/Split-radix_FFT_algorithm
|
# Split-radix FFT algorithm
The split-radix FFT is a fast Fourier transform (FFT) algorithm for computing the discrete Fourier transform (DFT), and was first described in an initially little-appreciated paper by R. Yavne (1968) and subsequently rediscovered simultaneously by various authors in 1984. (The name "split radix" was coined by two of these reinventors, P. Duhamel and H. Hollmann.) In particular, split radix is a variant of the Cooley-Tukey FFT algorithm that uses a blend of radices 2 and 4: it recursively expresses a DFT of length N in terms of one smaller DFT of length N/2 and two smaller DFTs of length N/4.
The split-radix FFT, along with its variations, long had the distinction of achieving the lowest published arithmetic operation count (total exact number of required real additions and multiplications) to compute a DFT of power-of-two sizes N. The arithmetic count of the original split-radix algorithm was improved upon in 2004 (with the initial gains made in unpublished work by J. Van Buskirk via hand optimization for N=64 [1] [2]), but it turns out that one can still achieve the new lowest count by a modification of split radix (Johnson and Frigo, 2007). Although the number of arithmetic operations is not the sole factor (or even necessarily the dominant factor) in determining the time required to compute a DFT on a computer, the question of the minimum possible count is of longstanding theoretical interest. (No tight lower bound on the operation count has currently been proven.)
The split-radix algorithm can only be applied when N is a multiple of 4, but since it breaks a DFT into smaller DFTs it can be combined with any other FFT algorithm as desired.
## Split-radix decomposition
Recall that the DFT is defined by the formula:
$X_k = \sum_{n=0}^{N-1} x_n \omega_N^{nk}$
where $k$ is an integer ranging from $0$ to $N-1$ and $\omega_N$ denotes the primitive root of unity:
$\omega_N = e^{-\frac{2\pi i}{N}},$
and thus $\omega_N^N = 1$.
The split-radix algorithm works by expressing this summation in terms of three smaller summations. (Here, we give the "decimation in time" version of the split-radix FFT; the dual decimation in frequency version is essentially just the reverse of these steps.)
First, a summation over the even indices $x_{2n_2}$. Second, a summation over the odd indices broken into two pieces: $x_{4n_4+1}$ and $x_{4n_4+3}$, according to whether the index is 1 or 3 modulo 4. Here, $n_m$ denotes an index that runs from 0 to $N/m-1$. The resulting summations look like:
$X_k = \sum_{n_2=0}^{N/2-1} x_{2n_2} \omega_{N/2}^{n_2 k} + \omega_N^k \sum_{n_4=0}^{N/4-1} x_{4n_4+1} \omega_{N/4}^{n_4 k} + \omega_N^{3k} \sum_{n_4=0}^{N/4-1} x_{4n_4+3} \omega_{N/4}^{n_4 k}$
where we have used the fact that $\omega_N^{m n k} = \omega_{N/m}^{n k}$. These three sums correspond to portions of radix-2 (size N/2) and radix-4 (size N/4) Cooley-Tukey steps, respectively. (The underlying idea is that the even-index subtransform of radix-2 has no multiplicative factor in front of it, so it should be left as-is, while the odd-index subtransform of radix-2 benefits by combining a second recursive subdivision.)
These smaller summations are now exactly DFTs of length N/2 and N/4, which can be performed recursively and then recombined.
More specifically, let $U_k$ denote the result of the DFT of length N/2 (for $k = 0,\ldots,N/2-1$), and let $Z_k$ and $Z'_k$ denote the results of the DFTs of length N/4 (for $k = 0,\ldots,N/4-1$). Then the output $X_k$ is simply:
$X_k = U_k + \omega_N^k Z_k + \omega_N^{3k} Z'_k.$
This, however, performs unnecessary calculations, since $k \geq N/4$ turn out to share many calculations with $k < N/4$. In particular, if we add N/4 to k, the size-N/4 DFTs are not changed (because they are periodic in k), while the size-N/2 DFT is unchanged if we add N/2 to k. So, the only things that change are the $\omega_N^k$ and $\omega_N^{3k}$ terms, known as twiddle factors. Here, we use the identities:
$\omega_N^{k+N/4} = -i \omega_N^k$
$\omega_N^{3(k+N/4)} = i \omega_N^{3k}$
to finally arrive at:
$X_k = U_k + \left( \omega_N^k Z_k + \omega_N^{3k} Z'_k \right),$
$X_{k+N/2} = U_k - \left( \omega_N^k Z_k + \omega_N^{3k} Z'_k \right),$
$X_{k+N/4} = U_{k+N/4} - i \left( \omega_N^k Z_k - \omega_N^{3k} Z'_k \right),$
$X_{k+3N/4} = U_{k+N/4} + i \left( \omega_N^k Z_k - \omega_N^{3k} Z'_k \right),$
which gives all of the outputs $X_k$ if we let $k$ range from $0$ to $N/4-1$ in the above four expressions.
Notice that these expressions are arranged so that we need to combine the various DFT outputs by pairs of additions and subtractions, which are known as butterflies. In order to obtain the minimal operation count for this algorithm, one needs to take into account special cases for $k = 0$ (where the twiddle factors are unity) and for $k = N/8$ (where the twiddle factors are $(1 \pm i)/\sqrt{2}$ and can be multiplied more quickly); see, e.g. Sorensen et al. (1986). Multiplications by $\pm 1$ and $\pm i$ are ordinarily counted as free (all negations can be absorbed by converting additions into subtractions or vice versa).
This decomposition is performed recursively when N is a power of two. The base cases of the recursion are N=1, where the DFT is just a copy $X_0 = x_0$, and N=2, where the DFT is an addition $X_0 = x_0 + x_1$ and a subtraction $X_1 = x_0 - x_1$.
These considerations result in a count: $4 N \log_2 N - 6N + 8$ real additions and multiplications, for N>1 a power of two. This count assumes that, for odd powers of 2, the leftover factor of 2 (after all the split-radix steps, which divide N by 4) is handled directly by the DFT definition (4 real additions and multiplications), or equivalently by a radix-2 Cooley–Tukey FFT step.
## References
• R. Yavne, "An economical method for calculating the discrete Fourier transform," in Proc. AFIPS Fall Joint Computer Conf. 33, 115–125 (1968).
• P. Duhamel and H. Hollmann, "Split-radix FFT algorithm," Electron. Lett. 20 (1), 14–16 (1984).
• M. Vetterli and H. J. Nussbaumer, "Simple FFT and DCT algorithms with reduced number of operations," Signal Processing 6 (4), 267–278 (1984).
• J. B. Martens, "Recursive cyclotomic factorization—a new algorithm for calculating the discrete Fourier transform," IEEE Trans. Acoust., Speech, Signal Processing 32 (4), 750–761 (1984).
• P. Duhamel and M. Vetterli, "Fast Fourier transforms: a tutorial review and a state of the art," Signal Processing 19, 259–299 (1990).
• S. G. Johnson and M. Frigo, "A modified split-radix FFT with fewer arithmetic operations," IEEE Trans. Signal Processing 55 (1), 111–119 (2007).
• Douglas L. Jones, "Split-radix FFT algorithms," Connexions web site (Nov. 2, 2006).
• H. V. Sorensen, M. T. Heideman, and C. S. Burrus, "On computing the split-radix FFT", IEEE Trans. Acoust., Speech, Signal Processing 34 (1), 152-156 (1986).
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 44, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9127281904220581, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/111868?sort=votes
|
## The use of Hall algebras in physics
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I once read a statement (not memorized precisely) that a certain physics quantity between two states of charge $d_1$ and $d_2$ respectively could be computed by running over the states of charge $d_1+d_2$ which is the extension of the original two states. Therefore we need to consider some Hall algebras on a moduli space.
I couldn't find that literature any more, so I am not sure that this statement is correct. Could anyone help me to make clear this sort of things? Thanks a lot!
My questions are:
1) What is the basic physics setting of this story?
2) Why is this "extension" important?
3) If this is not correct, what is the correct statement/why do physicists care about Hall algebras?
-
arxiv.org/abs/hep-th/9510182 arxiv.org/abs/hep-th/9609017 – Alexander Chervov Nov 9 at 12:54
## 3 Answers
In supersymmetric field theories and string theories there are special states called BPS states which are annihilated by some of the supercharges and whose mass is determined in terms of their charges by the supersymmetry algebra. The study of these states and how they behave as various moduli are varied has been one of the main tools physicists have used to find evidence for various kinds of dualities, particularly S-duality which relates weakly coupled theories to strongly coupled theories.
One particularly rich example of an S-duality involves a duality between the heterotic string on $K3 \times E$ with a particular choice of $E_8 \times E_8$ gauge bundle where $E$ is an elliptic curve and Type II string theory on Calabi-Yau manifolds which have the form of K3 surfaces fibered over rational curves. In the first paper mentioned by Alexander Chervov, Greg Moore and I computed certain one-loop integrals on the heterotic string side of this story and found two interesting facts. First, that these integrals were determined purely by the spectrum of BPS states, and second that the answers involved denominator formulae for Generalized Kac Moody algebras of the type studied previously by Borcherds. Given this fact it was natural to think that there was an algebraic structure that one could define on the BPS states that would explain'' why were getting the denominator formula for a GKM algebra. This was the physics motivation for the introduction of the algebra of BPS states. However we did not achieve our original goal in that we were not able to find a direct connection between the BPS algebra and the GKM denominator formulae. In spite of this failure the idea that there should be an algebraic structure on the space of BPS states seems to have some merit.
If you want to look at more recent developments you might have a look at arXiv:1102.1821 which finds a more direct relation between a particular Borcherds algebra and one-loop integrals. On the mathematical side there is arXiv:1006.2706 by Kontsevich and Soibelman where Hall algebras appear. They mention the idea of an algebra of BPS states as motivation for their construction.
I must confess though that I do not have the level of mathematical sophistication needed to understand this paper and unfortunately my colleague Greg Moore, who does, is not on MO.
-
Thank you very much! Additional question: what motivated you to use that correspondence conjecture to define the product of BPS state? – Xiao Xinli Nov 11 at 3:19
1
I can't offer much beyond what we say in the paper. The main motivations were 1) Nakajima's (and others) work on instantons on ALE spaces and Kac-Moody algebras which used a similar construction 2) comparison to the algebra of BPS states on the heterotic side which could be computed using conformal field theory techniques (or VOA's if you prefer) and 3) some physical intuition related to the behavior of bound states in quantum mechanics. N=2 string duality is not something that is yet understood in a mathematically precise way, so I'm not sure how helpful 2) is to mathematicians. – Jeff Harvey Nov 11 at 15:01
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Let me add a small comment to Jeff Harvey's nice answer. Some years after the original BPS-algebra papers of Harvey-Moore, there was a paper by Fiol-Marino, hep-th/0006189, which analyzed the algebra of BPS-states for local compactifications where the BPS-states can be described by quivers. They gave a rather explicit realization of Harvey and Moore's "correspondence conjecture", and they also noted that the algebra of BPS-states is essentially that of a Ringel-Hall algebra.
-
The papers by J. Harvey and G. Moore seems relevant to subj.
http://arxiv.org/abs/hep-th/9609017 On the algebras of BPS states
http://arxiv.org/abs/hep-th/9510182 Algebras, BPS States, and String
See for example pages 24-25 section 7.2 of the second paper. The "correspondence conjecture" (page 25 top) seems to be exactly the fact that physically defined BPS-state algebra for the particular situation of the IIA string on K3 is a Hall algebra related to the sheaves on K3.
conjecture, motivated by the work of Nakajima, and of Ginzburg et. al., [56][57], is the following. Suppose first that the three vectors Qi in (7.2) represent positive BPS states. Recall that the charges are Chern characters of sheaves. There is only one natural way that the three sheaves E1, E2, E3 can be related and satisfy (7.2). They must fit into an exact sequence: 0 → E1 → E3 → E2 → 0 (7.3) or 0 → E2 → E3 → E1 → 0 (7.4) The ambiguity between (7.3) and (7.4) is resolved by the requirement that E3 be semistable: since Chern characters are additive the inequality (5.2) cannot hold for both F = E1 and F = E2. 23 We define the correspondence region to be the subset of M(Q1) ×M(Q2) ×M(Q3) defined by the set of triples: C+++(Q1,Q2;Q3) = {(E1, E2, E3) : 0 → E1 → E3 → E2 → 0}
The correspondence conjecture given after that claims that the left hand-side product (physical definition of BPS-state algebra) equals to mathematical definition which is Hall algebra (actually for its analogue for cohomologies, i.e. for Chern characters of sheaves, while true Hall algebra would be for sheaves themselves).
PS
Seems one of the authors is sometimes on MO, may be notify him to get expert's answer.
-
Thank you very much! However, what I expected is something more physics. Those two papers are two mathematics. – Xiao Xinli Nov 10 at 4:23
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.947168231010437, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/1501/does-projectiveness-descend-along-field-extensions
|
## Does projectiveness descend along field extensions?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Background: Properness is a much more robust notion than projectiveness. For example, properness descends along arbitrary fpqc covers (see, for example, Vistoli's Notes on Grothendieck topologies, fibered categories and descent theory, Proposition 2.36). This is far from true for projectiveness. In fact, projectiveness doesn't even descend along Zariski covers! The standard example of a proper non-projective morphism is locally projective over the base. (pictured in an appendix of Hartshorne; page 443, unfortunately not on Google books)
So how delicate is projectivity? Suppose X is a scheme over a field k, and suppose K is a field extension of k such that XK is projective over K. Does it follow that X is projective over k?
The obvious thing to do (I think) is to pick a very ample line bundle L on XK and try to descend it to X. If K is a finite extension of k, then the norm of L will be an ample line bundle on X (if I haven't misunderstood EGA II, section 6.5 and Corollary 6.6.2). But could it be that there is an infinite extension K of k such that XK is projective over K, but X is not projective over k?
-
## 2 Answers
For finite extensions, this is stated explicitly as Corollary 6.6.5 in EGA II and it is also stated there that the result is true for arbitrary extensions. One may reduce the general case to the finite case as follows:
First, we may asume that K is finitely generated since any projective scheme is defined by finitely many equations. This allows us to find a finitely generated integral domain A over k with quotient K and a projective scheme Y over Spec(A) with generic fibre equal isomorphic to X_K. Let Y' = X x Spec(A) and p the projection from Y' to Spec(A). By construction the generic fibre of the two schemes above over Spec(A) are isomorphic. By properness it easily follows that there is an open set U of Spec(A) such that the two schemes become isomorphic over U. U has rational points over a finite extensions of k so we reduce to the finite extension case.
-
Why does properness imply that there's a dense open over which the two are isomorphic? – Anton Geraschenko♦ Oct 21 2009 at 6:01
Since the generic fibres are isomorphic, there are open sets V and V' of Y and Y' to which any isomorphism extends. The complements of these, say Z and Z', are closed sets whose images in Spec(A) are closed (by properness). Since the generic point is not contained in the image, it follows that the complement in Spec(A) is non-empty and open. (Properness is not really essential; one can use constructibility of images instead.) – ulrich Oct 21 2009 at 6:40
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The entirety of sections 8--12 (apart from 10) of EGA IV is devoted to fleshing out in remarkably exhaustive and elegant generality the entire yoga of "spreading out and specialization" of which this question is but a special case. Highly recommended reading; it is used (implicitly, if not explicitly) all the time when people prove theorems in algebraic geometry by specialization. This includes proving results over $\mathbf{C}$ by "reduction to the case of positive characteristic or finite fields" (e.g., Mori, Deligne-Illusie) as well as construction of moduli spaces of stable curves by digging out subschemes of Hilbert schemes, etc.
Here is a nifty little exercise to test one's understanding of the EGA formalism: if $X$ and $Y$ are schemes of finite type over a field $k$ and if there is an extension field $K/k$ such that there is a $K$-morphism $f:X _K \rightarrow Y _K$ with any "reasonable" property $\mathbf{P}$ then there is such a morphism with $K/k$ a finite extension; here, "reasonable" can be lots of things: isomorphism, surjective, open immersion, closed immersion, finite flat of degree 42, a semistable curve fibration, smooth, proper and flat with geometric fibers having 12 irreducible components which intersect according to such-and-such configuration and dimensions, and so on. The point is that the initial $f$ is certainly not descended to a finite subextension of the initial $K/k$, and if you made the construction over such an extension and extended scalars back up to the original $K$ then it has absolutely nothing to do with the original $f$.
On the topic of specialization for morphisms, I can't resist mentioning a useful fact which is not a formal consequence of that general stuff: if $A$ and $B$ are abelian varieties over a field $k$ then there exists a finite (even separable) extension $k'/k$ such that (loosely speaking) "all homomorphisms from $A$ to $B$" are defined over $k'$. This means that if $K/k'$ is any extension field whatsoever, then every $K$-homomorphism $A_K \rightarrow B_K$ is defined over $k'$. (Quick proof: the locally finite type Hom-scheme has finitely generated group of geometric points, and is unramified by functorial criterion, so it is \'etale since we are over a field.) There is nothing like this for general (even proper smooth) varieties; just think about automorphisms of projective space.
-
3
Another proof of the abelian variety fact: Each homomorphism is (loosely speaking, again) defined over a separable closure k^s since it maps a Zariski dense subset of A(k^s) into B(k^s), namely the prime-to-the-characteristic torsion points. Since the group Hom(A,B) is finitely generated, this is enough. – Bjorn Poonen Feb 16 2010 at 5:50
In fact, it suffices to take k' = k(A[3]) (when char(k) is not 3): see Fields of definition for homomorphisms of abelian varieties, A. Silverberg, Journal of Pure and Applied Algebra 77 (1992), 253-262. – Bjorn Poonen Feb 16 2010 at 5:52
1
Getting to a finite algebraic extension requires only the elementary 16.9 of my algebraic geometry notes. Silverberg uses a theorem of Chow to get to a separable extension. Here's a modern elementary argument. We have to show that an abelian variety $A$ over $k$ doesn't acquire extra endomorphisms when we pass from $k$ to $k[a]$, $a^p\in k$. By descent for $k\to k[a]$, it suffices to check that $A$ doesn't acquire extra endomorphisms when we pass from $k$ to $k[e]$, $e^p=0$, but this follows from (an almost trivial case of) the rigidity lemma. – JS Milne Feb 16 2010 at 12:57
This is a very nice discussion. I wanted to insert a warning about characteristic p weirdness (quoted from memory; please correct if necessary): on the other hand, there is an abelian variety A over an algebraically closed field of characteristic p and a positive integer N (= p^a for some a, presumably) such that the space of order N closed subgroup schemes of A varies in moduli: in particular, you get more by enlarging the algebraically closed base field. – Pete L. Clark Feb 16 2010 at 13:59
I personally like Jim's argument best of all, but I decided against using that descent/rigidity method above since the Hom-scheme argument fit in one sentence. Pete, for your comment see Theorem 3.18 in my expository paper on Chow trace. – BCnrd Feb 17 2010 at 2:53
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9307437539100647, "perplexity_flag": "head"}
|
http://physics.stackexchange.com/questions/tagged/momentum+education
|
# Tagged Questions
1answer
75 views
### Can electromagnetic momentum be introduced at pre-university level as for electromagnetic energy?
Electromagnetic energy is introduced at pre-university level, starting with static electric energy followed by static magnetic energy. But the introduction of electromagnetic momentum usually has to ...
4answers
225 views
### Applications of recoil principle in classical physics
Are there any interesting, important or (for the non physicist) astonishing examples where the recoil principle (as special case of conservation of linear momentum) is applied beside rockets and guns? ...
5answers
221 views
### Usefullness of an only qualitative understanding of momentum?
A few days ago I had a discussion with a friend who wants to become a physics teacher (in Germany). He told me that from a pedagogical/didactial point of view it seems to be a good idea to introduce ...
3answers
1k views
### Does conservation of momentum really imply Newton's third law?
I often heard that conservation of momentum is nothing else than Newton's third law. Ok, If you have only two interacting particles in the universe, this seems to be quite obvious. However if you ...
2answers
3k views
### Difference between momentum and kinetic energy
From a mathematical point of view it seems to be clear what's the difference between momentum and $mv$ and kinetic energy $\frac{1}{2} m v^2$. Now my problem is the following: Suppose you want to ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9509135484695435, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/28911/number-of-solutions-of-a-linear-equation-in-a-small-subset
|
## Number of solutions of a linear equation in a small subset.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $p$ be a prime. Let $F_p$ be the finite field of $p$ elements. Let $A$ be a subset of $F_p$ of size $s$. Assume that $s > 2$ is polylogarithmic in $p$.
Suppose that we want to count number of solutions of $$x_1 + x_2 + ... + x_k = t (\mod\ p)$$ under the restriction that $x_i \in A$ for all $1\leq i\leq k$. The expectation seems to be $s^k/p$ if $k$ is large, independent of $t$.
My question is: Does there exist $A$ such that the expectation is known to be quite sharp (namely that the error term is poly(k p) in absolute value)?
Thanks a lot
-
I am a little be confused by your terminology: what does "$s$ polylogarithmic in $p$" mean? – Wadim Zudilin Jun 21 2010 at 4:19
"$s$ is polylogarithmic in $p$" generally means $s = \Theta(\log^c p)$ for a fixed constant $c > 0$. – Ryan Williams Jun 21 2010 at 4:23
But then I can't follow the claim about the expectation to be independent of $t$: take $A$ to be $\lbrace 0,1,\dots,s-1\rbrace$, so that only $t<ks$ can be expressed in the required form... By the way for this form of $A$ one can easily count the number of solutions. – Wadim Zudilin Jun 21 2010 at 4:29
In my question k is large. In fact, the case most interesting to me is when k = p. So the expectation of number of solutions is independent of t, I guess. Error term will be dependent on t of course. Qi – Qi Jun 21 2010 at 6:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9133456945419312, "perplexity_flag": "head"}
|
http://en.m.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis)
|
# Decomposition of spectrum (functional analysis)
The spectrum of a linear operator $T$ that operates on a Banach space $X$ (a fundamental concept of functional analysis) consists of all scalars $\lambda$ such that the operator $T-\lambda$ does not have a bounded inverse on $X$. The spectrum has a standard decomposition into three parts:
• a point spectrum, consisting of the eigenvalues of $T$
• a continuous spectrum, consisting of the scalars that are not eigenvalues but make the range of $T-\lambda$ a proper dense subset of the space;
• a residual spectrum, consisting of all other scalars in the spectrum
This decomposition is relevant to the study of differential equations, and has applications to many branches of science and engineering. A well-known example from quantum mechanics is the explanation for the discrete spectral lines and the continuous band in the light emitted by excited atoms of hydrogen.
## Definitions
### For bounded Banach space operators
Let X be a Banach space, L(X) the family of bounded operators on X, and T ∈ L(X). By definition, a complex number λ is in the spectrum of T, denoted σ(T), if T − λ does not have an inverse in L(X).
If T − λ is one-to-one and onto, then its inverse is bounded; this follows directly from the open mapping theorem of functional analysis. So, λ is in the spectrum of T if and only if T − λ is either not one-to-one or not onto. One distinguishes three separate cases:
1. T − λ is not injective. That is, there exist two distinct elements x,y in X such that (T − λ)(x) = (T − λ)(y). Then z = x − y is a non-zero vector such that T(z) = λz. In other words, λ is an eigenvalue of T in the sense of linear algebra. In this case, λ is said to be in the point spectrum of T, denoted σp(T).
2. T − λ is injective, and its range is a dense subset R of X; but is not the whole of X. In other words, there exists some element x in X such that (T − λ)(y) can be as close to x as desired, with y in X; but is never equal to x. It can be proved that, in this case, T − λ is not bounded below (i.e. it sends far apart elements of X too close together). Equivalently, the inverse linear operator (T − λ)−1, which is defined on the dense subset R, is not a bounded operator, and therefore cannot be extended to the whole of X. Then λ is said to be in the continuous spectrum, σc(T), of T.
3. T − λ is injective but does not have dense range. That is, there is some element x in X and a neighborhood N of x such that (T − λ)(y) is never in N. In this case, the map (T − λ) x → x may be bounded or unbounded, but in any case does not admit a unique extension to a bounded linear map on all of X. Then λ is said to be in the residual spectrum of T, σr(T).
So σ(T) is the disjoint union of these three sets,
$\sigma(T) = \sigma_p (T) \cup \sigma_c (T) \cup \sigma_r (T).$
#### Spectrum of dual operator
If X* is the dual space of X, and T* : X* → X* is the adjoint operator of T, then σ(T) = σ(T*).
Theorem For a bounded operator T, σr(T) ⊂ σp(T*) ⊂ σr(T) ∪ σp(T).
Proof The notation <·, φ> will denote an element of X*, i.e. x → <x, φ> is the action of a bounded linear functional φ. Let λ ∈ σr(T). So Ran(T - λ) is not dense in X. By the Hahn–Banach theorem, there exists a non-zero φ ∈ X* that vanishes on Ran(T - λ). For all x ∈ X,
$\langle (T - \lambda)x, \varphi \rangle = \langle x, (T^* - {\lambda}) \varphi \rangle = 0.$
Therefore (T* - λ)φ = 0 ∈ X* and λ is an eigenvalue of T*. The shows the former inclusion. Next suppose that (T* - λ)φ = 0 where φ ≠ 0, i.e.
$\forall x \in X,\; \langle x, (T^* - \lambda) \varphi \rangle = \langle (T - \lambda) x, \varphi \rangle = 0.$
If Ran(T − λ) is dense, then φ must be the zero functional, a contradiction. The claim is proved.
In particular, when X is a reflexive Banach space, σr(T*) ⊂ σp(T**) = σp(T).
### For unbounded operators
The spectrum of an unbounded operator can be divided into three parts in exactly the same way as in the bounded case.
### Examples
#### Multiplication operator
Given a σ-finite measure space (S, Σ, μ), consider the Banach space Lp(μ). A function h: S → C is called essentially bounded if h is bounded μ-almost everywhere. An essentially bounded h induces a bounded multiplication operator Th on Lp(μ):
$(T_h f)(s) = h(s) \cdot f(s).$
The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts.
If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies Tg · Th − λ = Th − λ · Tg = I. So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets {Sn = h−1(B1/n(λ))}. Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly
$\| (T_h - \lambda) f_n \|_p ^p = \| (h - \lambda) f_n \|_p ^p = \int_{S_n} | h - \lambda \; |^p d \mu \leq \frac{1}{n^p} \; \mu(S_n) = \frac{1}{n^p} \| f_n \|_p ^p.$
This shows Th − λ is not bounded below, therefore not invertible.
If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th: Pick an open ball Bε(λ) that contains only λ from the essential range. Let f be the characteristic function of h−1(Bε(λ)), then
$\forall s \in S, \; (T_h f)(s) = \lambda f(s).$
Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum or in the resolvent spectrum of Th. To show this is to show that Th − λ has dense range for all such λ. Given f ∈ Lp(μ), again we consider the sequence of sets {Sn = h−1(B1/n(λ))}. Let gn be the characteristic function of S − Sn. Define
$f_n(s) = \frac{1}{ h(s) - \lambda} \cdot g_n(s) \cdot f(s).$
Direct calculation shows that fn ∈ Lp(μ), and, by the dominated convergence theorem,
$(T_h - \lambda) f_n \rightarrow f$
in the Lp(μ) norm.
Therefore multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum.
#### Shifts
In the special case when S is the set of natural numbers and μ is the counting measure, the corresponding Lp(μ) is denoted by lp. This space consists of complex valued sequences {xn} such that
$\sum_{n \geq 0} | x_n |^p < \infty.$
For 1 < p < ∞, l p is reflexive. Define the left shift T : l p → l p by
$T(x_1, x_2, x_3, \dots) = (x_2, x_3, x_4, \dots).$
T is a partial isometry with operator norm 1. So σ(T) lies in the closed unit disk of the complex plane.
T* is the right shift (or unilateral shift), which is an isometry, on l q where 1/p + 1/q = 1:
$T^*(x_1, x_2, x_3, \dots) = (0, x_1, x_2, \dots).$
For λ ∈ C with |λ| < 1,
$x = (1, \lambda, \lambda ^2, \dots) \in l^p$
and T x = λ x. Consequently the point spectrum of T contains the open unit disk. Invoking reflexivity and the theorem given above, we can deduce that the open unit disk lies in the residual spectrum of T*.
The spectrum of a bounded operator is closed, which implies the unit circle, { |λ| = 1 } ⊂ C, is in σ(T). Also, T* has no eigenvalues, i.e. σp(T*) is empty. Again by reflexivity of l p and the theorem given above, we have that σr(T) is also empty. Therefore, for a complex number λ with unit norm, one must have λ ∈ σp(T) or λ ∈ σc(T). Now if |λ| = 1 and
$T x = \lambda x, \; i.e. \; (x_2, x_3, x_4, \dots) = \lambda (x_1, x_2, x_3, \dots),$
then
$x = x_1 (1, \lambda, \lambda^2, \dots),$
which can not be in l p, a contradiction. This means the unit circle must be the continuous spectrum of T.
For the right shift T*, σr(T*) is the open unit disk and σc(T*) is the unit circle.
For p = 1, one can perform a similar analysis. The results will not be exactly the same, since reflexivity no longer holds.
↑Jump back a section
## Self adjoint operators on Hilbert space
Hilbert spaces are Banach spaces, so the above discussion applies to bounded operators on Hilbert spaces as well, although possible differences may arise from the adjoint operation on operators. For example, let H be a Hilbert space and T ∈ L(H), σ(T*) is not σ(T) but rather its image under complex conjugation.
For a self adjoint T ∈ L(H), the Borel functional calculus gives additional ways to break up the spectrum naturally.
### Borel functional calculus
Further information: Borel functional calculus
This subsection briefly sketches the development of this calculus. The idea is to first establish the continuous functional calculus then pass to measurable functions via the Riesz-Markov representation theorem. For the continuous functional calculus, the key ingredients are the following:
1. If T is self adjoint, then for any polynomial P, the operator norm
$\| P(T) \| = \sup_{\lambda \in \sigma(T)} |P(\lambda)|.$
2. The Stone-Weierstrass theorem, which gives that the family of polynomials (with complex coefficients), is dense in C(σ(T)), the continuous functions on σ(T).
The family C(σ(T)) is a Banach algebra when endowed with the uniform norm. So the mapping
$P \rightarrow P(T)$
is an isometric homomorphism from a dense subset of C(σ(T)) to L(H). Extending the mapping by continuity gives f(T) for f ∈ C(σ(T)): let Pn be polynomials such that Pn → f uniformly and define f(T) = lim Pn(T). This is the continuous functional calculus.
For a fixed h ∈ H, we notice that
$f \rightarrow \langle h, f(T) h \rangle$
is a positive linear functional on C(σ(T)). According to the Riesz-Markov representation theorem that there exists a unique measure μh on σ(T) such that
$\int_{\sigma(T)} f \, d \mu_h = \langle h, f(T) h \rangle.$
This measure is sometimes called the spectral measure associated to h. The spectral measures can be used to extend the continuous functional calculus to bounded Borel functions. For a bounded function g that is Borel measurable, define, for a proposed g(T)
$\int_{\sigma(T)} g \, d \mu_h = \langle h, g(T) h \rangle.$
Via the polarization identity, one can recover (since H is assumed to be complex)
$\langle k, g(T) h \rangle.$
and therefore g(T) h for arbitrary h.
In the present context, the spectral measures, combined with a result from measure theory, give a decomposition of σ(T).
### Decomposing the spectrum
Let h ∈ H and μh be its corresponding spectral measure on σ(T) ⊂ R. According to a refinement of Lebesgue's decomposition theorem, μh can be decomposed into three mutually singular parts:
$\, \mu = \mu_{\mathrm{ac}} + \mu_{\mathrm{sc}} + \mu_{\mathrm{pp}}$
where μac is absolutely continuous with respect to the Lebesgue measure, and μsc is singular with respect to the Lebesgue measure, and μpp is a pure point measure.
All three types of measures are invariant under linear operations. Let Hac be the subspace consisting of vectors whose spectral measures are absolutely continuous with respect to the Lebesgue measure. Define Hpp and Hsc in analogous fashion. These subspaces are invariant under T. For example, if h ∈ Hac and k = T h. Let χ be the characteristic function of some Borel set in σ(T), then
$\langle k, \chi(T) k \rangle = \int_{\sigma(T)} \chi(\lambda) \cdot \lambda^2 d \mu_{h}(\lambda) = \int_{\sigma(T)} \chi(\lambda) \; d \mu_k(\lambda).$
So
$\lambda^2 d \mu_{h} = d \mu_{k}\,$
and k ∈ Hac. Furthermore, applying the spectral theorem gives
$H = H_{\mathrm{ac}} \oplus H_{\mathrm{sc}} \oplus H_{\mathrm{pp}}.$
This leads to the following definitions:
1. The spectrum of T restricted to Hac is called the absolutely continuous spectrum of T, σac(T).
2. The spectrum of T restricted to Hsc is called its singular spectrum, σsc(T).
3. The set of eigenvalues of T are called the pure point spectrum of T, σpp(T).
The closure of the eigenvalues is the spectrum of T restricted to Hpp. So
$\sigma(T) = \sigma_{\mathrm{ac}}(T) \cup \sigma_{\mathrm{sc}}(T) \cup {\bar \sigma_{\mathrm{pp}}(T)}.$
### Comparison
A bounded self adjoint operator on Hilbert space is a bounded operator on a Banach space.
Unlike the Banach space formulation, the union
$\sigma(T) = {\bar \sigma_{\mathrm{pp}}(T)} \cup \sigma_{\mathrm{ac}}(T) \cup \sigma_{\mathrm{sc}}(T).$
need not be disjoint. It is disjoint when the operator T is of uniform multiplicity, say m, i.e. if T is unitarily equivalent to multiplication by λ on the direct sum
$\oplus _{i = 1} ^m L^2(\mathbb{R}, \mu_i)$
for some Borel measures $\mu_i$. When more than one measure appears in the above expression, we see that it is possible for the union of the three types of spectra to not be disjoint. If λ ∈ σac(T) ∩ σpp(T), λ is sometimes called an eigenvalue embedded in the absolutely continuous spectrum.
When T is unitarily equivalent to multiplication by λ on
$L^2(\mathbb{R}, \mu),$
the decomposition of σ(T) from Borel functional calculus is a refinement of the Banach space case.
### Physics
The preceding comments can be extended to the unbounded self-adjoint operators since Riesz-Markov holds for locally compact Hausdorff spaces.
In quantum mechanics, observables are, not necessarily bounded, self adjoint operators and their spectra are the possible outcomes of measurements. Absolutely continuous spectrum of a physical observable corresponds to free states of a system, while the pure point spectrum corresponds to bound states. The singular spectrum correspond to physically impossible outcomes. An example of a quantum mechanical observable which has purely continuous spectrum is the position operator of a free particle moving on a line. Its spectrum is the entire real line. Also, since the momentum operator is unitarily equivalent to the position operator, via the Fourier transform, they have the same spectrum.
Intuition may induce one to say that the discreteness of the spectrum is intimately related to the corresponding states being "localized". However, a careful mathematical analysis shows that this is not true. The following $f$ is an element of $L^2(\mathbb{R})$ and increasing as $x \to \infty$.
$f(x) = \begin{cases} n & \text{if }x \in \left[n, n+\frac{1}{n^4}\right], \\ 0 & \text{else.} \end{cases}$
However, the phenomena of Anderson localization and dynamical localization describe, when the eigenfunctions are localized in a physical sense. Anderson Localization means that eigenfunctions decay exponentially as $x \to \infty$. Dynamical localization is more subtle to define.
Sometimes, when performing physical quantum mechanical calculations, one encounters "eigenvectors" that do not lie in L2(R), i.e. wave functions that are not localized. These are the free states of the system. As stated above, in the mathematical formulation, the free states correspond to the absolutely continuous spectrum. Alternatively, if it is insisted that the notion of eigenvectors and eigenvalues survive the passage to the rigorous, one can consider operators on rigged Hilbert spaces.
It was believed for some time that singular spectrum is something artificial. However, examples as the almost Mathieu operator and random Schrödinger operators have shown, that all types of spectra arise naturally in physics.
↑Jump back a section
## See also
• Essential spectrum, spectrum of an operator modulo compact perturbations.
↑Jump back a section
## References
• N. Dunford and J.T. Schwartz, Linear Operators, Part I: General Theory, Interscience, 1958.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, Academic Press, 1972.
↑Jump back a section
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 41, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8936375379562378, "perplexity_flag": "head"}
|
http://www.physicsforums.com/showpost.php?p=292015&postcount=3
|
Thread: Christoffel symbol as tensor View Single Post
Blog Entries: 47
Recognitions:
Gold Member
Homework Help
Science Advisor
Here's a quote from Wald. I have highlighted what I think are the important parts.
From Wald, p. 33. Thus, we have shown that $\tilde\nabla_a - \nabla_a$ defines a map of dual vectors at p (as opposed to dual vector fields defined in a neighborhood of p) to tensors of type (0, 2) at p. By property (1), this map is linear. Consequently $\tilde\nabla_a - \nabla_a$ defines a tensor of type (1,2) at p, which we will denote as $C^c{}_{ab}$. Thus, we have shown that given any two derivative operators $\tilde\nabla_a$ and $\nabla_a$ there exists a tensor field $C^c{}_{ab}$ such that $\nabla_a \omega_b= \tilde\nabla_a \omega_b - C^c{}_{ab}\omega_c$ (3.1.7) ...snip... Continuing in a similar manner, we can derive the general formula for the action of $\nabla_a$ on an arbitrary tensor field in terms of $\tilde\nabla_a$ and $C^c{}_{ab}$. For $T \in {\cal T}(k,l)$ we find $\nabla_a T^{b_1 \cdots b_k}{}_{c_1 \cdots c_l}= \tilde\nabla_a T^{b_1 \cdots b_k}{}_{c_1 \cdots c_l} + \sum_i C^{b_i}{}_{ad} T^{b_1 \cdots d \cdots b_k}{}_{c_1 \cdots c_l} -\sum_j C^{d}{}_{ac_j} T^{b_1 \cdots b_k}{}_{c_1 \cdots d \cdots c_l}$ (3.1.14) Thus, the difference between the two derivative operators $\textcolor{red}{\nabla_a}$ and $\textcolor{red}{\tilde\nabla_a}$ is completely characterized by the tensor field $\textcolor{red}{C^c{}_{ab}}$.
From Wald, p. 34. The most important application of equation (3.1.14) arises from the case where $\tilde\nabla_a$ is an ordinary derivative operator $\partial_a$. In this case, the tensor field $C^c{}_{ab}$ is denoted $\Gamma^c{}_{ab}$ and called a Christoffel symbol. Thus, for example, we write $\nabla_a t^b = \partial_a t^b + \Gamma^b{}_{ac} t^c$ (3.1.15) Since we know how to compute the ordinary derivative associated with a given coordinate system, equation (3.1.15) (and, more generally, eq. [3.1.14] with $\partial_a$ and $\Gamma^b{}_{ac}$ replacing $\tilde\nabla_a$ and $C^b{}_{ac}$) tells us how to compute the derivative $\nabla_a$ if we know $\Gamma^b{}_{ac}$. Note that, as defined here, a Christoffel symbol is a tensor field associated with the derivative operator $\nabla_a$ and the coordinate system used to define $\textcolor{red}{\partial_a}$. However, if we change coordinates, we also change our ordinary derivative operator from $\partial_a$ to $\partial'_a$ and thus we change our tensor $\Gamma^c{}_{ab}$, to a new tensor $\Gamma'^c{}_{ab}$. Hence the coordinate components of $\textcolor{red}{\Gamma^c{}_{ab}}$, in the unprimed coordinates will not be related to the components of $\textcolor{red}{\Gamma'^c{}_{ab}}$ in the primed coordinates by the tensor transformation law, equation (2.3.8), since we change tensors as well as coordinates.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9248823523521423, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/56356/for-examples-in-probability/56357
|
## for examples in probability [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Give an example satisfying the following conditions: give out a sequence of random variables defined on a probability space, and a sub sigma algebra: the sequence converges almost surely to a limit and it is also uniformly integrable, but the sequence of the conditional expectation of the random variable sequence on the sub sigma algebra does not converges almost surely to the limit, which is the conditional expectation of the limit of the random variable sequence on the sub sigma algebra.
-
This isn't really a research level question (see the faq: mathoverflow.net/faq#whatquestions). Actually, it's not a question, rather, its a command. – George Lowther Feb 23 2011 at 3:30
math.stackexchange is probably a more suitable site although, if you post it there, it should still be written as a question. – George Lowther Feb 23 2011 at 3:31
## 1 Answer
Consider independent random variables $X_n, Y_n, n\in\mathbb N$, such that $\mathbb P(X_n = 1) = 1-\mathbb P(X_n = 0) = 1/n$, $\mathbb P(Y_n = n) = 1-\mathbb P(Y_n = 0) = 1/n$. Set $Z_n = X_nY_n$. Then, $Z_n$ is uniformly integrable and $Z_n\to 0$ as $n\to\infty$ almost surely, by Borel-Cantelli Lemma. However, set ${\cal A} = \sigma(X_n:n\in\mathbb N)$. Then $\mathbb E(Z_n\mid{\cal A}) = X_n\mathbb E(Y_n\mid {\cal A}) = X_n$, which converges to 0 in $L^1$ but not almost surely.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9243501424789429, "perplexity_flag": "head"}
|
http://mathhelpforum.com/math-challenge-problems/45963-russian-roulette-based-puzzle.html
|
# Thread:
1. ## Russian Roulette Based Puzzle
The mafia has kidnapped you and says that if you answer a question of theirs correctly that they will let you go free:
There are 4 guns on a table in front of you and another is being held to your head. Of the 5, one gun has 1 chamber, one has 2 chambers, one has 3, another 4 and the last has 5 chambers.
The person holding the gun to your head says that he chose it at random amongst the other 4 and put a random amount of bullets in it from 1 to the amount of chambers that were available.
The question you have to answer to avoid risking death is: If he pulls the trigger once, what is the chance that you will be shot and die regarding the information you have been presented with?
2. actually I think that you have 50% chance if he picks a gun with more than one round in it and 100% chance if he picks the gun with just the one chamber.
So that would mean the total odds would be: $1\cdot \frac{1}{5} + .5\cdot \frac{4}{5}$
So I say that you have a 3/5 chance of dying, or 60%
3. Originally Posted by Quick
actually I think that you have 50% chance if he picks a gun with more than one round in it
Say he picks the gun with 2 rounds/chambers, he'll fill it with either 1 or 2 bullets (not 0) which would lead to one of three outcomes:
-1 bullet loaded, loaded chamber fired (1/4 chance)
-1 bullet loaded, unloaded chamber fired (1/4 chance)
-2 bullets loaded, loaded chamber fired (1/2 chance)
Meaning a 3/4 chance of death.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9652565121650696, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/5904/must-basis-of-an-euclidean-space-be-ordered/5926
|
# Must Basis of an Euclidean Space Be Ordered
Does the basis of an Euclidean space have to be ordered by definition? Or can be left unordered?
I was also wondering about what is the morphism (i.e. the mapping that can preserve all the structures) on Euclidean spaces? Is it Euclidean transformation (rigid transformation), consisting of rotation, translation and reflection? Or the reflection is not part of it, because the basis of a Euclidean space must be ordered.
Thanks and regards!
-
1
From the answers below it appears the answer to your question is yes and no.:) A basis can be ordered or unordered, it depends on what you're using it for as to whether or not you want it ordered. For example, for representing linear maps I use the phrase "ordered basis" rather than the term "basis" to make it clear what conditions I need on the basis. – Ryan Budney Oct 3 '10 at 16:01
@Ryan: If what you're using it for intrinsically requires a basis order --- as opposed to it being merely convenient for presentation to label the basis vectors with an index set, which has a conventional ordering --- then I'm not sure that what you're studying is just "geometry" or "linear algebra" any more. Similarly, someone studying group theory, who finds that their problem simply requires that some notion of "distance" be continuously deformed, is not just working on "group theory", but algebraic topology. Do group actions have to preserve open sets? – Niel de Beaudrap Oct 8 '10 at 6:08
## 4 Answers
Bases must be ordered, otherwise you couldn't speak of coordinates as ordered tuples. And this could make things pretty messy. For instance, if the standard basis for $\mathbb{R}^2$ could be either $e_1, e_2$ or $e_2, e_1$ simultaneously, then it would be difficult to speak of "the point of coordinates $(2,1)$".
Reflexions send bases to bases, because they are isomorphisms. But this doesn't mean they necessarily send a particular basis to the same particular basis. So they can change the order of the vectors of the basis. They don't preserve orientation, but this is an extra piece of structure, not included in the definition of Euclidean space.
-
2
Yes, a basis should be thought of as an ordered list of vectors. However, one should be aware that some textbooks define a basis (incorrectly, IMO) as a set of vectors; as far as I can see, there is no advantage to this. – Darsh Ranjan Oct 3 '10 at 4:31
1
@Darsh. It is in fact standard to define a basis as a set of vectors rather than as an ordered tuple. Most of the time we are interested in linear combination of basis vectors where order does not matter because vector addition is commutative. – Jyotirmoy Bhattacharya Oct 3 '10 at 5:41
### On bases
Bases do not have to be ordered. I respectfully present the following as a counterpoint to Agustí Roig's response.
First of all, a basis is simply a set of vectors, by definition; and sets are unordered. If you wish to enumerate the vectors in a basis, you will of course do so in some order, but this order is arbitrary.
This arbitrary order of enumeration is the reason why we describe vectors as tuples. The order of the coefficients in a tuple matters only in as much as it must be in agreement with the arbitrary order which was selected for the basis vectors. To wit: let x, y be two arbitrary linearly independent vectors. The tuple [ 7 5 ] with respect to the (enumeration of the) basis v1 = x, v2 = y represents the same vector as the tuple [ 5 7 ] with respect to the (different enumeration of the same) basis v2 = x, v1 = y. The tuple is just a representation of a vector, relative to an arbitrarily chosen order for the basis.
Even more foundationally: "tuples" can be regarded as functions from the integers (e.g. the indices '1', '2', etc.) to the reals, complex numbers, or whichever set you draw your coefficients from. So [ 7 5 ] can be thought of as 'really being' the function mapping
'1' $\mapsto$ 7,
'2' $\mapsto$ 5.
Our enumeration of the basis has a similar role: choosing the enumeration v1 = x, v2 = y is equivalent to defining a mapping
'1' $\mapsto$ x,
'2' $\mapsto$ y.
Why do I have '1' and '2' in quotes? Because they're just labels; any other labels would do just as well. For instance, I could replace '1' and '2' with the vectors x and y themselves. Then we could define vectors by coefficient functions such as
x $\mapsto$ 7,
y $\mapsto$ 5;
that is, the coefficient 7 is associated to the vector x, and the coefficient 5 is associated to y, to represent the vector 7x + 5y. This is what we really mean anyway; never mind any ordering of the vectors in your basis. With this, the arbitrary ordering of the basis disappears, leaving nothing but what it is we really mean by it all.
So: when we order our bases, its only to make it easier to present things — it is not part of the definition, or part of the structure of the objects we really care about.
### On morphisms
A morphism on a Euclidean space ought to preserve all of the structures of Euclidean geometry. In particular, it should map circles to other circles, to preserve the structures guaranteed by the third postulate; so it must be an isothety (a rigid transformation up to scaling). This includes, but is not restricted to, the rigid transformations.
Can you make any argument why the morphisms should consist only of rigid transformations (excluding isothetic maps such as v $\mapsto$ 2v)?
-
This began as a reply to Jyotirmoy's comment on Agustí's answer, but it became way too long.
As I stated in my comment on Agustí's answer, some textbooks define a basis as a set, and I think this is wrong. Obviously, some people disagree (just read some of the other answers and comments). I'll try to explain why a basis should never be thought of as a set.
Do the columns of the following matrix form a basis for $\mathbb{R}^2$? $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1\end{bmatrix}$$ The answer is "yes" if a basis is a set, but obviously the right answer is "no." This makes it clear that whatever sort of collection to which we apply concepts like "linear [in]dependence" or "basis", it must allow repeated vectors; otherwise, a lot of basic theorems from linear algebra would require extra clauses to deal with matrices with repeated rows or columns. This rules out sets as our collections of choice.
To find out what our collection of choice should be, let's revisit the definition of a basic concept like linear dependence (in somewhat imprecise "collection-agnostic" language (whose meaning should nevertheless be clear)):
The vectors $v_1,\ldots,v_n$ are linearly dependent if there are scalars $c_1,\ldots,c_n$, not all zero, such that $c_1v_1+\cdots+c_nv_n=0$.
It's obvious here that the natural object to apply such a concept to is an indexed tuple of vectors (indexed by $\{1,\ldots,n\}$ above, but we could formulate the concept equally well for an arbitrary finite index set): we begin with an indexed tuple of vectors and then consider a tuple of scalars indexed by the same set, and then we apply a function to these two tuples that is now completely specified (taking the linear combination). We could have started with a completely unindexed, unordered collection of vectors with repetition allowed (i. e., a "multiset"), but then to consider a linear combination, we would have had to choose some indexing of the multiset first and then to consider some scalars indexed by our arbitrary index set in order to form a linear combination. The core point is that the natural input to the "linear combination" operation is a tuple of vectors and a tuple of scalars indexed by the same finite set. This is the fundamental operation on a vector space, so to me, it makes sense to formulate everything in terms of indexed tuples of vectors whenever possible.
I talked about indexing lists of vectors in the previous paragraph, but indexing isn't the same as ordering. When I say "order," though, I'm actually being a bit imprecise. It's usually not the order relation itself that matters, but the choice of index set does matter. For example, if you want to write down the coordinate matrix of a vector or the matrix of a linear transformation relative to a basis, then your index set pretty much has to be $\{1,...,n\}$ (or $\{0,...,n-1\}$ if you've ever programmed in a language other than Matlab or Fortran) because whatever you're using to index the elements of the basis, you're also using to index the rows or columns of your matrices. Since one of the main uses of bases is to reduce things to matrices, the sets $\{1,...,n\}$ are very convenient for indexing. In this very common context, all the basic concepts like "basis," "linear independence," "spanning," etc. would apply to finite sequences of vectors. In no context, however, do they naturally apply to finite sets of vectors.
-
1
Your complaint is not so much that we should be using sequences instead of sets, as it is that we should be using bags (i.e. multi-sets, sets where membership has multiplicity) rather than sets. This obviously applies for the columns of the matrix you describe; and the case of the definition of linear independence hilights this, where even in the sum the particular order does not matter, nor does the choice of indexing scheme, so long as it is a bijection between some index-set and the vector-set (or vector-bag) under consideration. – Niel de Beaudrap Oct 3 '10 at 11:58
1
Anyway: I would readily concede that bags are a better choice than sets for describing collections of vectors for the purpose of describing linear independence, etc. However, I think that details such as a specific choice of indexing into the definitions only cloud details, and introduce distinctions which are on the whole totally uninteresting. – Niel de Beaudrap Oct 3 '10 at 12:02
@Niel, I think the definitions should be chosen to make the surrounding theory as natural and simple as possible, and as such, sequences are still better than multisets. Again, all of the definitions are about linear combinations, and "linear combination" is an operation on a tuple of vectors and a tuple of scalars. Moreover, while the property of "being a basis" is indeed independent of the indexing, it's still useful to talk about permuting a basis, e. g., when discussing orientation. – Darsh Ranjan Oct 3 '10 at 19:32
well, it's also useful to talk about permuting a basis when talking about permutation matrices, or anything where permutations play a role. (For instance, the sign of the cross product can be seen as taking the 'sign' of a permutation, rather than saying anything about the geometry per se.) — I respectfully disagree that 'sequences' make the theory more natural, or simple; and if it makes it pedagogically easier, this can just as easily be an accident of the fact that we teach sequences, but not multisets, in grade-school. – Niel de Beaudrap Oct 4 '10 at 6:32
Orientation does not play any role in Euclidean geometry. So it is most natural to define a Euclidean space as a finite-dimensional vector space with an inner product. Lengths, distances and angles are then defined in terms of this inner product. We are interested in transformations which preserve lengths and angles, which are precisely the orthogonal transformations including reflections. There is no need to explicitly choose a basis at any point.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9407557845115662, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/21383/applications-of-cauchys-arm-lemma/21400
|
## Applications of Cauchy’s Arm Lemma
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Cauchy's Arm Lemma is used in the proof of Cauchy's Rigidity Theorem for convex polyhedra. The Lemma states that in the plane or on the sphere that if all but one of the side lengths of two convex polygons $P$ and $P'$ are the same, and the angles formed by the remaining sides of P are less than or equal to those of the remaining sides of $P'$, then the ommitted side length from $P$ is less than or equal to the omitted side length for $P'$, with equality occurring iff the angles are all the same. I know of a couple of extensions of this theorem (a nice presentation of this sort of thing is available in O'Rourke's paper [O'R01]). The Lemma can also be used to show that convex linkages may be (unsurprisingly) straightened.
I'm seeking other applications of this Lemma, particularly those that might be suitable for use as exercises in an advanced undergraduate course, or other applications in the theory of polyhedra.
[O’R01] Joseph O’Rourke, An extension of Cauchy’s arm lemma with application to curve development, Discrete and computational geometry (Tokyo, 2000), Lecture Notes in Comput. Sci., vol. 2098, Springer, Berlin, 2001, pp. 280–291. MR MR2043660 (2004m:52052)
-
## 2 Answers
The question is a little too general to have a single answer. Cauchy's arm lemma is a basic technical result in rigidity theory. One reason it has a name is because it is intuitively obvious, but the most natural proof Cauchy originally came up with is false.
Now, other than Cauchy theorem, it has few direct consequences. A.D. Alexandrov developed a whole family of results similar in nature (say, with angles preserved but polygon lengths extended), which proved new results on rigidity. This was, e.g. his way of extending the Minkowski uniqueness theorem in $\Bbb R^3$ to polytopes with (say) equal normals and perimeters of faces. You can read it in Alexandrov's famous monograph. Sabitov made a historical study behind the original Cauchy lemma and proved a number of related results as well.
Let me emphasize that from a modern point of view, the Arm Lemma is a "relative" version of the discrete four vertex theorem, which is a fundamental result in its own right. I explain the background, connections and variations in my book, sections 21-23. Note that this is not the only "relative result" of this type: in addition to Alexandrov's lemmas, see Tabachnikov's theorem in 21.6. See also numerous historical and other references mentioned there. In conclusion, much of this is relatively easy but not straigtforward. I think at least some of these results fit well to an undergrad course.
-
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
If I may add one explicit example as a late-comer, even though it is along the lines with which you are already familiar: Cauchy's arm lemma may be used to prove that the curve that is the intersection of a plane with a convex polyhedron develops on a plane without self-intersection. This is described in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, p.377ff.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9359049797058105, "perplexity_flag": "head"}
|
http://unapologetic.wordpress.com/2011/04/11/the-tangent-bundle-of-a-euclidean-space/?like=1&_wpnonce=401286028f
|
The Unapologetic Mathematician
The Tangent Bundle of a Euclidean Space
Let’s look at the tangent bundle to a Euclidean space. That is, we let $V$ be a finite-dimensional real vector space with its standard differentiable structure and see what these constructions look like.
Well, if $V$ has dimension $n$, then we know $V$ is an $n$-manifold, and thus we know that $\mathcal{T}_vV$ is an $n$-dimensional vector space for every point $v\in V$. Now, of course all $n$-dimensional vector spaces are equivalent, but I say that $\mathcal{T}_vV$ can be made canonically equivalent to $V$ itself.
First of all, let’s set up an isomorphism between $V$ and $\mathcal{T}_0V$. Given a vector $v\in V$, we can set up the curve $c(t)=tv$. Then we just define a vector in $\mathcal{T}_0V$ by picking the tangent vector to $c$ at $c(0)=0\in V$. This correspondence is clearly linear. For instance, if $v$ and $w$ are two vectors in $V$, then $tv+tw = t(v+w)$. The linearity of the derivative shows that the tangent vector to $t(v+w)$ is the sum of the tangent vectors to $tv$ and $tw$.
In terms of components, pick a basis $\{e_i\}$ of $V$ and use it to get a coordinate map on all of $V$. We also get a basis of coordinate vectors for $\mathcal{T}_vV$ at each point $v\in V$, and in particular at $v=0$. What does this isomorphism look in terms of these coordinates?
Well, the $i$th component of the tangent vector at $c(0)=0$ is the derivative of the $i$th component of the curve written out in terms of coordinates. And this component is $tv^i$, where $v^i$ is the $i$th component of $v$ in the $\{e_i\}$ basis, so the derivative in question is just $v^i$. That is, if we use a particular basis of $V$ and the basis of coordinate vectors it induces on $\mathcal{T}_0V$, we get the exact same components for each vector $v\in V$ and its corresponding vector in $\mathcal{T}_0V$.
In fact, there’s nothing particularly special about $0\in V$ here. We can do pretty much the exact same thing at any other point $p\in V$; just replace the curve $c(v)=tv$ with the curve $c(v)=p+tv$. That is, we use the same line translated (slid around) parallel to itself by adding an offset of $p$ to every point. This is such a fundamental technique in Euclidean spaces that we have a name for this method of comparing vectors in $\mathcal{T}_pV$ with vectors in $\mathcal{T}_0V$ — and thus with vectors in any other $\mathcal{T}_qV$ — “parallel translation”.
Parallel translation works as simply as this for two fundamental reasons: we can cover the entire manifold $V$ by a single coordinate patch, and the vector space structure allows us to sensibly define an operation of “add $p$ to each point”. In more general manifolds, this parallel translation operation is not possible, or at least not so straightforwardly. But Euclidean spaces are, of course, special.
In fact, they’re so special that they’re basically all that most people study in multivariable calculus. And in that situation everything works so smoothly that people often conflate many vector spaces which are really only equivalent. It’s common to automatically compare tangent vectors rooted at different points, parallel translating them with impunity. It’s even common to cavalierly identify the tangent space $\mathcal{T}_vV$ with the space $V$ itself, identifying a point $p$ with its position vector — that which points from $0$ to $p$.
But to move forward in differential geometry it is absolutely essential to unlearn this identification. When working in a Euclidean space, it’s okay to use it to simplify matters, but in general it pays off to be careful whether we’re talking about a point in a manifold, a tangent vector at one point in a manifold, or a tangent vector at a different point in the manifold. They’re just not all the same thing.
Like this:
Posted by John Armstrong | Differential Topology, Topology
3 Comments »
1. [...] has dimension , and thus they’re all isomorphic. Worse, when working over Euclidean space there is a canonical identification between a tangent space and the space itself, and thus between any two tangent spaces. But when [...]
Pingback by | May 23, 2011 | Reply
2. [...] matrix is the same as the tangent space to at . And since is (isomorphic to) a Euclidean space, we can identify with using the canonical isomorphism . In particular, we can identify it with the tangent space [...]
Pingback by | June 9, 2011 | Reply
3. [...] At any point the induced distribution is the subspace , which is the image of under the standard identification of with [...]
Pingback by | July 1, 2011 | Reply
« Previous | Next »
About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 59, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9219138622283936, "perplexity_flag": "head"}
|
http://physics.stackexchange.com/questions/20403/pair-correlation-function-for-a-inhomogeneous-laughlin-droplet
|
# Pair correlation function for a inhomogeneous Laughlin droplet
Pair correlation function for the usual Laughlin droplet is defined as $g(\vec{r})$: $$\rho_0 g(\vec{r})=\frac{1}{N}\langle\sum_i^N \sum_{j \neq i}^N \delta(\vec{r}-\vec{r_i}+\vec{r_j})\rangle$$, where $\rho_0=\frac{\nu}{2\pi l_0^2}$. So if now the Laughlin droplet is inhomogeneous, meaning that the density function $\rho(\vec{r})$ is direction dependent, then should the pair correlation function still be defined the same as above, or do we need to change $\rho_0$ to something else?
-
## 1 Answer
If the system is anisotropic, you'll need to consider expanding the pair correlation function to accommodate the anisotropy - typically this is done using an expansion in spherical harmonics.
-
Any paper on that I can read about? – huyichen Feb 2 '12 at 18:08
1
I'll look for some others, but for the case applied to metallic glasses try: prl.aps.org/abstract/PRL/v105/i20/e205502, or prb.aps.org/abstract/PRB/v35/i5/p2162_1 Also, "Underneath the Bragg Peaks" by Egami has a section on the anisotropic pair distribution function. – Jen Feb 2 '12 at 20:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8570681810379028, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/15902/one-dimensional-phi-gamma-modules-in-char-p/15940
|
One dimensional (phi,Gamma)-modules in char p
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I would like to better understand the simplest case of the correspondence between Galois representations and (phi,Gamma)-modules. Namely, consider 1-dimensional Galois representations of $G_{Q_p}$ over $F_p$ which are in correspondence with 1-dimensional etale (phi,Gamma)-modules over $F_p((T))$.
There are finitely many such Galois representations. Moreover, their associated (phi,Gamma)-modules are very simple -- the action of phi and Gamma can be described (on some basis element) as scaling by an element of $F_p$ (as opposed to $F_p((T))$).
My question: can one see this directly on the (phi,Gamma)-module side? That is, given a 1-dimensional etale (phi,Gamma)-module $D$ over $F_p((T))$, find a $D'$ isomorphic $D$ such that $D'$ has a basis in which the matrices for phi and elements of Gamma are in $F_p^\times$.
-
I added the etale condition per Brian's correction. – Robert Pollack Feb 20 2010 at 20:33
Hey Rob---I once failed to do that exercise too :-) I could see all the phi-Gamma modules associated to the Galois reps but couldn't prove from first principles that there weren't any more. – Kevin Buzzard Feb 20 2010 at 20:45
3 Answers
OK...I think I see how to do this now. In the end, I am seeing $(p-1)^2$ distinct $(\phi,\Gamma)$-modules which matches well with the Galois side.
To do this, let $D$ be any 1-dimensional etale $(\phi,\Gamma)$-module. Let $e$ be a basis, and set $\phi(e)=h(T)e$ with $h(T) \in F_p((T))^\times$. Write $h(T) = h_0 T^a f(T)$ with $h_0 \in F_p^\times$ and $f(T) \in F_p[[T]]$ with $f(0)=1$.
Changing basis from $e$ to $u(T)e$ with $u(T) \in F_p((T))^\times$ gives $$\phi(u(T)e) = u(T^p)h(T)e = \frac{u(T^p)}{u(T)} h(T) (u(T)e).$$ I claim one can find $u(T)$ such that $u(T)/u(T^p)$ equals any element of $1+TF_p[[T]]$. Indeed, for such an element $g(T)$, the infinite product $\prod_{j=1}^\infty \phi^j(g(T))$ (which hopefully converges since $g(0)=1$) works.
Thus, we can change basis so that $\phi$ has the form $\phi(e) = h_0 T^a e$ -- i.e. we can kill off the $f(T)$ term. Further, by making a change of basis of the form $e$ goes to $T^b e$, we may assume that $0 \leq a < p-1$.
Now, we use the fact that the $\phi$ and $\Gamma$ actions commute (which is a strong condition even in dimension 1). Namely, let $\gamma$ be a generator of $\Gamma$, and set $\gamma e = g(T) e$. Then $\gamma \phi e = \phi \gamma e$ implies $$((1+T)^{\chi(\gamma)}-1)^a g(T) = g(T^p) T^a.$$ Comparing leading coefficients, we see this is only possible if $a=0$ and $g(T)$ is a constant.
Thus, $D$ has a basis $e$ so that $\phi(e) = h_0 e$ and $\phi(e) = g_0 e$ with $h_0,g_0 \in F_p^\times$ as desired.
Does this look okay? Any takers for the 2-dimensional case?
-
1
Yes, it looks good. This same style of calculation comes up when using Dieudonne theory to work out p-torsion commutative groups of order p^2, as well as Fontaine-Laffaille modules (as in the original paper of Fontaine-Laffaille). The fun part is to dig out the Galois action in the rings to identify exactly which character corresponds to the parameters (p_0, g_0) (which one can guess in advance, up to some inversions perhaps). I should revise the CMI lecture notes to address the general p-torsion rank-1 case (in the form of an exercise...)! As for the irreducible 2-dimensional case...oy. – BCnrd Feb 21 2010 at 1:36
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The correspondence requires the $(\phi,\Gamma)$-module to have the \'etale property for its underlying $\phi$-module, and this plays an essential role in the proof of the correspondence (see Fontaine's original articles, or the CMI summer school lecture notes on $p$-adic Hodge theory, for example). So I assume you mean to impose the \'etale property. But even for the trivial representation (of dimension 1) the actions of $\phi$ and $\Gamma$ are more complicated than $\mathbf{F}_p^{\times}$-scalings relative to a suitable basis. Do you mean to ask why in a suitable basis the $(\phi,\Gamma)$-modules associated to powers of the mod $p$ cyclotomic character are related to the case of the trivial character by a "Tate twist" on the $(\phi, \Gamma)$-module side? If so, see Example 13.6.6 in the CMI summer school lecture notes (which handles $\mathbf{Z} _p(r)$ more generally, for any $p$-adic field $K$, and can pass to mod $p$ version there in the same way to get to the case you ask about).
Looking back at the current draft of the notes, I now see a typo in the notes there (I think $\chi^t$ at the end should be $\chi^r$). I'd better fix it for the final draft, so thanks for asking the question, Rob!
-
Yes. I certainly meant to include etale. In this case (i.e. 1-dimensional in char p), this is simply asserting that $\phi$ is non-zero, yes? Regarding the trivial representation, its associate $(\phi,\Gamma)$-module must just be $F_p((T))$ with the standard $\phi$ and $\Gamma$ actions. Since $\phi(1)=1$ and $\gamma(1)=1$ for all $\gamma \in \Gamma$, with respect to the basis 1" all of these operators have matrix 1 (certainly in $F_p^\times$). What am I missing? – Robert Pollack Feb 20 2010 at 20:29
The etale property in this rank-1 case mod p is indeed the same as phi being nonzero on some basis vector. Concerning the 2nd part of your comment, nothing is being missed; I was just expressing (perhaps in an unclear way) that there's a lot of nontrivial action on the coefficients. When reading the question I misread it as not accounting for that (but I see it was implicit when you wrote "(on some basis element)"). Anyway, see the Example 13.6.6, which I hope will clear things up. – BCnrd Feb 20 2010 at 20:42
Hey Rob! You may want to take a look at Colmez's "Représentations triangulines de dimension 2" (here). He's working with (φ, Γ)-modules over Qp (by which I mean the Robba ring of Qp) not Fp((T)) though. But he shows that every one-dimensional (φ, Γ)-module over Qp is given by a p-adic character δ of $\mathbf{Q}_p^\times$ with the action of φ on a basis vector given by δ(p) and the action of $\gamma\in\Gamma$ given by $\delta(\chi(\gamma))$ (with $\chi$ the $p$-adic cyclotomic character). See section 2 and proposition 3.1 in particular of Colmez's paper (in which he also describes the $H^1$'s of these one-dimensional $(\phi,\Gamma)$-modules).
EDIT: looking at Colmez's paper, it looks like the way that he knows that this is all one-dimensional $(\phi,\Gamma)$-modules is by using the equivalence of catgeories with Galois representations, so I guess this doesn't really answer your question as it doesn't address the $(\phi,\Gamma)$-modules intrinsically.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 83, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9298526644706726, "perplexity_flag": "head"}
|
http://stats.stackexchange.com/tags/media/hot
|
# Tag Info
## Hot answers tagged media
33
### How exactly does one “control for other variables”?
There are many ways to control for variables. The easiest, and one you came up with, is to stratify your data so you have sub-groups with similar characteristics - there are then methods to pool those results together to get a single "answer". This works if you have a very small number of variables you want to control for, but as you've rightly discovered, ...
29
### Article about misuse of statistical method in NYTimes
I will answer the first question in detail. With a fair coin, the chances of getting 527 or more heads in 1,000 flips is less than 1 in 20, or 5 percent, the conventional cutoff. For a fair coin the number of heads in 1000 trials follows the binomial distribution with number of trials $n=1000$ and probability $p=1/2$. The probability of getting ...
28
### Why law of large numbers does not apply in the case of Apple share price?
Humorously enough, I just wrote a blog post on this very subject: http://confounding.net/2012/03/12/thats-not-how-the-law-of-large-numbers-works/ Essentially, the Law of Large Numbers is that as the number of trials of a random process increases, the mean of those trials will approach the actual mean (or expectation, for more complex distributions). So ...
20
### How exactly does one “control for other variables”?
1. Introduction I like @EpiGrad's answer (+1) but let me take a different perspective. In the following I am referring to this PDF document: "Multiple Regression Analysis: Estimation", which has a section on "A 'Partialling Out' Interpretation of Multiple Regression" (p. 83f.). Unfortunately, I have no idea who is the author of this chapter and I will refer ...
11
### Why law of large numbers does not apply in the case of Apple share price?
Here is the rub: Apple is so big, it’s running up against the law of large numbers. Also known as the golden theorem, with a proof attributed to the 17th-century Swiss mathematician Jacob Bernoulli, the law states that a variable will revert to a mean over a large sample of results. In the case of the largest companies, it suggests that high ...
11
### Can one leave out data from research because it is not significant?
That sentence does not actually make sense and is clearly in error. Data cannot be statistically significant or insignificant. Only relationships between data, the product of statistical tests, can be spoken about in these terms. If the question is: Can we drop data from our analyses because the inclusion of that data means we cannot reject the null ...
9
### Equations in the news: Translating a multi-level model to a general audience
"Your teaching score depends on how well your students did compared to a prediction made based on What they knew beforehand, as measured by a pretest, How well we think the students can learn based on what we know about them individually (their "characteristics"), And how well students do on average in your district, school, and classroom (if there are ...
9
### Equations in the news: Translating a multi-level model to a general audience
Here's one possibility. Assessing teacher performance has traditionally been difficult. One part of this difficulty is that different students have different levels of interest in a given subject. If a given student gets an A, this doesn't necessarily mean that teaching was excellent -- rather, it may mean that a very gifted and interested student did his ...
6
### Can one leave out data from research because it is not significant?
In the report cited in whuber's comment, it says on page 104 [pg 114 in the pdf]: The survey succeeded in activating the participation of approximately 8,900 doctoral candidates from more than 30 countries... Then, spanning pages 104-105, it says: While conducting data cleaning procedures, the Eurodoc survey experts' team decided to run a power ...
4
### How to interpret the margin of error in a poll?
The claim that the margin of error is $4.9$% follows from assuming that the poll was conducted as if a box had been filled with tickets--one for each member of the entire population (of "hardcore Republican voters")--thoroughly mixed, $400$ of those were blindly taken out, and each of the associated $400$ voters had written complete answers to all the poll ...
3
### How to interpret the margin of error in a poll?
I won't try to deliver my own answer, but I would refer you to the "What Is a Survey?" booklet compiled by the Survey Research Methods Section of the American Statistical Association. (Fritz Scheuren endorsing it on the title page is a former President of ASA from about five years ago. He used to be a high profile statistician in federal agencies such as the ...
2
### How to interpret the margin of error in a poll?
To answer your question: It is possible to extrapolate from a sample of 400 to the views of all 700,000. This is contingent on the sample being random. Statistical Power is the topic you'd want to look into to confirm this. If I ask 400 of my closest friends, this doesn't work. To get a truly random sample, I'd have to get the list of all 700,000 people, ...
2
### Equations in the news: Translating a multi-level model to a general audience
There is just nothing to understand here. Well, ok, it is just a standard linear regression model. It assumes that the score of a student can be described as a linear function of several factors, including school and teacher efficiency coefficients -- thus it shares all the standard problems of linear models, mainly the fact that it is a great approximation ...
Only top voted, non community-wiki answers of a minimum length are eligible
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9544830918312073, "perplexity_flag": "middle"}
|
http://www.impan.pl/cgi-bin/dict?contradiction
|
## contradiction
[see also: contrary, otherwise]
To obtain a contradiction, we suppose that ......
Suppose, to derive a contradiction, that ......
Striving for a contradiction, suppose that ......
Aiming for a contradiction, suppose that ......
Suppose, towards a contradiction $\langle$for contradiction$\rangle$, that ......
......, which is a contradiction.
Now we have the required contradiction since ......
This leads to the contradiction that $0<a<b=0$.
......, in contradiction with Lemma 2.
This is a clear contradiction of the fact that ......
Go to the list of words starting with: a b c d e f g h i j k l m n o p q r s t u v w y z
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8501849174499512, "perplexity_flag": "head"}
|
http://physics.stackexchange.com/questions/tagged/drag
|
# Tagged Questions
The drag tag has no wiki summary.
1answer
60 views
### In tennis, why does topspin serve bounce higher than flat serve?
When receiving servers (while playing tennis), I've noticed that the tennis ball seems to bounce up higher on me when the server uses a topspin serve than when the server hits a flat serve. Why is ...
2answers
74 views
### How to calculate time for ball to reach plane while considering mass, gravity and wind resistance?
The scenario that I'm having is such that a ball of radius $15mm$ is thrown from a location point $\vec{p}=(2, 5, 2)$ in a direction of $\vec{d}=(3, 0, 4)$. The initial velocity is $30m/s$. There were ...
2answers
115 views
### Could some design of a propeller be used in both air and water?
Propellers in water are smaller in diameter. They also move more slowly. On the other hand, aircraft propellers are larger in diameter, have narrower blades and operate at very high speeds. An ...
3answers
90 views
### Parachute in vacuum
Consider that the earth is vacuum. Consider a person of weight 100 kg is falling from sky with an parachute . He is free falling at height of 3000 m. When will he reach the ground? What would happen ...
0answers
89 views
### Why did increasing the Ackermann geometry in my race car make it faster in corners?
Ackermann geometry is used to account for the different radius arcs that the front tires follow when the steering wheel is turned from center. It's often expressed as a percentage: e.g. 25% Ackermann, ...
1answer
92 views
### Concerning drag on a flow past a cylinder
I am wondering about the drag coefficient for a flow past a cylinder. I am reading this article. I understand why the drag is high to begin with (point 2), when the boundary layer separates and the ...
4answers
467 views
### What causes a soccer ball to follow a curved path?
Soccer players kick the ball in a linear kick, though you find it to turn sideways, not even in one direction. Just mid air it changes that curve's direction. Any physical explanation? Maybe this ...
2answers
234 views
### Explanation that air drag is proportional to speed or square speed?
A falling object with no initial velocity with mass $m$ is influenced by a gravitational force $g$ and the drag (air resistance) which is proportional to the object's speed. By Newton´s laws this can ...
1answer
152 views
### Drag on a spinning ball in fluid
I am a physics newbie (high school level) and I am wondering what happens when a spherical object is spinning on the spot in a bunch of gas (no gravity here, just an imaginary physics sandbox). Am I ...
2answers
128 views
### Why is simple harmonic motion called so?
Is the motion of a simple pendulum, a simple harmonic motion? It stops vibrating after sometime.
1answer
112 views
### Skiing downhill
The other day on skiing holiday we've been arguing about whether an adult has weight advantage over a child when skiing downhill. I was claiming that gravity is a constant regardless of object's ...
3answers
143 views
### If an electric car were to drive without having to stop, would the range be greatly affected by the speed at which the vehicle is moving?
Of course aerodynamics factors into this question, and the faster you are moving the more air you have to push out of your way, the more energy you use. But would the difference be only a small ...
2answers
180 views
### Forces on an aircraft - thrust, lift, drag, weight
I'm extremely sceptical about the wikipedia page on aircraft flight mechanics. When describing 'straight climbing flight', it says: lift is unable to alter the aircraft's potential energy or ...
1answer
321 views
### Does the metal foam “whiffleball” orbital reentry idea make any sense?
Planetary Resources is a company speaking publicly about mining near Earth asteroids for resources. I'm interested the physicality of a claim they make in a video here: ...
0answers
83 views
### How much force is required to hold an umbrella?
Assuming a small umbrella (e.g. http://www.amazon.com/ShedRain-WindPro-Umbrella-Close-Size/dp/B001DL5WN0). What are the factors that need to be considered? I assume where I hold the umbrella might ...
1answer
143 views
### How would you use the Euler-Lagrange equation to predict the motion of projectiles with linear (Stokes) drag (but no wind)?
My first instinct would be to use the force $$\vec{F} =- \alpha \vec{v}$$ and therefore V(\vec{r}) = \alpha \int_C \vec{v}\cdot d\vec{s} = \alpha \int_C \vec{v}\cdot \vec{v} dt = \alpha \int_C ...
2answers
165 views
### Finding optimal angle for projectile, taking into account linear (Stokes) drag
If you throw a projectile from the ground at a certain angle, it's not hard to see that, assuming we're in a vacuum, throwing it at 45º from the ground will always make it go farthest before it hits ...
2answers
110 views
### Estimating atmospheric friction by measuring the change in velocity of a ball thrown straight upwards
Imagine I throw a ball straight upwards with some velocity $v_1$, and filming the ball with a camera, I can estimate a velocity $v_2$ (along the same vector) after the ball has moved a distance $D$. ...
2answers
328 views
### Work done by the air resistance [closed]
A ball of mass 0.37 kg is thrown upward along the vertical with a initial speed of 14 m / s, and reaches a maximum height of 8.4 m. a) What is the work done by air resistance on the ball? b) ...
1answer
390 views
### Equation for the trajectory of a frisbee?
I'm the lead programmer on a FIRST robotics team, and this year's competition is about throwing Frisbees. I was wondering if there was some sort of "grand unified equation" for Frisbee trajectory that ...
2answers
86 views
### Estimating drag due to wind - ribbon-shaped sail
In the countless calculations and discussions concerning the "space rope" I've never found any addressing its capability to resist winds. Consider, as in most current works, it's a 1m wide ribbon, ...
1answer
127 views
### Is resistance to motion directly proportional to the speed of a moving object?
Power is known to be equal to the force x velocity (P=FV). Im solving a question that states the following : Car with engine working at 32 kW, mass of 1 tonne, travels at a constant speed of 40m/s ...
1answer
203 views
### Limitations of drag equation
The magnitude of the air resistance for objects with Reynolds numbers greater than 1000 is given by the formula: Why it does not hold for objects with lower Reynolds numbers? Can I use this ...
2answers
156 views
### Minimum size of an asteroid to actually impact earth
From what I understand, an object entering the atmosphere will start to burn up from the tremendous resistance of the atmosphere. Presumably, for asteroids under a certain size, they will burn up ...
0answers
73 views
### Integration question from book “e: The Story of a Number” by Eli Maor [closed]
In the book 'e: The Story of a Number', a derivation of a parachutist problem is given on pg. 109-110. A parachutist jumps from a plane and at $t=0$ opens his chute. At what speed will he reach the ...
2answers
104 views
### Is air drag equation in term of momentum still valid?
This is the known equation of air drag: $$m{\bf a}=mg-\mathcal D=mg-b{\bf v}.$$ Considering this, is air drag equation in term of momentum still valid? $$m{\bf v}=mv_g-b{\bf r}.$$
4answers
968 views
### Acceleration of two falling objects with identical form and air drag but different masses
I have a theoretical question that has been bugging me and my peers for weeks now - and we have yet to settle on a concrete answer. Imagine two balloons, one is filled with air, one with concrete. ...
1answer
393 views
### Projectile Motion with Air Resistance and Wind
I am wondering how the general kinematics equations would change in the following situation. If an object were fired out of a cannon, or some sort of launcher, so that it had both an initial velocity ...
0answers
129 views
### Torque required to spin a disk along its diameter
How would I calculate (or simulate) this? I am only interested in the aerodynamic drag caused by the surface moving, not any other forces. As far as I know, the only variables needed are the drag ...
2answers
104 views
### What was the muzzle velocity of a home made gun launch strait in the air if air time was 8.2sec?
I built a potato gun and wanted to calculate the muzzle velocity. I remember from physics that I could run the numbers by calculating time from launch until landing. After pointing strait into the air ...
0answers
65 views
### I'm having trouble understanding the dimensions of this falling body problem
I'm working with this: A 50 kg mass is shot from a cannon straight up with an initial velocity of 10m/s off a bridge that is 100 meters above the ground. If air resistance is given by 5v ...
4answers
1k views
### After what speed air friction starts to heat up an object?
I understand that air friction cools off an object at low speeds. For example, if you blow on a spoon of hot soup, it cools off. Or if you swing a hot frying pan in the air, it cools off faster. But ...
1answer
125 views
### How much runway does the space shuttle need to land?
A textbook question proposes the following scenario: NASA needs to plan emergency landing sites for the Space Shuttle (RIP). There are a number of things to consider. The shuttle touches down no ...
1answer
227 views
### Finding the diameter a parachute must be to land a probe on Mars at specific speed
Hi I'm trying to solve this textbook example but I don't know where to begin; NASA has decided to send an experimental probe to Mars. Its weight on earth is $40 kg_f$. When the probe is near ...
2answers
116 views
### Kinetic Energy of Stone
If a stone is launched upward, of which is subject to gravity and air resistance, which of the following will have a greater kinetic energy? The stone at a point on its way up, or the stone at the ...
2answers
169 views
### Parabolic motion and air drag
Are this equations correct, in order to calculate the parabolic motion of an arrow with the computation of the drag with the air? \begin{cases} x(t)=\left(v_0-\frac{1/2C_DA\rho ...
1answer
410 views
### Friction due to air drag at high speeds
I am trying to set up this problem, but I am not sure how to go about doing so. (From University Physics, Young & Freedman): You throw a baseball straight up. The drag force is proportional to ...
1answer
247 views
### How does Baumgartner slow his velocity so that the parachute can open up?
How does Baumgartner slow his velocity so that the parachute can open up? What is Baumgartner's speed just before his parachute opens? What causes his speed to slow down just before the chute opens? ...
6answers
22k views
### Could someone jump from the international space station and live?
Felix Baumgartner just completed his breathtaking free-fall skydiving jump from $120,000\,\text{feet} = 39\,\text{km}$ above the Earth, breaking the speed of sound during the process. I was wondering ...
4answers
2k views
### How Felix Baumgartner has reached the speed of sound quickly
I have watched Felix Baumgartner freefall; but I wonder how Felix has reached the speed of sound quickly, in a matter of some seconds, then we had no idea of its speed? Any explanation please.
1answer
89 views
### Time of a ball going up and down with air resistance
A ball is shot directly upward, and then it comes back to the place where it was shot. Suppose we have air resistance. Suppose $t_1$ is the time period from the moment that the ball was shot to the ...
2answers
181 views
### Does air resistance ever slow a particle down to zero velocity?
If a particle moves in a place with air resistance (but no other forces), will it ever reach a zero velocity in finite time? The air resistance is proportional to some power of velocity - $v^\alpha$, ...
3answers
514 views
### Why would a fat skydiver fall first in free fall?
I was having one of those obnoxious conversations with a friend where he was arguing that a fat skydiver would reach the ground faster than a skinny skydiver. To me it seemed as obvious that the world ...
1answer
156 views
### How can the Air friction force be applied to an object?
Suppose we have an object and we throw it straight upward in the air. How do we apply the Air friction force to this object while moving upwards and after that downwards? Sorry if it's easy because ...
3answers
426 views
### Is it possible to find out the distance traveled by a car if the force applied on it is given?
Say you have car which produces $F$ amount of force which is transferred to the wheels directly. Now assuming that there is air friction which is causing a retarding force proportional to the ...
2answers
1k views
### Equations for an object moving linearly but with air resistance taken into account?
I know (from Kinematics) that for an object moving linearly with an acceleration and without air resistance the following equations can be used to determine v(velocity) or x(position of the object) at ...
2answers
223 views
### Bicycle Wheel Drag in Slipstream
I was recently driving behind a car that had a bicycle mounted on a carrier over the rear bumper. The bicycle wheels were not bound so they were rotating in the slip-stream of the car. I wonder, the ...
1answer
122 views
### What would be an appropiate shape for a parachute?
I'm trying to design a parachute that minimizes the descent velocity, but I'm not sure what shape I should use. From what I've read, ellipse-shaped parachutes are too aerodynamic and minimize drag, ...
2answers
411 views
### Free fall of two spheres made of same materials, different masses, with air resistance
Two people falling with the same parachute will gain different speeds if their masses are different. The upward air drag will needed to be bigger for a heavier person, since gravitational force is ...
2answers
212 views
### At what g is terminal velocity not terminal?
How weak would gravity need to be in order for a human to reliably survive the terminal velocity of falling through air? (Context: watching scifi on a space station with a variety of artificial ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9415584206581116, "perplexity_flag": "middle"}
|
http://mathoverflow.net/users/19949?tab=favorites
|
# Ronnie Brown
5,209 Reputation
2331 views
## Registered User
Name Ronnie Brown
Member for 3 years
Seen 2 hours ago
Website
Location
Age 78
I was a research student of Henry Whitehead till 1960, when he died suddenly at Princeton; then Michael Barratt, as new supervisor, suggested the problem of the homotopy type of function spaces $X^Y$, by induction on the Postnikov system of $X$. A particular problem was calculating $k^Y$ where $k$ is a cohomology operation. This worked out well. In doing this, I came across lots of exponential laws, essentially monoidal closed categories, and this led me to the notion of convenient category for topology, and my first two papers.
As a result of giving courses at BSc and MSc level at Liverpool, I got involved in writing a text published as "Elements of Modern Topology" with McGraw Hill in 1968, and the 3rd edition is now available as "Topology and Groupoids" from amazon, see my web page. It was writing this book, and trying to clarify certain points, such as the fundamental group of the circle, that got mew into the area of groupoids; this suggested the area of higher groupoids; research on this got going in the 1970s, and has been a major area in my work, with fortunate collaborations with Chris Spencer (1971-73), Philip Higgins (1974-2005), and Jean-Louis Loday (1981-1985), and excellent contributions from research students.
Since 2001 I was engaged in writing a Tract giving an exposition in one place of my work since 1974 with Philip Higgins and a number of others on (strict) higher homotopy groupoids and their applications. A first step was to make available my out of print book on Topology, and this was made available in 2006 under the title "Topology and Groupoids" see my web page. The higher dimensional work was published in August, 2011, as a European Mathematical Society Tract, Vol 15, under the title "Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids" with co-authors P.J. Higgins and R. Sivera.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9497795701026917, "perplexity_flag": "middle"}
|
http://mathhelpforum.com/calculus/145182-integration-product-two-exponential-functions.html
|
# Thread:
1. ## Integration of product of two exponential functions
Hello,
I apologise in advance if this is a naive question (which it probably is) but my maths is very rusty and my old notes and text books don't seem to cover this particular problem.
What is the correct approach to integrating the product of two exponential functions of the form below?
I have current, i(t) = 10e^-5000t
voltage, v(t) = 50(1-e^-5000t).
Power, p(t) = i(t).v(t)
Energy for t is greater than equal to zero is therefore the integral of p(t) for t = 0 to infinity.
I know that the answer is 50 mJ (or 0.050 joules) from calculating p(t) in a spreadsheet for small increments of t until p(t) tends to zero but cannot demonstrate it mathematically.
I've tried integrating by parts but this does not seem to close the deal as I always end up with a integral of the same or similar form to evaluate.
What is the blindingly obvious thing that I am missing ? Am i on the right track with integrating by parts or is there an easier reduction of the problem given the similarities between the two functions?
Many thanks
2. Originally Posted by nortyrich
Hello,
I apologise in advance if this is a naive question (which it probably is) but my maths is very rusty and my old notes and text books don't seem to cover this particular problem.
What is the correct approach to integrating the product of two exponential functions of the form below?
I have current, i(t) = 10e^-5000t
voltage, v(t) = 50(1-e^-5000t).
Power, p(t) = i(t).v(t)
Energy for t is greater than equal to zero is therefore the integral of p(t) for t = 0 to infinity.
I know that the answer is 50 mJ (or 0.050 joules) from calculating p(t) in a spreadsheet for small increments of t until p(t) tends to zero but cannot demonstrate it mathematically.
I've tried integrating by parts but this does not seem to close the deal as I always end up with a integral of the same or similar form to evaluate.
What is the blindingly obvious thing that I am missing ? Am i on the right track with integrating by parts or is there an easier reduction of the problem given the similarities between the two functions?
Many thanks
According to what you wrote, you have the improper integral:
$\int\limits^\infty_010e^{-5,000t}\cdot 50(1-e^{-5,000t})\,dt=$ $500\int\limits_0^\infty\left(e^{-5,000t}-e^{-10,000t}\right)\,dt=$ $500\lim_{b\to\infty}\left[-\frac{1}{5,000}\,e^{-5,000t}+\frac{1}{10,000}\,e^{-10,000t}\right]^b_0=$ $\frac{1}{20}\lim_{b\to\infty}\left(-\frac{2}{e^{5,000b}}+2e^0+\frac{1}{e^{10,000b}}-e^0\right)=$
$=\frac{1}{20}=0.05$
Tonio
3. Multiply the functions together before integrating, what do you get?
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9394372701644897, "perplexity_flag": "head"}
|
http://physics.stackexchange.com/questions/52891/what-if-f-neq-fracdpdt
|
# What if $F\neq \frac{dp}{dt}$?
I was thinking of this idea that maybe there are esoteric cases where the force is not given in classical mechanics as $F=dp/dt$ but as some function of $F=F(p,q,\dot{p},\dot{q})$
E.g, something like: $k\cdot \frac{dp}{dq}$ with a suitable constant $k$, or any other sort of function of p,q and its time derivatives.
Are there any toy models that theorists suggest on this idea?
-
## 1 Answer
If you're using $q$ and $p$ then you're implicitly using Hamiltonian formalism. Then $\dot{p}=u(q,p)$ and $\dot{q}=v(q,p)$ so $F=F(p,q,\dot{p},\dot{q})\equiv F(q,p)$. In your particular proposal $k\displaystyle\frac{dq}{dp}=0$ because $q$ and $p$ are treated as independent variables. In classical mechanics the choice of the independent variables is of capital importance.
Let's take point particle electromagnetism. The Hamiltonian is:
$$H=\frac{1}{2}\left(\vec{p}-e\vec{A} \right)^2+e\phi$$
and Hamilton's equation for the momentum reads:
$$\dot{p}_{i}=-\frac{\partial H}{\partial q_{i}}=0$$
but $\vec{F}\neq 0$. I think the problem is mixing terms about Newtonian Mechanics (i.e. Force) and terms about Hamiltonian Mechanics (momentum).
-
I don't think choosing $q$ and $p$ as independent variable means Hamiltonian formalism. It is only Hamiltonian formalism when $-\partial H/\partial q=\dot{p}$ and $\partial H/\partial p=\dot{q}$. – Sankaran Feb 2 at 22:37
Isn't $F = dp/dt$ the definition of force? – Paul J. Gans Feb 3 at 1:05
– Manishearth♦ Feb 3 at 14:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9332588911056519, "perplexity_flag": "head"}
|
http://en.wikipedia.org/wiki/Analytic_function
|
# Analytic function
Not to be confused with analytic expression or analytic signal.
In mathematics, an analytic function is a function that is locally given by a convergent power series. There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others. Functions of each type are infinitely differentiable, but complex analytic functions exhibit properties that do not hold generally for real analytic functions. A function is analytic if and only if its Taylor series about x0 converges to the function in some neighborhood for every x0 in its domain.
## Definitions
Formally, a function ƒ is real analytic on an open set D in the real line if for any x0 in D one can write
$\begin{align} f(x) & = \sum_{n=0}^\infty a_{n} \left( x-x_0 \right)^{n} \\ & = a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + a_3 (x-x_0)^3 + \cdots \end{align}$
in which the coefficients a0, a1, ... are real numbers and the series is convergent to ƒ(x) for x in a neighborhood of x0.
Alternatively, an analytic function is an infinitely differentiable function such that the Taylor series at any point x0 in its domain
$T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^{n}$
converges to f(x) for x in a neighborhood of x0 pointwise (and uniformly). The set of all real analytic functions on a given set D is often denoted by Cω(D).
A function ƒ defined on some subset of the real line is said to be real analytic at a point x if there is a neighborhood D of x on which ƒ is real analytic.
The definition of a complex analytic function is obtained by replacing, in the definitions above, "real" with "complex" and "real line" with "complex plane."
## Examples
Most special functions are analytic (at least in some range of the complex plane). Typical examples of analytic functions are:
• Any polynomial (real or complex) is an analytic function. This is because if a polynomial has degree n, any terms of degree larger than n in its Taylor series expansion must immediately vanish to 0, and so this series will be trivially convergent. Furthermore, every polynomial is its own Maclaurin series.
• The exponential function is analytic. Any Taylor series for this function converges not only for x close enough to x0 (as in the definition) but for all values of x (real or complex).
• The trigonometric functions, logarithm, and the power functions are analytic on any open set of their domain.
Typical examples of functions that are not analytic are:
• The absolute value function when defined on the set of real numbers or complex numbers is not everywhere analytic because it is not differentiable at 0. Piecewise defined functions (functions given by different formulas in different regions) are typically not analytic where the pieces meet.
• The complex conjugate function z → z* is not complex analytic, although its restriction to the real line is the identity function and therefore real analytic, and it is real analytic as a function from R² to R².
• See here for another example of a non-analytic smooth function.
## Alternative characterizations
If ƒ is an infinitely differentiable function defined on an open set D ⊂ R, then the following conditions are equivalent.
1) ƒ is real analytic.
2) There is a complex analytic extension of ƒ to an open set G ⊂ C which contains D.
3) For every compact set K ⊂ D there exists a constant C such that for every x ∈ K and every non-negative integer k the following bound holds[citation needed]
$\left | \frac{d^k f}{dx^k}(x) \right | \leq C^{k+1} k!$
The real analyticity of a function ƒ at a given point x can be characterized using the FBI transform.
Complex analytic functions are exactly equivalent to holomorphic functions, and are thus much more easily characterized.
## Properties of analytic functions
• The sums, products, and compositions of analytic functions are analytic.
• The reciprocal of an analytic function that is nowhere zero is analytic, as is the inverse of an invertible analytic function whose derivative is nowhere zero. (See also the Lagrange inversion theorem.)
• Any analytic function is smooth, that is, infinitely differentiable. The converse is not true; in fact, in a certain sense, the analytic functions are sparse compared to all infinitely differentiable functions.
• For any open set Ω ⊆ C, the set A(Ω) of all analytic functions u : Ω → C is a Fréchet space with respect to the uniform convergence on compact sets. The fact that uniform limits on compact sets of analytic functions are analytic is an easy consequence of Morera's theorem. The set $\scriptstyle A_\infty(\Omega)$ of all bounded analytic functions with the supremum norm is a Banach space.
A polynomial cannot be zero at too many points unless it is the zero polynomial (more precisely, the number of zeros is at most the degree of the polynomial). A similar but weaker statement holds for analytic functions. If the set of zeros of an analytic function ƒ has an accumulation point inside its domain, then ƒ is zero everywhere on the connected component containing the accumulation point. In other words, if (rn) is a sequence of distinct numbers such that ƒ(rn) = 0 for all n and this sequence converges to a point r in the domain of D, then ƒ is identically zero on the connected component of D containing r.
Also, if all the derivatives of an analytic function at a point are zero, the function is constant on the corresponding connected component.
These statements imply that while analytic functions do have more degrees of freedom than polynomials, they are still quite rigid.
## Analyticity and differentiability
As noted above, any analytic function (real or complex) is infinitely differentiable (also known as smooth, or C∞). (Note that this differentiability is in the sense of real variables; compare complex derivatives below.) There exist smooth real functions which are not analytic: see non-analytic smooth function. In fact there are many such functions.
The situation is quite different when one considers complex analytic functions and complex derivatives. It can be proved that any complex function differentiable (in the complex sense) in an open set is analytic. Consequently, in complex analysis, the term analytic function is synonymous with holomorphic function.
## Real versus complex analytic functions
Real and complex analytic functions have important differences (one could notice that even from their different relationship with differentiability). Analyticity of complex functions is a more restrictive property, as it has more restrictive necessary conditions and complex analytic functions have more structure than their real-line counterparts.[1]
According to Liouville's theorem, any bounded complex analytic function defined on the whole complex plane is constant. The corresponding statement for real analytic functions, with the complex plane replaced by the real line, is clearly false; this is illustrated by
$f(x)=\frac{1}{x^2+1}.$
Also, if a complex analytic function is defined in an open ball around a point x0, its power series expansion at x0 is convergent in the whole ball (analyticity of holomorphic functions). This statement for real analytic functions (with open ball meaning an open interval of the real line rather than an open disk of the complex plane) is not true in general; the function of the example above gives an example for x0 = 0 and a ball of radius exceeding 1, since the power series 1 − x2 + x4 − x6... diverges for |x| > 1.
Any real analytic function on some open set on the real line can be extended to a complex analytic function on some open set of the complex plane. However, not every real analytic function defined on the whole real line can be extended to a complex function defined on the whole complex plane. The function ƒ(x) defined in the paragraph above is a counterexample, as it is not defined for x = ±i. This explains why the Taylor series of ƒ(x) diverges for |x| > 1, i.e., the radius of convergence is 1 because the complexified function has a pole at distance 1 from the evaluation point 0 and no further poles within the open disc of radius 1 around the evaluation point.
## Analytic functions of several variables
One can define analytic functions in several variables by means of power series in those variables (see power series). Analytic functions of several variables have some of the same properties as analytic functions of one variable. However, especially for complex analytic functions, new and interesting phenomena show up when working in 2 or more dimensions. For instance, zero sets of complex analytic functions in more than one variable are never discrete.
## References
• Conway, John B. (1978). . Springer-Verlag. ISBN 0-387-90328-3.
• Krantz, Steven; Parks, Harold R. (2002), A Primer of Real Analytic Functions (Second ed.), Birkhäuser, ISBN 0-8176-4264-1
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9021451473236084, "perplexity_flag": "head"}
|
http://www.reference.com/browse/Gelfand_representation
|
Definitions
# Gelfand representation
In mathematics, the Gelfand representation in functional analysis (named after I. M. Gelfand) has two related meanings:
• a way of representing commutative Banach algebras as algebras of continuous functions;
• the fact that for commutative C*-algebras, this representation is an isometric isomorphism.
In the former case, one may regard the Gelfand representation as a far-reaching generalization of the Fourier transform of an integrable function. In the latter case, the Gelfand-Naimark representation theorem is one avenue in the development of spectral theory for normal operators, and generalizes the notion of diagonalizing a normal matrix.
## Historical remarks
One of Gelfand's original applications (and one which historically motivated much of the study of Banach algebras) was to give a much shorter and more conceptual proof of a celebrated lemma of Norbert_Wiener (see the citation below), characterizing the elements of the group algebras $L^1\left(\left\{mathbb R\right\}\right)$ and $ell^1\left(\left\{mathbb Z\right\}\right)$ whose translates span dense subspaces in the respective algebras.
## The model algebra
For any locally compact Hausdorff topological space X, the space C0(X) of continuous complex-valued functions on X which vanish at infinity is in a natural way a commutative C*-algebra:
• The structure of algebra over the complex numbers is obtained by considering the pointwise operations of addition and multiplication.
• The involution is pointwise complex conjugation.
• The norm is the uniform norm on functions.
Note that A is unital if and only if X is compact, in which case C0(X) is equal to C(X), the algebra of all continuous complex-valued functions on X.
## The Gelfand representation of a commutative Banach algebra
Let A be a commutative Banach algebra, defined over the field $\left\{mathbb C\right\}$ of complex numbers. A non-zero algebra homomorphism $phi:Ato \left\{mathbb C\right\}$ is called a character of $A$; the set of all characters of A is denoted by $Phi_A$.
It can be shown that every character on A is automatically continuous, and hence $Phi_A$ is a subset of the space $A^*$ of continuous linear functionals on A; moreover, when equipped with the relative weak-* topology, $Phi_A$ turns out to be locally compact and Hausdorff. (This follows from the Banach-Alaoglu theorem.) The space $Phi_A$ is compact (in the topology just defined) if and only if the algebra A has an identity element.
Given $ain A$, one defines the function $widehat\left\{a\right\}:Phi_Ato\left\{mathbb C\right\}$ by $widehat\left\{a\right\}\left(phi\right)=phi\left(a\right)$. The definition of $Phi_A$ and the topology on it ensure that $widehat\left\{a\right\}$ is continuous and vanishes at infinity, and that the map $amapsto widehat\left\{a\right\}$ defines a norm-decreasing, unit-preserving algebra homomorphism from A to $C_0\left(Phi_A\right)$. This homomorphism is the Gelfand representation of A, and $widehat\left\{a\right\}$ is the Gelfand transform of the element $a$. In general the representation is neither injective nor surjective.
In the case where A has an identity element, there is a bijection between $Phi_A$ and the set of maximal proper ideals in A (this relies on the Gelfand-Mazur theorem). As a consequence, the kernel of the Gelfand representation $A to C_0\left(Phi_A\right)$ may be identified with the Jacobson radical of A. Thus the Gelfand representation is injective if and only if A is semisimple.
### Examples
In the case where $A=L^1\left(\left\{mathbb R\right\}\right)$, the group algebra $\left\{mathbb R\right\}$, then $Phi_A$ is homeomorphic to $\left\{mathbb R\right\}$ and the Gelfand transform of $fin L^1\left(\left\{mathbb R\right\}\right)$ is the Fourier transform $tilde\left\{f\right\}$.
In the case where $A=L^1\left(\left\{mathbb R\right\}_+\right)$, the L1-convolution algebra of the real half-line, then $Phi_A$ is homeomorphic to $\left\{ zin\left\{mathbb C\right\} : Re \left(z\right) geq 0 \right\}$, and the Gelfand transform of an element $fin L^1\left(\left\{mathbb R\right\}\right)$ is the Laplace transform $\left\{mathcal L\right\}f$.
## The C*-algebra case
As motivation, consider the special case $A=C_0\left(X\right)$. Given x in X, let $varphi_x in A^*$ be pointwise evaluation at x, i.e. $varphi_x\left(f\right) = f\left(x\right)$. Then $varphi_x$ is a character on $A$, and it can be shown that all characters of $A$ are of this form; a more precise analysis shows that we may identify $Phi_A$ with X, not just as sets but as topological spaces. The Gelfand representation is then an isomorphism $C_0\left(X\right)to C_0\left(Phi_A\right)$.
### The spectrum of a commutative C*-algebra
The spectrum or Gelfand space of a commutative C*-algebra A, denoted Â, consists of the set of non-zero *-homomorphisms from A to the complex numbers. Elements of the spectrum are called characters on A. (It can be shown that every algebra homomorphism from A to the complex numbers is automatically a *-homomorphism, so that this definition of the term 'character' agrees with the one above.)
In particular, the spectrum of a commutative C*-algebra is a locally compact Hausdorff space: In the unital case, i.e. where the C*-algebra has a multiplicative unit element 1, all characters f must be unital, i.e. f(1) is the complex number one. This excludes the zero homomorphism. So  is closed under weak-* convergence and the spectrum is actually compact. In the non-unital case, the weak-* closure of  is  ∪ {0}, where 0 is the zero homomorphism, and the removal of a single point from a compact Hausdorff space yields a locally compact Hausdorff space.
Note that spectrum is an overloaded word. It also refers to the spectrum σ(x) of an element x of an algebra with unit 1, that is the set of complex numbers r for which x - r 1 is not invertible in A. For unital C*-algebras, the two notions are connected in the following way: σ(x) is the set of complex numbers f(x) where f ranges over Gelfand space of A. Together with the spectral radius formula, this shows that  is a subset of the unit ball of A* and as such can be given the relative weak-* topology. This is the topology of pointwise convergence. A nets {fk}k of elements of the spectrum of A converges to f if and only if for each x in A, the net of complex numbers {fk(x)}k converges to f(x).
If A is a separable C*-algebra, the weak-* topology is metrizable on bounded subsets. Thus the spectrum of a separable commutative C*-algebra A can be regarded as a metric space. So the topology can be characterized via convergence of sequences.
Equivalently, σ(x) is the range of γ(x), where γ is the Gelfand representation.
### Statement of the commutative Gelfand-Naimark theorem
Let A be a commutative C*-algebra and let X be the spectrum of A. Let $gamma:A to C_0\left(X\right)$ be the Gelfand representation defined above.
Theorem. The Gelfand map γ is an isometric *-isomorphism from A onto C0(X).
See the Arveson reference below.
The spectrum of a commutative C*-algebra can also be viewed as the set of all maximal ideals m of A, with the hull-kernel topology (see the earlier remarks for the general, commutative Banach algebra case). For any such m the quotient algebra A/m is one-dimensional (by the Gelfand-Mazur theorem), and therefore any a in A gives rise to a complex-valued function on Y.
In the case of C*-algebras with unit, the spectrum map gives rise to a contravariant functor from the category of C*-algebras with unit and unit-preserving continuous *-homomorphisms, to the category of compact Hausdorff spaces and continuous maps. This functor is one half of a contravariant equivalence between these two categories (its adjoint being the functor that assigns to each compact Hausdorff space X the C*-algebra $C_0\left(X\right)$). In particular, given compact Hausdorff spaces X and Y, then C(X) is isomorphic to C(Y) (as a C*-algebra) if and only if X is homeomorphic to Y.
The 'full' Gelfand–Naimark theorem is a result for arbitrary (abstract) noncommutative C*-algebras A, which though not quite analogous to the Gelfand representation, does provide a concrete representation of A as an algebra of operators.
## Applications
One of the most significant applications is the existence of a continuous functional calculus for normal elements in C*-algebra A: An element x is normal if and only if x commutes with its adjoint x*, or equivalently if and only if it generates a commutative C*-algebra C*(x). By the Gelfand isomorphism applied to C*(x) this is *-isomorphic to an algebra of continuous functions on a locally compact space. This observation leads almost immediately to:
Theorem. Let A be a C*-algebra with identity and x an element of A. Then there is a *-morphism f → f(x) from the algebra of continuous functions on the spectrum σ(x) into A such that
• It maps 1 to the multiplicative identity of A;
• It maps the identity function on the spectrum to x.
This allows us to apply continuous functions to bounded normal operators on Hilbert space.
## References
• W. Arveson (1981). An Invitation to C*-Algebras. Springer-Verlag. ISBN 0-387-90176.
• Frank F. Bonsall, John Duncan (1973). Complete Normed Algebras. Springer-Verlag, New York.
• N. Wiener (1932). "Tauberian theorems". Ann. of Math. (2) 33 1-100.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 42, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8840477466583252, "perplexity_flag": "head"}
|
http://mathhelpforum.com/pre-calculus/3389-find-domain-print.html
|
# Find Domain
Printable View
• June 11th 2006, 06:44 PM
Kiwigirl
Solve equations
1. Give the range of the function y = tan^-1 x
2. Give the value of tan^-1 5
• June 11th 2006, 06:56 PM
ThePerfectHacker
1 Attachment(s)
Quote:
Originally Posted by Kiwigirl
$y\in (-\pi/2,\pi/2)$
The second way. The function $\tan^{-1}(x)$ is defined as the inverse function of the function,
$\tan x \mbox{ on } -\pi/2<x<\pi/2$. Now there is a rule that the range of the inverse function is the domain of the orignal function. Since the domain of the original function is $(-\pi/2,\pi/2)$ thus the range of $\tan^{-1}(x)$ is $(-\pi/2,\pi/2)$
There is actually a third way (much more advanced). Find all such $y$ such as the equation $y=\tan^{-1}(x)$ has a solution.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.914145827293396, "perplexity_flag": "middle"}
|
http://stats.stackexchange.com/questions/tagged/forecasting+machine-learning
|
# Tagged Questions
2answers
211 views
### Predicting time series with NNs: should the data set be shuffled?
Suppose I'm trying to predict time series with a neural network. The data set is created from a single column of temporal data, where the inputs of each pattern are ...
5answers
384 views
### How to handle online time series forecast?
I have been dealing with the following problem. I have kind of a real time system and every time frame I read its current value, creating a time series (such as 1, 12, 2, 3, 5, 9, 1, ...). I'd like to ...
1answer
125 views
### Lemonade consumption forecast
Some guys, decided to make some money, by selling lemonade in different public places. Each guy has a fix spot on which she sell the lemonade. In every morning, they go to a lemonade maker warehouse ...
6answers
738 views
### What algorithm could be used to predict consumables usage given data from past purchases?
Thinking about a supposedly simple but interesting problem, I'd like to write some code to forecast consumable I'll need in the near future given the full history of my previous purchases. I'm sure ...
2answers
390 views
### Machine learning techniques for time series estimation - forecasting price
Can anyone recommend any machine learning techniques for time series estimation? I have a series of times $t_{1}...t_{n}$, each having a set of associated features $f_{1}...f_{m}$, and a value $x$. ...
1answer
395 views
### Documented/reproducible examples of successful real-world applications of econometric methods?
This question might sound very broad, but here is what I am looking for. I know there are many excellent books about econometric methods, and many excellent expository articles about econometric ...
2answers
155 views
### How to predict Ozone concentration in few years time?
my friend is a chemist and his problem is to predict the level of ozone concentration in a single site. We have the data for the last 12 years. We want to predict the concentration for the coming ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9466308355331421, "perplexity_flag": "middle"}
|
http://motls.blogspot.com/2012/06/why-matrix-theory-contains-membranes.html?m=0
|
# The Reference Frame
## Thursday, June 07, 2012
... /////
### Why M(atrix) theory contains membranes
A week ago, I wrote a long article on Matrix theory. I planned a few shorter additions and this is the first one.
M-theory may stand for "mother", "magic", "mystery", "matrix", upside down "W" for "Witten" (this interpretation was discovered by Sheldon Glashow), but it may also stand for "membrane". And Matrix theory actually contains states that look like membranes. It has to contain such states because M-theory contains membranes and Matrix theory should be physically equivalent to M-theory.
The BFSS, discoverers of Matrix theory. I am only able to identify Willy Fischler, a part-time paramedic, in the middle.
In this text, I want to explain why the matrix model contains configurations (and quantum states) that look like two-dimensional sheets of a supersmooth atomless Lorentz-invariant charged paper. We will focus on membranes of spherical and toroidal topology but the conclusion is more general.
The insight that the matrix model describes membranes wasn't quite new in 1996. In fact, the matrix model was encountered a decade earlier in efforts to "discretize" a theory of membranes in order to make it as well-defined as a theory of strings. However, only in 1996, BFSS were able to find out that the matrix model is actually an exact description of the M-theoretical membranes as well as everything else. They also resolved some puzzling issues about degenerate shapes of membranes and their topology change.
All this discussion is linked to noncommutative geometry, a subdiscipline of mathematics whose physically meaningful part has been tightly incorporated (by Nature) into string/M-theory. The particular constructions of noncommutative geometry resembling a sphere and a torus are known as the fuzzy sphere and fuzzy torus, respectively.
The Hamiltonian of Matrix theory
We will begin with a particular form of the Matrix theory Hamiltonian:\[
P^- \equiv H = R\cdot {\rm Tr} \zav { (\Pi^i)^2 - ([X_i,X_j])^2}, \quad P^+ = \frac{N}{R}
\] I am neglecting coefficients of order one and I am neglecting Yukawa-like terms with the fermions. The fermionic matrices $\theta^a$ where $a$ takes 16 possible values give you new degrees of freedom; but these degrees of freedom are also necessary to cancel some terms that would cause inconsistencies such as the "zero-point energies" arising from the off-diagonal modes – as I mentioned in the first blog entry on Matrix theory. So if you do a quantum calculation involving the bosons only and you run into some pathological quantum effect that shouldn't be there, chances are that this effect would be cancelled if you added the fermions.
Note that the light-like components of the spacetime momentum, $P^+$ and $P^-$, include the factors of $1/R$ and $R$, respectively. The quantity $R$ may be rescaled by a multiplicative factor which is nothing else than a Lorentz boost: if we increase $P^+$ by a factor and reduce $P^-$ by the same factor, it's the same operation as if we Lorentz transform or boost the $P^0$-$P^{10}$ plane.
Ultimately we want to describe objects with finite, continuous values of $P^+$ and $P^-$ in an eleven-dimensional spacetime. So we have to take $R$ to infinity to achieve the continuity while $N/R$ is kept fixed. Because the Hamiltonian $P^-$ has to be finite as well, the trace in it has to scale as $1/N$ for $P^+ P^-$ to stay finite. So only the states of the large $N$ matrix models whose energy scales like $1/N$ i.e. drops appropriately with $N$ as $N\to\infty$ are relevant for the decompactified M-theory.
Because we have understood the factor $R$ or $1/R$ in various quantities, we may ignore it and effectively set $R=1$, a full-fledged light-like compactification in which the periodicity of the spacetime coordinate $X^-$ may a priori be very visible. The mathematical essence of the matrix model involves\[
P^- \equiv H = {\rm Tr} \zav { (\Pi^i)^2 - ([X_i,X_j])^2}, \quad P^+ = N
\] That's very simple: the kinematic, quantized longitudinal light-like momentum $P^+$ is simply $N$, the size of the matrices, while the dynamical light-like momentum $P^-$ that we treat as the Hamiltonian is the trace of the kinetic term and the quartic potential term without any additional factors (except the factors of order one that we neglected). And we're only interested in the large $N$ models and low-energy states of $P^-$ that scale like $1/N$. The states we care about must kind of exist for each value of $N$ and their energy has to go down with an increasing $N$. Quite generally, the dualities mapping the model to M-theory imply that in the large $N$ limit, the model has to become $N$-independent up to the simple scalings.
A detail: note that I wrote a minus sign in front of the quartic, commutator term. That's because the commutator (computed only from matrices, regardless of whether the entries are classical or quantum observables!) of two Hermitian matrices is antihermitian and its square (which we want to sum over $i,j$) is negatively definite. The minus sign is needed for the positivity.
Minimizing the Hamiltonian
Look at the Hamiltonian above classically. How do we make it small for large $N$? In the quantum theory, we're constrained by the uncertainty principle so if $X^i$ are too sharply defined, then $\Pi^i$ is highly uncertain and we will get a high contribution from the kinetic energy $\Pi^2$. To minimize the total energy, we really have to make both terms small, including the potential energy (think about the virial theorem and the way how the Hydrogen atom minimizes the energy in a "balanced way").
I just wanted to say something that is intuitively obvious. The commutator term $[X^i,X^j]^2$ has to be small, too. How do we make it small? Note that it is the sum of the traces of all the squared commutators of pairs of those nine $X^i$ matrices. In fact, the trace may be obtained as follows: compute all the $9\times 8/2\times 1 = 36$ commutators of the matrices and sum the squared absolute values in all these $36$ commutator-matrices.
Well, a simple way to reduce this term is to make all the commutators vanish,\[
\forall i,j:\quad [X^i,X^j] = 0
\] It means that all the matrices $X^i$ may be simultaneously diagonalized. But such a state has a simple interpretation. We have seen in the previous article that block-diagonal matrices describe states with several independent, separated objects. Fully diagonal matrices are an extreme example which is composed from gravitons carrying the minimal unit value of $P^+$; gravitons with other, larger values of $P^+=N$ may be interpreted as larger blocks in which $X^i$ are proportional to the unit matrix, i.e. conglomerates of coincident and overlapping minimal gravitons. (In the quantum theory, there exists exactly one zero-energy ground state wave function for the $SU(N)$ problem describing all the "relative" coordinates between the D0-branes. There also exist lots of nonzero energy states that are similarly localized and they generically describe black hole microstates.)
But that's not new. We have already discussed the decomposition into blocks. We want some nontrivial solution in which the commutators aren't strictly zero. We want a new solution in which the configuration of matrices is irreducible; it can't be decomposed to smaller objects via the block diagonal decomposition. Can we find a solution? Yes. A beautiful class of such configurations is physically identified as membranes, 2-dimensional submanifolds floating in the 9-dimensional transverse space parameterized by the coordinates $X^i$.
(The locations of the points on the membrane in the remaining, tenth "spatial" coordinate $X^-$ is obtained by a Fourier transform because we know how the complementary momentum $P^+$ is uniformly divided among the bits of the membrane. I don't want to get into this technicality here but only the nine "purely transverse" coordinates are truly independent and physical here.)
Fuzzy torus
We really want to find values of matrices $X^i$ such that they're of order one but they're large matrices whose size is $N$ and all the commutators of these matrices, while nonzero, naturally scale like $1/N$. Let me just immediately give you a solution. Consider the matrices\[
U = \pmatrix{1&0&0&\cdots& 0 \\
0&\omega&0&\cdots &0\\
0&0&\omega^2&\cdots &0\\
\vdots & \vdots&\vdots & \ddots& \vdots\\
0&0&0&\cdots&\omega^{N-1}
}, \quad \omega\equiv e^{2\pi i/N}
\] and \[
V = \pmatrix{0&1&0&\cdots& 0 \\
0&0&1&\cdots &0\\
0&0&0&\cdots &0\\
\vdots & \vdots&\vdots & \ddots& \vdots\\
1&0&0&\cdots&0
}.
\] The matrix $U$ is diagonal and the diagonal entries are the $N$-th roots of unity. All of the possible $N$ roots appear on the diagonal, in a kind of nicely ordered way, much like when a clock hand is showing the current time. As you circle around the diagonal, the complex unit is going from one midnight to another midnight (or is it a noon?). That's why $U$ is known as the clock operator.
Analogously, $V$ is the cyclic permutation matrix acting on the $N$ basis vectors, it's the so-called shift operator. If you think about it, the eigenvalues of $V$ are the $N$-th roots of unity, too. The eigenvectors will remind you of the (discrete) Fourier transform. All these comments mean that $U$ and $V$ are actually similar to each other. There exists a unitary matrix $M$ such that $U=MVM^{-1}$. This matrix $M$ is actually the defining matrix of the discrete Fourier transform and you may pick\[
M_{ab} = \frac{\omega^{ab}}{\sqrt{N}}.
\] Pure phases bilinearly depending on both indices $a,b=1,2,\dots, N$. Sorry if there should be $-ab$ in the exponent. I didn't have to add the $1/\sqrt{N}$ factor but it's helpful if you want $M^{-1}$ to be similar to $M$, including the normalization.
A funny thing is that $U,V$ don't commute with one another but they're damn close to commuting. Why don't they commute? Well, if you first decorate $N$ coordinates of a complex vector by the clock phases and then you cyclically permute them, it's different from permuting them at first and then decorating them with the clock phases. It's because you need the clock phases shifted by one i.e. by the factor of $\omega$.
At any rate, if you understand the previous sentence or if you compute the products explicitly, you will find out that\[
UV = VU \cdot \omega
\] Sorry if there should be $\omega^{-1}$ here; I don't want to waste a minute here, it's an exercise for you. The products are the same up to a factor of $\omega$. The previous relationship may be rewritten as\[
UV-VU =(\omega-1) VU.
\] But $\omega=\exp(2\pi i/N)$ and for a large value of $N$, it's very close to a one. Keeping the first subleading term in the expansion of the exponential, we see that\[
UV-VU \sim \frac{2\pi i}{N} VU
\] Here, $VU$ on the right hand side is "of the same order" as the normal matrices and doesn't depend on $N$. However, there's an extra factor of $1/N$. That's exactly what we need for the low-energy states in Matrix theory. All the previous insights mean that if all the matrices $X^i$ are simple functions of $U,V$ and their inverses, the commutators will be small. They will scale as $1/N$ which, by the way, will also produce $1/N$ terms after we square the commutators – that will change $1/N$ to $1/N^2$ – and trace over the matrices – that will add a factor of $N$ again, thus boosting $1/N^2$ to $1/N$.
That's great. What the form of the matrices is? It is\[
X^i = \sum_{k,l=-T}^{T} \xi^i_{k,l} U^k V^l
\] with lots of coefficients $\xi^i_{k,l}$. I have truncated the sum to an interval of integers between $-T$ and $T$. To decompose the most general matrix $X^i$, we need $T$ to be approximately $N/2$ so that each sum over $k$ and $l$ goes exactly over $N$ possible values. In such a case, the sum will have $N^2$ independent terms which is exactly what you need to reconstruct a general $N\times N$ matrix.
However, imagine that $T$ is taken to be much smaller than $N/2$. Imagine that $N$ is one million but we will only take $T\sim 1,000$, summing a much smaller number of terms.
Now, what is the physical interpretation of the polynomial formula for $X^i$ above? The interpretation is easily obtained if we make the following replacement or identification:\[
U\equiv \exp(i\sigma_1), \quad V\equiv \exp(i\sigma_2)
\] Note that the eigenvalues of $U,V$ are almost "any" complex numbers whose absolute value equals one (if $N$ is large) so it is legitimate to say that they may be written as the complex exponential of a phase. The angles $\sigma_1,\sigma_2$ are two independent variables with periodicity $2\pi$. They clearly parameterize a torus. In the normal geometry, they commute with each other. In our geometry, the commutator is nonzero but small.
So the formula for $X^i$ written in terms of $U,V$ is nothing else than a Fourier decomposition of a function $X^i(\sigma_1,\sigma_2)$ on a torus! If the configuration of $X^i$ matrices is close to our Ansatz, the matrices know exactly everything about the shape of a membrane in a 9-dimensional space! The membrane has a toroidal topology. In some sense, it resembles a periodic phase space with $N^2$ phase cells.
If you rewrite the Hamiltonian for the matrices $X^i$ as a Hamiltonian for the non-matrix functions on the torus, $X^i(\sigma_1,\sigma_2)$, you will get nothing else than a Hamiltonian for a membrane (a supersymmetric membrane if you also include the fermions and do the analogous operations for them). This Hamiltonian for the membrane will look like a higher-dimensional generalization of the Hamiltonians for a string. The usual quadratic Hamiltonians for strings may be obtained as a clever rewriting of the Nambu-Goto action (the proper area of the world sheet); and in the same way, the Hamiltonian we get for the membranes are also physically identical to a higher-dimensional generalization of the Nambu-Goto action (the volume of the world volume).
(Instead of the conformal gauge, i.e. the condition that the string's world sheet metric is flat up to a position-dependent Weyl rescaling, the natural gauge condition we automatically get in the membrane case is the condition that the determinant of the induced 2+1-dimensional membrane metric is constant. So no scaling or conformal symmetry is allowed for the membranes. In this way, the gauge-fixed Hamiltonian contains quartic terms, and not just "free quadratic" terms such as the stringy ones, and many things about the membrane are more complicated. They're still pretty important and mathematically natural – after all, the BFSS matrix model may have been obtained from the maximally supersymmetric gauge theories etc.)
Membranes have been considered and for a while, people thought that you could treat them just like strings, obtaining a higher-dimensional generalization of string theory. However, this "membrane theory" didn't quite enjoy the same remarkable properties as string theory. The higher-dimensional world volume theory was less well-behaved at short distances; the topology of the membranes could degenerate more uncontrollably than the strings' topology; the constant quantifying the strength of the interactions between the membranes wasn't really adjustable so it couldn't be made small.
I will return to these questions but before I do, let's ask: is that just the toroidal topology or can we explicitly describe membranes of another shape, e.g. spherical ones? The answer is Yes.
Fuzzy sphere
In the fuzzy torus case, we made the identification\[
U\equiv \exp(i\sigma_1), \quad V\equiv \exp(i\sigma_2)
\] of two unitary matrices $U,V$ with two coordinates which are complex numbers of absolute value one (equivalently, two independent angular variables). The commutator of the matrices was scaling like $1/N$ which was fine, and so on. Can we do something similar for the sphere? Yes, the answer is the fuzzy sphere.
We will describe the $r=1$ unit sphere not by $\theta,\phi$, the spherical coordinates, but by three coordinates $x,y,z$ that just satisfy\[
x^2+y^2+z^2=1.
\] Because we will require the relationship above to hold automatically, one of the coordinates will actually fail to be quite independent from the other two (except for its adjustable sign). By using $x,y,z$, we will avoid the problems with the singular poles, $\theta=0$ and $\theta=\pi$, where new checks and balances would have to be imposed.
Now, $x,y,z$ play exactly the same role as $\exp(i\sigma_1)$ and $\exp(i\sigma_2)$ in the fuzzy torus case. In the fuzzy torus case, we identified them with two matrices. So the analogous operation for the fuzzy sphere must find $N\times N$ matrices representing the coordinates $x,y,z$. Moreover, these matrices have to satisfy $x^2+y^2+z^2=1$ automatically. Can we find them?
Yes, we can. Just make the following identification:\[
(x,y,z) = \frac{1}{\sqrt{J(J+1)}} (J_x,J_y,J_z).
\] Here, $J=(N-1)/2$ so that $N=2J+1$ gives us the right size of the matrices and $J_x,J_y,J_z$ are simply matrices of the angular momentum in an $N$-dimensional irreducible representation, i.e. in a representation with a large value of the spin $J$. Note that $x^2+y^2+z^2=1$ is identically satisfied if $x,y,z$ represent the associated matrices simply because \[
J_x^2+J_y^2+J_z^2 = J(J+1)\cdot 1_{N\times N}
\] is the well-known "Casimir" that is proportional to the unit matrix. Note that in this case, we associated Hermitian, and not unitary, matrices to $x,y,z$ because unlike $\exp(i\sigma_{1,2})$, the variables $x,y,z,$ are real.
Now, you may verify that for a large $N=2J+1$, the commutators such as $[x,y]$ go like $1/N$ times $iz$ because $\sqrt{J(J+1)}\sim J$ for a large $J$ or $N$ so we pick $1/J$ from each $x$ and $y$, getting $1/J^2$ in total which is equal to $1/J$ times $z$ because $z$ contains one $1/J$ by itself.
Just like we could write $X^i$ as a simple Fourier-transform-like function of $U,V$ in the toroidal case, we may write each of the nine $X^i$ matrices as a simple polynomial of the matrices that represent $x,y,z$. In this fashion, you may approximate any spherical harmonic or any function on the $x^2+y^2+z^2=1$ surface, on the two-sphere.
So the matrices $X^i$ may also naturally and beautifully encode a function on the two-sphere, via a Fourier-like decomposition that is closely associated with the decomposition into spherical harmonics. For a finite $N$, it is not a perfect approximation: we only allow spherical harmonics up to $L\sim N$ or so. Also, the multiplication is a bit noncommutative. If you analyze the maths carefully, you will find out that all the refusal of the matrices to commute may be explained by a totally equivalent feature of the multiplication: the multiplication of the matrices representing spherical harmonics simply eliminates all the spherical harmonics in the product that have $L\gt N$ or so.
Dynamics, degeneration, topology change
I have mentioned that one may rewrite the Matrix theory Hamiltonian as a 2+1-dimensional Hamiltonian for functions such as $X^i(\sigma_1,\sigma_2)$ in the toroidal case or $X^i(\theta,\phi)$ in the spherical case, if I rewrite $x,y,z$ using the polar coordinates again. The local integrand of these Hamiltonians (which dictates how the waves are propagating along the membranes, and related things) will be the same – for any topology, we will derive the same local dynamics on a piece of a membrane – so it's not hard to guess that the Hamiltonian of the matrix model actually allows orientable membranes of any topology.
Only the torus and the sphere can be written down very explicitly; that's because of their isometries, $U(1)\times U(1)$ and $SO(3)$, respectively. However, membranes of all topologies are allowed in the matrix model (and with some more abstract maths, we may describe them somewhat explicitly, too). It's not hard to figure out that the matrix model allows the topology to change. After all, the space of matrices is completely connected; it isn't really decomposed into several isolated "sectors with a different topology".
The same is true for single-membrane and multi-membrane states.
Imagine that someone in the 1980s wrote the continuous Hamiltonian – a 2+1-dimensional quantum field theory – for a membrane of a spherical topology. Yes, several people did. And now imagine that they had the creativity to regulate the membranes using the matrices. Well, they even did that. What they found was a problematic model because it didn't prevent the spherical membrane from developing sharp spikes on the surface and it was plausible that these spikes get connected and change the topology of the previously spherical membrane into a torus or a higher-genus topology. Or the sphere could get pinched near the equator and the thin tube connecting the previous hemispheres could have broken, thus producing a pair of spherical membranes.
Those folks in the 1980s were not able to deduce what was really happening in these extreme limits and whether the membrane was allowed to do so; these questions boiled down to some short-distance properties of the 2+1-dimensional model that weren't well-defined because 2+1 dimensions is already too high for a world volume theory to be fully consistent in the ultraviolet.
But if you rewrite the spherical-membrane Hamiltonian in terms of the matrix model, you will be able to find the answer to the question whether the degeneration processes above are allowed etc. They are allowed. The reason is that the correct and correctly evaluated and interpreted matrix regularization of the single-spherical-membrane model actually describes membranes of arbitrary orientable topologies – and an arbitrary number of disconnected membranes, too!
This is just another manifestation of the remarkable feature of the matrix models that gave the title to the first blog entry on Matrix theory. In quantum field theory using the second quantization, the multi-particle and other multi-object states have to be created "out of pieces", i.e. as a product of several independent creation operators acting on the vacuum state. But Matrix theory erases the qualitative difference between single-particle and multi-particle objects – and between single-membrane and multi-membrane states. All of them are represented by certain configurations of matrices (or, quantum mechanically, by wave functions whose support is located near these classical configurations). The multi-object states correspond to classical configurations that may be simultaneously block-diagonalized.
(Matrix string theory, a matrix model that also describes strings in string theory, also erases the qualitative difference between one-string and multi-string states, a feature I had believed to be fundamentally necessary in a complete formulation of string theory at least since 1992 when I read the first Green-Schwarz light-cone-gauge superstring articles from the early 1980s. Matrix theory just came to me as a marvelous framework in which my "dreams of the teen age" could have been transformed into a working reality which was a reason why I could explain how strings were included in Matrix theory much more quickly than anyone else, within a week of thinking about Matrix theory. The feeling when you discover something relatively important – the first nonperturbatively complete definition of type II strings' dynamics and why it agrees with the previous approximate descriptions – before anyone else in the world and when you're sure that it actually works is special. I will postpone the discussion of matrix string theory and some thoughts I had when I discovered it to a future blog entry.)
However, the realistic wave functions in the quantum mechanics are nonzero for configurations of diagonalizable as well as non-diagonalizable matrices. So the interactions between the membranes, gravitons, and other objects – including the splitting of membranes into pieces and changing the membrane topology – is always predicted to occur with a nonzero probability. You can't eliminate them from the picture. In perturbative string theory, the string interactions are uniquely determined once you specify the single-string dynamics; the local character of the world sheet remains the same and you just allow more complicated topologies.
Now we see that something analogous holds for membranes, too. String/M-theory, in this case its matrix model definition, uniquely dictates how the membranes should split, merge, degenerate, and simplify once you write a consistent regularization of the theory for a single membrane of any topology. In as consistent and robust theories as string/M-theory – and string/M-theory is probably the only element in this set – learning about a small piece of the structure (e.g. the free limit describing a single object) uniquely determines everything else (including interactions). Once you find a small (but not too small) piece of a mosaic, the whole mosaic may be "holographically reconstructed" out of it.
Noncommutative geometry and noncommutative field theories: promotion of a future text
The fuzzy torus and the fuzzy sphere above are the most canonical and simplest examples of noncommutative geometry. Studied in very abstract mathematics by folks such as Alain Connes, the noncommutative geometry was found to be a natural exact description of some pretty elementary configurations in string theory. There exist whole quantum field theories in which the supporting spacetime is noncommutative; they may be obtained by replacing the ordinary commutative point-wise product of classical fields by the star-product.
The Feynman diagrams for these theories have extra phases associated with vertices; these theories relate $d$-dimensional theories (and branes) with $(d-2)$-dimensional and $(d-2k)$-dimensional theories (and branes) in natural ways; and – which is related – they have very natural and interesting solutions (solitons) that use some mathematical tricks we know from phase spaces in quantum mechanics (although their interpretation is different). This text has gotten pretty long so I will postpone the discussion of noncommutative geometry to a future blog entry.
Posted by Luboš Motl
|
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 176, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9303049445152283, "perplexity_flag": "head"}
|
http://math.stackexchange.com/questions/173530/unprovable-statements-in-zf
|
# Unprovable statements in ZF [duplicate]
Possible Duplicate:
Advantage of accepting the axiom of choice
Advantage of accepting non-measurable sets
As you all know, Banach-Tarski paradox is solely a consequence of Axiom of Choice, and I think it is just absurd.
I'm trying to take ZF as my axiomatic model rather than ZFC. I wonder if there are some theorems really important understanding our 'Number system' that can be proved in ZFC, but not in ZF. (i.e., every vector space has a basis is equivalent to axiom of choice, but i think we actually don't need this strong theorem since we are always working on finite dimensions (maybe not! Please tell me if there are some branches of mathematics studying infinite dimension))
I wonder how many risks should I take when I'm removing AC.
-
– Gerry Myerson Jul 21 '12 at 8:41
2
– Martin Sleziak Jul 21 '12 at 8:43
4
Function spaces are (generally) infinite-dimensional. – Gerry Myerson Jul 21 '12 at 8:44
1
@Katlus: When you use a "standard" mathematical result, then, in many fields, there is a significant risk that you will use a result that in fact uses AC. And, as happened to some analysts who were early opponents of AC, you may end up using AC without being aware of it. It really is a "natural" principle of reasoning. But apart from this, there are no risks. The theorems one proves are just less general. – André Nicolas Jul 21 '12 at 20:32
## marked as duplicate by Asaf Karagila, Zhen Lin, Gerry Myerson, Henning Makholm, t.b.Jul 21 '12 at 23:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
The axiom of choice, while seemingly having counterintuitive results is needed to ensure that infinite sets are well-behaved.
Of course if you would only like to work with finite sets then the axiom of choice is not needed, however some things which you think hold immediately would fail without the axiom of choice, and they may fail badly.
• The union of countably many pairs may not be countable.
• The real numbers may be a countable union of countable sets.
• We may be able to partition a set into more parts than elements, in particular this set might be the real numbers.
• In the real numbers continuity by sequences and by $\varepsilon$-$\delta$ are no longer equivalent.
• Topology breaks down in acute manners, to horrid to begin to describe.
These three are far far more disturbing to me than Banach-Tarski, and these would hold in several models of ZF without the axiom of choice.
What more? Let's see, what else can fail badly
• There might be no free ultrafilters, on any set.
• In turn some fields might not have an algebraic closure; others could have two non-isomorphic closures, for example the rationals.
• There could be a tree that every point has a successor, but there is no $\omega$-branch.
• There could be a vector space which has two bases of different cardinality.
• Functional analysis may stop working due to lack of Hahn-Banach, Krein-Milman, Banach-Alaoglu theorems.
If you wish to do some set theory, perhaps, it also becomes hard:
• Cardinal arithmetics can fail for infinite sums and products.
• In forcing the mixing lemma fails.
• The partial ordering of cardinalities is not necessarily well-founded.
• There may be no canonical representative for $|A|$, namely a function which returns a particular set of the cardinality of $A$ (like the $\aleph$ numbers).
This list can be made really quite infinite. Why? Because modern mathematics is very much about infinitary objects and for those to be well-behaved we really need the axiom of choice, or else a lot of bad things may occur.
It is also the case that most people are educated by choice-using people, so the basic intuitions about mathematics actually use the axiom of choice a lot more than you would think.
However there are still merits to working without the axiom of choice. For those, see my recent post: Is trying to prove a theorem without Axiom of Choice useless?
To read even more:
-
I couldn't hold myself, despite voting to close - I still felt I had more to say. – Asaf Karagila Jul 21 '12 at 9:16
I suppose you could have added it to the earlier question. – Gerry Myerson Jul 21 '12 at 9:18
@Gerry: Which one? :-) With these sort of answers I usually stop at some point, otherwise I'll write a book in the answer. – Asaf Karagila Jul 21 '12 at 9:27
I was surprised that you left out that a product of nonempty sets might be empty. – MJD Jul 21 '12 at 12:36
@Mark, well that is actually less useful in a direct way to mathematics, since most sets you encounter are such that they ensure someone the product is non-empty (e.g. vector spaces; open sets; etc.) furthermore the question suggested "fine, we can do without basis for vector spaces" so writing "products of sets may be empty" is kinda moot in my eyes. – Asaf Karagila Jul 21 '12 at 12:38
show 9 more comments
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9506241679191589, "perplexity_flag": "middle"}
|
http://stats.stackexchange.com/questions/46826/influential-residual-vs-outlier
|
# Influential residual vs. outlier
First, I should state that I have searched on this site for the answer. I either didn't find a question that answered my question or my knowledge level is so low I didn't realize I already read the answer.
I am studying for the AP Statistics Exam. I have to learn linear regression and one of the topics is residuals. I have a copy of Introduction to Statistics and Data Analysis on page 253 it states.
Unusual points in a bivariate data set are those that fall away from most of the other points in the scatterplot in either the $x$ direction or the $y$ direction
An observation is potentially an influential observation if it has an $x$ value that is far away from the rest of the data (separated from the rest of the data in the $x$ direction). To determine if the observation is in fact influential, we assess whether removal of this observation has a large impact on the value of the slope or intercept of the least-square line.
An observation is an outlier if it has a large residual. Outlier observation fall far away from the least-square line in the $y$ direction.
Stattreck.com states four methods to determine an outlier from residuals:
Data points that diverge in a big way from the overall pattern are called outliers. There are four ways that a data point might be considered an outlier.
1. It could have an extreme X value compared to other data points.
2. It could have an extreme Y value compared to other data points.
3. It could have extreme X and Y values.
4. It might be distant from the rest of the data, even without extreme X or Y values.
These two sources seem to conflict each other. Could anyone help clear up my confusion. Also, how does one define extreme. The AP Statistics uses the rule if the data point is outside of (Q1-1.5IQR,Q3+1.5IQR) the it is an outlier. I don't know how to apply that from just a graph off the residuals.
-
## 2 Answers
The stattrek site seems to have a much better description of outliers and influential points than your textbook but you've only quoted a short passage that may be misleading. I don't have that particular book so I cannot examine it in context. Keep in mind though, that the textbook passage you quoted says, "potentially". It's not exclusive either. Keeping those points in mind, stattrek and your book don't necessarily disagree. But it does appear that your book is misleading in the sense that it implies (from this short passage) that the only difference between outliers and influential points is whether they deviate on x or y axis. That is incorrect.
The "rule" for outliers varies depending on context. The rule you cite is just a rule of thumb and yes, not really designed for regression. There are a few ways to use it. It might be easier to visualize if you imagine multiple y-values at each x and examining the residuals. Typical textbook regression examples are too simple to see how that outlier rule might work, and in most real cases it is quite useless. Hopefully, in real life, you collect much more data. If it's necessary that you may be applying the quantile rule for outliers to a regression problem then they should be providing data for which it is appropriate.
-
Thanks for the answer, it just gets annoying that different books try to state these rules without really saying it honestly depends on the data, as you are saying. – MaoYiyi Jan 2 at 16:23
Actually, I stated it wrong as well... it depends on theory, method, and data... the entire study. – John Jan 2 at 17:58
I agree with John. Here are a few more points. An influential observation is (strictly) one that influences the parameter estimates. A small deviation in the Y value gives a big change in the estimated beta parameter(s). In simple regression of 1 variable against another, influential variables are precisely those whose X value is distant from the mean of the X's. In multiple regression (several independent variables), the situation is more complex. You have to look at the diagonal of the so called hat matrix $X(X'X)^{-1}X'$, and regression software will give you this. Google "leverage".
Influence is a function of the design points (the X values), as your textbook states.
Note that influence is power. In a designed experiment, you want influential X values, assuming you can measure the corresponding Y value accurately. You get more bang for the buck that way.
To me, an outlier is basically a mistake - that is, an observation that does not follow the same model as the rest of the data. This may occur because of a data collection error, or because that particular subject was unusual in some way.
I don't much like stattrek's definition of an outlier for several reasons. Regression is not symmetric in Y and X. Y is modelled as a random variable and the X's are assumed to be fixed and known. Weirdness in the Y's is not the same as weirdness in the X's. Influence and outliership mean different things. Influence, in multiple regression, is not detected by looking at residual plots. A good description of outliers and influence for the single variable case should set you up to understand the multiple case as well.
I dislike your textbook even more, for the reasons given by John.
Bottom line, influential outliers are dangerous. They need to be examined closely and dealt with.
-
Your dislike of the stattrek regression explanation is appropriate if you come from a background where true experiments are the norm. Your reasons all apply there. But if you come from a background where quasi-experimental designs are more common then the stattrek site has more relevance. In those cases both x and y values are often just random samples. – John Jan 3 at 5:52
@John how about the background of wanting to pass the AP Statistics Exam? What is quasi-experimental design? Is that using a random number table for a simulation? – MaoYiyi Jan 3 at 7:41
I don't know anything about the AP statistics exam. True experiments are ones where you manipulate the predictor variable and make groups to test multiple hypothesis or control and experimental groups, etc. Quasi-experimental designs are pretty much anything else that looks like an experiment. So, imagine a regression where the x value is weight and the y value is some sport skill. You don't manipulate either variable, you randomly sample both. So, Placidia's criticisms of stattrek are quite valid for true experiments but not as much so for quasi. – John Jan 3 at 12:29
@John ... I do come from a background where designed experiments are seen as the gold standard. In practice, I know that X and Y are often both random samples, which begs the question of why regression is being used, and not some form of latent variable analysis. – Placidia Jan 3 at 14:17
When you've only got two variables... :) Sometimes you have good theory to suggest one thing predicts another, for example, height and probability of getting into the NBA... both random samples. In cases with one, or a few (especially uncorrelated) linear relationships regression is good. – John Jan 3 at 14:49
show 3 more comments
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9369058609008789, "perplexity_flag": "middle"}
|
http://physics.stackexchange.com/questions/4404/time-dilation-what-happens-when-you-bring-the-observers-back-together?answertab=votes
|
# Time Dilation - what happens when you bring the observers back together?
I'm having trouble getting my head around the time dilation paradox.
Observer A and B are at the same "depth" in a gravity well. Observer B then descends into the well. A will observe B's time as going slower than their own. B will observe A's time as going faster than their own.
What happens if B were to ascend the well back to A's depth, would B's local time speed back up to the same rate as A's, but B would be younger (relative to A)?
What about the paradox caused by relative motion (ignoring gravity)? If A is moving relative to B, A and B will both observe the other's time as going slower. If A and B were together initially, then B moves away and returns, do their clocks agree? they can't both be younger than each other :s (i get thats the paradox, but what explanation resolves it?)
Thanks
-
## 3 Answers
Calculations show that younger will be the observer who suffered accelerations/decelerations.
-
is there a lay explanation as to why? – Andrew Bullock Feb 1 '11 at 22:19
I do not think so. Such is our reality. – Vladimir Kalitvianski Feb 1 '11 at 22:23
1
I wish we could vote down comments! Yes, accelerated twin will be younger, but don't physics would be ridiculous if there was no explanation for that! I guess you can find your answer in any standard text on special relativity! Just go and look at the chapter on accelerated reference frames and see how time dilation goes there! – iii Feb 2 '11 at 1:26
1
@Sina Salek: there is an explanation, but I don't know how well it can be qualified as a "lay explanation". In a way, Vladimir is right, it's just the way nature is. But he could have added what it is that is more fundamental than the passage of time. Well, turns out that there is a speed that is invariant. It is this fact combined with the relativity of motion that forces us to the conclusion that time is not absolute and that different observers will record different times. It is a fact borne out by experience and the most basic explanationis invariant speed $c$ + relativity. – Raskolnikov Feb 2 '11 at 10:07
– Andrew Bullock Feb 2 '11 at 16:30
show 1 more comment
Answer to the first question: yes, B will be younger than A.
-
As I've been brought to understand this - the person who has descended down the gravity and then returned has not been accelerating at a constant rate throughout the journey (the change in direction requires a change in acceleration), and this causes the differential aging.
This is the same thing for the original twin paradox where one of them travels at near light speed and then returns. This may be oversimplifying things - but you seem to be after a lay explanation.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9625318050384521, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/77152/increasing-union-of-contractible-cw-complexes/77155
|
## Increasing union of contractible CW complexes
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let X be CW complex. I'm trying to prove (using Zorn's lemma) that there is maximal contractible subcomplex. Problem is that I'm not able to show that increasing union of contractible subcomplexes has to be contractible itself.
-
## 1 Answer
By various standard lemmas, a CW complex $X$ is contractible if and only if every map $u:S^{n-1}\to X$ (for any $n>0$) can be extended over $B^n$. In this context $u(S^{n-1})$ is compact and therefore (by another standard lemma) contained in some subcomplex with only finitely many cells. If $X$ is the union of some totally ordered family of subcomplexes $X_\alpha$, it follows that $u(S^{n-1})\subseteq X_\alpha$ for some $\alpha$. This is enough to prove what you want.
-
1
Calling Whitehead's theorem a standard lemma is a slight understating :) – Mariano Suárez-Alvarez Oct 4 2011 at 18:13
Saying "Whitehead's theorem" is almost as misleading (I believe Henry had proved at least two reasonably well-known results). – Igor Rivin Oct 4 2011 at 20:42
So for short: $\pi_n$ commutes with increasing unions in consideration and detect contractibility by Whitehead. – Martin Brandenburg Oct 4 2011 at 21:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.919773280620575, "perplexity_flag": "middle"}
|
http://www.physicsforums.com/showthread.php?t=511174
|
Physics Forums
Blog Entries: 1
Recognitions:
Gold Member
Science Advisor
Staff Emeritus
## What is Heat?
Written by Hootenanny. Edited by berkeman and Kurdt. Feel free to contact me with any questions, comments or corrections.
What is Heat?
One of the most frustrating misconceptions in Thermodynamics centres on the actual definition of heat. Many science/engineering students often refer to a body as to be possessing heat, but as we shall see, to do so is completely nonsensical.
We start by examining the first law and defining each of it’s terms. The first law is simply a statement of the principle of conservation of energy and is often stated thus;
$$\Delta U = Q + W$$
Where $\Delta U$ is the change in internal energy, Q is the heat added to the system and W is the work done on the system. So what do all these terms actually mean?
Internal Energy
We define the internal energy as the energy associated with the microscopic energies of system, that is with the energy associated with the random motion of the molecules within a system. So for a general fluid, the internal energy of a system is the sum of the translational kinetic energies, the rotational kinetic energies, the vibrational kinetic energies and the potential energies of all the molecules in that system. The internal energy of a system is often erroneously referred to as the heat of a system and we shall see why this is incorrect later. One important point to note here is that the internal energy is a state variable, that is, the change in internal energy between any two states is independent of the path taken.
Work
Well, if you're reading this I assume that you know the definition of work; in thermodynamics work is usually associated with a transfer of energy into or out of a system. An example of work specific to thermodynamics would be the application of a force to a piston, which would then compress the gas within the cylinder, thus doing work on the gas. Since work is being done on the gas the W term in our expression would be positive. If we assume that the walls of the cylinder are adiabatic (no heat transfer) then all the work done would be converted to internal energy. Suppose that after we have compressed the piston, we release it. Intuitively, we would expect the piston to recoil back, and this is exactly what happens; the gas expands and does [an equal amount of] work on the piston against atmospheric pressure. In this case, since it is the gas that is doing work, our W term would be negative.
Heat
So we have defined the internal energy of a system and we can quantify the work done on the system, but what about heat? First let us examine temperature. One useful definition of temperature is often called kinetic temperature and is derived from kinetic theory. Using kinetic theory the temperature of a system is taken to be a measure of the average translational kinetic energy associated with the random motion of the molecules with the system. It should be noted that although related to internal energy, temperature is not directly proportional to internal energy since internal energy also involves the rotational and vibrational kinetic energies and the potential energies of the constituent molecules.
So, we come to the definition of heat. If we examine the first law, we can see that we can increase the internal energy of a system either by doing work on it, or adding heat to it. Consider a piston and a cylinder filled with gas, we can increase the internal energy of the system by either compressing the gas by applying a force to the piston (work) or by fixing the piston and placing the cylinder in a flame (heat). We can compress and heat the gas in such a way that after the operation all the macroscopic properties (pressure, volume & temperature) are identical, that is the two cylinders are in identical states. Suppose we take two identical cylinders (but not necessarily in identical initial states) filled with a gas at 373K, one of which we compress and the other of which we heat such that both cylinders are at 473K and all their macroscopic quantities are identical, that is the final states of the two cylinders are identical. If we were to now examine the final states of the two cylinders, we have no way of knowing which was compressed and which was heated; the only conclusion we can draw is that their internal energies have increased. In this way we can consider heat as the microscopic analogy of work (macroscopic). I therefore, offer you a formal definition of heat:
"Heat is the non-mechanical exchange of energy between the system and surroundings as a result of a difference in temperature"
Both work and heat can be considered as methods of transferring energy within or between systems. It should now be apparent why the statement "a body posses heat" is nonsensical. To say that a body posses heat is analogous to stating that a "body has work", which you must agree is utter rubbish. Rather, one transfers energy to a body by doing work on that body and one transfers energy to a body by heating or adding heat to that body. Similarly, it is incorrect to state that a body's heat has increased, rather it's internal energy has increased.
Some texts make use of the term "thermal energy" when discussing the "translational kinetic energy" of the molecules, I personally find that the term "thermal energy" only serves to confuse discussions further.
Further Reading
Thermal Physics, 2nd Edition, C.B.P. Finn
Heat@Hyperphysics
Internal Energy@Hyperphysics
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Thread Tools
| | | |
|------------------------------------|-------------------------------|---------|
| Similar Threads for: What is Heat? | | |
| Thread | Forum | Replies |
| | Advanced Physics Homework | 4 |
| | Advanced Physics Homework | 1 |
| | Introductory Physics Homework | 3 |
| | Introductory Physics Homework | 1 |
| | Introductory Physics Homework | 8 |
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9509781002998352, "perplexity_flag": "head"}
|
http://math.stackexchange.com/questions/tagged/turing-machines
|
# Tagged Questions
This tag is suited for questions involving Turing machines. Not to be confused with finite state machines and finite automata.
learn more… | top users | synonyms
1answer
36 views
### Complement of halting set is not r.e.
suppose we don't know that Halting problem is not recursive. I want to prove that complement of halting set is not r.e. then we can find halting problem is not recursive. Can you direct prove that ...
2answers
97 views
+200
### Decidability and undecidability of a set or language
I want to find out whether the following sets are decidable or not. Generally speaking, what exactly should be done about it? Doing some research, I think a language or set is decidable if a Turing ...
1answer
39 views
### Rice's theorem_Theory of computation
Is there any body tell me, where is wrong in this proof Problem: The set of number of turing machine that has 5 state is decidable or not? Answer: The set is obviously 'Set of partial computable ...
0answers
27 views
### Describing a multitape Turing Machine that enumerates the set of $i$ such that $w_i$ is accepted by $M_i$
I am having trouble with this problem. It regards the theory of Turing Machines. Describe a multitape Turing Machine that enumerates the set of $i$ such that the word $w_i$ is accepted by the ...
0answers
36 views
### A question about the analogy between formal systems and Turing machines
It is well known the analogy between formal systems and Turing machines. If I am not wrong, you can code any formal system of language L in first order logic into a Turing machine, and there is a ...
1answer
123 views
### Injection from computable numbers into natural numbers
Each Turing machine which writes an infinite sequence of 1 and 0 can be regarded as representing a (computable) real number (and of course each Turing machine represents a natural number by its ...
0answers
55 views
### Understanding this specific turing machine, NOT HOMEWORK
In regards to the turing machine below, what is it's language? What is the final string if the input is 0011 and in general. what kind of 4 symbol strings does the turing machine accept? I am trying ...
0answers
68 views
### Turing Machines!!! Please check! [closed]
I was having a hard time understanding and solving this question that wants me to show the final tape and figuring out if whether or not the turning machine accepts it or not. I have a list of 20 ...
0answers
31 views
### Turing machine question [duplicate]
I was having a hard time understanding and solving this question that wants me to show the final tape and figuring out if whether or not the turning machine accepts it or not. I have a list of 20 ...
0answers
127 views
### Turing Machine question, this is NOT HW
I was having a hard time understanding and solving this question that wants me to show the final tape and figuring out if whether or not the turning machine accepts it or not. I have a list of 20 ...
0answers
16 views
### Automata. Concatenation of Turing Machine
Given TMs T1 = (Q1,S1, R1, q1, δ1) and T2 = (Q2,S2, R2, q2, δ2), with R1 ⊆ S2, give a precise definition of the TM T1T2 = (Q,S,R, q0, δ). Say precisely what Q, S,R , q0, and δ are.
1answer
57 views
### Showing this language is not decidable by rice theorem or reduction
Consider this language: L = {<M1,M2> : M1 and M2 are TMs and L(M1) contained in L(M2) contained in {1}*} Intuition says that it's undecidable, though can ...
1answer
65 views
### Decidability/Undecidability Question
Could someone please help me with this question? I'm really having a hard time understanding reductions and decidability. Prove that the language L = \{\langle M,N \rangle \mid M,N\text{ are Turing ...
1answer
78 views
### Decidability of a Turing machine always halting in at most ten steps
I've exam comping up soon and I need help with this. Consider the problem: Given a Turing machine $M$, determine if $M$ halts in at most ten steps on every input. Is this decidable? Prove your ...
0answers
124 views
### Turing Machine that accepts the language { www : w ∈ {a, b}*} [closed]
Turing Machine that accepts the language { www : w ∈ {a, b}*} I'm trying to solve this but I'm not sure how to start! I understand how to solve it if it only has ...
1answer
25 views
### Writing multiple symbols on a Turing Machine
Just a quick question: is it possible to write multiple symbols in succession onto a tape of a Turing Machine at once? For example, I'm trying to make a Turing machine that will accept the language ...
1answer
170 views
### Describe a Turing Machine that accepts the language of all non-negative decimal integers that are multiples of 3.
I have exam coming up and I need help with this: Describe a Turing Machine that accepts the language of all non-negative decimal integers that are multiples of 3 Thank you :)
1answer
41 views
### Decidability Turing Machine Problem
$L=\{G|G$ is a context free grammar over ${a,b}$ and $L\{G\}$ contains at least one string $w$ such that the number of $a$'s in $w$ is a multiple of $5\}$ Show that L is decidable by ...
1answer
35 views
### Decidablility Turing machine
Is it decidable whether a Turing machine takes more than 481 steps for some input? This was asked in one of the exams. I found some solutions but are not clear to me.
2answers
108 views
### Why is propositional logic not Turing complete?
According to 1 (probably not the most relevant source), propositional logic is not Turing complete. Aren't all computations in computers performed using logic gates, which can be represented as ...
4answers
74 views
### Non-Deterministic Turing Machine Algorithm
I'm having trouble with this question: Write a simple program/algorithm for a nondeterministic Turing machine that accepts the language: L = \left\{\left. xw w^R y \right| x,y,w \in \{a,b\}^+, ...
1answer
145 views
### Is this undecidable language recognizable?
Is this language: $L = \{\langle M\rangle : \text{$M$is a Turing machine and$L(M)$is decidable}\}$ which I know that is undecidable, turing-recognizable? Is its complement recognizable? ...
1answer
72 views
### Turing Machine Variation
Hi i'm trying to figure out this question: Give a formal definition of multihead-multitape Turing machine. Then show how such a machine can be simulated by a standard Turing machine Can someone ...
2answers
89 views
### Turing machine for balancing parentheses on a two letter alphabet
How to construct a Turing machine $M=(Q,\Gamma,b,\Sigma,\delta,q_0,F)$ which decides if a sting on the alphabet $\{(,)\}$ is ''balanced'' (e.g. $(()())$ is balanced and $))(($ or $()(($ is not) with ...
1answer
60 views
### How many digits of Chaitin's $\Omega$ constant would we know if we had a $\Sigma_1$-Oracle?
According to Wikipedia (and it seems intuitive from the definition itself), $\Omega$ is Turing equivalent to the halting problem and thus at level $\Delta_2^0$ of the arithmetical hierarchy. Do this ...
2answers
173 views
### Why is showing a language is Turing recognizable trickier than showing Turing decidable?
I have written a proof to show that a Turing Decidable languages are closed under union (amongst other things). Later, I have written a proof to show that Turing Recognizable languages are closed ...
1answer
39 views
### Prove or disprove: Superlanguages of Turing-recognizable languages are themselves Turing-recognizable.
Consider the following claim: Prove or disprove: If $L_a$ is Turing-recognizable and $L_b$ contains (or equal to) La, then $L_b$ is recognizable. I'd love to get a hint or a direction Thanks ...
0answers
80 views
### Equivalence of 2-dimensional Turing machine and a standard Turing machine
I'm asked to prove that a two-dimensional TM (one with 2-dim tape that has the upper-left end, and downwards and to the right it goes infinitely) is equivalent to a standard TM. Can I please get a ...
3answers
638 views
### Proving that a Turing Machine that only accepts even length strings is undecidable
I need to prove that a Turing Machine that only accepts even length strings in undecidable. The proof I was thinking is explaining the following: Given an input that contains even length strings, if ...
1answer
43 views
### Proving a language is Turing recognizable
Turing Machine M with a wait option has the option to make the machine's head wait where it is, until a case comes along where ...
1answer
87 views
### Where is the flaw in the following proof?
Where is the flaw in the following proof, that if a language is Turing recognizable then we can enumerate it? Proof Let $TM1$ be a Turing machine for language $L$. We can create an enumerator $E$ ...
1answer
102 views
### Computability of busy-beaver sequence? [closed]
We can draw a parallel between cellular automata and busy-beaver numbers. For example the initial case occupies some kxk square in the plane,leaving all the other cells emty, after how many ...
0answers
96 views
### What turmite runs the longest before becoming predictable?
When looking at 2D Turing machines, many of them eventually become predictable. For example, Langton's Ant, the champion 2-color 1-state turmite, develops a highway after 10,000 steps. Predictable ...
2answers
79 views
### Are there known natural problems of intermediate degrees of unsolvability?
I know there exist intermediate degrees of unsolvability, i.e. there are undecidable problems which can be reduced to the Halting Problem, but not vice versa. Are there any "natural" problems known or ...
1answer
106 views
### Turing machines, halting problem
Let's assume there exists hardware that is able to compute the halting function H(n). That is, if you give it the number of a Turing Machine program/input combination, it will output a 1 if the TM ...
1answer
57 views
### Explain why if the language A is recursive, then A is reducible to 0*1*
I'm in a theory of computation class and there is a problem that I think I am way overthinking. Can anyone point me in the right direction with the following: Give a short justification of the fact ...
4answers
412 views
### Is the set of PA theorems the same as the set of solvable halting problems?
I am not sure if this is a trivial question. By Post's theorem we know that every PA (first order logic) theorem is equivalent to stating that a given input C in a given Turing machine halts or ...
0answers
35 views
### Simulation time of Universal turing machine
How to Design a UTM in which the simulation time of TM M is only O($log |T| +log |Q|$) times more than the original execution time of M. where T & Q denotes the tape alphabet & states of M ...
0answers
79 views
### Implement a Turing machine, Solve the system of logical equations
№ 1 Implement a Turing machine: The input is a sequence of 0 and 1. The machine should be replaced every second 0 to 1. Example: 000111 replaced to 010111. To demonstrate the correctness of your ...
1answer
48 views
### Where does the input x in Turing Machine subroutines come from in solving reductions to undecidable problems?
I'm taking an introduction to computation theory class and we went over the chapter on undecidable problems and proving undecidability through reductions. I can't seem to grasp some of the simplest ...
1answer
90 views
### The set of Turing machines that recognize $\{00, 01\}$ is undecidable
$L =\big\{\langle T\rangle \mid T\text{ is a Turing machine that recognizes }\{00, 01\}\big\}$. Prove $L$ is undecidable. I am really having difficulties even understanding the reduction to use ...
1answer
102 views
### Question about $\Sigma_n$-soundness
According to wikipedia (http://en.wikipedia.org/wiki/%CE%A9-consistent_theory#Definition): "$\Sigma_n$-soundness has the following computational interpretation: if the theory proves that a program C ...
2answers
132 views
### I do not understand why the Turing computable sets of N are exactly the sets at level $\Delta_1^0$ of the arithmetical hierarchy
The reason I don't understand it is this. Take for example the twin primes conjecture, which is $\Pi_2^0$. The set of twin primes is computable right? (there is a Turing machine that enumerates all of ...
1answer
62 views
### Seeking Alternate Proof Regarding Closure Of Recursively Enumerable Languages Under Shrink
So I would like to show that the class of Recursively Enumerable languages are closed under the shrink operation. In other words, $\mathrm{shrink}_a(L) = \{x \mid x=\mathrm{shrink}_a(w), w\in L\}$ and ...
1answer
110 views
### Baker-Gill-Solovay theorem
I have been trying to understand the proof of Baker-Gill-Solovay theorem as described in Complexity Theory: Modern Approach. I think I do understand most of it, but what troubles me is that let's say ...
1answer
352 views
### Language that is recursively enumerable, but not recursive
I have a problem with this task: Show that this language is recursive enumerable, but not recursive: $L = \{ w \in \{0,1\}^* | M_w(x)\; \text{converges for some input}\; x \}$ (where $M$ is turing ...
1answer
70 views
### Understanding of working of Turing Machine for $\{0^k1^k\}$
I try to learn Computation Complexity by Sipser's textbook "Introduction to the Theory of Computation". The problem is I have a lack in understanding how Turing Machine is working. Example from the ...
1answer
150 views
### How can a turing machine solve the element distinctness problem?
I am reading an example in Sipser's famous book on the theory of computation. In this example, Sipser creates a turing machine M to solve the element distinctness problem. M is given a list of strings ...
2answers
256 views
### How can Busy beaver($10 \uparrow \uparrow 10$) have no provable upper bound?
This wikipedia article claims that the number of steps for a $10 \uparrow \uparrow 10$ state (halting) Turing Machine to halt has no provable upper bound: "... in the context of ordinary ...
0answers
32 views
### Would adding a stack to a 2-stack Turing machine allow it to recognize more languages?
I don't think it should because a third stack would be superfluous. The machine could just reuse the first stack after it uses the second right? I'm just beginning to learn about Turing machines, so ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9139069318771362, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/177658/a-number-theory-question-about-a-double-infimum
|
# A number theory question about a “double infimum”
Let $x_1,x_2,x_3,\ldots,x_S$ be numbers with $x_i>-1$ for all $i$ and $x_k<0$ for some $k$.
How can one show that \begin{equation} \inf_{s\in[1,S]}\inf_{t\in[1,s]}\prod_{i=t}^s (1+\frac{1}{2}x_i) < \inf_{s\in[1,S]}\inf_{t\in[1,s]}\prod_{i=t}^s (1+\frac{1}{4}x_i) \end{equation} This seems to instinctively be obvious, because the "most destructive path" surely must be a bit less destructive when we reduce the "destruction" from 1/2 to 1/4, but I'm not sure how to formalize this thought.
Update: Try also to generalize this for not just 1/2 and 1/4, but for any number $q$ and any other number $p<q$, with $0<p,q<1$.
-
– Did Aug 15 '12 at 10:30
## 2 Answers
The observation that $1+\frac{1}{2}x \le \left(1 + \frac{1}{4}x\right)^2$ for all $x$, and that the inequality is strict whenever $x\neq 0$, is all you need. Then $$\prod_{i=t}^{s}\left(1 + \frac{1}{2}x_i\right) < \left(\prod_{i=t}^{s}\left(1 + \frac{1}{4}x_i\right)\right)^2$$ for any $t \le s$ as long as $x_i \neq 0$ for some $t\le i \le s$. The double infinimum of the latter product over $s$ and $t$ must be less than $1$, since we're given that at least one $x_i$ is negative, and hence squaring it makes it even smaller.
-
Do you know how the result could be made more general, using not just 1/2 vs. 1/4, but rather any number $q$ vs. any other number $p<q$? (with $0<p,q<1$). – godel68 Aug 1 '12 at 23:10
For convenience, define $f(s,t) = \prod_{t=i}^s \left(1 + \frac{1}{4} x_i\right)$ and $g(s,t) = \prod_{t=i}^s \left(1 + \frac{1}{2} x_i\right)$. Note that since $x_i \leq 0$, both $f(s,t)$ and $g(s,t)$ are monotonically decreasing in $s$ and monotonically increasing in $t$. Therefore, $$f(S,1) = \inf_{s \in [1,S]} \inf_{t \in [1,s]} f(s,t) \quad \text{and} \quad g(S,1) = \inf_{s \in [1,S]} \inf_{t \in [1,s]} g(s,t)\enspace.$$ Since $x_j \leq 0$ for all $j$, we have $$\prod_{i \in A} \left( 1 + \frac{1}{4} x_i \right) \geq \prod_{i \in A} \left( 1 + \frac{1}{2} x_i \right) \enspace,$$ for any set $A \subset ( \mathbb{N} \cap [1,S] )$. To get a strict inequality, pick any $x_k$ such that $x_k < 0$ (and $k \in [1,S]$) and define $A = ((\mathbb{N} \cap [1,S]) \setminus \{x_k\})$. Clearly, \begin{align*} f(S,1) = \prod_{i=1}^S \left( 1 + \frac{1}{4} x_i \right) & = \left( 1 + \frac{1}{4} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{4} x_i \right) \\ & \geq \left( 1 + \frac{1}{4} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{2} x_i \right) \\ & > \left( 1 + \frac{1}{2} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{2} x_i \right) = \prod_{i=1}^S \left( 1 + \frac{1}{4} x_i \right) = g(S,1) \enspace. \end{align*} And we are done.
-
I never mentioned that $x_j\le 0$ for all $j$..., just for some $j$. – godel68 Aug 1 '12 at 23:10
Very true. I'm sorry, I misread. – MLS Aug 1 '12 at 23:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.849065363407135, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/67968/how-many-minimal-pair-wise-coverings-can-there-be
|
## How many minimal pair-wise coverings can there be?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose I have a set of finite sets: $X = \{V_1, V_2, V_3\}$ where each set called $V_i$ in $X$ contains a number of symbols (i.e. $V_1 = \{a,b,c\}$). $Z$ contains all of the Cartesian products of pairs of sets chosen from $X$. So in this case $Z = \{V_1 \times V_2, V_1 \times V_3, V_2 \times V_3\}$. A "word" is a set with each character at position $i$ drawn from the corresponding $V_i$. The set of all words is $W = V_1 \times V_2 \times V_3$. A "sentence" is an ordered set of "words" taken from $W$.
How many "sentences" of minimal length can I make such that every pair from $Z$ is in that "sentence" at least once? Minimal means fewest possible number of "words."
-
Are $V_1,V_2,V_3$ disjoint sets? – Hsueh-Yung Lin Jun 16 2011 at 16:24
yes... I wouldn't consider pairs made from elements within a single $V_i$ to be valid. For example if $V_1 = \\{a,b,c\\}$ and $V_2 = \\{1,2,3\\}$, "words" starting with $\\{a,1,...\\}$ would be valid but "words" like ${a,a,...}$ or ${1,1,...}$ would not be since a "word" is an ordered set where each character at position $i$ may only be drawn from the corresponding $V_i$. – Misha Jun 16 2011 at 17:41
It is not clear to me what it means for a pair from $Z$ to be "in" a sentence. Your sentence is an ordered $n$-tuple of ordered triples; what do you mean by asking that a pair be in such an $n$-tuple? Maybe an example of sets $V_1,V_2,V_3$ together with a minimal sentence would help. – Gerry Myerson Jun 16 2011 at 23:41
Here's an example of what I mean: $V_1=\{a,b\}$, $V_2=\{c,d\}$, $V_3=\{e,f\}$. $V_1 \times V_2 \times V_3 = \{ace,acf,ade,adf,bce,bcf,bde,bdf\}$ There are 12 total pairs: $Z=\{\{ac,ad,bc,bd\},\{ae,af,be,bf\},\{ce,cf,de,df\}\}$. A minimal acceptable sentence is: $S=\{acf,ade,bce,bdf\}$. ...another is $\{ace,adf,bcf,bde\}$ – Misha Jun 17 2011 at 3:53
@Misha, thanks, now I think I understand the question (but regret I don't have a solution). Can you answer the question of how many minimal sentences there are in your example? – Gerry Myerson Jun 17 2011 at 12:29
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9480363726615906, "perplexity_flag": "head"}
|
http://math.stackexchange.com/questions/94690/is-anybody-researching-ternary-groups
|
# Is anybody researching “ternary” groups?
As someone who has an undergraduate education in mathematics, but didn't take it any further, I have often wondered something.
Of course mathematicians like to generalize ideas. i.e. it is often better to define and write proofs for a wider scope of objects than for a specific type of object. A kind of "paradox" if you will - the more general your ideas, often the deeper the proofs (quoting a professor).
Anyway I used to often wonder about group theory especially the idea of it being a set with a list of axioms and a binary function $a\cdot b = c$. But has anybody done research on tertiary (or is that trinary/ternary) groups? As in, the same definition of a group, but with a $\cdot (a,b,c) = d$ function.
Is there such a discipline? Perhaps it reduces to standard group theory or triviality and is provably of no interest. But since many results in Finite Groups are very difficult, notably the classification of simple groups, has anybody studied a way to generalize a group in such a way that a classification theorem becomes simpler? As a trite example: Algebra was pretty tricky before the study of imaginary numbers. Or to be even more trite: the Riemann $\zeta$ function wasn't doing much before it was extended to the whole complex plane.
EDIT: Just to expand what I mean. In standard groups there is an operation $\cdot :G\times G\to G$ I am asking about the case with an operation $\cdot :G\times G\times G\to G$
-
Could you elaborate on your definition? It reminds me of vectors: take $a,b,c\in\mathbb{R}$ and $(a,b,c)$ a vector with $a$ a scalar. Maybe $d$ comes from taking the magnitude? Are we trying to talk about a grouplike thing that operates on itself and 3-tuples of itself in a different way? – Tony Dec 28 '11 at 14:27
3
– Chris Taylor Dec 28 '11 at 14:33
7
– lhf Dec 28 '11 at 14:36
@Chris Taylor. Thanks for that. I'll have a look at it. It is COMMUTATIVE, which boils down to commutative binary groups, so I'm not sure it would be helpful in dealing with simple groups. – JJG Dec 28 '11 at 14:41
2
– Zhen Lin Dec 28 '11 at 15:05
show 1 more comment
## 1 Answer
One problem with the idea is that the most obvious generalization to ternary operations really adds nothing new:
Proposition: Let $f:G\times G\times G:\to G$, and for brevity write $[abc]$ for $f(a,b,c)$. Suppose that there is an identity element $e\in G$ such that $[aee]=[eae]=[eea]$ for all $a\in G$. Suppose further that the operation is associative in the following sense: $$\big[[abc]de\big]=\big[a[bce]d\big]=\big[ab[cde]\big]$$ for all $a,b,c,d,e\in G$. Then there is an associative binary operation $\otimes$ on $G$ such that $[abc]=a\otimes b\otimes c$ for all $a,b,c\in G$, and $e$ is the $\otimes$-identity.
The proof is easy. Define $\otimes:G\times G\to G$ by $a\otimes b=[abe]$. Then
$$\begin{align*} (a\otimes b)\otimes c&=[(a\otimes b)ce]=\big[[abe]ce\big]=\big[ab[ece]\big]=[abc]\\ &=\big[[abc]ee\big]=\big[a[bce]e\big]=[a(b\otimes c)e]=a\otimes(b\otimes c)\;, \end{align*}$$
and $a\otimes e=[aee]=a=[eae]=e\otimes a$ for all $a,b,c\in G$. Note that this does not require any kind of generalized commutativity for the ternary operation.
I remember noticing this as an undergraduate in the late 60s. My roommate was looking at a less obvious generalization of the associative law. The generalization, as I recall, was $$\big[[abc]de\big]=\big[a[bde][cde]\big]\;.\tag{1}$$ The idea is that if $\otimes$ is a binary operation on a set $G$, one can think of each element $a\in G$ as defining a function $f_a:G\to G:x\mapsto a\otimes x$, and associativity of $\otimes$ is then the statement that $$\operatorname{Comp}(f_a;f_b)=f_{f_a(b)}\;,\tag{2}$$ where $\operatorname{Comp}$ is the composition operator. In the ternary setting think of $a\in G$ as defining a function $$f_a:G\times G\to G:\langle b,c\rangle\mapsto [abc]\;;$$ then $(1)$ becomes $$\operatorname{Comp}(f_a;f_b,f_c)=f_{f_a(b,c)}\;,$$ generalizing $(2)$.
If I remember correctly, this approach produced more interesting structures, but I no longer remember the details.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9620222449302673, "perplexity_flag": "head"}
|
http://math.stackexchange.com/users/16192/gt6989b?tab=activity&sort=comments
|
gt6989b
reputation
310
bio
location New York, NY
age 34
member for 1 year, 8 months
seen 12 hours ago
profile views 150
| | | bio | visits | | |
|------------|------------------|--------------|-------------------------|--------------|------------------|
| | 2,688 reputation | website | linkedin.com/in/gt6989b | member for | 1 year, 8 months |
| 310 badges | location | New York, NY | seen | 12 hours ago | |
307 Comments
| | | |
|------|---------|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| 15h | comment | Proof for an integral identity+1 - more general than my answer - I didn't want to bother with the integrals :) |
| 15h | comment | Proof for an integral identity@robjohn thanks, making the change. |
| 16h | comment | How can I find all the solutions of $\sin^5x+\cos^3x=1$Substitute $\sin x = \sqrt{1 - \cos x}$ and solve the result in terms of $\cos x$. You could try to reduce $\cos x$ in terms of $\sin x$ as well. |
| 16h | comment | Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$You are asing to show that $\sqrt{x}$ is concave. |
| 17h | comment | Formula for Sum of Logarithms $\ln(n)^m$@DonAntonio $\ln \left(n^{\ln n}\right) = \ln n \cdot \ln n = \ln^2 n$ |
| 18h | comment | Formula for Sum of Logarithms $\ln(n)^m$Perhaps an idea - not sure if it will help. $$\sum \ln^2 n = \sum \ln n^{\ln n} = \ln \left( \prod n^{\ln n} \right).$$ |
| 18h | comment | Proof for an integral identityDo you mean $A=B$? Setting $f(x) = f(y) = 1$ as in the above hint yields $AB = A^2$... |
| May7 | comment | Need an algorithm to compute number of elements in sample space@YevgeniyRozhkov for this case, i would condition on $z$ - it's either there or not, and if yes, choose a place for it. Now you reduced it to 2 elements and 2 or 3 places. 2 places are easy and for 3 you condition again. |
| May7 | comment | Evaluate using Riemann sums $\int_a^b \frac{1}{x^2}\mathrm{d}x$@darenn You're right, looks pretty nasty. |
| May7 | comment | Evaluate using Riemann sums $\int_a^b \frac{1}{x^2}\mathrm{d}x$What are you asking? How to evaluate the integral using riemann sums? |
| May7 | comment | A problem on calculating rank of a matrixIf all except one are $0$, then you get a $0$-matrix with one non-zero diagonal entry, which will have deteminant $0$. |
| May7 | comment | Simple fractions question@user2008436 I see, but a bit weird. My daughter learned solving 1st degree equations in 3rd grade (in the Russian math program), and I've heard the Singapore one is more advanced. Interesting. |
| May6 | comment | Finding the error of the Taylor expansion of $\log(1 + x)$Write out the first couple of terms in the Taylor expansion of $\log(1+x)$ at $x=0$. Find $R_{0,k}$ and show that $R_{0,k} \to 0$ as $k \to \infty$. |
| May6 | comment | In order to factor we must find its zeros?You can proceed to factor $(x-1)^2-2$ using $a^2-b^2=(a+b)(a-b)$ with $a=x-1$. (What is $b$?) As was pointed out by @TheChaz2.0, factoring and finding zeroes is equivalent. |
| May3 | comment | Numerical estimation of simple integralI got the question. That is the definition of expected value of a function. |
| May3 | comment | Numerical estimation of simple integralYou pick the function in such a way that $I = \mathbb{E}[f(U)]$, and then use Monte-Carlo, for example, to compute the expected value... |
| May3 | comment | Considering the linear system Y'=AYTo take @julien's answer one step further, note that $AX=\lambda X$ is equivalent to $AX - \lambda X = 0$, which simplifies to $(A-\lambda I)X = 0$, which requires $det(A-\lambda I)=0$ if you want $X \neq 0$... |
| May3 | comment | How to find equation of tangent line to $x^2 = 2y$ at $(-3, 9/2)$+1, this is clever :) |
| May3 | comment | How to find equation of tangent line to $x^2 = 2y$ at $(-3, 9/2)$@Steven Please see the edit which explains the concept of the discriminant. |
| May2 | comment | Overlapping AreasIf $a+b > c$, by Pigeonhole principle they overlap. Other than that, it is impossible to say - you may have $a = b = c/3$ and they overlap or not overlap, both are possible. Or perhaps I did not understand the problem? |
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9217972755432129, "perplexity_flag": "middle"}
|
http://mathhelpforum.com/advanced-algebra/156803-prime-numbers.html
|
# Thread:
1. ## Prime Numbers
Here is my problem:
Let a,b be nonzero integers. Prove that (a,b) = 1 if and only if (a+b, ab) = 1.
Any suggestion on where to start?
2. You could try playing around with Bezout's identity.
(From right to left) Suppose that $\gcd(a+b,ab)=1$. There exist integers $x,y$ such that $(a+b)x+(ab)y=1$. Distributing the terms, $ax+bx+aby=1$, and $a(x+by)+bx=1$. This shows that $\gcd(a,b)$ must divide 1. Therefore, $\gcd(a,b)=1$.
You can try a similar argument for the other direction. (You may need to insert some terms into the equation by adding and subtracting the same thing.)
3. Alternatively:
$<br /> \gcd(a+b,ab) = 1 \Leftrightarrow \gcd(a+b,a) = \gcd(a+b,b) = 1 \Leftrightarrow \gcd(a,b)=1<br />$
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.895769476890564, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/32296/periodicity-theorems-in-generalized-cohomology-theories/32314
|
## Periodicity theorems in (generalized) cohomology theories
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
It is well-known that topological K-theory is blessed with the Bott periodicity theorem, which specifies an isomorphism between $K^2(X)$ and $K^0(X)$ (where $K^n$ is defined from $K^0$ by taking suspensions). I am wondering if other generalized cohomology theories have their own periodicity theorems, and if there is a general framework for conceptualizing them. I am interested in any substantive answer to this question, but there are two specific avenues for generalization that I am particularly curious about.
The first avenue begins with the Clifford algebra approach to Bott periodicity. This approach relates periodicity in K-theory to a certain natural periodicity present in the theory of complex Clifford algebras, and it generalizes the 8-fold periodicity of real K-theory (corresponding to an 8-fold periodicity in real Clifford algebras). Can one fruitfully generalize the notion of a Clifford algebra, associate to it a generalized cohomology theory, and analogously produce a periodicity theorem?
The second avenue involves Cuntz's proof of Bott periodicity for C*-algebras (which in particular implies topological Bott periodicity by specializing to commutative C*-algebras). Cuntz proves Bott periodicity for any functor from the category of C*-algebras to the category of Abelian groups which is stable (i.e. insensitive to tensoring with the C*-algebra of compact operators on Hilbert space), half exact, and homotopy invariant. The proof uses topological properties of Toeplitz algebras in an essential way. Because of the generality of his approach, I am left wondering if the essential features of his argument can be translated into more general contexts.
Any ideas are welcome!
-
4
Unfortunately, unlike K-theory it happens that most of the time we might know that a cohomology theory is periodic, or we might know a geometric interpretation like in terms of vector bundles, but rarely do we have both. – Tyler Lawson Jul 17 2010 at 18:57
## 3 Answers
In the spirit of first approach, there is a conjecture for a Clifford-algebra type proof of the 576-fold periodicity of TMF. This is a generalized cohomology theory constructed by piecing together all the elliptic cohomology theories together in a suitable way. I heard about this conjecture from Andre Henriques, who is working on a geometric approach to TMF using conformal nets.
The idea is that the free fermion conformal net (a introduction can be found in this article by Bartels, Douglas and Henriques) is to TMF as the Clifford algebras are to K-theory. For a suitable sense of equivalence, i.e. some generalization of Morita equivalence, $Free(n)$ and $Free(n+576)$ should be equivalent. I believe people are still far from a proof, but the motivation comes from looking at orientations: a manifold is orientable for K-theory if the frame bundle (a principal $SO(n)$-bundle) lifts to a principal $Spin(n)$-bundle. The $Spin(n)$ groups can be defined as a group in the Clifford algebra. For TMF, a manifold is orientable if the frame bundle extends to a principal $String(n)$-bundle, which can be obtained from $Spin(n)$ by killing $\pi_3$, just like $Spin(n)$ is obtained from $SO(n)$ by killing $\pi_1$. There is a way to define $String(n)$ using the free fermion conformal nets.
The only reference I know for these ideas is the following summary.
-
This sounds fascinating! I can't wait to find out where the number 576 comes from... – Paul Siegel Jul 18 2010 at 2:57
576 is $24^2$. It is supposed to come from points in the moduli stack of elliptic curves with additional automorphisms: those corresponding to j=0 and j=12^3 have Z/6Z and Z/4Z respectively. However, the proof uses spectral sequence arguments at primes 2 and 3: math.mit.edu/conferences/talbot/2007/tmfproc/… – skupers Jul 18 2010 at 12:19
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
In this video of a lecture given at Atiyah's 80th Birthday Conference, Mike Hopkins gives a description of the solution to the Kervaire Invariant Problem which relies heavily on some periodicity in a cohomology theory created for this problem. He mentions Clifford algebras and Bott Periodicity at one point towards the end of the talk... I don't know a whole lot about this topic so I didn't understand too much, but maybe it will be helpful for you!
-
Amusingly enough I was actually there in the audience, and I definitely remember some discussion of generalized periodicity theorems (though I guess I was pretty lost by the end). In any event, it seems I am not the first to wonder about this sort of thing. – Paul Siegel Jul 18 2010 at 3:00
Cohomology theories constructed by the Landweber exact functor theorem are periodic, and there is also a periodization of cohomology theories.
-
Is this really the case? all genera lead to periodic theories? – Sean Tilson Aug 8 2010 at 18:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9217488169670105, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/93772/triangulations-of-polytopes-and-tilings-of-zonotopes
|
## Triangulations of polytopes and tilings of zonotopes
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider a set $A = \{ a_1,a_2,\ldots, a_n \}$ of vectors in $\mathbb{R}^d$, which lie in a common affine hyperplane. Two convex polytopes may be obtained from $A$, namely the convex hull of the vectors in $A$, $conv(A)$ and the zonotope generated by vectors in $A$, $Z(A)$. Both polytopes can be viewed as the projection of a polytope $P$ which sends the unit vectors in $\mathbb{R}^n$ to the columns of $A$, where $P$ is the standard $n-1$-simplex in the first case and the unit $n$-cube in the second. The general study of affine projections of convex polytopes is developed in this paper of Billera and Sturmfels entitled "Fiber Polytopes"
The first question may be common knowledge, but is there some relation between the triangulations of $conv(A)$ and cubical tilings of $Z(A)$ which arise from this projection? I haven't seen this explicitly written down, but in the light of fiber polytopes both should be related concepts.
Question: Is there a reason why the triangulations of $conv(A)$ should be considered in some sense "more natural" than the cubical tilings of $Z(A)$ (both subdivisions seen as arising from the projection)?.
The reason I ask this is because I have seen many instances in the literature of toric ideals where triangulations of a convex polytope are used to characterize some algebraic construction but, as far as I remember, none performing similar characterizations in terms of tilings of zonotopes. As an example, a theorem of Sturmfels here characterizes the radicals of the initial ideals of a toric variety (associated to a matrix $A\in \mathbb{Z}^{d\times n}$) as the radical of the Stanley-Reisner ideals of regular triangulations of the convex hull of the columns of $A$.
However, when it comes to the combinatorial information of a vector configuration, the zonotope associated to it seems to relate more directly to the oriented matroid of the vector configuration. Recall, for instance the Bohne-Dress theorem relating the set of zonotopal tilings of the zonotope generated by the vector configuration and the one-element liftings of its oriented matroid.
I would be very satisfied with answers which, say,
• give pointers to characterization of algebraic objects (e.g. Groebner bases, minimal free resolutions) from the theory of toric ideals in terms of some associated zonotopes.
• give pointers to a general relation between triangulations of $conv(A)$ and cubical tilings of $Z(A)$,
• indicate why, and in which context, one of both concepts should be more natural than the other.
p.s. may be someone reputed enough would like to create the tag oriented-matroids?
-
1
Maybe the connection equivalence of unimodular triangulations and square-free initial ideals is already the best argument why triangulations are 'better'. – Thomas Kahle Apr 11 2012 at 15:53
For me this would rather be a disadvantage, since it means that triangulations can only be associated with square-free initial ideals (in the general case, one would have to go to the radical to get a triangulation). – Camilo Sarmiento Apr 12 2012 at 8:51
## 1 Answer
Triangulations of polytopes are "more fundamental" than cubical tilings of zonotopes. By the Cayley trick, every cubical tiling of a zonotope can be seen as a triangulation of the Cayley lifting of the segments defining it. The latter is equal to their Lawrence lifting.
This can be seen as an addition to the Bohne-Dress theorem, relating tilings of a zonotope to liftings of the associated oriented matroid.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9322925806045532, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/7933/differentiable-function-germs-on-differentiable-manifolds
|
## Differentiable function germs on differentiable manifolds
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello everyone, I was wondering if anyone knew how to prove that the map from $C^{\infty}(M)$ to $\xi (p)$, that is, from the infinitely differentiable functions on a manifold M to the space of (once)-differentiable function germs, where the map is associating to each f in $C^{\infty}(M)$ its class in $\xi (p)$ is onto. By the way, since you ask, the reason I'm interested in this is because its a question that WAS on my final for differential topology, I've tried to work it out since then but no luck so far, this is not homework it's just curiosity now, hope its ok ill have to check the post regultaions, sorry, if not just tell me and i'll delete the question...
-
3
1. Your question's well-formed, but nearly impossible to parse. Edit? 2. I don't know any differential geometry, but is this homework? It reads kind of like it might be, and it's gotten downvoted. If it's homework, don't bother editing. – Harrison Brown Dec 6 2009 at 0:37
1
Assuming you mean $C^{\infty}$ germs, then this is basically the existence of a cutoff function. – Akhil Mathew Dec 6 2009 at 2:12
Upvoted since reasonableness should be encouraged. :) – Harrison Brown Dec 6 2009 at 2:21
## 1 Answer
I think your question is mis-stated, because this map is not onto:
There is no smooth function on the reals whose germ at $0$ is the germ of $x|x|$, a once-differentiable function germ.
Assuming this is not what you wanted, please restate the question, and include an explanation of why you are interested in it ;)
-
Thanks, I'll look into your counterexample! – unknown (yahoo) Dec 6 2009 at 1:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9536365866661072, "perplexity_flag": "middle"}
|
http://mathhelpforum.com/advanced-statistics/175983-expected-ranked-position-set-normal-distributed-random-variables.html
|
# Thread:
1. ## Expected ranked position of a set of normal-distributed random variables
Given several normal-distributed random variables X1 .. Xn with means m1,m2,m3 .. mn and variance s1,s2,s3 .. sn respectively, how can the ranking (when sorted from highest to lowest) of a particular sample can be predicted in general?
For example, if X1 ~ N(1,1) , X2 ~ N(2,1) and X3 ~ N(1,2), what will be the average "ranked position" (after sorting) of the X1 sample, the X2 sample and the X3 sample?
I tried to research the problem, but all I came up was a solution for the case with two variables.
Thanks in advance for help and hints!
2. This probably isn't the most efficient way of doing this, but I guess you can get the probability of each of the orderings and work with that (i.e. $P(X_1 \le X_2 \le X_3)$ and so forth, which I assume you know how to get). Probably that would get out of control for large n though. At the very least you can streamline those sort of calculations - you would be working with a bunch of multivariate normals - but then you would have to do $n!$ of these.
I don't see any reason to think this is an easy problem in practice if you really want to get exact answers. You could Monte Carlo this if it is a practical problem.
At least this takes care of the example you gave though. You only need to calculate 6 probabilities to know the distributions of the rankings.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9470511674880981, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/42113/a-small-collection-of-large-subsets-covering-all-small-subsets/42115
|
## A small collection of large subsets covering all small subsets.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $r,s,n$ be positive integers with $r < s < n$. Let $U = \{1,\ldots,n\}$.
Let $S$ contain $s$-element subsets of $U$ (of our choosing). What is that smallest we can make $S$ such that every $r$-element subset of $U$ is a subset of some element in $S$?
I'm curious what the best known upper and lower bounds are on the smallest we can make $S$, and I am especially interested in the case where $r << s << n$. I am more worried about asymptotic behavior than exact bounds.
This can also be viewed as a special case of a set-cover problem that has a lot of symmetry. I'm afraid/hoping there's something simple from graph theory that solves my problem.
EDIT: Thanks to @Gerhard, I see this is a well known problem called covering numbers / covering designs.
-
3
www.ccrwest.org/cover.html , the La Jolla covering repository, should be your first stop. Let us know what it does not answer. Gerhard "Ask Me About System Design" Paseman, 2010.10.13 – Gerhard Paseman Oct 14 2010 at 5:23
## 1 Answer
What you are looking for is an `$(r,s,n)$`-covering design. A good starting point might be the La Jolla Covering Repository or in fact, any book on design theory. In general, the smallest possible size for your covering design is given by `$\binom{n}{r}/\binom{s}{r}$` in which case every $r$ -element subset is contained in exactly one $s$-element subset and the design is then called a `$(r,s,n)$`-Steiner system. There are necessary conditions for Steiner systems to exist such as `$\binom{s-i}{r-i} | \binom{n-i}{r-i}$` for all $i$. Quite a lot is known about the existence of `$(2,3,n)$`-Steiner systems, and `$(3,4,n)$`-Steiner systems, but in general their existence is an open question.
-
Thank you -- this is perfect. It seems Erdos and Spencer give an upper bound of (1+\ln(C(s,r)))C(n,r)/C(s,r), which is perfect for my purposes! If I had enough reputation, I would upvote your answer : – Lev Reyzin Oct 14 2010 at 5:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.934317409992218, "perplexity_flag": "head"}
|
http://physics.stackexchange.com/questions/26890/principle-behind-fidelity-balance-in-quantum-cloning?answertab=votes
|
# Principle behind fidelity balance in quantum cloning
If we do optimal state estimation on an unknown qubit, we can recreate a state with fidelity $F_c=2/3$ with respect to the original. Let us define the "quantum information content" $I_q=1-2/3=1/3$ as the "amount of fidelity lost" in this measurement procedure.
If we, instead of measuring, decide to clone the qubit using an optimal cloning machine, we can obtain two imperfect copies with fidelity $F_q=5/6$ each. The "quantum information content" of the two qubits is now $I_q=2 \times (5/6-2/3)=1/3$. Note that the value is the same as in the measurement procedure above.
This conservation of "quantum information content" holds more generally: it is true for symmetric, $N \to M$ system cloning, for systems of any dimensionality (see reference [1]). The question then is: is there a deeper principle or operational justification that can be invoked to justify this curious fidelity balance result? I originally raised this question in my PhD thesis (ref. [2] below, section 4.3.4).
References:
[1] M. Keyl and R. F. Werner. Optimal cloning of pure states, testing single clones. J. Math. Phys., 40(7):3283–3299 (1999).
[2] E. F. Galvão, PhD thesis, http://arxiv.org/abs/quant-ph/0212124
-
## 1 Answer
This may seem something of an obvious answer, and may well have already occured to you, but perhaps there is something interesting in it for you.
The input system has Holevo information of $\chi = \log_2 D$, where $D$ is the dimensionality of the system to be cloned. Applying the cloning procedure does not change this in any way, it is simply spreading this information out over a number of systems. You appear to be implicitly assuming that the input state is chosen uniformly at random from the pure states (since otherwise you can obtain $I_q = 0$ by taking the distribution only over orthogonal states). As the Holevo information is defined as $\chi = S(\sum_i p_i \rho_i) + \sum_i p_i S(\rho_i)$, and $p_i=p$ is fixed and $\rho_i$ are pure, you have $$\chi = S(p \sum_i \rho_i) = S(\bar{\rho})$$ where the bar denotes the average density matrix. Presumably any cloning scheme which is optimal must not decrease the Holevo information of the joint system, and so the entropy is necessarily preserved. When you make a projective measurement of all of final system you project onto a product state, and hence the entropy is zero, independent of the size of the system. Thus, the change in entropy is identical whether or not optimal cloning has been preformed. The reason that you can express this as a condition on the fidelity of the system, is then simply a consequence of the fact that fixing the joint entropy of the system imposes a constraint on the maximum fidelity of the clones.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9156796932220459, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/68367/are-almost-commuting-hermitian-matrices-close-to-commuting-matrices-in-the-2-nor/68379
|
## Are almost commuting hermitian matrices close to commuting matrices (in the 2-norm)?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I consider on $M_n(\mathbb C)$ the normalized $2$-norm, i.e. the norm given by $\|A\|_2 = \sqrt{Tr(A^* A)/n}$.
My question is whether a $k$-uple of hermitian matrices that are almost commuting (with respect to the $2$-norm) is close to a $k$-uple of commuting matrices (again with respect to the $2$-norm). More precisely, for an integer $k$, is the following statement true?
For every $\varepsilon>0$, there exists $\delta>0$ such that for any $n$ and any matrices $A_1,\dots, A_k\in M_n(\mathbb C)$ satisfying $0\leq A_i\leq 1$ and $\|A_iA_j - A_j A_i\|_2 \leq \delta$, there are commuting matrices $\tilde A_1,\dots,\tilde A_k$ satisfying $0\leq \tilde A_i\leq 1$ and such that $\|A_i - \tilde A_i\|_2 \leq \varepsilon$.
The important point is that $\delta$ does not depend on $n$.
I could not find a reference to this problem in the litterature. However, this question with the $2$-norm replaced by the operator norm is well-studied. And the answer is known to be true if $k=2$ (a result due to Lin) and false for $k=3$, and hence $k\geq 3$ (a result of Voiculescu).
-
## 3 Answers
There is a recent paper by Glebsky titled "Almost commuting matrices with respect to normalized Hilbert-Schmidt norm" which shows that this is indeed true for any $k$ for Hermitian matrices (and in fact also unitary and normal matrices).
-
Thanks a lot for your answer. I did not yet have time to read the whole proof, but this seems to completely answer my question. I just wonder why this paper was posted in the algebraic geometry section of arXiv. – Mikael de la Salle Jun 21 2011 at 14:31
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The answer is yes, and much more is true. Any hyperfinite von Neumann algebra (with separable predual) has a unique embedding (up to conjugacy) into the ultra-product of the hyperfinite $II_1$-factor.
This implies in particular, that almost commuting matrices in Hilbert-Schmidt are close to commuting matrices. The proofs goes by contradiction; assume that there is a sequence of counterexamples and construct non-conjugate embeddings. Since any abelian von Neumann algebra is hyperfinite, this yields a contradiction.
Kenley Jung showed that uniqueness of the embedding also implies that the algebra is hyperfinite.
-
Andreas, is your argument presupposing that the matrices almost commute in HS-norm and are uniformly bounded in operator norm? – Yemon Choi Jun 21 2011 at 21:16
Thanks Andreas. I already accepted Ashley's answer, but your answer is more what I was hoping for (Glebsky's proof has however the advantage of giving an explicit bound on $\delta$ depending on $k$ and $\varepsilon$). – Mikael de la Salle Jun 22 2011 at 6:42
I guess that the precise statement is :"Given a non principal ultrafilter $\mathcal U$, any two embeddiing of a hyperfinite von Neumann algebra into $\prod_{\mathcal U} M_n$ (or $R^{\mathcal U}$) are conjugate". (unless I missed something, the ultrapower might depend on $\mathcal U$). By the way, do you have a reference for this statement? – Mikael de la Salle Jun 22 2011 at 6:57
Yemon: yes, just as in the question. – Andreas Thom Jun 22 2011 at 7:32
2
Mikael: you are right, this might depend on the ultrafilter. I would try looking at the paper by Kenley Jung (Math. Ann. 2007 vol. 338 (1) pp. 241-248) and see whether he gives a reference. This is the easy part anyway, it is based on the fact that a full matrix-algebras embeds (unitally) in a unique way (up to conjugacy) into each other. The same holds if you take block-sums of matrix-algebras and remember their relative multiplicities. – Andreas Thom Jun 22 2011 at 7:36
show 1 more comment
I just found the discussion. In the paper there is the better estimates then the ones of mine and it contains citations on proofs using von Neumann algebras.
(It was a surprise for mi too why my paper is in Algebraic Geometry. Probably it is my error. I have not found an easy way to fix it.)
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9298096895217896, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/70348/if-two-ideal-quotients-are-isomorphic-are-the-ideals-isomorphic
|
If two ideal quotients are isomorphic are the ideals isomorphic? [closed]
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose $J_1$ and $J_2$ are two ideals in a ring both containing another ideal $I$. If $J_1/I \cong J_2/I$ then is $J_1 \cong J_2$?
-
2
Reading the FAQ will explain why this question is not really well-suited to this site (in short, it is not a question about "research math"). The FAQ also suggests other places, like math.stackexchange.com, where your question will be much more at home. Good luck! – Mariano Suárez-Alvarez Jul 14 2011 at 16:36
2
What notion of isomorphism of ideals is everyone implicitly working with here? – Qiaochu Yuan Jul 14 2011 at 18:03
@Qiaochu: I had a similar question, as I think of ideals as things that can be equal to each other, not isomorphic. One possibility is to use "isomorphic as $R$-modules"; this is Neil's interpretation below. But an ideal is really an $R$-module with a monomorphism to the rank-$1$ free module, and the category thereof is a poset. – Theo Johnson-Freyd Jul 14 2011 at 18:28
@Theo: the category thereof is that if you want it to be. In some conexts, you don't, though. For example, the ideal class group of a Dedekind domain is usefully seen as the set of isomorphism classes of ideals (as modules). Fractional ideals are, from that point of view, a kludge one uses to contruct the operation in the group. – Mariano Suárez-Alvarez Jul 14 2011 at 23:47
1 Answer
Take $R=\mathbb{Z}[x,y]/(x^2,xy,y^2,4x,4y)$ and $J_1=(x)$ and $J_2=(2x,2y)$ and $I=(2x)$. Then $J_1/I\simeq J_2/I\simeq R/(2,x,y)$ as $R$-modules, but $2J_2=0$ and $2J_1\neq 0$, so $J_1\not\simeq J_2$.
On the other hand, if $R$ is a principal ideal domain, then the answer to your question is positive.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9307258725166321, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/161659/integral-as-a-limit-of-a-sum?answertab=active
|
# Integral as a limit of a sum
How do I get $$\int_a^b \frac{1}{x}dx = \ln\left(\frac{b}{a}\right)$$ as a limit of sum. The constant width partition of the interval $(a,b)$ doesn't seem to work.
-
What makes you think the fixed-width Riemann sum doesn't work in the limit? – anon Jun 22 '12 at 14:14
2
Typically the logarithm is defined $\log t:= \int_1^t\frac{1}{x}dx$, so your question follows nearly by definition. – nullUser Jun 22 '12 at 14:15
@Anon I tried it but couldn't get to the answer. – Prakash Gautam Jun 22 '12 at 14:25
@PrakashGautam Write up what you tried below your question. – rschwieb Jun 22 '12 at 14:29
## 4 Answers
Take $x_i = a \left( \frac{b}{a} \right)^{i/n}$ then:
$$\lim\limits_{n \to \infty} \sum_{i = 0}^{n-1} \frac{x_{i+1} - x_i}{x_i} = \lim\limits_{n \to \infty} \frac{ \left(\frac{b}{a}\right)^{\frac{1}{n}} - 1 }{\frac{1}{n}} = \ln \frac{b}{a}$$
-
(+1), but I guess it is good to append that $$\log x = \lim\limits_{k \to 0} \frac{x^k-1}{k}$$ – Peter Tamaroff Jun 24 '12 at 22:43
$$\int _a^b {f(t) dt}=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n f(\frac{k(b-a)}{n}+a)$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{1}\sum \limits_{k=1}^n \frac{1}{k(b-a)+na}$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{(b-a)}\sum \limits_{k=1}^n \frac{1}{k+\frac{na}{b-a}}$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{(b/a)-1}}$$
$b/a=x$
$$f(x-1)=\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x-1}}$$
$$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$
$$f(x)=\lim_{n\to\infty}\frac{x}{n} \sum \limits_{k=1}^n \frac{1}{1+\frac{kx}{n}}=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n (1-\frac{kx}{n}+\frac{k^2x^2}{n^2}-\frac{k^3x^3}{n^3}+....)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....$$
$$f(x)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....=\lim_{n\to\infty} \frac{x}{n} n -\lim_{n\to\infty} \frac{x^2}{n^2} (\frac{n^2}{2}+\frac{n}{2})+\lim_{n\to\infty} \frac{x^3}{n^3} (\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})-\lim_{n\to\infty} \frac{x^4}{n^4}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+....$$
$$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants.where aj are constants. More information about summation http://en.wikipedia.org/wiki/Summation
After solving limits. We get:
$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$
$$f(x-1)=\ln(x)$$
$b/a=x$
$$\int _a^b \frac{1}{t} dt=f(x-1)=\ln(x)=\ln(b/a)$$
-
The constant width partition works fine. No need for fancy partitions. To wit:
Let $x_k = a+\frac{k}{N}(b-a)$, with $k = 0,...,N$. We have, of course, $x_0=a$, $x_N = b$.
Using the Taylor series expansion of $x \mapsto \ln x$, and the mean value theorem, we obtain the bounds, for $x \geq y$:
$$\ln y + \frac{x-y}{y} - \frac{(x-y)^2}{2 y^2} \leq \ln x \leq \ln y + \frac{x-y}{y}.$$ Letting $y=x_k$, $x=x_{k+1}$, we get: $$\ln x_k + \frac{1}{N x_k} - \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} \leq \ln x_k + \frac{1}{N x_k}.$$
Rearranging gives: $$- \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} - \ln x_k - \frac{1}{N x_k} \leq 0.$$ Now sum the inequality over $k = 0,...,N-1$, to get: $$- \frac{1}{2 N^2} \sum_{k=0}^{N-1}\frac{1}{x_k^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$
Since $x_k\geq a$, we obtain: $$- \frac{1}{2 N a^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$
Now let $N \to \infty$, to obtain the desired result, $$\lim_{N \to \infty} \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} = \int_a^b \frac{1}{t} dt = \ln \frac{b}{a}.$$
-
Well, fine is obviously not the same as "nicely". – Peter Tamaroff Jun 24 '12 at 22:42
No symbols where none intended. – copper.hat Jun 24 '12 at 23:34
I don't follow. – Peter Tamaroff Jun 24 '12 at 23:36
Unlike mathematics, it loses meaning when explained; it is the last line of Beckett's idiosyncratic novel Watt. Seemed an appropriate response to your enigmatic comment... – copper.hat Jun 24 '12 at 23:47
What I mean is that if one chooses an appropriate "fancy" partition the solution is much manageable (i.e. qoqozs solution or Amihai's), which I guess you're refering to. – Peter Tamaroff Jun 25 '12 at 0:09
show 2 more comments
Lets do it for $\int\limits_1^b\frac{1}{x}dx$:
Consider the following partition of our interval $\,[1,b]$:
$$P_n=\{1, \sqrt[n]{b},\sqrt[n]{b^2},\dots,\sqrt[n]{b^n}=b\}$$
Now, we need to calculate the upper and lower Darboux sums in respect to the $P_n$'s. Notice that $\frac{1}{x}$ is monotonic decreasing on the interval so the $\sup$ and the $\inf$ are obtained on the edges:
$$\mathcal{U}\left(\frac{1}{x},P_n\right)=\sum_{k=1}^n b^{-\frac{k}{n}}(b^{\frac{k}{n}}-b^{\frac{k-1}{n}})=\sum_{k=1}^n 1-b^{-\frac{1}{n}}=n(\sqrt[n]{b}-1)$$
define $g(x)=b^x$:
$$n(\sqrt[n]{b}-1)=\frac{g(\frac{1}{n})-g(0)}{\frac{1}{n}-0}\to_{n\to\infty}g'(0)=b^x\ln b|_{x=0}=\ln b$$
I'll leave it to you to figure out the lower sums as it's practically the same procedure.
-
Well, how do you define $\ln b$? This seems circular. THe derivative of $b^x$ is defined using that of $e^x$, which might or might not be defined in terms of the $\log x$. – Peter Tamaroff Jun 22 '12 at 16:05
Why can't I define the logarithm as the inverse of the $\exp$? – Amihai Zivan Jun 23 '12 at 17:36
I said "might or might not" in the sense you can choose that method, but there are others available. – Peter Tamaroff Jun 24 '12 at 0:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 24, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8913530111312866, "perplexity_flag": "middle"}
|
http://physics.stackexchange.com/questions/tagged/fluid-dynamics%20water
|
# Tagged Questions
0answers
37 views
### Finding Surface Tension of water at certain Temperature and Pressure
So the question is: Using the Young-Laplace Equation (if applicable), find the surface tension (dynes/cm) for water at 20 degrees Celsius with 2.5 psi. Round to the nearest tenth. ...
2answers
146 views
### Why water in the sink follow a curved path?
When you fill the sink with water and then allow the water to be drained, the water forms a vortex.. And then it starts to follow a curved path downwards by effects of gravity.. Why this phenomena ...
2answers
103 views
### Best shape to reduce the splash of a droplet?
Our coffee machine catches the last couple of droplets, after your cup is removed on a shape to reduce plash of the coffee droplets. These shapes are placed inside the spill reservoir. The shape ...
2answers
108 views
### Pumping water from a closed column. Will the column get empty or just circualte?
As a picture is worth a thousand words, here is my problem: It is a closed water column with a bit of air in the top section. If I run the pump and make the water flow from IN to OUT, Will the ...
1answer
126 views
### How fast does water fall in the middle of a very very thick waterfall?
Let me create a very artificial experimental set up. Take a bathtub the size of Delaware and suspend it a mile above the ground. Fill it with water (though I'm not sure to what depth - and it might ...
3answers
181 views
### How can I keep a smaller water reservoir's water level at half available when being fed from a larger reservoir? [closed]
I'm trying to create my own ultrasonic humidifier. I ordered the misting part which works great but it only functions correctly in shallow water. So I'd like to feed from a large water reservoir to a ...
1answer
80 views
### Measure maximum volume of liquid of a water tap
I need to know the consumption of a water tap in Liters. I'm able to know the cross-sectional area of the pipe and how long the tap is open. With these data, how can I know the consumption/volume of ...
1answer
161 views
### Can you measure the speed of water coming out of a hose by its arc?
Water comes out of a horizontally stationed hose and creates an arc as it heads towards the ground. Can I determine the speed the water was traveling in when it exited the hose by the measuring the ...
1answer
205 views
### Pump pressure required to fill water tower
I have a water pump at a distance of x meters from a water tower located h meters above the pump through a pipe with diameter d. I am not concerned with the exact value just want to make sure my pump ...
4answers
418 views
### How far can water rise above the edge of a glass?
When you fill a glass with water, water forms a concave meniscus with constant contact angle $\theta$ (typically $\theta=20^\circ$ for tap water): Once you reach the top of the glass, the water-air ...
2answers
225 views
### Will the siphoning effect help a system pump water upwards if the water's entry and exit points are at the same height?
I am looking to pump water from a pool up to a roof for solar heating (black plastic tubing) and then back into the pool with the original source water. Does the gravitational force of the water ...
2answers
302 views
### Why water is not superfluid?
My question is in the title. I do not really understand why water is not a superfluid. Maybe I make a mistake but the fact that water is not suprfluid comes from the fact that the elementary ...
2answers
155 views
### Fluid Dynamics - Water Bottle Drink Mix: Air or No Air?
If you take an ordinary sized plastic water bottle full of water and pour a packet of powdered (or liquid) drink mix into it, Will shaking the bottle with the cap screwed on to dissolve the mix into ...
0answers
128 views
### Reduction in velocity in pipe
I have two pipes divided from main pipe. If I add one more pipe, what would be the amount of velocity in each pipes. I know the velocity of main pipe, then diameter and radius of each pipe. Edit: 1) ...
5answers
313 views
### Zigzag flow of water along a vertical glass window
I've observed this behavior many times. When it rains, the rainwater will form vertical channels along a glass window. The flow of water is mostly confined within these vertical channels and the ...
1answer
75 views
### how to stop water reaching up inside an under water structure (see video)
in this video at arround 0:47 they deliver the 'dry pots' to the under water lab, bellow the water and they seem to swim up into an opening where it is completly dry.. how is this dry, surly the water ...
2answers
2k views
### Flow of liquid among branches
If water is flowing through big pipe is branched into 4 branches of small pipe. Lets say the flow is around 4 m/sec. I have the following questions: What will be the flow rate in each of the ...
1answer
115 views
### How would this rapid depressurization of water really behave?
An in-development game(Starbound) recently released a demo of their water physics. At about 39 seconds into the video, a volume of water in a container is released when the cap on the container is ...
1answer
713 views
### Why does water flow out of an upside-down bottle? (Rayleigh Taylor Instability)
I am currently reading the excellent book An Indispensable Truth: How Fusion Power Can Save the Planet by Francis F. Chen and I came across this explanation. The Rayleigh–Taylor Instability ...
1answer
408 views
### How much water must flow trough canal to maintain a constant water deep?
In order to maintain a constant water deep in canal, how much water must flow trought the pipe ? As shown on picture, canal have a rectangular shape. I don't know if canal length have an influence. ...
2answers
2k views
### Why does water pouring from a glass sometimes travel down the side of the glass?
If you have a glass of water, say, three quarters full and you pour it at an angle of say, $45^{\circ}$ with respect to the the table, the water comes out of the glass and goes directly down towards ...
3answers
7k views
### Relation between water flow and pressure
Is there any equation that states the relation between pressure and water flow. I.e. Let's say that in 1 hour with 8mca (water collum meters) pressure I obtain 50m3. What if (giving the same ...
2answers
2k views
### Before a once-warm lake starts to freeze, must its temperature be 4°C throughout at some point?
This is a problem I just started puzzling over, and I felt this would be a good forum to check my reasoning. So here are the relevant observations followed by my question: Water achieves its maximum ...
1answer
554 views
### static flow of water
The title, I don't know whether it's correct or not, but I came across a video in youtube, http://youtu.be/_PkgQQqpH2M. The author of video used the title and hence I used the same.. The video ...
2answers
107 views
### Wave with mass transport?
I don't think this question has been asked on this forum before (at least I didn't find it). In the case of a tsunami, an earthquake generates a wave which will travel with the sea/ocean as the ...
1answer
219 views
### How to model water flow with and without friction against other objects?
I am interested in modeling water flow including boundary flow (with friction) against other objects in the path of the water flow. I'm not a physicist but I am interested to learn the relevant ...
1answer
175 views
### I want to recreate a phenomenon I've noticed from our water tank
I'm trying to recreate a phenomenon I have noticed from our water tank at home. Ok so we have a deep well pump that fills our water tank everyday. But something odd always happen everytime the water ...
1answer
240 views
### Calculating spacecraft water leak flow rate
A large container of heated water is floating in space. How can I calculate leak rate if a small puncture is opened? I'm assuming the water is kept at 298K and in the liquid state inside the chamber. ...
1answer
99 views
### Clouds in closed hydrosphere
Is possible estimate the needed size of an geodesic dome (like in the Eden project) for creating an real hydrosphere - especially clouds (and rain)? With other words, under what circumstances can ...
2answers
658 views
### Why doesn't water come out of tap/faucet at high pressure when I turn it on?
(tap=faucet) When I turn a tap on full and then put my thumb over the spout covering, say, 90% of it, then the water spurts out. If I turn it on to, say 10%, then the water dribbles out. What's the ...
2answers
206 views
### Water and superconductors
If you have seen my previous questions then this may make better sense. If you put an insanely powerful superconducter on the outside layer of a ROV that is underwater, would it form an outer layer ...
1answer
250 views
### Tear drop shape
i heard that a tear drop shap is the most aerodynamic shape possible or the best is this true? If this isn't true what is since i need to make a fast ROV? Also since i need to have a propeller to ...
3answers
197 views
### can Superconductivity be used to repel water?
Can water be repeled with the force from superconductivity gathering current on the surface cuased by Faraday law of induction? I'm trying to make a ROV that can run underwater that uses ...
1answer
148 views
### Surface Tension of a Liquid - When a liquid body is acclerating
As far as I understand it (which admittedly isn't very far), surface tension forces are made up by the tension-attractive forces of molecules at the liquid-gas/vacuum interface, such as those between ...
5answers
336 views
### What can wavy patterned sand tell about the Fluid that formed it?
Sand on the bottom of the ocean as well as sand on the low-tide beach often forms wavy patterns. Do the parameters of these wavy patterns have any relation to the water and waves that formed them? If ...
2answers
611 views
### How to calculate concentration of vapor at the surface of a water drop
I'm reading a paper that examines the evaporation rates of water. In the final formula, it has the following constant: $c_s - c_\infty$ where $c_s$ is the concentration of the vapor at the sphere ...
1answer
383 views
### Water droplet evaporation due to humidity and temperature
I am conducting an experiment that requires comparison to models for a water droplet's change in size due to evaporation. I've found plenty of papers, but they all seem to require measurement of the ...
2answers
345 views
### the form of a kettle
What is the best form for a kettle, that is, to have the water boil the fastest? I am particularly interested in the following case: for one given kettle (whose volume is constant) containing ...
1answer
739 views
### water flow in a sink
When one turns on the tap in the kitchen, a circle is observable in the water flowing in the sink. The circle is the boundary between laminar and turbulent flow of the water (maybe this is the wrong ...
1answer
168 views
### Resonance modes of a cubic box of water which we shake
I don't know hydrodynamics, but I wonder how one would compute resonance modes of a cubic box of water which we shake. I believe the waves would directly depend on the height of water and the width ...
3answers
2k views
### Why does the water level equalize in a series of tubes?
Say I have a series of tubes (not the internet) looking like this, where w represents water: ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9430021643638611, "perplexity_flag": "middle"}
|
http://www.physicsforums.com/showthread.php?p=4252844
|
Physics Forums
## Space Dependence of Electric potential
1. The problem statement, all variables and given/known data
The space dependence of an electric potential V($\vec{r}$) = V(x,y,z)=V0ln((sqrt{x2 + y2})/a)
1. What is the electric field at position $\vec{r}$ = <x,y,z>?
2. Explain how the electric field looks in general. Make a sketch.
3. What object would produce an electric field like this?
3. The attempt at a solution
I haven't really tried much yet. What really confuses me about this question is the part in the prompt that comes after the equals sign. I'm not really sure what I'm supposed to do with it. To find the electric field do they just want me to take the gradient of the potential or something? That doesn't seem right because there's only a constant V0. I'm a bit lost....
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Mentor
Quote by PhysicsRob 1. The problem statement, all variables and given/known data The space dependence of an electric potential V($\vec{r}$) = V(x,y,z)=V0ln((sqrt{x2 + y2})/a) 1. What is the electric field at position $\vec{r}$ = <x,y,z>? 2. Explain how the electric field looks in general. Make a sketch. 3. What object would produce an electric field like this? 3. The attempt at a solution I haven't really tried much yet. What really confuses me about this question is the part in the prompt that comes after the equals sign. I'm not really sure what I'm supposed to do with it. To find the electric field do they just want me to take the gradient of the potential or something? That doesn't seem right because there's only a constant V0. I'm a bit lost....
Yes, the electric field is the (negative) gradient of the electric potential. What do you mean "in the prompt?" In any case, V is certainly not constant. It varies with x and y.
Well I understand the idea that V($\vec{r}$) varies with x and y, but isn't V0 a constant? That's what's confusing me. Generally when things have the subscript "0", they're looked at as a constant. If this is true, then how would I take the gradient of the voltage? And when I say "In the prompt", I mean the information they give you before they ask the questions, aka: "The space dependence of an electric potential V(r⃗ ) = V(x,y,z)=V0ln((sqrt{x2 + y2})/a)"
Mentor
## Space Dependence of Electric potential
Uhh...yes V0 is a constant, but it is being multiplied by something that is a function of x and y. The V(x,y,z) that you have been given in this problem is not constant.
Ohhhh, wait a second. I see what I'm doing now. So by doing the gradient I got that: Ex = -(V0x)/(x2 + y2) Ey = -(V0y)/(x2 + y2) Ez = 0 And if you model sqrt{x2 + y2} as a vector "r", you get that the field has an inverse dependence on r and that the object producing the field would probably be a line of charge, correct?
Recognitions:
Homework Help
Science Advisor
Quote by PhysicsRob Ohhhh, wait a second. I see what I'm doing now. So by doing the gradient I got that: Ex = -(V0x)/(x2 + y2) Ey = -(V0y)/(x2 + y2) Ez = 0 And if you model sqrt{x2 + y2} as a vector "r", you get that the field has an inverse dependence on r and that the object producing the field would probably be a line of charge, correct?
Well, sqrt(x^2+y^2) is the scalar "r", but yes, I'd say it looks like a line of charge. Can you say where the line of charge is and what direction it's pointing?
Well... if you were to move along the z-axis while keeping the same x and y coordinates, your change in voltage would be 0 since the voltage doesn't depend on z. So the line of charge would be along the z axis and just points in the $\hat{k}$ direction, right?
Recognitions:
Homework Help
Science Advisor
Quote by PhysicsRob Well... if you were to move along the z-axis while keeping the same x and y coordinates, your change in voltage would be 0 since the voltage doesn't depend on z. So the line of charge would be along the z axis and just points in the $\hat{k}$ direction, right?
Right!
Tags
electric field, gradient, potential, vectors
Thread Tools
Similar Threads for: Space Dependence of Electric potential
Thread Forum Replies
Precalculus Mathematics Homework 2
Introductory Physics Homework 5
Introductory Physics Homework 0
Calculus & Beyond Homework 2
General Physics 0
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9444499015808105, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/87030/proving-sum-limits-i-0n-i-2i-1-n1-2n-1-by-induction
|
# Proving $\sum\limits_{i=0}^n i 2^{i-1} = (n+1) 2^n - 1$ by induction
I'm trying to apply an induction proof to show that $((n+1) 2^n - 1$ is the sum of $(i 2^{i-1})$ from $0$ to $n$.
• the base case: L.H.S = R.H.S
• we assume that $(k+1) 2^k - 1$ is true.
• we need to prove that $(k+2) 2^{k+1} - 1$
My try to prove 3 is as follows:
$(k+2) 2^{k+1} - 1$
$(k+2) (2^k * 2) - 1$ , from 2: $2^k = 1/(k+1)$
$(k+2) (2 / (k+1)) - 1$
$(k+1) 2 - 1$
My question is, how could I get to $2^k$ from the last line to prove this formula is right?
-
I don't think you have the right formula. It does not hold for $n=1$. – David Mitra Nov 30 '11 at 11:45
Two similar questions: math.stackexchange.com/questions/53496/… and math.stackexchange.com/questions/11464/… (But I do not think that this should be closed as duplicate - since OP is asking about mistake in his approach and he is interested only in the proof by induction. I just wanted to point him to similar question.) – Martin Sleziak Nov 30 '11 at 12:02
## 2 Answers
You have a typo in your statement. You want to show $$\sum_{i=0}^n i 2^{i-1} = n 2^n - 2^n + 1$$
Base case $n = 0$:
$$\sum_{i=0}^0 i 2^{i-1} = 0 = 0 - 1 + 1$$
Assume that $$\sum_{i=0}^n i 2^{i-1} = n 2^n - 2^n + 1$$
Then make an induction step from $n$ to $n+1$. This means you want to show that given your assumption you can show
$$\sum_{i=0}^{n+1} i 2^{i-1} = (n+1) 2^{n+1} - 2^{n+1} + 1$$
Do this as follows:
$$\sum_{i=0}^{n+1} i 2^{i-1} = \sum_{i=0}^{n} i 2^{i-1} + (n+1)2^n = n 2^n - 2^n + 1 + (n+1)2^n = n 2^{n+1} + 1 = (n+1)2^{n+1} - 2^{n+1} + 1$$
Which is what you wanted to show. Hope this helps.
-
2
Is this right? $(n+1)2^n+(n+1 )2^n-1 = 2(n+1)2^n-1=(n+1)2^{n+1}-1$... The OP's formula does not hold for $n=1$. – David Mitra Nov 30 '11 at 11:43
Darn, you're right! – Matt N. Nov 30 '11 at 11:44
@DavidMitra: Let's assume it's a typo. Let's see whether I can fix this. – Matt N. Nov 30 '11 at 11:45
@DavidMitra: Thanks, David. Fixed. – Matt N. Nov 30 '11 at 11:57
Thanks! very clear.. – Sosy Nov 30 '11 at 18:25
show 1 more comment
HINT $\$ The RHS should be $\rm\:(n-1)\ 2^n + 1\:.\:$ As always, by telescopy, the inductive step arises from equating the first difference of the LHS and RHS. Here that yields the trivially proved identity
$$\rm (n+1)\ 2^n\ =\ n\ 2^{n+1} - (n-1)\ 2^n$$
which, combined with the trivial proof of the base case $\rm\:n=0\:,\:$ completes the proof by induction.
REMARK $\$ Note that absolutely no ingenuity is required. The proof by telescopy is so mechanical that it can be done by a machine. Indeed, cancelling the factor of $\rm\:2^n\:,\:$ said inductive step reduces to proving the polynomial identity $\rm\ n+1\ =\ 2\:n - (n-1)\:.\:$ For much further discussion see my many posts on telescopy.
-
sounds very helpful, but hmmm what is the first difference of the LHS and RHS? at what value (k or k+1)? – Sosy Nov 30 '11 at 18:05
@Sosy The first difference of $\rm\:f(n)\:$ wrt $\rm\:n\:$ is $\rm\:\Delta_n f(n) := f(n+1) - f(n)\:.\:$ The displayed equation above is $\rm\: \Delta\: LHS(n) = \Delta\: RHS(n)\:.\:$ This method works to mechanically "discover" the induction step on all problems of this type. See the linked posts on telescopy for background. – Gone Nov 30 '11 at 19:10
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9408401846885681, "perplexity_flag": "middle"}
|
http://crypto.stackexchange.com/questions/1921/secure-multiparty-computation-for-multiplication?answertab=active
|
# secure multiparty computation for multiplication
Suppose there are $N$ parties $p_j$, each with a binary $b_j\in{\{0,1\}}$. The problem needs to compute the multiplication of number of ones times that of zeros, that is, $R=(\sum{b_j})\times(N-\sum{b_j})$.
The computation should be secure in the sense that no party can know more than the final result $R$. For example, it's not ok to perform secure sum, because then $\sum{b_j}$ is known, which is sensitive in my problem. So, is there any existing secure computation protocol that fits my demand?
-
What if N = 6 and it turns out that R = 9? – Henrick Hellström Feb 24 '12 at 19:56
good point!... but that's just fine to my problem. actually, $\sum{b_j}$ is sensitive only when it's big, like equal to 6. – Richard Feb 24 '12 at 20:10
## 2 Answers
I'm no expert, but homomorphic encryption may work for this. I believe you can use a system that supports unlimited additions and one multiplication.
-
Well, I am unaware of any published work on this precise problem, however I believe I do have a solution.
First of all, I would note that your problem is equivalent to the problem "if each party has a bit $b_i$, can they securely compute a value $k$ with either $k = \sum b_i$ or $k = N - \sum b_i$ (without giving any further information, including which of those values it might be)". So, we can use a secure sum as part of the solution, as long as we're careful not to give any information about which sum it is.
Here is how my protocol works; it takes 4 rounds and $4N$ messages between party members. In each round, the party members arrange themselves in a cycle; each party member takes a value from the previous party member (or the data from the start of the cycle for the first party member), does some computation on the values, and then passes it along to the next party member. I also do not assume that the message between party members are private.
In an initialization phase, the party members agree on a group where the Diffie-Hellman problem is hard, and two group members $A$ and $B$ such that $A \neq 2^k \cdot B$ and $B \neq 2^k \cdot A$ for any value of $0 \leq k \leq N$ (note: I'm using additive notation for the group, if you're using a multiplicative group, that would normally be written $A \neq B ^ {2^k}$ and $B \neq A ^ {2^k}$)
Now, round 1: we start with the data consisting of the pair $(A, B)$. Each party member takes the pair that they get $(X, Y)$, selects an integer $c_i$, randomly selects either $(c_i \cdot X, c_i \cdot Y)$ or $(c_i \cdot Y, c_i \cdot X)$, and sends that pair to the next party member.
• The point of this round is that the output from the final party will be either $(c \cdot A, c \cdot B)$ or $(c \cdot B, c \cdot A)$ (for some blinding factor $c$) without anyone being able to determine which.
Now, round 2: we start with the data from the last round. Each party takes the pair they get $(X, Y)$, selects another integer $d_i$, and then if their bit $b_i$ is zero, computes $(2 \cdot d_i \cdot X, d_i \cdot Y)$, and if their bit $b_i$ is one, computes $(d_i \cdot X, 2 \cdot d_i \cdot Y)$, and sends that pair to the next party member
• The point of this round is to gather the votes; the first of the pair collects the zero vots, and second of the pair collects the one votes (again, with a blinding factor so no one can see what someone else voted).
Now, round 3: we start with the data from the last round. Each party member takes the pair that they get $(X, Y)$, selects an integer $e_i$, randomly selects either $(e_i \cdot X, e_i \cdot Y)$ or $(e_i \cdot Y, e_i \cdot X)$, and sends that pair to the next party member (except the last party member; they random select either $(e_N \cdot X)$ or $(e_N \cdot Y)$ and sends that
• The point of this round is to disguise which of the pair was the '0' vote, and which was the '1' vote; the last member returns one of them, but no one knows which one it has.
Now, round 4: we start with the data from the last round. Each party member takes the value that they get $(X)$ and removes the blinding factors they applied; that is, computes $((c_i \cdot d_i \cdot e_i) ^ {-1} \cdot X)$, and passes that to the next part member.
The result of the last party member will be either $2^k \cdot A$ or $2^k \cdot B$; we can recover the value $k$ in $\sqrt N$ time, and that is our result.
Note that:
• It will be $2^k \cdot A$ if an even number of party members decided to swap during round 1, and $2^k \cdot B$ if an odd number decided to swap. Because each party member selected this randomly and independently of anything else, tthis is unbiased, and uncorrelated to anything.
• The computed $k$ will be $\sum b_i$ if an odd number of party members decided to swap during round 3, and $N - \sum b_i$ if an even number decided to swap (and for the last party member, we count him as swapping if he selected $(e_N \cdot Y)$. Again, because each party member selected this randomly and independently of anything else, this is unbiased, and uncorrelated to anything.
-
Thinking about this, I realized that while this protocol works in the "honest but curious" model, it is vulnerable to cheaters which don't follow the protocol. For example, if someone in round 2 gave their output as $(x \cdot X, y \cdot Y)$ (for values of $x$ and $y$ they knew), then the could deduce the value of $\sum b_i$ for everyone else at the end of the protocol. Whether or not this matters would depend on whether this attack model is relevent. – poncho Feb 24 '12 at 22:49
'honest but curious' model is just fine to the problem. – Richard Feb 24 '12 at 22:57
this does not work! 'cause first party can simply know $k$ by comparing the result of round 1 and that of round 2. – Richard Feb 26 '12 at 7:45
There doesn't appear to be a problem. If N=3 (for example), and if the result of round 1 was $(X,Y)$, then the result of round two is $(2^{3-k}d_1 d_2 d_3 \cdot X, 2^k d_1 d_2 d_3 \cdot Y)$, where the first party doesn't know the values $d_2, d_3$. Verifying a guessed value of k would involve solving a decisional Diffie-Hellman problem, and we assumed that they picked a group where that problem was hard. – poncho Feb 26 '12 at 17:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 59, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9573104977607727, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/151984/find-all-graphs-with-valency-sequence-4-2-2-2-1-1
|
# Find all graphs with valency sequence $(4, 2, 2, 2, 1, 1)$
Test the following sequences using to see if they are graphical or not: (if they are graphical draw an exemplar graph having that valency sequence)
-
1
The title and the question don’t agree; are you interested in finding one such graph, or in finding all of them? – Brian M. Scott May 31 '12 at 11:34
What following sequences? Can you do something so the title and the body agree, and make sense? – Gerry Myerson May 31 '12 at 11:35
5
Perhaps you should start by considering the vertex of degree 4. Draw it and the 4 neighbors. What can you do next? – John Engbers May 31 '12 at 11:38
## 1 Answer
There is an algorithm to easily verify valency sequences.
Enumering all possible graphs is a slightly more difficult problem. As there are only a few graphs with 6 nodes you can however figure out that there are only 2 possible graphs with that sequence: (you can modify the algorithm to give you all possible graphs)
Graphs 1 and 2:
-
2
– jp26 Jun 1 '12 at 15:50
@jp26 neat to see that you are using stackexchange too. I found your description of the algorithm very straight forward and easy to understand, so I linked it. – Listing Jun 1 '12 at 20:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9412073493003845, "perplexity_flag": "head"}
|
http://unapologetic.wordpress.com/2008/11/12/representations-of-hopf-algebras-i/?like=1&source=post_flair&_wpnonce=bcc517a348
|
# The Unapologetic Mathematician
## Representations of Hopf Algebras I
We’ve seen that the category of representations of a bialgebra is monoidal. What do we get for Hopf algebras? What does an antipode buy us? Duals! At least when we restrict to finite-dimensional representations.
Again, we base things on the underlying category of vector spaces. Given a representation $\rho:H\rightarrow\hom_\mathbb{F}(V,V)$, we want to find a representation $\rho^*:H\rightarrow\hom_\mathbb{F}(V^*,V^*)$. And it should commute with the natural transformations which make up the dual structure.
Easy enough! We just take the dual of each map to find $\rho(h)^*:V^*\rightarrow V^*$. But no, this can’t work. Duality reverses the order of composition. We need an antiautomorphism $S$ to reverse the multiplication on $H$. Then we can define $\rho^*(h)=\rho(S(h))^*$.
The antiautomorphism we’ll use will be the antipode. Now to make these representations actual duals, we’ll need natural transformations $\eta_\rho:\mathbf{1}\rightarrow\rho^*\otimes\rho$ and $\epsilon_\rho:\rho\otimes\rho^*\rightarrow\mathbf{1}$. This natural transformation $\epsilon$ is not to be confused with the counit of the Hopf algebra. Given a representation $\rho$ on the finite-dimensional vector space $V$, we’ll just use the $\eta_V$ and $\epsilon_V$ that come from the duality on the category of finite-dimensional vector spaces.
Thus we find that $\epsilon_\rho$ is the pairing $v\otimes\lambda\mapsto\lambda(v)$. Does this commute with the actions of $H$? On the one side, we calculate
$\begin{aligned}\left[\left[\rho\otimes\rho^*\right](h)\right](v\otimes\lambda)=\left[\rho\left(h_{(1)}\right)\otimes\rho^*\left(h_{(2)}\right)\right](v\otimes\lambda)\\=\left[\rho\left(h_{(1)}\right)\otimes\rho\left(S\left((h_{(2)}\right)\right)^*\right](v\otimes\lambda)\\=\left[\rho\left(h_{(1)}\right)\right](v)\otimes\left[\rho\left(S\left(h_{(2)}\right)\right)^*\right](\lambda)\end{aligned}$
Then we apply the evaluation to find
$\begin{aligned}\left[\left[\rho\left(S\left(h_{(2)}\right)\right)^*\right](\lambda)\right]\left(\left[\rho\left(h_{(1)}\right)\right](v)\right)=\lambda\left(\left[\rho\left(S\left(h_{(2)}\right)\right)\right]\left(\left[\rho\left(h_{(1)}\right)\right](v)\right)\right)\\=\lambda\left(\left[\rho\left(h_{(1)}S\left(h_{(2)}\right)\right)\right](v)\right)=\lambda\left(\left[\rho\left(\mu\left(h_{(1)}\otimes S\left(h_{(2)}\right)\right)\right)\right](v)\right)\\=\lambda\left(\left[\rho\left(\iota\left(\epsilon(h)\right)\right)\right](v)\right)=\epsilon(h)\lambda(v)\end{aligned}$
Which is the same as the result we’d get by applying the “unit” action after evaluating. Notice how we used the definition of the dual map, the fact that $\rho$ is a representation, and the antipodal property in obtaining this result.
This much works whether or not $V$ is a finite-dimensional vector space. The other direction, though, needs more work, especially since I waved my hands at it when I used $\mathbf{FinVect}$ as the motivating example of a category with duals. Tomorrow I’ll define this map.
## 3 Comments »
1. [...] Representations of Hopf Algebras II Now that we have a coevaluation for vector spaces, let’s make sure that it intertwines the actions of a Hopf algebra. Then we can finish showing that the category of representations of a Hopf algebra has duals. [...]
Pingback by | November 14, 2008 | Reply
2. [...] Category of Representations of a Hopf Algebra It took us two posts, but we showed that the category of representations of a Hopf algebra has duals. This is on [...]
Pingback by | November 18, 2008 | Reply
3. [...] algebra. Thus the category of representations of is monoidal — symmetric, even — and has duals. Let’s consider these structures a bit more [...]
Pingback by | November 21, 2008 | Reply
« Previous | Next »
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 21, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8988432288169861, "perplexity_flag": "head"}
|
http://mathhelpforum.com/geometry/153727-finding-volume-surface-area-right-prism.html
|
# Thread:
1. ## Finding volume and surface area of right prism
How to find the volume and surface area of right prisms in the attachments ?
2. $\mbox{Volume} = \mbox{Base} * \mbox{Height} * \mbox{Depth}$
$\mbox{Surface area (Of each face)} = \mbox{Base} * \mbox{Height}$
Divide the prisms into many simpler shapes (eg. rectangular cubes for volume, or rectangles/squares for surface area) so it is easier to work with and find the surface area and volume seperately. Afterwards add them together to get the final surface area or volume of each prism.
For the wedge, use:
$\mbox{Volume} = \dfrac{\mbox{Base} * \mbox{Height} * \mbox{Depth}}{2}$ for the volume
$\mbox{Surface area (Of each triangular face)} = \dfrac{\mbox{Base} * \mbox{Height}}{2}$ for the surface area of the triangular faces
3. I think I'll put more detail for the volume of a prism.
$Volume\ =\ Area\ of\ Cross\ Section\ \times\ Depth$
You need to first identify the cross section of the solid, the area which 'repeats' itself through the solid. In the first one, the cross section is a cross, in the second, it's this 'L' shape and in the last, it's a right angled triangle. Once you get their cross sectiona; area, you'll be able to find the volume by multiplying it by the depth of the solid.
4. For attachment 1 Volume=12*16*20=3840 ?
and area=12*16=192
Is this the right answer ?
5. Please solve the 1st one with ur details so that i can understand.
6. First one:
$\mbox{Total Volume} = (4 * 5 * 3) + (4 * 5 * 3) + (14 * 3 * 4)$
$\mbox{Total Volume} = 288~ \mbox{units}^3$
$\mbox{Surface Area} = (5*4*(8)) + (4*4*(2)) + (4*3*(4)) + (5*3*(8))$ (The single numbers in brackets () are how many faces of each there are.)
$\mbox{Surface Area} = 360~\mbox{units}^2$
7. The area of the cross can be obtained by dividing the cross into 5 parts; 4 rectangles and 1 central square.
Each rectangle has area 5*4 = 20.
Area of 4 rectangles = 20*4 = 80.
Area of central square = 4*4 = 16.
Total area of cross section = 16 + 80 = 96.
Volume of Prism = 96 x 3 = 288 units^3
8. Very difficult question to understand.
9. Unit^2 or unit^3 ?
10. Oh, sorry, a typo from my part. It's units^3
Here is a picture to help you understand. The rectangles are reddish and the central square is grey.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9006372690200806, "perplexity_flag": "middle"}
|
http://physics.stackexchange.com/questions/22480/nuclear-reactor-control-rods
|
Nuclear reactor control rods
What is the relation between control rods surface exposed into a nuclear reactor and neutron energy? Is it linear? I mean, how do neutron absorbing rate change with the progressive immersion of control rods into the core?
-
Gut feeling says linear, as it would seem that its proportional to surface area. But, like so many things in nuclear physics, it's rarely that simple. – Manishearth♦ Mar 17 '12 at 17:25
– anna v Mar 17 '12 at 19:22
2 Answers
What is the relation between control rods surface exposed into a nuclear reactor and neutron energy?
You mention control rod surface. Why? Do you understand how control rods work? They're inserted into the core and absorb neutrons. Now, why would the surface be a metric that matters? Perhaps this was just the most obvious thing to you at the time, so let me get into the complications here.
Cross sections are higher at thermal versus fast energies. The fast and thermal neutron flux both matter a great deal for typical light water reactors. The main difference is that the average path length of a neutron (before interacting) is much longer for fast neutrons than thermal neutrons. Now, since we've made this distinction, we can ask if the control rods absorb a significant fraction of the neutrons incident on its surface. The answer is mostly "yes" for thermal energies and "no" for fast energies. See, because the thermal flux has a much smaller mean free path, it is much bumpier because of the presence of different materials, including fuel, moderator, and absorbers. It would be mostly correct to suppose that some significant fraction of a thermal neutron beam is absorbed after entering the control rod surface in terms percentage. As an order of magnitude guidance, it would be more than 1%. The same is probably not true for the fast flux. Furthermore, according to transport/diffusion physics, if you were to reduce the fast flux by a significant percentage, you would necessarily suppress the fast flux in the entire region of the core.
Anyway, let me address your question by saying that under the unphysical assumption that a control rod absorbed all neutrons incident on its surface, its reactivity contribution would be proportional to its surface area, relying on a few other assumptions as well. Notably, if you have too many control rods in the same small area they will depress the flux around each other so its not 100% valid. As I've already pointed out, this doesn't fit current reactors.
Let's look at the other extreme. Say, for instance, that the control rods absorbed a negligible fraction of the flux relative to the total flux. This is much closer to normal reactor conditions. In this case, the reactivity contribution would be proportional to the volume of the control rods times the local flux value, and this is shown mathematically in several nuclear engineering text books using perturbation theory.
Is it linear?
Again, use perturbation theory and it's linear for small movements. But what does this physically correlate to? My answer is that it is with respect to the total amount of absorbing material times the local flux where it exists. This is because the control rods don't significantly affect the shape of the flux. In the thermal energies, however, this is a mediocre assumption, which is why people use actual computer codes to design reactors.
(math warning) variables:
• Neutron flux $\phi$
• Absorption cross section $\sigma$
• Absorber atom number density $N$
• Rod cross-sectional area $A$
• Linear distance rod is moved $\Delta l$
Let's speak of an isotropic neutron flux of a single energy. Then the flux will be of a certain physical shape $\phi(\vec{r})$ and the entire point of using the words "perturbation theory" is to refer to the assumption that $\phi(\vec{r})$ doesn't change with the insertion of the control rod. If the end of a control rod is at some given $\vec{r}$ in the core, then the change in reactivity due to movement of the rod will be $\phi(\vec{r}) N \sigma A \Delta l$ which will come out to units of $1/s$, representing the number of neutrons absorbed per second from the volume of the absorber element introduced. This affects the power of the core through a concept called the multiplication factor, which is how much the neutron chain reaction grows or declines for each neutron generation.
$$k = \frac{ \text{Fission neutrons created} }{\text{Neutrons born} }$$
The absorber removes neutrons from the population that could cause a fission and thus create move neutrons, so it can be said to subtract from the numerator of that equation. The reactor changes flux (and thus power) over time as:
$$\phi(t) \propto e^{t \frac{k-1}{\Lambda} }$$
You need not concern yourself with the details other than the fact that inserting absorbers causes the flux to decrease over time.
(end math)
I mean, how do neutron absorbing rate change with the progressive immersion of control rods into the core?
This is the most objective part of the question.
According to my prior arguments, the differential reactivity contribution depends on the flux value at the end of the rod being inserted.
So let me go back to the linearity question. The reactivity contribution from control rods is highly nonlinear with respect to control rod position, because the flux in the core changes dramatically with vertical position. AdamRedwine points out, correctly, that the flux is well-smoothed by core design, but that statement is specific to the xy-plane. This is not true for the z-direction, where it is nearly a cosine shape for a PWR, something else for the complicated thermal-hydraulic and neutronic feedback of the BWR.
Here is an example of a differential control rod worth curve.
It is greater than zero at the top and bottom of the core because some neutrons do leave the core, bounce of Hydrogen, and then enter the core again to cause fission.
For any small control rod movement it's linear. Sure. That follows from the fact that the above graph is continuous. The fact also implies that the integral control rod worth curve is smooth.
-
It's never proportional to the surface area, but to the cross section, averaged over the neutron directions. This doesn't matter, as the end result is that the absorption of relevant neutrons will be approximately proportional to the length of immersed portion of the rod. This is only modified to the extend that higher parts of other rods are in the neutron shadow cast by the other rods, so the answer is too complicated. – Ron Maimon May 15 '12 at 7:16
1
Proportional to the insertion length is a poor approximation because of the axial profile, which goes to nearly zero at the ends of the active fuel region aside from the extrapolation length, which admittedly can be over a foot while the active fuel region is 12 or 14 ft. So we're talking about a differential worth at the beginning of insertion being 5-10% what it is when the rod-end is in the middle of the reactor. – AlanSE May 15 '12 at 12:03
I see--- you mean that the outgoing neutrons don't matter, so absorbing them does nothing. You are right. Then you can give a simple function of the height which tells you how many neutrons you absorb at each depth, and integrate this. It should be zero near the ends and increasing linearly to a constant value, and the integral of this absorption per unit length (no shadows) is the number of active neutrons absorbed (as you said in the answer, but it isn't so easy to extract), +1, but try to make it simpler--- you only need one function of height. – Ron Maimon May 15 '12 at 16:42
I've left my reactor physics texts at work, but I'll at least give a general answer.
Nuclear reactor fuel loading is designed to keep the scalar flux as flat as possible. In other words, they try to avoid having the center of the reactor be much hotter than the edges. Because reactors tend toward cylindrical, you can usually work with horizontal cross sections when analyzing. When you insert the control rods, (bottom up for BWRs and top down for PWRs), you, more or less, insert them all the same distance; again, this is to keep the flux profile flat. In other words, as the rods are inserted, the neutron flux is reduced (absorption is increased) near the rods, but not in the rest of the reactor.
If you are asking just about how much of the neutron flux is absorbed by the rods, like any complex system, it depends. The amount of boron in the rods plays a part, as does the location of the particular rod under consideration. If you consider a single rod, with a known composition, in the center of a reactor, and wonder how much the flux is reduced by a move of $\Delta d$, it is close to linear. If that same rod is near the top or bottom, it is much less so. There is a graph of the integral control rod worth on page 51 of this reference.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9453656077384949, "perplexity_flag": "middle"}
|
http://stats.stackexchange.com/questions/28406/is-cosine-similarity-a-classification-or-a-clustering-technique/28414
|
# Is cosine similarity a classification or a clustering technique?
In document classification, is cosine similarity considered a classification or a clustering technique? But you need training data with the cosine similarity for creation of the centroid right?
-
## 5 Answers
No.
Cosine similarity can be computed amongst arbitrary vectors. It is a similarity measure (which can be converted to a distance measure, and then be used in any distance based classifier, such as nearest neighbor classification.)
$$\cos \varphi = \frac{a\cdot b}{||a|| ||b||}$$
Where $a$ and $b$ are whatever vectors you want to compare.
If you want to do NN classification, you would use $a$ as your new document, and $b$ as your known sample documents, then classify the new document based on the most similar sample(s).
Alternatively, you could compute a centroid for a whole class, but that would assume that the class is very consistent in itself, and that the centroid is a reasonable estimator for the cosine distances (I'm not sure about this!). NN classification is much easier for you, and less dependent on your corpus to be very consistent in itself.
Say you have the topic "sports". Some documents will talk about Soccer, others about Basketball, others about American Football. The centroid will probably be quite meaningless. Keeping a number of good sample documents for NN classification will likely work much better.
This happens commonly when one class consists of multiple clusters. It's an often misunderstood thing, classes do not necessarily equal clusters. Multiple classes may be one big cluster when they are hard to discern in the data. And on the other hand a class may well have multiple clusters if it is not very uniform.
Clustering can work well for finding good sample documents from your training data, but there are other more appropriate methods. In a supervised context, supervised methods will always perform better than unsupervised.
-
If the cosine btw two vectors is 0, they are considered to be similar. Isn't cosine similarity a distance measure then? – bayerj May 14 '12 at 12:58
2
It doesn't satisfy the triangle equality, so it is not a distance function. It also is not well-defined for the vector 0. – Anony-Mousse May 14 '12 at 14:11
you said No to "But you need training data with the cosine similarity for creation of the centroid right?"?. If so how do you create the centroid? – user601 May 16 '12 at 3:03
– Anony-Mousse May 16 '12 at 9:54
yes it is defined on two vectors. In my case since I want to determine how much a new document (vector) is relevant to the topic of my interest, for this I have to keep one vector (topic vector) constant all the time right? And that vector was what I referring to as the centroid. You initially need data (I referred as training data) in order to build that 'centroid' right? Am I referring a different approach here? – user601 May 17 '12 at 17:22
show 2 more comments
I think you have not yet understood the difference between clustering and classification.
Document classification (or supervised learning) requires a set of documents and a class information for each document (example: the topic of the document). The goal of classification is to build a model which predicts the class for documents where the class (in this example the topic) is not known. When models are applied on documents where the class is known, they can be evaluated by comparing the predicted with the true class (hence supervised). The data used for training but not evaluating the model is called training data.
Document clustering (or unsupervised learning) requires a set of documents but not a class information. The goal is to find groups / clusters in the data, so that
• documents which are similar according to a specified distance function are in one cluster. Example: Documents which roughly contains the same keywords.
• documents which are not similar according to a specified distance function are in different clusters
The resulting clusters cannot be evaluated like a classification model, because the true clusters are not known (hence unsupervised). Hence there is no such thing as training data, you simply use all data to build the clusters.
See also: Classification vs clustering
Now the connection between both techniques, and imho the source of your confusion: By defining the clusters generated by document clustering as class, one can train a classification model on the data. Example: If you cluster documents by words, you may detect that the resulting clusters are indeed describing topics. Now you can build a classification model for that automatically derived class.
Finally, as put by Anony-Mousse et al., the cosine similarity can be used both for
• clustering, by defining 1-cosine as distance function (which may not be a metric). Maybe you want use the loosely related Jaccard distance instead
• classification, by using it in e.g. k-nearest-neighbor
-
A cosine similarity function returns the cosine between vectors. A cosine is a cosine, and should not depend upon the data. However, how we decide to represent an object, like a document, as a vector may well depend upon the data.
Often, we represent an document as a vector where each dimension corresponds to a word. If the word does not appear we assign a value of 0 to that dimension. If the word does appear the value corresponds to the number of times that word appears in the document normalized by the number of times that word appears in all the documents with our data. This is the general idea behind TF/IDF. Since different sets of documents will have a different distribution of words, the TF/IDF vector representations of documents depend upon the particular document set you are working with.
Many classification and clustering methods depend upon some measure of distance and similarity or distance between objects. If they do, then they can use cosine similarity.
-
Then why cosine similarity comes as a similarity measure for text document clustering? – user601 May 13 '12 at 15:42
Because documents are often represented as vectors. Cosine similarity is a measure of whether these vectors are pointing in the same direction. In practice, this often works as a measure of whether documents are about the same thing. – fgregg May 13 '12 at 17:41
Similarity measures are not machine learning algorithm per se, but they play an integral part. After features are extracted from the raw data, the classes are selected or clusters defined implicitly by the properties of the similarity measure.
It might help to consider the Euclidean distance instead of cosine similarity. Is the Euclidean distance a learning algorithm? No, but you can use it to define one.
-
I'm doing some tests in document clustering in weka with Euclidean distance as the distance function. But my problem is even if I give a mix of unrelated (topically) and related instances during tests (even some empty instances), they all get clustered into the same group. What could be the reason for this? – user601 May 15 '12 at 15:59
You are not getting good separation in the feature space? Can you see the clusters with PCA or does it look like a jumble? – Emre May 15 '12 at 16:07
I'm using weka. I checked it but there is no option for that. Do you know of a different tool kit with those options? So if I test the model with same instances from the train set and empty instances shouldn't they get clustered into different groups? – user601 May 15 '12 at 16:26
– Emre May 15 '12 at 16:29
Thanks for the link. I did a try but it was taking so long time when I stopped. I'm wondering if I have some empty instances in my test set shouldn't they get clustered into a different cluster? – user601 May 17 '12 at 17:17
Yes. In document clustering problem, cosine similarity sometimes get better result. Check this paper: "concept decompositions for large sparse text data using clustering", Dhillon, Modha, 2000.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.928558886051178, "perplexity_flag": "middle"}
|
http://cogsci.stackexchange.com/questions/3259/what-is-the-standard-error-of-measurement-for-teacher-made-multiple-choice-tests
|
# What is the standard error of measurement for teacher made multiple choice tests?
Assume a teacher constructs a four-choice multiple choice test. Each item has only one correct response. The test is scored from 0 to 100 representing the percentage of items answered correctly.
I want to have some rules of thumb that could be informative regarding how many items are required to achieve a given standard error of measurement. For example, it would be nice to be able to advice teachers who write their own multiple choice exams "if you have 100 items that are reasonably well worded, you can expect a standard error of measurement of 2.5".
The standard error of measurement is often defined as:
$$s_e=s_x \sqrt{1-r_{xx}}$$
where $s_x$ is the standard deviation and $r_{xx}$ is the reliability.
Furthermore internal consistency reliability can be calculated from the number of items $k$ and the mean inter-item correlation $\bar{r}_{ij}$ (i.e., average correlation between item $i$ and item $j$ for all $k$ items where $i\neq j$):
$$r_{xx}=\frac{k(\bar{r}_{ij})} {1 + (k -1) \bar{r}_{ij}}$$
However, I'm wanting to translate the above information into meaningful recommendations for teachers. Thus, this assumes that I have some empirical estimate of typical values of $\bar{r}_{ij}$ and that I have an estimate of $s_x$. It then requires application of the formulas to calculate standard errors of measurement for likely numbers of items $k$. In particular, I was thinking about numbers of items equal to: 10, 20, 50, 80, 100, 120, 150, and 200.
Thus, I was wondering whether there are any published estimates of the standard error of measurement teacher constructed multiple choice tests.
-
I'm only just learning statistics, so my question stems from my ignorance: What does the standard error in a multiple choice test mean? You are not drawing a random sample from a population, but testing the complete population (the population being the class taught by the teacher). I always assumed the standard error referred to the difference between the sample and the population that it is supposed to represent. – user1196 Mar 10 at 17:19
1
Standard error of measurement is the standard deviation that would be obtained if you were able to repeatedly obtain a measure for a particular individual under hypothetical identical circumstances. I.e., it's a measure of the uncertainty you have about a measure you have obtained on a person. – Jeromy Anglim♦ Mar 10 at 23:31
1
This is awesome! It'd be great if I received all my exams and assignments back with error bars. – Artem Kaznatcheev Mar 14 at 12:48
A major issue here is how related are the items? In particular, I would worry that the item similarity increases as the number increases --- a 20 item test may be rather different, but a 200 item test may have substantial redundancy. – Joshua Mar 15 at 18:39
## 2 Answers
Harvill mentions an estimate by Lord (1959). Lord (1959) presents some data for the standard error of measurement for some moderately difficult cognitive measures. While there are many caveats (e.g., the estimate of the standard error is most accurate for scores around 50% and the estimates are based on tests that are neither particularly easy or particularly difficult with means in the .35 to .75 range), Lord provides a simple formula that can be used as a rule of thumb for predicting standard error of measurement in his sample of cognitive measures which performed quite well.
$$\hat{s}_e = .432 \sqrt{k}$$
where $k$ is the number of items. Alternatively, if you are interested in the mean correct on a 0 to 100 scale rather than the total correct, you can divide by $k$ and multiply by 100.
$$\hat{s}_e = \frac{.432 \sqrt{k}}{k} \times 100$$
When I plugged this into R for some sample values I obtained:
````> lord_approximation <- function(k) 0.432 * sqrt(k) /k * 100
> k <- c(10, 20, 50, 80, 100, 120, 150, 200)
> cbind(k, sem=round(lord_approximation(k), 2))
k sem
[1,] 10 13.66
[2,] 20 9.66
[3,] 50 6.11
[4,] 80 4.83
[5,] 100 4.32
[6,] 120 3.94
[7,] 150 3.53
[8,] 200 3.05
````
Of course, not all of this reduction in standard error of measurement is due to greater accuracy. Some of it comes from the smaller standard deviation in true scores that occurs when you take the mean of more items. Furthermore, these estimates are based on relatively well designed cognitive measures. Teacher designed tests may have slightly lower reliability and thus larger SEM.
### References
• Harvill, L. M. (1991). Standard error of measurement. Educational Measurement: Issues and Practice, 10(2), 33-41. PDF
• Lord, F. M. (1959). Tests of the same length do have the same standard error of measurement. Educational and Psychological Measurement, 19, 233-239.
-
1
interesting, but this seems to assume that each question is independent of the others. However, in an actual test there is extremely high correlations between questions. In particular, a valid measure should at least account for questions coming from some fixed number of relatively-independent units (i.e. chapters or topics) with high question-question correlation within the units. You could also give teachers a software tool to infer typical question-question correlations by doing statistics on student results on their previous exams. – Artem Kaznatcheev Mar 14 at 12:52
To me the most natural solution is to just use item response theory (IRT). IRT has been around for a few decades, so it is well established, implemented in a variety of software packages and provides a sensible, extensible framework for this type of problem.
Essentially, one assumes an underlying latent construct of interest, values of which should drive responses on the test. For multiple choice where the answer is "right" or "wrong", you can use a series of logit (canonically) or probit models. Then for each student, you can estimate the score on the underlying latent variable and that will naturally come with some estimate of its quality/variability.
Issues that are automatically handled:
• If everyone (or nearly everyone) gets an answer correct, it contributes very little information
• Corollary to #1, if almost no one gets an answer correct and someone does, it should be weighted more heavily. Essentially, item difficulty is automatically handled.
• Inter-dependence among items is accounted for. Asking the same item 10 times in a row will not artificially decrease your measurement error.
If you take a Bayesian view, for each student, you could use the model and their test responses to calculate a posterior distribution for the latent construct of interest, which would allow both a point estimate (e.g., posterior mean, median, or mode), as well as estimates of variability (e.g., standard deviation; 95% high posterior density region).
This sort of thing is essentially what big nation wide tests and testing services do. It is actually not too hard to do, but probably enough effort most teachers who already feel overworked do not adopt them.
-
Solid answer, maybe teachers would use this if you made a plug-and-play version of this that directly takes the results from scantron sheets (are those still in use?) and does the statistics for them? – Artem Kaznatcheev Mar 15 at 18:50
+1 I agree IRT (bayesian IRT in particular) provides a great framework for modelling multiple choice tests. In my particular case, I'm looking for heuristics that teachers might use to guide their decision about how many items to include in a test. I'd like to be able to give teachers a rough sense of how much more accurate their test could be if they for example increased their test from 80 to 100 items. – Jeromy Anglim♦ Mar 17 at 23:55
In addition to difficulties in implementation, I think IRT is not applied in teacher constructed test settings because there is often an implicit contract with the students. That is, a student's mark is directly related to the proportion of items answered correctly. Thus, in IRT there is still the question of how thetas will be mapped on to exam grades. Of course this can be overcome, but I think it presents another obstacle to uptake, particularly where the test is used to make judgements about absolute performance and not just normative performance. – Jeromy Anglim♦ Mar 18 at 0:54
@JeromyAnglim I agree with that, with respect to absolute performance. However, the implicit assumption there is that teachers actually have a valid measure of absolute performance. I think that is a much stricter assumption than that they have a valid measure of relative performance (I am open to being argued out of that). Regarding the accuracy of the test, I am not sure how reasonable any heuristic could be. That is dependent on the relative (in)dependence of items, right? In the extreme, can a teacher ask the same question another 20 times to get to 100 and still be more accurate? – Joshua Mar 24 at 4:29
With regards to heuristics: Any heuristic would be based on the assumption that the teacher would be writing a set of additional items of roughly comparable quality, difficulty, and variety as the existing items. I acknowledge that some teachers might differ in their skill in doing this, but at the same time, a heuristic could be frame in terms of a ball park or a typical range. – Jeromy Anglim♦ Mar 24 at 12:05
show 1 more comment
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9404118061065674, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/77806/eigenvalues-in-the-semiclassical-limit/77809
|
## Eigenvalues in the semiclassical limit
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider the Schrödinger operator $H_\hbar = -\hbar^2\Delta + V$ on $M=\mathbb{R}^n$, where $V$ is a potential that behaves well in a certain sense ($C^\infty$, bounded from below, going to infinity for large $|x|$, ...).
In the case $V = x^2$, $n=1$, the eigenfunctions are $\psi_n(x/\sqrt{\hbar})$ where the $\psi_n$ are hermite functions, and the corresponding eigenvalues are $\hbar(2n+1)$.
My question is: Is it a well-known theorem, that in the semiclassical limit $\hbar \rightarrow 0$, the eigenvalues tend to the minimum or minima of $V$ and the corresponding eigenvectors behave asymptotically like delta peaks? Can you give me references?
What about if M is a (possibly compact) riemannian manifold and $\Delta$ the Laplace-Beltrami operator?
-
## 2 Answers
The canonical reference is:
Introduction to spectral theory: with applications to Schrödinger operators by Hislop and Sigal.
Your statement about the semiclassical behavior of eigenvalues seems to be proved by Barry Simon in:
http://archive.numdam.org/ARCHIVE/AIHPA/AIHPA_1983__38_3/AIHPA_1983__38_3_295_0/AIHPA_1983__38_3_295_0.pdf
(where he calls it a "folk theorem").
-
Thank you very much! – Kofi Oct 11 2011 at 12:20
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Another reference is Helffer and several Coauthors (Sjöstrand, Nier, Klein, Garyard, ...) using the Witten-Laplace approach.
An introduction is given in the book Semiclassical analysis, Witten Laplacians, and statistical mechanics
Later sharp asymptotics for the low lying spectra in the case where $V$ consists of several minima were obtained. Some lecture note on this topic Low lying eigenvalues of Witten Laplacians and metastability (after Helffer-Klein-Nier and Helffer-Nier).
If you are only interessted in the Schrödinger Operator, maybe the book Semi-Classical Analysis for the Schrödinger Operator and Applications is the most interesting for you. There are also lecture notes available Semiclassical Analysis for Schrödinger Operators, Laplace Integrals and Transfer Operators in large dimension: an introduction.
-
Thanks, these are really nice references! – Igor Rivin Oct 11 2011 at 13:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8635799884796143, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/62683?sort=newest
|
## fgf = f, gfg = g, fg not necessarily identity, what was that called?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A very simple question, I just totally forgot how it was called, and google is not helping.
There's a pair of functions $f:X\to Y$, $g:Y\to X$.
$fgf = f$, $gfg = g$, but $fg$ and $gf$ don't need to be identities (and usually are not in interesting cases).
A simple example would be $f(a,b,c)=(a,b)$, $g(a,b)=(a,b,0)$
What were $f$ and $g$ called?
-
The notion of a von Neumann inverse or of a von Neumann regular element has some resemblance to what you're looking for. – Matthias Künzer Apr 26 2011 at 8:56
## 2 Answers
It is called "generalized inverse". In that case $fg$ and $gf$ are idempotents. In particular, if you have a semigroup of maps $X\to X$ (i.e. a set of maps closed under composition) such that every $f$ has a generalized inverse, the semigroup is called regular. If the generalized inverse is unique, the semigroup is called inverse. See Clifford and Preston "Algebraic theory of semigroups".
-
Yes, $fg$ and $gf$ would be idempotents. I'm sure the name was something else than "generalized inverse" back when I learned it. It was in context of computer science and program analysis, $f$ would normally project to a simpler domain, $g$ would expand back. If $fgf=f$ and $gfg=g$, this allows every function of tuples of $X$ to be extended to tuples of $Y$. Or something like that. Does it help? – t-a-w Apr 23 2011 at 1:25
3
I do not know the application you have in mind, unfortunately. In the case of matrices (i.e. linear maps from ${\mathbb R}^n$ to ${\mathbb R}^m$), it is sometimes called pseudoinverse (see, say, en.wikipedia.org/wiki/…). – Mark Sapir Apr 23 2011 at 2:02
In your case, by the way, you have two linear transformations $(a,b,c)\to (a,b)$ and $(a,b)\to (a,b,0)$ which are Moore-Penrose pseudoinverses of each other. – Mark Sapir Apr 23 2011 at 3:41
2
@t-a-w: In domain theory you'd have seen Galois connections - is that it? – Colin McQuillan Apr 23 2011 at 9:19
Thanks, "galois connections" is just the name I've been looking for. Too bad I cannot accept a comment as answer, so I'll accept this one and upvote the comment. – t-a-w Apr 23 2011 at 13:30
show 2 more comments
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Linear case
In the linear case, these identities are part of the definition of the Moore-Penrose pseudo-inverse, which exists and is unique. Given $A\in M_{p\times q}(\mathbb C)$, its MPpi is the matrix $A^\dagger\in M_{q\times p}(\mathbb C)$ that satisfies $$AA^\dagger A=A,\qquad A^\dagger AA^\dagger=A^\dagger,\qquad(AA^\dagger)^H=AA^\dagger,\qquad(A^\dagger A)^H=A^\dagger A,$$ where the superscript $H$ stands for the Hermitian adjoint.
If $A\in GL_n(\mathbb C)$, then $A^\dagger=A^{-1}$. But otherwise, $AA^\dagger$ and $A^\dagger A$ are only unitary projections.
Nonlinear case
The situation where $f=g$ is amazing: one looks at functions $h$ such that $h\circ h\neq {\rm id}$, whereas $h\circ h\circ h=h$. Then we have $h^{(2k)}=h^2$ and $h^{(2k-1)}=h$ for every $k\ge1$.
Such an $h$ can be obtained by the following construction, when we are given $f,g$ such that $fgf=f$, $gfg=g$ and at least one of $fg$ or $gf$ is not the identity. Just define $h(x,y)=(f(x),g(y))$ on the cartesian product.
Application: take for $f$ the backward shift on $\ell^p({\mathbb N})$ and for $g$ the forward shift.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9452020525932312, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/tagged/monads
|
## Tagged Questions
1answer
238 views
### When do reflexive coequalizers preserve weak equivalences?
In my work I've run into the following situation. In a model category, I have two reflexive coequalizers $A_i \stackrel{\to}{\to} B_i \to C_i$ and a map of diagrams which is level …
1answer
354 views
### Reference for my monads?
I'm looking for a reference for a certain pair of monads on $Cat$. One problem is that I don't know the modern way of thinking about some basic things, so excuse me if my presentat …
1answer
189 views
### The crude monadicity theorem
In order to test the monadicity of a functor, there is a precise monadicity theorem (PM) as well as a crude monadicity theorem (CM), see the nlab. In CM, the forgetful functor shou …
1answer
170 views
### Free cocommutative commutative Hopf monoids
I have some questions about generalizations of abelian groups, relative to symmetric monoidal categories. 1) Let $C$ be a cocomplete cartesian monoidal category with equalizers. I …
1answer
252 views
### Applications of the Giry monad in probability and statistics
In another thread, I asked about the $M$ endofunctor on the category $\operatorname{Meas}$ of measurable spaces, which sends a space $X$ to its space of measures $M(X)$. Will Saw …
1answer
137 views
### Seems like Reader monad composed with a strong monad produces a monad, am I right?
Take a Cartesian (or monoidal) closed category; define Reader monad for a given object $E$ as $X \mapsto X^E$; and take a strong monad $M$ (strong means preserves product or tensor …
0answers
82 views
### Compatibility between strength and costrength of a monoidal monad
Let $C$ be a closed monoidal category, and let $T : C \to C$ be a monad on the underlying category. Let $\sigma$ be a tensorial strength of $T$ and let $\sigma'$ be a cotensorial s …
0answers
236 views
### Can we describe equivariant vector bundles of free group action in terms of descent theory (Barr-Beck theorem)?
It is well known that for a compact topological group $G$ acts (say, from the right) freely on a compact space $X$. Then the category of equivariant complex vector bundles on $X$, …
2answers
484 views
### “Functors between monads”: what are these really called?
Let $(S,\eta,\mu)$ be a monad on a category $C$, and $(T,\eta,\mu)$ a monad on a category $D$. The following kind of gadget is ubiquitous: a functor $F:D\to C$, together with a na …
2answers
348 views
### When do functors induce monadic adjunctions of presheaf categories
For a category $C$, let $C-Set$ denote the category of set-valued functors $\delta\colon C\to Set$. Given categories $C$ and $D$, and a functor $F\colon C\to D$, composition with \$ …
1answer
237 views
### Coproducts of modules over an algebraic monad
Coproducts of modules over an algebraic monad $\Sigma$ are described in Section 4.16.14/15 in Durov's thesis. It is claimed there that for $\Sigma$-modules $M,N$, the set \$M \copro …
0answers
139 views
### Coalgebra for a comonad for a really tiny category [closed]
I have a category with only two objects. These objects are just sets, each set has two elements in it. I take the morphisms as all endo-maps of the sets. There are no morphisms …
1answer
361 views
### Monadicity theorem in homotopy theory.
Let $\mathbf{C}$ be a cofibrantly generated model category (assume for simplicity that all objects are fibrant) and $\mathbf{C}^{\mathrm{T}}$ the category of $\mathrm{T}$-algebras …
1answer
615 views
### The reflexive free-category comonad-resolution is a cofibrant replacement of the discrete simplicial category associated with an ordinary category in the Bergner model structure on the category of small simplicial categories?
Let $X$ be the category of reflexive quivers, and let $Cat$ be the category of small categories. There exists an evident forgetful functor $U:Cat\to X$ sending a category $A$ to i …
3answers
768 views
### Relation between monads, operads and algebraic theories
I've begun to interest in algebraic theories and their categorical models: in particular monads, generalized multicategories and operads, lawvere theories and their generalization. …
15 30 50 per page
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8729142546653748, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/57279/question-about-all-the-homomorphisms-from-mathbbz-to-mathbbz/104640
|
# Question about all the homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$
An exercise in "A first course in Abstract Algebra" asked the following: Describe all ring homomorphisms from the ring $\mathbb{Z},+,\cdot$ to itself.
I observed that for any such ring homomorphism the following has to hold: $$\varphi(1) = \varphi(1\cdot 1) = \varphi(1) \cdot \varphi(1)$$ In $\mathbb{Z}$ only two numbers exists so that their square equals itself: 0 and 1.
When $\varphi(1) = 0$ then $\varphi = 0$ hence $\forall n \in \mathbb{Z}$: $\varphi(n) = \varphi(n \cdot 1) = \varphi(n) \cdot \varphi(1) = \varphi(n) \cdot 0 = 0$.
Now, when $\varphi(1) = 1$ I showed that $\varphi(n) = n$ using induction
Base case: $n = 1$, which is true by our assumption
Induction hypothesis: $\varphi(m) = m$ for $m < n$
Induction step: $\varphi(n) = \varphi((n-1) + 1) = \varphi(n-1) + \varphi(1) = n-1 + 1 = n$
Now I wonder whether you could show that $\varphi(n) = n$ when $\varphi(1) = 1$ without using induction, which seems overkill for this exercise.
EDIT: Forgot about the negative n's. Hence $\varphi$ is also a group homomorphism under $\mathbb{Z},+$, we know that $\varphi(-n) = -\varphi(n)$. Thus, $$\varphi(-n) = -\varphi(n) = -n$$
-
4
I would say that you need to do the induction at least once while learning. As you said, it is relatively straightforward, so you can skip it the next time it shows up (at least while self-learning). But you also need to check the negative numbers, and another trick is needed for that (induction takes care of the positive side only)! Also many (but not all) algebraists require that a ring homomorphism must map 1 to 1. – Jyrki Lahtonen Aug 13 '11 at 19:23
oh you are right I forgot about -n – Dimitri Surinx Aug 13 '11 at 19:25
1
Authors often require $1$ to map to $1$ under a ring homomorphism, so check your text (the zero ring has $0 = 1$, so there's no issue there). As for the induction, I think it's pretty rare for induction to be overkill. Most people would write $\varphi(n) = \varphi(\sum_{i = 1}^n 1) = \sum_{i = 1}^n \varphi(1)$, but at some point in the logical chain there is induction, even in these obvious equalities. – Dylan Moreland Aug 13 '11 at 19:27
2
@Dimitri: The part about negative numbers looks good to me. You may also consider typing your completed argument as an answer. If you are happy with your understanding of this problem, you can accept that answer. The purpose of that exercise is twofold: 1) some of us may want to give you upvotes, and (more importantly) 2) the question won't be lingering in the ranks of unanswered ones (helping the system a bit). I don't know the proper protocol for this. May be it is polite to wait a little while in case somebody wants to add something that we may have overlooked before accepting? – Jyrki Lahtonen Aug 13 '11 at 20:17
2
Dear Dimitri, I'd say that the answer is no because $\mathbb Z$ itself is characterized by a property which involves induction. So, you may be able to hide induction, but not to remove it. – Pierre-Yves Gaillard Aug 14 '11 at 3:39
show 3 more comments
## 2 Answers
I assume you are talking about Fraleigh's book. If so, he does not require that a ring homomorphism maps the multiplicative identity to itself. Follow his hint by concentrating on the possible values for $f(1)$. If $f$ is a (group) homomorphism for the group $(\mathbb{Z},+)$ and $f(1)=a$, then $f$ will reduce to multiplication by $a$. For what values of $a$ will you get a ring homomorphism? You will need to have $(mn)a=(ma)(na)$ for all pairs $(m,n)$ of integers. What can you conclude about the value of $a$? You still won't have a lot of homomorphisms.
-
If you are working in the category of unitary associative rings, the morphisms $\varphi : R\to S$ must satisfy $\varphi (1)=1$. In this category the ring $\mathbb{Z}$ is an initial object, that is, for any ring $R$ there is exactly one morphism (i.e. ring homomorphism) $\chi : \mathbb{Z}\to R$ (which defines the characteristic of the ring $R$). In particular there is exactly one ring homomorphism $\mathbb{Z}\to\mathbb{Z}$, which is the identity map.
-
2
Sorry, -1: I think this a) Clearly not a suitable response for this asker, and b) Essentially hiding all of the essential mathematics behind a veil of language. (How do you prove that $\mathbb{Z}$ is an initial object?) – Cam McLeman Feb 1 '12 at 16:03
OK, I forgot to comment that @Dimitri 's proof by induction works well to prove that $\mathbb{Z}$ is an initial object. The statement to prove is: for each $n \in\mathbb{Z}$, $\varphi (n)=n1_R$ ($n$ times the identity of $R$). – Loronegro Feb 1 '12 at 18:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 47, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9473157525062561, "perplexity_flag": "head"}
|
http://mathhelpforum.com/statistics/6645-please-help.html
|
# Thread:
1. ## please help
here is a probability question:
if two of the four expressions (x+y), (x+5y), (x-y) and (5x-y) are selected at random, what is the probability of their product being of the form x^2-(by)^2, where b is an integer? thanks in advance
2. Originally Posted by helpjoe
here is a probability question:
if two of the four expressions (x+y), (x+5y), (x-y) and (5x-y) are selected at random, what is the probability of their product being of the form x^2-(by)^2, where b is an integer? thanks in advance
Probability is not my forte so someone please correct me if I'm wrong.
1. $(x+y)(x+5y) = x^2 + 6xy + 5y^2$
2. $(x+y)(x-y) = x^2 - y^2$
3. $(x+y)(5x-y) = 5x^2 + 4xy - y^2$
4. $(x+5y)(x-y) = x^2 + 4xy - 5y^2$
5. $(x+5y)(5x-y) = 5x^2 -5y^2$
6. $(x-y)(5x-y) = 5x^2 - 6xy +y^2$
These are all the different possibilities. Of these only one pair produces the required form, #2. So we need to pick two choices, of which there are only two elements we can choose. So by the counting principle we have 2 choices for the first factor, then we are left with only 1 possibility for the second choice. There are 4*3 ways to make a product of any two of the factors. So:
$P = \frac{2*1}{4*3} = \frac{1}{6}$.
So we have a 1 in 6 chance of getting the required product.
-Dan
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9422880411148071, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/105474?sort=newest
|
## Proving that a specific function is quasiconvex
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello all,
Assume we have a sequence of quasiconcave functions (in $X$) denoted by $f_{i,j}(X)$ for $i,j = 1,\ldots,n$. Denote by $F(X)$ the $n\times n$ matrix whose $(i,j)$ entry is the function $f_{i,j}(X)$.
Assuming that $F\succ0$ (positive definite for all $X$), I want to prove (or disprove) that the function $g(X)=a^TF^{-1}a$, where $a\in\mathbb{R}_+^{n\times 1}$, is quasiconvex.
Someone have any idea?
Thank you!
Correction: $f_{i,j}(X)$ are quasiconcave and not quasiconvex. Credit to Robert.
-
1
What do you mean by $F \succ 0$? Positive definite for all $X$? What about the $1 \times 1$ example $f_{1,1}(X) = 1+x^2$? – Robert Israel Aug 26 at 22:17
Hmm, this is exactly what I meant. Oops!! I meant that $f_{i,j(X)}$ are QUASICONCAVE !! Thank you for the correction! – Josh Aug 27 at 5:53
## 1 Answer
It's not true.
Consider the $2 \times 2$ matrix $$F(X) = \pmatrix{f(X) & 0\cr 0 & f(X-2)\cr}$$ where $f$ is an even function, everywhere $> 0$, and decreasing on $[0,\infty)$. Take $a = (1,1)^T$. Then $g(X) = a^T F(X)^{-1} a = 1/f(X) + 1/f(X-2)$. In particular $g(0) = g(2) = 1/f(0) + 1/f(2)$ while $g(1) = 2/f(1)$.
If $2/f(1) > 1/f(0) + 1/f(2)$, $g$ is not quasiconvex. For example, we could take $f(0) = 4$, $f(1) = 2$, $f(2) = 1$.
-
Great example. Thanks again – Josh Aug 27 at 18:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8804900050163269, "perplexity_flag": "head"}
|
http://www.reference.com/browse/crumbling+power
|
Definitions
# Power-law index profile
For optical fibers, a power-law index profile is an index of refraction profile characterized by
$n\left(r\right) =$
begin{cases} n_1 sqrt{1-2Deltaleft({r over alpha}right)^g} & r le alpha n_1 sqrt{1-2Delta} & r ge alpha end{cases} where $Delta = \left\{n_1^2 - n_2^2 over 2 n_1^2\right\},$
and $n\left(r\right)$ is the nominal refractive index as a function of distance from the fiber axis, $n_1$ is the nominal refractive index on axis, $n_2$ is the refractive index of the cladding, which is taken to be homogeneous ($n\left(r\right)=n_2 mathrm\left\{ for \right\} r ge alpha$), $alpha$ is the core radius, and $g$ is a parameter that defines the shape of the profile. $alpha$ is often used in place of $g$. Hence, this is sometimes called an alpha profile.
For this class of profiles, multimode distortion is smallest when $g$ takes a particular value depending on the material used. For most materials, this optimum value is approximately 2. In the limit of infinite $g$, the profile becomes a step-index profile.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8080602288246155, "perplexity_flag": "head"}
|
http://mathforum.org/mathimages/index.php?title=Divergence_Theorem&diff=6618&oldid=6609
|
Divergence Theorem
From Math Images
(Difference between revisions)
| | | | |
|----------|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|----------|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| | | | |
| Line 38: | | Line 38: | |
| | We now turn to the right side of the equation, the integral of flux. | | We now turn to the right side of the equation, the integral of flux. |
| | :'''Step 3:''' We first parametrize the parts of the surface which have non-zero flux. | | :'''Step 3:''' We first parametrize the parts of the surface which have non-zero flux. |
| - | :{{hide|1= :Notice that the given vector field has vectors which only extend in the x-direction, since each vector has zero y and z components. Therefore, only two sides of our cube can have vectors normal to them, those sides which are perpendicular to the x-axis.<math> test </math> }} | + | :{{hide|1= :Notice that the given vector field has vectors which only extend in the x-direction, since each vector has zero y and z components. Therefore, only two sides of our cube can have vectors normal to them, those sides which are perpendicular to the x-axis. Furthermore, the side of the cube perpendicular to the x axis and with all points at x = 0 cannot have any flux, since all vectors in this surface are zero vectors. |
| | | + | |
| | | + | We are thus only concerned with one side of the cube since only one side has non-zero flux. This side is parametrized using |
| | | + | |
| | | + | <math>\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix} = \begin{bmatrix} 2 \\ u\\ v\\ \end{bmatrix}</math>, for u \in (0,2) and v \in (0,2) </math>}} |
| | | | |
| | | | |
Revision as of 15:13, 30 June 2009
Fountain Flux
The water flowing out of a fountain demonstrates an important theorem for vector fields, the Divergence Theorem.
Fountain Flux
Field: Calculus
Created By: Brendan John
Basic Description
Consider a fountain like the one pictured, particularly its top layer. The rate that water flows out of the fountain's spout is directly related to the amount of water that flows off the top layer. Because something like water isn't easily compressed like air, if more water is pumped out of the spout, then more water will have to flow over the boundaries of the top layer. This is essentially what The Divergence Theorem states: the total the fluid being introduced into a volume is equal to the total fluid flowing out of the boundary of the volume.
A More Mathematical Explanation
Note: understanding of this explanation requires: *Some multivariable calculus
[Click to view A More Mathematical Explanation]
The Divergence Theorem in its pure form applies to Vector Fields. Flowing water can be considere [...]
[Click to hide A More Mathematical Explanation]
The Divergence Theorem in its pure form applies to Vector Fields. Flowing water can be considered a vector field because at each point the water has a position and a velocity vector. Faster moving water is represented by a larger vector in our field. The divergence of a vector field is a measurement of the expansion or contraction of the field; if more water is being introduced then the divergence is positive. Analytically divergence of a field $F$ is
$\nabla\cdot\mathbf{F} =\partial{F_x}/\partial{x} + \partial{F_y}/\partial{y} + \partial{F_z}/\partial{z}$,
where $F _i$ is the component of $F$ in the $i$ direction. Intuitively, if F has a large positive rate of change in the x direction, the partial derivative with respect to x in this direction will be large, increasing total divergence. The divergence theorem requires that we sum divergence over an entire volume. If this sum is positive, then the field must indicate some movement out of the volume through its boundary, while if this sum is negative, the field must indicate some movement into the volume through its boundary. We use the notion of flux, the flow through a surface, to quantify this movement through the boundary, which itself is a surface.
The divergence theorem is formally stated as:
$\iiint\limits_V\left(\nabla\cdot\mathbf{F}\right)dV=\iint\limits_{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf F\;\cdot\mathbf n\,{d}S .$
The left side of this equation is the sum of the divergence over the entire volume, and the right side of this equation is the sum of the field perpendicular to the volume's boundary at the boundary, which is the flux through the boundary.
Example of Divergence Theorem Verification
The following example verifies that given a volume and a vector field, the Divergence Theorem is valid.
Consider the vector field $F = \begin{bmatrix} x^2 \\ 0\\ 0\\ \end{bmatrix}$.
For a volume, we will use a cube of edge length two, and vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2), (2,2,0), (2,0,2), (0,2,2), (2,2,2). This cube has a corner at the origin and all the points it contains are in positive regions.
We begin by calculating the left side of the Divergence Theorem.
Step 1: Calculate the divergence of the field:
$\nabla\cdot F = 2x$
Step 2: Integrate the divergence of the field over the entire volume.
$\iiint\nabla\cdot F\,dV =\int_0^2\int_0^2\int_0^2 2x \, dxdydz$
$=\int_0^2\int_0^2 4\, dydx$
$=16$
We now turn to the right side of the equation, the integral of flux.
Step 3: We first parametrize the parts of the surface which have non-zero flux.
Notice that the given vector field has vectors which only extend in the x-direction, since each vector has zero y and z components. Therefore, only two sides of our cube can have vectors normal to them, those sides which are perpendicular to the x-axis. Furthermore, the side of the cube perpendicular to the x axis and with all points at x = 0 cannot have any flux, since all vectors in this surface are zero vectors.
We are thus only concerned with one side of the cube since only one side has non-zero flux. This side is parametrized using
$\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix} = \begin{bmatrix} 2 \\ u\\ v\\ \end{bmatrix}$, for u \in (0,2) and v \in (0,2) </math>
Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.903596043586731, "perplexity_flag": "head"}
|
http://physics.stackexchange.com/questions/tagged/boundary-condition?sort=unanswered&pagesize=30
|
Tagged Questions
The boundary-condition tag has no wiki summary.
2answers
248 views
Maxwell equation boundary conditions on a conducting sheet
I'm having difficulties solving boundary conditions for an infinitely thin conducting layer in a presence of an alternating field. I use the Maxwell equations: $\nabla \cdot \mathbf B = 0$ \$\nabla ...
1answer
275 views
Do spacelike junctions in the Thin-Shell Formalism imply energy nonconservation and counterintuitive wormholes?
The Thin Shell Formalism (MTW 1973 p.551ff) is used to properly paste together different vacuum solutions to the Einstein equations. At the junction of the two solutions is a hypersurface of matter – ...
1answer
140 views
What's the average position of oscillating particles in a box with periodic boundary conditions?
Imagine an open box repeating itself in a way that a if a particle crossing one of the box boundary is "teleported" on the opposite boundary (typical periodic boundary position in 3D). Now put a ...
0answers
25 views
Interface condition for heat exchange
I would like to compute the heat distribution of a piece of metal with some surrounding material. The heat is assumed to propagate by diffusion, so inside the metal piece and also on the outside, the ...
0answers
99 views
Quantization and natural boundary conditions
The Euler-Lagrange equations follow from minimizing the action. Usually this is done with fixed (e.g. vanishing) boundary conditions such that we do not have to worry about any boundary terms. ...
0answers
78 views
How do I simulate a constant velocity flow in porous media
I am modelling gas combustion in porous media. Most contemporary models assume that the pressure drop from the porous media is small enough to disregard, but I want to include that in my ...
0answers
262 views
What is a boundary condition for capacitors/dielectrics?
I am extremely confused about what boundary conditions are. One minute ago I was solving easy capacitor questions and the next minute I am being asked boundary condition questions and there is no such ...
0answers
8 views
Periodic boundary condition on a Wave Function of a Particle in a Box
Until now solving the Schrodinger Equation for a particle in a box was relatively easy because the boundaries conditions imposed zero value on the wave function at the boundaries. But now I must find ...
0answers
41 views
equilibrium intensity Helmoltz equation
Helmoltz equation describes the evolution of the bulk electromagnetic field, even when doing scalar optics as an approximation. Beam Propagation Method is a common approximation that assumes certain ...
0answers
37 views
Non reflecting boundaries in waveguides
Can someone please explain the Sommerfeld radiation condition and what is the alternative non-reflecting boundary conditions for waveguides of general geometries?
0answers
153 views
boundary conditions
let be the operator $-i\hbar x\frac{df(x)}{dx}-i\hbar \frac{f(x)}{2}=E_{n}f(x)$ what kind of bundary conditions can i put ?? i have tried to find a function so for every integer 'n' i get \$ ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8961793184280396, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/65557/random-versions-of-deterministic-problems/65558
|
Random versions of deterministic problems
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
What are the examples of situations where "randomizing" a problem (or some part of it) and analyzing it using probabilistic techniques yields some insight into its deterministic version?
An example of what I have in mind: it is a well-known conjecture that the Hausdorff dimension of the graph of Weierstrass function (everywhere continuous, nowhere differentiable) is given by a certain simple formula, depending on the amplitudes and phases of the cosines in the series. This is still open; however, in the paper "The Hausdorff Dimension of Graphs of Weierstrass Functions" Hunt proved that if you add a uniformly distributed independent random "noise" to each phase, the conjectured formula holds with probability 1. So while the "randomized" approach does not solve the original problem, it somehow lends credibility to the original conjecture and thus gives us some insight about the problem.
-
1
If you assume that the primes are random, you get a whole bunch of results. I am actually going to post a question in this vein as a follow-up. – Steve Huntsman May 20 2011 at 19:39
...mathoverflow.net/questions/65584 – Steve Huntsman May 20 2011 at 19:54
11 Answers
There are several applications of probability to attack deterministic problems. Generally, this approach is called the 'probabilistic method' or the probabilistic method of Erdos, who first applied it to solve a wide class of problems. For a survey of the technique, check out Noga Alon and Joel Spencer's book "The Probabilistic Method".
A specific example, due to Erdos, is an answer to Sidon's problem. Sidon asked whether or not one can find a set $B \subset \mathbb{N}$ such that $|B \cap [1, N]| = N^{1/2 + o(1)}$ for all $N$, and such that $2B =$ {$b_1 + b_2 : b_1, b_2 \in B$ is equal to $\mathbb{N}$. Erdos answered this question in the positive by using probabilistic arguments to show that such $B$'s exist. In particular, if one considers the random set $B$ defined by $\displaystyle p(x \in B) = \min\left(1, 10 \sqrt{\frac{\log x}{x}}\right)$, then $B$ satisfies Sidon's condition with probability 1. To date, no explicit example of such a $B$ is known to exist.
-
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Neudecker and Williams proved that an analogue of the Riemann hypothesis holds with probability one for numbers drawn from a randomised version of the sieve of Eratosthenes.
-
This surely isn't what you had in mind either, but just one word: Collatz. (OK, more than one word: Assume that the 3n+1|n/2 process is completely random. Odd->even, even->1/2*even+1/2*odd. Now it is obvious the random process should converge, since the Markov chain equilibrium is 2/3:1/3 and 3^(1/3)<2^(2/3). Not much of an insight, I admit...)
-
3
The most interesting questions related to the $3n + 1$ function (e.g. Are there any other cycles? Do sequences exist that tend to infinity?) are not addressed by this approach. There could be infinitely many cycles or diverging paths, which does not even contradict the convergence of the random process. It only tells us that the set of these counterexamples is (probably) not dense in $\mathbb{N}$. – TMM May 20 2011 at 17:38
There are many examples in fractal geometry/dynamical systems (including of course the one in the OP).
For example, in general it is very hard to compute the dimension (any kind of fractal dimension) of a set invariant under a nonconformal dynamical system. However adding randomness makes the situation much easier. See for example the paper "Hausdorff dimension for randomly perturbed self-affine attractors" by Jordan, Pollicott and Simon.
A related example concerns Bernoulli convolutions and more general self-similar measures with overlaps. Very little is known for specific cases, but the more randomness one adds the easier the situation becomes. See for example the paper "Absolute continuity for random iterated function systems with overlaps" by Peres, Simon and Solomyak, in which they establish absolute continuity of a family of random measures, whose deterministic counterpart includes a measure whose absolute continuity was asked by Sinai motivated by connections with the Collatz conjecture. Namely, Sinai asked about the absolute continuity of the distribution $\mu_a$ of the random series $$1+Z_1+Z_1 Z_2+Z_1 Z_2 Z_3+\ldots,$$ where $P(Z_i=1+a)=P(Z_i=1-a)=1/2$ for some $a\in (0,1)$. (The authors of the cited paper show that if $a>\sqrt{3}/2$ then $\mu_a$ is singular, while if $a<\sqrt{3}/2$, then a variant of $\mu_a$ in which $Z_i$ has a multiplicative random error, is absolutely continuous.)
-
I'm sure you're looking for something deeper, but one example I like where using a rough probabilistic argument actually gives the right answer is when considering the number of derangements (fixed-point free permutations) of ${1,2,\ldots, n}.$ Putting the uniform probablity distribution on `$S_n$`, the probablity that a given symbol $i$ is fixed by a random permutation $\tau$ is `$1 - \frac{1}{n}$`. Therefore, you might expect the probability that a given permutation should fix no symbol would be around `$(1 - \frac{1}{n})^{n}$`, so around `$e^{-1}$`. But all sorts of unjustified independence assumptions have been made here. However, if we do a more precise analysis, an inclusion-exclusion argument shows that the number of permutations in `$S_n$` fixing no symbol is exactly `$\sum_{j=0}^{n-1} (-1)^{j}$` " n choose j" `$|S_{n-j}|$`. Dividing by $n!$, we are left with the $n-1$-th degree Taylor polynomial for `$e^{-x}$` at $x =1$, which soon gets very close to `$e^{-1}$`. The unjustified independence assumptions seem to be compensated for by the fact that `$(1- \frac{1}{n})^{n}$` is only an approximation to $e^{-1}$. In fact, the proportion of derangments clearly approaches $e^{-1}$ much faster than `$(1- \frac{1}{n})^{n}$` does.
In another direction, there are situations in finite group theory where it is possible to show that the probability that a given pair of elements with given properties generate the group is close to $1$, yet it may be very difficult to exhibit an explicit pair of such generators. See for example the work of Martin Liebeck and various co-authors on this topic.
-
This is heuristic so it may not be what you are asking for. There is a probabilistic argument for the hook length formula. Let $\lambda$ be a partition: then the number of standard tableaux of shape $\lambda$ is $$\frac{n!}{\prod_{(i,j)\in\lambda}h(i,j)}$$ Here $(i,j)$ runs over the coordinates of the boxes of the Ferrers diagram and $h(i,j)$ is the hook length of the box $(i,j)$.
Then the usual definition of a standard tableau is that it is a filling of the boxes by $1,2,\ldots ,n$ such that the entries increase along each row and down each column. However this is equivalent to saying that for all boxes the entry in $(i,j)$ is the smallest entry in the hook for $(i,j)$.
Now there are $n!$ ways to fill the boxes and the probability of the condition holding for box $(i,j)$ is $\frac1{h(i,j)}$. If these probabilities were independent (which they are not!) this would prove the hook length formula.
-
While it is not known that $P \neq NP$, it is known that for most oracles $A \subseteq \omega$ we have $P^A \neq NP^A$. This might be interpreted as "evidence" for the conjecture when $A = \emptyset$.
-
1
Given that there are also *A*'s for which $P^A=NP^A$, I'm not sure we should read too much into this. – Thierry Zell May 20 2011 at 20:38
See [The Probabilistic Method][1] by Noga Alon and Joel H. Spencer.
Somehow the link above was broken. Anyway, I intended to link to
http://amzn.com/0470170204
-
One problem that I am aware of is the problem of finding the Poisson boundary of a manifold, that is, the bounded harmonic functions. There is a probabilistic way of approaching this problem using Brownian motion (c.f. here) which has evolved out of the probabilistic proof of Liouville's theorem.
If we let $B=(B_t:t \geq 0)$ be a Brownian motion and $Inv(B)$ be the sigma-algebra of events of the form ${B_t \in A}$ iff ${B_{t+s}\in A}$ for all $s \geq 0$, then there is a one-to-one correspondence between $Inv(B)$ measurable random variables and bounded harmonic functions.
Liouville's theorem holds on $\mathbb{R}^d$ because if we can start two Brownian motions at $x,y \in \mathbb{R}^d$ and couple them in a finite time, so the invariant sigma algebra is trivial. On manifolds where Liouville's theorem does not hold, then by looking at the invariant algebra, one can work out all of the bounded harmonic functions.
-
This is a nice answer to a different question. – Ori Gurel-Gurevich Oct 7 2011 at 17:43
This may not be quite what you had in mind, but: suppose you were trying to compute the absolute value of a Gauss sum $\sum_{a=0}^{p-1} \zeta^{a^2}$ where $\zeta = e^{ \frac{2 \pi i}{p} }, p$ an odd prime. Intuitively you might guess that quadratic residues are uniformly distributed and thus this sum should behave like a sum of $p$ random unit vectors. The expected value of the square root of the length of such a sum is just $p$, so one might expect that the absolute value of a Gauss sum should be about $\sqrt{p}$. In fact it is exactly $\sqrt{p}$!
-
Lets say you want to know if $\sum_n^\infty sin(a_n)$ converges for some set of values $a_n$. In many cases it might be really hard to prove (if not impossible) that $a_n\mod \pi$ eventually is not really close to integers. Even though a_n has no obvious relation to Pi. Then we might want to assume a_n mod Pi behaves somewhat randomly (but perfectly normal with regards to this property) and the divergence is trivial.
For instance $a_n=2^{1/3}*\zeta(2n+1)*e^{1/e}$
Or $a_n=2^{1/2}*(the-n'th-prime-number)$
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 65, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9316201210021973, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/57945?sort=oldest
|
## Markov chain: Obtaining transition matrix from recurrence probabilities
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider a markov chain with finite space { 0,1,..n} with transition probability matrix whose entries are $P_{ij}$. Let
$f_{ij}^n$ = probability that starting from state $i$ it goes to state $j$ first time.
Question 1. What are the necessary and sufficient condition arbitrary $a_{ij}^{n}$ needs to satisfy to be valid $f_{ij}^{n}$ of some markov chain ?
Qustion 2. If existence of such a markov chain is shown, can we calculate $P_{ij}$ given valid $f_{ij}^n$ ? ( I mean, is there any algorithm to calculate?)
-
Does the answer below correspond to what you were asking for? – Didier Piau Apr 11 2011 at 17:27
## 1 Answer
Re 2, for $n=1$, $P_{ij}=f^1_{ij}$ for every $i\ne j$ and $1-P_{ii}$ is the sum of $f^1_{ij}$ over $j\ne i$, hence one recovers trivially $P$ from $f^1=(f^1_{ij})_{ij}$.
Re 2 again, on the contrary, there is no hope to recover $P$ from $f^n=(f^n_{ij})_{ij}$ in general, for any given $n\ge2$. For example, the two (deterministic) one-step rotations on the discrete circle with $2n$ vertices, clockwise and counterclockwide, yield the same matrix $f^n$ although their transition matrices $P$ are different.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9055821895599365, "perplexity_flag": "head"}
|
http://www.openwetware.org/index.php?title=User:Timothee_Flutre/Notebook/Postdoc/2011/11/10&diff=672771&oldid=672770
|
# User:Timothee Flutre/Notebook/Postdoc/2011/11/10
### From OpenWetWare
(Difference between revisions)
| | | | |
|-----------------------------|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|-----------|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| m ( use beta instead of x) | | m () | |
| Line 354: | | Line 354: | |
| | <math>\frac{\partial f}{\partial \beta} = - \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N (y_i - p_i ) \frac{\partial X_i^TB}{\partial \beta}</math> | | <math>\frac{\partial f}{\partial \beta} = - \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N (y_i - p_i ) \frac{\partial X_i^TB}{\partial \beta}</math> |
| | | | |
| - | When setting <math>\frac{\partial f}{\partial \beta}(\beta^\star) = 0</math>, we observe that <math>\beta^\star</math> is present not only alone but also inside the sum, in the <math>p_i</math>'s: indeed <math>p_i>/math> is a non-linear function of <math>B</math>. This means that an iterative procedure is required, typically [http://en.wikipedia.org/wiki/Newton_method_in_optimization Newton's method]. | + | When setting <math>\frac{\partial f}{\partial \beta}(\beta^\star) = 0</math>, we observe that <math>\beta^\star</math> is present not only alone but also inside the sum, in the <math>p_i</math>'s: indeed <math>p_i</math> is a non-linear function of <math>B</math>. This means that an iterative procedure is required, typically [http://en.wikipedia.org/wiki/Newton_method_in_optimization Newton's method]. |
| | | | |
| | To use it, we need the second derivatives of <math>f</math>: | | To use it, we need the second derivatives of <math>f</math>: |
## Revision as of 18:41, 3 February 2013
Project name Main project page
Previous entry Next entry
## Bayesian model of univariate linear regression for QTL detection
This page aims at helping people like me, interested in quantitative genetics, to get a better understanding of some Bayesian models, most importantly the impact of the modeling assumptions as well as the underlying maths. It starts with a simple model, and gradually increases the scope to relax assumptions. See references to scientific articles at the end.
• Data: let's assume that we obtained data from N individuals. We note $y_1,\ldots,y_N$ the (quantitative) phenotypes (e.g. expression levels at a given gene), and $g_1,\ldots,g_N$ the genotypes at a given SNP (encoded as allele dose: 0, 1 or 2).
• Goal: we want to assess the evidence in the data for an effect of the genotype on the phenotype.
• Assumptions: the relationship between genotype and phenotype is linear; the individuals are not genetically related; there is no hidden confounding factors in the phenotypes.
• Likelihood: we start by writing the usual linear regression for one individual
$\forall i \in \{1,\ldots,N\}, \; y_i = \mu + \beta_1 g_i + \beta_2 \mathbf{1}_{g_i=1} + \epsilon_i \; \text{ with } \; \epsilon_i \; \overset{i.i.d}{\sim} \; \mathcal{N}(0,\tau^{-1})$
where β1 is in fact the additive effect of the SNP, noted a from now on, and β2 is the dominance effect of the SNP, d = ak.
Let's now write the model in matrix notation:
$Y = X B + E \text{ where } B = [ \mu \; a \; d ]^T$
This gives the following multivariate Normal distribution for the phenotypes:
$Y | X, \tau, B \sim \mathcal{N}(XB, \tau^{-1} I_N)$
Even though we can write the likelihood as a multivariate Normal, I still keep the term "univariate" in the title because the regression has a single response, Y. It is usual to keep the term "multivariate" for the case where there is a matrix of responses (i.e. multiple phenotypes).
The likelihood of the parameters given the data is therefore:
$\mathcal{L}(\tau, B) = \mathsf{P}(Y | X, \tau, B)$
$\mathcal{L}(\tau, B) = \left(\frac{\tau}{2 \pi}\right)^{\frac{N}{2}} exp \left( -\frac{\tau}{2} (Y - XB)^T (Y - XB) \right)$
• Priors: we use the usual conjugate prior
$\mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B | \tau)$
A Gamma distribution for τ:
$\tau \sim \Gamma(\kappa/2, \, \lambda/2)$
which means:
$\mathsf{P}(\tau) = \frac{\frac{\lambda}{2}^{\kappa/2}}{\Gamma(\frac{\kappa}{2})} \tau^{\frac{\kappa}{2}-1} e^{-\frac{\lambda}{2} \tau}$
And a multivariate Normal distribution for B:
$B | \tau \sim \mathcal{N}(\vec{0}, \, \tau^{-1} \Sigma_B) \text{ with } \Sigma_B = diag(\sigma_{\mu}^2, \sigma_a^2, \sigma_d^2)$
which means:
$\mathsf{P}(B | \tau) = \left(\frac{\tau}{2 \pi}\right)^{\frac{3}{2}} |\Sigma_B|^{-\frac{1}{2}} exp \left(-\frac{\tau}{2} B^T \Sigma_B^{-1} B \right)$
• Joint posterior (1):
$\mathsf{P}(\tau, B | Y, X) = \mathsf{P}(\tau | Y, X) \mathsf{P}(B | Y, X, \tau)$
• Conditional posterior of B:
$\mathsf{P}(B | Y, X, \tau) = \frac{\mathsf{P}(B, Y | X, \tau)}{\mathsf{P}(Y | X, \tau)}$
Let's neglect the normalization constant for now:
$\mathsf{P}(B | Y, X, \tau) \propto \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B)$
Similarly, let's keep only the terms in B for the moment:
$\mathsf{P}(B | Y, X, \tau) \propto exp(B^T \Sigma_B^{-1} B) exp((Y-XB)^T(Y-XB))$
We expand:
$\mathsf{P}(B | Y, X, \tau) \propto exp(B^T \Sigma_B^{-1} B - Y^TXB -B^TX^TY + B^TX^TXB)$
We factorize some terms:
$\mathsf{P}(B | Y, X, \tau) \propto exp(B^T (\Sigma_B^{-1} + X^TX) B - Y^TXB -B^TX^TY)$
Importantly, let's define:
$\Omega = (\Sigma_B^{-1} + X^TX)^{-1}$
We can see that ΩT = Ω, which means that Ω is a symmetric matrix. This is particularly useful here because we can use the following equality: Ω − 1ΩT = I.
$\mathsf{P}(B | Y, X, \tau) \propto exp(B^T \Omega^{-1} B - (X^TY)^T\Omega^{-1}\Omega^TB -B^T\Omega^{-1}\Omega^TX^TY)$
This now becomes easy to factorizes totally:
$\mathsf{P}(B | Y, X, \tau) \propto exp((B^T - \Omega X^TY)^T\Omega^{-1}(B - \Omega X^TY))$
We recognize the kernel of a Normal distribution, allowing us to write the conditional posterior as:
$B | Y, X, \tau \sim \mathcal{N}(\Omega X^TY, \tau^{-1} \Omega)$
• Posterior of τ:
Similarly to the equations above:
$\mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \mathsf{P}(Y | X, \tau)$
But now, to handle the second term, we need to integrate over B, thus effectively taking into account the uncertainty in B:
$\mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \int \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) \mathsf{d}B$
Again, we use the priors and likelihoods specified above (but everything inside the integral is kept inside it, even if it doesn't depend on B!):
$\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} \tau^{N/2} exp(-\frac{\tau}{2} B^T \Sigma_B^{-1} B) exp(-\frac{\tau}{2} (Y - XB)^T (Y - XB)) \mathsf{d}B$
As we used a conjugate prior for τ, we know that we expect a Gamma distribution for the posterior. Therefore, we can take τN / 2 out of the integral and start guessing what looks like a Gamma distribution. We also factorize inside the exponential:
$\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} exp \left[-\frac{\tau}{2} \left( (B - \Omega X^T Y)^T \Omega^{-1} (B - \Omega X^T Y) - Y^T X \Omega X^T Y + Y^T Y \right) \right] \mathsf{d}B$
We recognize the conditional posterior of B. This allows us to use the fact that the pdf of the Normal distribution integrates to one:
$\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} exp\left[-\frac{\tau}{2} (Y^T Y - Y^T X \Omega X^T Y) \right]$
We finally recognize a Gamma distribution, allowing us to write the posterior as:
$\tau | Y, X \sim \Gamma \left( \frac{N+\kappa}{2}, \; \frac{1}{2} (Y^T Y - Y^T X \Omega X^T Y + \lambda) \right)$
• Joint posterior (2): sometimes it is said that the joint posterior follows a Normal Inverse Gamma distribution:
$B, \tau | Y, X \sim \mathcal{N}IG(\Omega X^TY, \; \tau^{-1}\Omega, \; \frac{N+\kappa}{2}, \; \frac{\lambda^\ast}{2})$
where $\lambda^\ast = Y^T Y - Y^T X \Omega X^T Y + \lambda$
• Marginal posterior of B: we can now integrate out τ:
$\mathsf{P}(B | Y, X) = \int \mathsf{P}(\tau) \mathsf{P}(B | Y, X, \tau) \mathsf{d}\tau$
$\mathsf{P}(B | Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}}}{(2\pi)^\frac{3}{2} |\Omega|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \int \tau^{\frac{N+\kappa+3}{2}-1} exp \left[-\tau \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right) \right] \mathsf{d}\tau$
Here we recognize the formula to integrate the Gamma function:
$\mathsf{P}(B | Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}} \Gamma(\frac{N+\kappa+3}{2})}{(2\pi)^\frac{3}{2} |\Omega|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right)^{-\frac{N+\kappa+3}{2}}$
And we now recognize a multivariate Student's t-distribution:
$\mathsf{P}(B | Y, X) = \frac{\Gamma(\frac{N+\kappa+3}{2})}{\Gamma(\frac{N+\kappa}{2}) \pi^\frac{3}{2} |\lambda^\ast \Omega|^{\frac{1}{2}} } \left( 1 + \frac{(B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY)}{\lambda^\ast} \right)^{-\frac{N+\kappa+3}{2}}$
We hence can write:
$B | Y, X \sim \mathcal{S}_{N+\kappa}(\Omega X^TY, \; (Y^T Y - Y^T X \Omega X^T Y + \lambda) \Omega)$
• Bayes Factor: one way to answer our goal above ("is there an effect of the genotype on the phenotype?") is to do hypothesis testing.
We want to test the following null hypothesis:
$H_0: \; a = d = 0$
In Bayesian modeling, hypothesis testing is performed with a Bayes factor, which in our case can be written as:
$\mathrm{BF} = \frac{\mathsf{P}(Y | X, a \neq 0, d \neq 0)}{\mathsf{P}(Y | X, a = 0, d = 0)}$
We can shorten this into:
$\mathrm{BF} = \frac{\mathsf{P}(Y | X)}{\mathsf{P}_0(Y)}$
Note that, compare to frequentist hypothesis testing which focuses on the null, the Bayes factor requires to explicitly model the data under the alternative. This makes a big difference when interpreting the results (see below).
$\mathsf{P}(Y | X) = \int \mathsf{P}(\tau) \mathsf{P}(Y | X, \tau) \mathsf{d}\tau$
First, let's calculate what is inside the integral:
$\mathsf{P}(Y | X, \tau) = \frac{\mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B)}{\mathsf{P}(B | Y, X, \tau)}$
Using the formula obtained previously and doing some algebra gives:
$\mathsf{P}(Y | X, \tau) = \left( \frac{\tau}{2 \pi} \right)^{\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} exp\left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY) \right)$
Now we can integrate out τ (note the small typo in equation 9 of supplementary text S1 of Servin & Stephens):
$\mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \frac{\frac{\lambda}{2}^{\frac{\kappa}{2}}}{\Gamma(\frac{\kappa}{2})} \int \tau^{\frac{N+\kappa}{2}-1} exp \left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY + \lambda) \right)$
Inside the integral, we recognize the almost-complete pdf of a Gamma distribution. As it has to integrate to one, we get:
$\mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{2} \right)^{-\frac{N+\kappa}{2}}$
We can use this expression also under the null. In this case, as we need neither a nor d, B is simply μ, ΣB is $\sigma_{\mu}^2$ and X is a vector of 1's. We can also defines $\Omega_0 = ((\sigma_{\mu}^2)^{-1} + N)^{-1}$. In the end, this gives:
$\mathsf{P}_0(Y) = (2\pi)^{-\frac{N}{2}} \frac{|\Omega_0|^{\frac{1}{2}}}{\sigma_{\mu}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY - \Omega_0 N^2 \bar{Y}^2 + \lambda}{2} \right)^{-\frac{N+\kappa}{2}}$
We can therefore write the Bayes factor:
$\mathrm{BF} = \left( \frac{|\Omega|}{\Omega_0} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{Y^TY - \Omega_0 N^2 \bar{Y}^2 + \lambda} \right)^{-\frac{N+\kappa}{2}}$
When the Bayes factor is large, we say that there is enough evidence in the data to support the alternative. Indeed, the Bayesian testing procedure corresponds to measuring support for the specific alternative hypothesis compared to the null hypothesis. Importantly, note that, for a frequentist testing procedure, we would say that there is enough evidence in the data to reject the null. However we wouldn't say anything about the alternative as we don't model it.
The threshold to say that a Bayes factor is large depends on the field. It is possible to use the Bayes factor as a test statistic when doing permutation testing, and then control the false discovery rate. This can give an idea of a reasonable threshold.
• Hyperparameters: the model has 5 hyperparameters, $\{\kappa, \, \lambda, \, \sigma_{\mu}, \, \sigma_a, \, \sigma_d\}$. How should we choose them?
Such a question is never easy to answer. But note that all hyperparameters are not that important, especially in typical quantitative genetics applications. For instance, we are mostly interested in those that determine the magnitude of the effects, σa and σd, so let's deal with the others first.
As explained in Servin & Stephens, the posteriors for τ and B change appropriately with shifts (y + c) and scaling ($y \times c$) in the phenotype when taking their limits. This also gives us a new Bayes factor, the one used in practice (see Guan & Stephens, 2008):
$\mathrm{lim}_{\sigma_{\mu} \rightarrow \infty \; ; \; \lambda \rightarrow 0 \; ; \; \kappa \rightarrow 0 } \; \mathrm{BF} = \left( \frac{N}{|\Sigma_B^{-1} + X^TX|} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY - Y^TX (\Sigma_B^{-1} + X^TX)^{-1} X^TY}{Y^TY - N \bar{Y}^2} \right)^{-\frac{N}{2}}$
Now, for the important hyperparameters, σa and σd, it is usual to specify a grid of values, i.e. M pairs (σa,σd). For instance, Guan & Stephens used the following grid:
$M=4 \; ; \; \sigma_a \in \{0.05, 0.1, 0.2, 0.4\} \; ; \; \sigma_d = \frac{\sigma_a}{4}$
Then, we can average the Bayes factors obtained over the grid using, as a first approximation, equal weights:
$\mathrm{BF} = \sum_{m \, \in \, \text{grid}} \frac{1}{M} \, \mathrm{BF}(\sigma_a^{(m)}, \sigma_d^{(m)})$
In eQTL studies, the weights can be estimated from the data using a hierarchical model (see below), by pooling all genes together as in Veyrieras et al (PLoS Genetics, 2010).
• Implementation: the following R function is adapted from Servin & Stephens supplementary text 1.
```BF <- function(G=NULL, Y=NULL, sigma.a=NULL, sigma.d=NULL, get.log10=TRUE){
stopifnot(! is.null(G), ! is.null(Y), ! is.null(sigma.a), ! is.null(sigma.d))
subset <- complete.cases(Y) & complete.cases(G)
Y <- Y[subset]
G <- G[subset]
stopifnot(length(Y) == length(G))
N <- length(G)
X <- cbind(rep(1,N), G, G == 1)
inv.Sigma.B <- diag(c(0, 1/sigma.a^2, 1/sigma.d^2))
inv.Omega <- inv.Sigma.B + t(X) %*% X
inv.Omega0 <- N
tY.Y <- t(Y) %*% Y
log10.BF <- as.numeric(0.5 * log10(inv.Omega0) -
0.5 * log10(det(inv.Omega)) -
log10(sigma.a) - log10(sigma.d) -
(N/2) * (log10(tY.Y - t(Y) %*% X %*% solve(inv.Omega)
%*% t(X) %*% cbind(Y)) -
log10(tY.Y - N*mean(Y)^2)))
if(get.log10)
return(log10.BF)
else
return(10^log10.BF)
}
```
In the same vein as what is explained here, we can simulate data under different scenarios and check the BFs:
```N <- 300 # play with it
PVE <- 0.1 # play with it
grid <- c(0.05, 0.1, 0.2, 0.4, 0.8, 1.6, 3.2)
MAF <- 0.3
G <- rbinom(n=N, size=2, prob=MAF)
tau <- 1
a <- sqrt((2/5) * (PVE / (tau * MAF * (1-MAF) * (1-PVE))))
d <- a / 2
mu <- rnorm(n=1, mean=0, sd=10)
Y <- mu + a * G + d * (G == 1) + rnorm(n=N, mean=0, sd=tau)
for(m in 1:length(grid))
print(BF(G, Y, grid[m], grid[m]/4))
```
• Binary phenotype: using a similar notation, we model case-control studies with a logistic regression where the probability to be a case is $\mathsf{P}(y_i = 1) = p_i$.
There are many equivalent ways to write the likelihood, the usual one being:
$y_i | p_i \; \overset{i.i.d}{\sim} \; Bernoulli(p_i)$ with the log-odds (logit function) being $\mathrm{ln} \frac{p_i}{1 - p_i} = \mu + a \, g_i + d \, \mathbf{1}_{g_i=1}$
Let's use $X_i^T=[1 \; g_i \; \mathbf{1}_{g_i=1}]$ to denote the i-th row of the design matrix X. We can also keep the same definition as above for $B=[\mu \; a \; d]^T$. Thus we have:
$p_i = \frac{e^{X_i^TB}}{1 + e^{X_i^TB}}$
As the yi's can only take 0 and 1 as values, the likelihood can be written as:
$\mathcal{L}(B) = \mathsf{P}(Y | X, B) = \prod_{i=1}^N p_i^{y_i} (1-p_i)^{1-y_i}$
We still use the same prior as above for B (but there is no τ anymore), so that:
$B | \Sigma_B \sim \mathcal{N}_3(0, \Sigma_B)$
where ΣB is a 3 x 3 matrix with $[\sigma_\mu^2 \; \sigma_a^2 \; \sigma_d^2]$ on the diagonal and 0 elsewhere.
As above, the Bayes factor is used to compare the two models:
$\mathrm{BF} = \frac{\mathsf{P}(Y | X, M1)}{\mathsf{P}(Y | X, M0)} = \frac{\mathsf{P}(Y | X, a \neq 0, d \neq 0)}{\mathsf{P}(Y | X, a=0, d=0)} = \frac{\int \mathsf{P}(B) \mathsf{P}(Y | X, B) \mathrm{d}B}{\int \mathsf{P}(\mu) \mathsf{P}(Y | X, \mu) \mathrm{d}\mu}$
The interesting point here is that there is no way to analytically calculate these integrals (marginal likelihoods). Therefore, we will use Laplace's method to approximate them, as in Guan & Stephens (2008).
Starting with the numerator:
$\mathsf{P}(Y|X,M1) = \int \exp \left[ N \left( \frac{1}{N} \mathrm{ln} \, \mathsf{P}(B) + \frac{1}{N} \mathrm{ln} \, \mathsf{P}(Y | X, B) \right) \right] \mathsf{d}B$
$\mathsf{P}(Y|X,M1) = \int \exp \left\{ N \left[ \frac{1}{N} \left( \mathrm{ln} \left( (2 \pi)^{-\frac{3}{2}} \, \frac{1}{\sigma_\mu \sigma_a \sigma_d} \, \exp\left( -\frac{1}{2} (\frac{\mu^2}{\sigma_\mu^2} + \frac{a^2}{\sigma_a^2} + \frac{d^2}{\sigma_d^2}) \right) \right) \right) + \frac{1}{N} \left( \sum_{i=1}^N \left( y_i \, \mathrm{ln} (p_i) + (1-y_i) \, \mathrm{ln} (1-p_i) \right) \right) \right] \right\} \mathsf{d}B$
Let's use f to denote the function inside the exponential:
$\mathsf{P}(Y|X,M1) = \int \exp \left( N \; f(B) \right) \mathsf{d}B$
The function f is defined by:
$f: \mathbb{R}^3 \rightarrow \mathbb{R}$
$f(B) = \frac{1}{N} \left( -\frac{3}{2} \mathrm{ln}(2 \pi) - \frac{1}{2} \mathrm{ln}(|\Sigma_B|) - \frac{1}{2}(B^T \Sigma_B^{-1} B) \right) + \frac{1}{N} \sum_{i=1}^N \left( y_i \, X_i^T B - \mathrm{ln}(1 + e^{X_i^TB}) \right)$
This function will then be used to approximate the integral, like this:
$\mathsf{P}(Y|X,M1) \approx N^{-3/2} (2 \pi)^{3/2} |H(B^\star)|^{-1/2} e^{N f(B^\star)}$
where H is the Hessian of f and $B^\star = [\mu^\star a^\star d^\star]^T$ is the point at which f is maximized.
We therefore need to find $B^\star$. As it maximizes f, we need to calculate the first derivatives of f. Let's do this the univariate way:
$\frac{\partial f}{\partial \beta} = - \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N \left(\frac{y_i}{p_i} - \frac{1-y_i}{1-p_i} \right) \frac{\partial p_i}{\partial \beta}$
where β is μ, a or d.
A simple form for the first derivatives of pi also exists when writing $p_i = e^{X_i^tB} (1 + e^{X_i^tB})^{-1}$:
$\frac{\partial p_i}{\partial \beta} = \left[ e^{X_i^tB} (1 + e^{X_i^tB})^{-1} + e^{X_i^tB} \left( -e^{X_i^tB} (1 + e^{X_i^tB})^{-2} \right) \right] \frac{\partial X_i^TB}{\partial \beta}$
$\frac{\partial p_i}{\partial \beta} = \left[ \frac{e^{X_i^tB} (1 + e^{X_i^tB}) - (e^{X_i^tB})^2}{(1 + e^{X_i^tB})^2} \right] \frac{\partial X_i^TB}{\partial \beta}$
$\frac{\partial p_i}{\partial \beta} = \left[ p_i (1 - p_i) \right] \frac{\partial X_i^TB}{\partial \beta}$
where $\frac{\partial X_i^TB}{\partial \beta}$ is equal to $1, \, g_i, \, \mathbf{1}_{g_i=1}$ when β corresponds respectively to $\mu, \, a, \, d$.
This simplifies the first derivatives of f into:
$\frac{\partial f}{\partial \beta} = - \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N (y_i - p_i ) \frac{\partial X_i^TB}{\partial \beta}$
When setting $\frac{\partial f}{\partial \beta}(\beta^\star) = 0$, we observe that $\beta^\star$ is present not only alone but also inside the sum, in the pi's: indeed pi is a non-linear function of B. This means that an iterative procedure is required, typically Newton's method.
To use it, we need the second derivatives of f:
$\frac{\partial^2 f}{\partial \beta^2} = - \frac{1}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N \left[ (-p_i(1-p_i)\frac{\partial X_i^TB}{\partial \beta}) + (y_i-p_i)\frac{\partial^2 X_i^TB}{\partial \beta^2} \right]$
The second derivatives of $X_i^TB$ are all equal to 0:
$\frac{\partial^2 f}{\partial \beta^2} = - \frac{1}{N \, \sigma_\beta^2} - \frac{1}{N} \sum_{i=1}^N p_i(1-p_i)\frac{\partial X_i^TB}{\partial \beta}$
Note that the second derivatives of f are strictly negative. Therefore, f is globally convex, which means that it has a unique global maximum, at $B^\star$. As a consequence, we have the right to use Laplace's method to approximate the integral around its maximum.
finding the maximums: iterative procedure, update equations or generic solver -> to do
implementation: in R -> to do
finding the effect sizes and their std error: to do
• Link between Bayes factor and P-value: see Wakefield (2008)
to do
• Hierarchical model: pooling genes, learn weights for grid and genomic annotations, see Veyrieras et al (PLoS Genetics, 2010)
to do
• Multiple SNPs with LD: joint analysis of multiple SNPs, handle correlation between them, see Guan & Stephens (Annals of Applied Statistics, 2011) for MCMC, see Carbonetto & Stephens (Bayesian Analysis, 2012) for Variational Bayes
to do
• Confounding factors in phenotype: factor analysis, see Stegle et al (PLoS Computational Biology, 2010)
to do
• Genetic relatedness: linear mixed model, see Zhou & Stephens (Nature Genetics, 2012)
to do
• Discrete phenotype: count data as from RNA-seq, Poisson-like likelihood, see Sun (Biometrics, 2012)
to do
• Multiple phenotypes: matrix-variate distributions, tensors
to do
• Non-independent genes: enrichment in known pathways, learn "modules"
to do
• References:
• Servin & Stephens (PLoS Genetics, 2007)
• Guan & Stephens (PLoS Genetics, 2008)
• Stephens & Balding (Nature Reviews Genetics, 2009)
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 85, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8756369948387146, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/tagged/idempotents
|
# Tagged Questions
For questions about elements which satisfy $x\cdot x=x$ where $\cdot$ is a composition law.
2answers
44 views
### Idempotents in a Quotient Ring
Let $R=\mathbb{Z}_p[x]/(x^p-x)$. Show that $R$ has exactly $2^p$ elements satisfying $r^2=r$. I know that for $f,g\in\mathbb{Z}_p[x]$, we have $f-g\in(x^p-x)$ if and only if $f(a)=g(a)$ for all ...
1answer
30 views
### Is there a name for a relationship like idempotence between two functions?
If $f(f(x)) = f(x) \quad \forall \space x$ then $f$ is idempotent. If $g(f(x)) = f(x) \quad \forall \space x$ then is there a term to describe the relationship between $g$ and $f$?
1answer
95 views
### Sum of idempotent matrices is Identity
[Ciarlet, Problem $1.1-10$] Let $A_k$, $1 \leq k\leq m$, be matrices of order $n$ satisfaying $$\sum_{k=1}^mA_k\ =\ I.$$ Show that the following conditions are equivalent. \$A_k = ...
1answer
44 views
### Formal Notation of Functions
Say i have a function $f$ with the following signature: $f: S \times S \to S$ and I wanted to formally represent that this function was idempotent and reflexive. How would i do right this? For ...
0answers
51 views
### Representation for idempotent semiring
I have a semi-ring whose multiplication is non-commutative, and addition is idempotent. That is, $ab \neq ba$ and $a + a = a$. The semi-ring is freely generated from a finite set $\Sigma$, the ...
0answers
88 views
### A special idempotent function
Usually an idempotent function is $f: S \rightarrow S$ such that $f \circ f = f$. But I need something slightly different: $f: S \times S \rightarrow S$, such that: For $x, y, z \in S$: \$f(x, y) ...
1answer
78 views
### Determine if the matrix is idempotent?
I am dealing with an example to show that the matrix($M = I − X(X'X)^{−1}X'$) is idempotent. X is a matrix with T rows and k columns and I the unit matrix of dimension T. And then to determine the ...
0answers
32 views
### Left continuous magmas with no fixed points
Let $X$ be a compact Hausdorff topological space, and $*: X^2\rightarrow X$ an associative map (so that $(X, *)$ is a semigroup) which is left continuous (for all $s\in X$, the map $t\mapsto ts$ is ...
0answers
32 views
### fully idempotent and co-idempotent
I want the proof of : For each prime number p, the sub module N = Zp ⊕0 of the Z-module M = Zp⊕Zp is not idempotent but N = Hom z (M;N)N. Z4 is a multiplication Z-module which is not fully ...
1answer
95 views
### Can a sum of idempotents vanish?
Let $A$ be a finite dimensional $\mathbb C$-algebra. Let $e_1,\ldots,e_r\in A$ be nonzero idempotents (with $r>0$), i.e. $e_i^2=e_i$. My question is: Can it happen that $e_1+\cdots+e_r=0$? I can't ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8933125734329224, "perplexity_flag": "middle"}
|
http://physics.stackexchange.com/questions/27445/constructing-a-cp-map-with-some-decaying-property?answertab=active
|
# Constructing a CP map with some decaying property
Given some observable $\mathcal O \in \mathcal H$ it is simple to construct a CP (completely positive) map $\Phi:\mathcal{H}\mapsto \mathcal{H}$ that conserves this quantity. All one has to observe is that $$\text{Tr}(\mathcal O \, \Phi[\rho]) = \text{Tr}(\Phi^*[\mathcal O] \rho).$$ Therefore, if we impose $\Phi^*[\mathcal O] = \mathcal O$, then $\text{Tr}(\mathcal O \, \Phi[\rho])=\text{Tr}(\mathcal O \rho), \; \forall \rho\in \mathcal H$. That amounts to impose that the Kraus operators of $\Phi^*$ should commute with $\mathcal O$.
I'd like, however, to construct a trace-preserving CP map for which the expectation value of $\mathcal O$ does not increase for any $\rho \in \mathcal H$. More explicitly, given $\mathcal O\in \mathcal H$, I want to construct $\Gamma:\mathcal H \mapsto \mathcal H$ such that $$\text{Tr}(\mathcal O\, \Gamma[\rho]) \le \text{Tr}(\mathcal O \rho), \; \forall \rho \in \mathcal H .$$
How would you go about that? Any ideas?
-
Do you want to construct $\Gamma$, or to characterize all possible $\Gamma$ ? – Frédéric Grosshans Feb 23 '12 at 11:22
## 3 Answers
I'll restrict myself to trace-preserving CP-maps.
One can rewrite $\mathcal O=\sum_{k,l}o_k|k,l\rangle\langle k,l|$, where the $o_k$ are in decreasing order. The non-increasing condition $\langle\mathcal O\rangle$ corresponds then to an non-increasing condition on $k$. Writing $\Gamma$ in terms of Kraus operators, one has $\Gamma(\rho)=\sum_i B_i\rho B_i^*$ with $\sum_iB_iB_i^*=\mathbb1$. The condition on $k$ given above is then translated into the following writing of the Kraus operators: $$B_i=\sum_{\substack{k,l,k',l'\\k\le k'}}B_i^{klk'l'}|k,l\rangle\langle k',l'|.$$ Another way to say the same thing is the condition $B_i^{klk'l'}=0$ if $k>k'$.
Then, of course, the normalization condition imposes $$\sum_{\substack{i,k',l'//k\le k'}}\left|B_i^{klk'l'}\right|^2=1, \forall k,l.$$
If you apply the same reasoning with a non-increasing and non-decreasing condition, you find that $k$ has to be conserved, and this is then equivalent to the commutativity condition you give in your question. In the same way, this answer is not general: you have operations which preserve $\langle\mathcal O\rangle$ without commuting with $\mathcal O$.
-
Thanks for the answer Frédéric, but I don't totally follow it. To start with, I believe you need just one index for the eigenbasis of $\mathcal O$, right? Assuming that, and decomposing the Kraus operators in the eigenbasis of $\mathcal O$, as you suggested, I get the condition: $\sum_i \sum_{k,s,t} o_k B_i^{ks} {B^*}_i^{tk} \langle s|\rho |t\rangle \le \sum_k o_k \langle k|\rho |k\rangle$. How do I conclude something from this? – Fernando Feb 23 '12 at 18:26
@Fernando : if the observable is non-degenerate, the index $l$ is indeed is useless. But if the observable is degenerate, i.e. if two or more state correspond to the same value $o_k$ of the observable, this second index helps to have a more general solution. – Frédéric Grosshans Feb 23 '12 at 18:31
@Fernando : the right hand term is $\mathrm{Tr}(\rho\mathcal O)$, and, if $s=t$ in the left-hand term, the left hand term becomes $\mathrm{Tr}(\Gamma(\rho)\mathcal O)$. – Frédéric Grosshans Feb 23 '12 at 18:34
Thanks for the quick reply! Indeed the r.h.s. is $\text{Tr}(\rho \mathcal O)$, and the l.h.s is $\text{Tr}(\Gamma[\rho]\mathcal O)$. The inequality is what I want to obtain by putting constraints on the $B_i$'s. (Note that the l.h.s is $\text{Tr}(\Gamma[\rho]\mathcal O)$ even without imposing $s=t$). Of course I could ask $B_i$ to be diagonal in the basis of $\mathcal O$, but still there is the problem that some $o_k$ can be negative. Am I loosing something? – Fernando Feb 23 '12 at 18:48
1
I got it. The idea is to shift "population" of $rho$, in the eigenbasis of $\mathcal O$, towards eigenvectors that will have smaller contributions to $\langle \mathcal O \rangle_\rho$. Thanks a lot. Just one last comment, I believe the trace-preserving condition should be $\sum_l B_l^* B_l = \idty$, which then imposes a couple of other constraints in $B_l$ (maybe already implied by your conditions, but not explicitly). Let me know if you want to change something in your answer, and I'll accept it right afterwards. Thanks. – Fernando Feb 24 '12 at 10:31
show 2 more comments
The formal condition---the correspondent of $\Phi^*[\mathcal{O}]=\mathcal{O}$ in the other case---is $\Gamma^*[\mathcal{O}]\leq \mathcal{O}$. For $\mathcal{O}>0$ (all eigenvalues strictly positive), if one multiplies on the right and on the left by $\mathcal{O}^{-1/2}$, one can see that this condition is equivalent to the map with Kraus operators $\mathcal{O}^{1/2}K_i\mathcal{O}^{-1/2}$ being trace non-increasing.
-
This is probably not exactly what you had in mind, but how about the channel that discards its input and always outputs the state corresponding the the minimum eigenvalue of $\mathcal{O}$?
-
Thanks Dan, but indeed that's not what I had in mind. I'd like something smoother than that. More a characterization of the channels than a single construction. – Fernando Feb 23 '12 at 18:15
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9313279390335083, "perplexity_flag": "head"}
|
http://math.stackexchange.com/questions/187985/example-of-a-cluster-algebra-of-geometric-type-and-ground-field
|
# Example of a cluster algebra of geometric type and ground field
In most definitions of a cluster algebra of geometric type, it is said to be the subalgebra of the field $\mathbb Q (x_1,\cdots,x_n)$ generated by the cluster variables.
I would like to compute such an algebra but I am not sure of the result. As an example, let's try with the cluster algebra of rank $2$ associated with the seed $((x,y),B)$ where $B$ is the skew-symetric matrix $\begin{pmatrix} 0 & 1 \\-1 & 0\end{pmatrix}$.
There is only a finite number of clusters : $x,y,\dfrac{1+y}{x},\dfrac{1+x}{y},\dfrac{1+x+y}{xy}$ but what is the associated cluster algebra ? What are elements of it ?
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9321068525314331, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/33304?sort=votes
|
## Marey’s problem: Generating all prime numbers in $[n_1,n_2]$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Is there are some references to algorithms that generate the set of prime numbers located between two given numbers n1 and n2?
I would like to consider the cases when n1 is large while n2-n1 is small or while n2-n1 is large.
If we consider these cases:
[1] n1=$2^{10}$ & n2=$2^{11}$;
[2] n1=$2^{40}$ & n2=$2^{45}$, (modified to [2a]);
[3] n1=$2^{100}$ & n2=$2^{101}$, (modified to [3a]);
[4] n1=$2^{1000}$ & n2=$2^{1001}$, (modified to [4a]);
Is there is a well known algorithm to generate the set of all primes p ∈ [n1,n2] without generating all primes p < n1?
By considering for example these cases:
[2a] n1=$2^{40}$ & n2=$2^{40}+2^{20}$;
[3a] n1=$2^{100}$ & n2=$2^{100}+2^{20}$;
[4a] n1=$2^{1000}$ & n2=$2^{1000}+2^{20}$.
-
2
How do $n_1$ and $n_2$ compare? I think the answers below are assuming that $n_1$ is significantly larger than $n_2-n_1$. – David Speyer Jul 25 2010 at 18:35
@David: Without this assumption he could just generate all primes less $n_1$ using Atkin's sieve (cr.yp.to/primegen.html) if $n_2$ is around $10^{15}$ or simply forget about it if $n_2$ is much bigger. – jp Jul 25 2010 at 20:48
1
What do you consider large and what small? E.g., is $n_1$ around 1000 bit long? And does the "large" for $n_1$ mean the same as the "large" for $n_2−n_1$? (I wouldn't try to find all primes in an interval of length $2^{1000}$, but finding primes of size $2^{1000}$ is no problem.) – Someone Jul 26 2010 at 8:44
1
Is there any reason why you don't want to generate the primes $<n_1$ for the cases [1] and [2]? Otherwise you can simply use the program from cr.yp.to/primegen.html . And in your cases [3] ($n_1=2^{100}$) and [4] ($n_2 = 2^{1000}$) with always $n_2 = 2*n_1$ the intervals are much too long. You'll get around $2^{100}/ln(2^{100}) =1.8..10^{28}$ rsp. $2^{1000}/ln(2^{1000})=1.5..10^{298}$ primes. – Someone Jul 26 2010 at 10:20
1
Do not use the Gries algorithm! It's actually slower than the Sieve of Eratosthenes. See Pritchard's "Improved incremental prime number sieves". – Charles Jul 26 2010 at 20:22
show 8 more comments
## 9 Answers
The fastest approach should be first to sieve the numbers by marking the numbers divisible by small primes (for this you should use only one long division per short prime), then to use a Fermat test to the base 2 (as it is more efficient since multiplication with 2 is a left shift) on all unmarked numbers. Finally apply a certain number of Miller-Rabin tests to all candidates passing the Fermat test to reduce your error probability to a level you can tolerate (e.g., \$2^{-100}).
-
1
That gives you a prime; he is looking for all the primes – BlueRaja Jul 25 2010 at 18:29
Thanks BlueRaja,I corrected my answer accordingly. – jp Jul 25 2010 at 20:34
Don't use a Fermat test; use a Miller-Rabin test for the initial scan. It's just as fast and slightly better. If you want an error level as low as $2^{-100}$ you should use one of the Frobenius tests like Grantham's (but not for the initial test, of course!) – Charles Jul 27 2010 at 6:37
@Charles: A Fermat test with base $2$ takes approx. 25% less time than a MR-test, as one can replace the Square-and-Multiply algorithm for the exponentiation by a Square-8-Times-Then-Multiply algorithm if the numbers have more than 256 bit. Per bit of the exponent one needs therefore only 9/8 operations instead of 3/2 in average. – jp Jul 27 2010 at 10:34
@Charles, part II: To get an error level of $2^{-100}$ one should use the results of Damgard, Landrock, Pomerance (math.dartmouth.edu/~carlp/PDF/paper88.pdf) instead of the weaker bound $4^{-k}$ for $k$ tests (DLP show that 9 tests for bit length 512 and 2 tests for bit length 768 are sufficient). I do not know of a similar analysis of Grantham's Frobenius test, but would be interested in it. – jp Jul 27 2010 at 10:46
show 6 more comments
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This was originally a comment on TonyK's answer, but it was too large to submit as a comment so I'm putting it here. I thought the numbers might have value to some.
Let's say that it takes $b^{\lg 6}/10^{10}$ seconds to test a $b$-bit number with the M-R test and $b^{\lg 24}/10^{13}$ seconds to prove primality with ECPP. Testing a range up to $n_2=2^b$ with M-R + ECPP would take about $$\left(n_2-n_1\right)\left(\frac{b^{\lg 6}}{10^{10}}+\frac{b^{\lg 24}}{10^{13}\ln n_2}\right)=\left(n_2-n_1\right)\left(\frac{b^{\lg 6}}{10^{10}}+\frac{b^{\lg 12}}{10^{13}\ln2}\right)$$ seconds. For $n_2-n_1=2^{20}$ and $n_2$ large, this is about $1.5(\lg n_2)^{\lg12}/10^7$ seconds.
On the other hand, suppose sieving $n_1$ to $n_2$ takes $\sqrt{n_2}/10^6+(n_2-n_1)/10^{8.5}$ seconds. For $n_2-n_1=2^{20}$ and $n_2$ large, this is about $\sqrt{n_2}$ seconds.
Equating the two suggests that, for intervals about a million wide, testing each member is superior to sieving beyond about $n_2>8\cdot10^8$. Only about 12 kB of memory are needed to store all the primes up to the square root of that limit, so the fourth power trick doesn't seem viable here at all -- by the time you run out of primary memory you shouldn't be sieving at all.
-
@Charles: may be your example is correct and valid when the interval $[n_1,n_2]$ is small compared with $n_2$, but when the interval is relatively wide, (for example when $n_1=n_2/2$), the same demonstration can be used to prove the contrary. – Mohammed Marey Jul 27 2010 at 18:57
I used the width you used. For any fixed interval width it can be shown that beyond some fixed point, standard sieving (Eratosthenes/Atkin) is less efficient than testing each member in the range. (If you implement the algorithms yourself, it's obvious.) But of course I intend for the argument to go both ways: when the interval is wide compared to its height, sieving is better -- often vastly better. – Charles Jul 27 2010 at 19:36
@Charles: Yes, you are true, the efficiency of the used algorithm depends on the value of $n_1$ and $n_2$. – Mohammed Marey Jul 28 2010 at 21:16
If $n_1$ and $n_2$ are not too big you can use Eratosthenes sieve (see wikipedia for this). If they are big enought you can use sieve to cross out numbers which have small prime divisors and after that check every number that wasn't crossed out and check if it is prime using any primality test (see http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test).
-
Unless you have very specialized needs, you can probably apply standard sieving techniques. Here are a couple references. First, hot off the press is Kjell Wooding's Calgary thesis [1], which includes a good overview of both software and hardware sieving techniques (e.g. FPGA-based sieves, Calgary's scalable sieve, etc). Its bibliography should prove useful as an entry point into the literature. Also, here [2] is a charming classic paper by D.H. Lehmer - one of the pioneers of computational number theory. It gives a nice succinct introduction to sieving in general. Such methods have a long venerable history that stimulated much development in both number theory and computer science, e.g. google "Lehmer sieve" and you'll discover many ingenious machines devised to carry out number theoretical computations long before the dawn of the modern digital computer, e.g. via bicycles chains, photoelectric devices, etc.
1 Wooding, Kjell. The Sieve Problem in One- and Two-Dimensions.
PhD Thesis. Calgary, Alberta. April, 2010
2 Lehmer, D.H. The sieve problem for all-purpose computers, MTAC, v. 7 1953, p. 6-14
-
@Dubuque: Thank you Dubuque.. – Mohammed Marey Jul 28 2010 at 21:13
You can get away with storing $n_2^{1/4}$ bits for the sieving data. Generate the primes $<= n_2^{1/4}$, and use them to generate the primes in successive intervals of length $n_2^{1/4}$ dynamically, up to $n_2^{1/2}$. The primes thus generated can be used to generate the primes in the interval $[n_1,n_2]$.
-
@TonyK: Is there is a good reference that explains generating primes in successive intervals dynamically. – Mohammed Marey Jul 27 2010 at 11:59
@Mohammed: If you have all the primes up to k, and an interval [a,b] with k < a < b < k^2, then you can sieve the interval [a,b], right? That's all I meant. – TonyK Jul 29 2010 at 12:45
You can use prime sieves.
-
I'm assuming you're asking if we can generate primes in [n1, n2] without generating the primes before that (using a sieve), to which the answer is, no, nothing aside from testing every number between n1 and n2 for primality using something like Rabin-Miller.
Note that this will be significantly slower than sieving unless the numbers are extremely large and the range [n1, n2] is comparatively small.
-
1
NB. From what I understand Miller-Rabin is still the preferred method for something like this in practice even though it's probabilistic; AKS is much much slower and Bernstein's probabilistic version of AKS is still slower than MR. – Steve Huntsman Jul 25 2010 at 19:04
Of course M-R doesn't prove primality, so in some cases it's inappropriate unless followed by a proof. For example, it would be a bad choice if you were enumerating the pseudoprimes like, e.g., Richard Pinch... chalcedon.demon.co.uk/rgep/carpsp.html – Charles Jul 25 2010 at 19:08
Yes, Rabin-Miller is used in practice in cryptography, usually 100 times to give a probability of failure of $1−1/2^{100}$ , which is close enough to certain for practical purposes. However, if this worries you too much, you can of course use something like AKS, which is deterministic. – BlueRaja Jul 25 2010 at 23:39
1
That is what i am asking about, to generate all primes in [n1,n2] without generating primes less than n1. – Mohammed Marey Jul 26 2010 at 8:23
Let $S=\sqrt{n_2}$. If S is bigger than your computer's memory in bits, then you're going to have trouble generating all the primes in that range; if you insist, break the range into pieces that fit into memory, mark off the ones with small divisors, and test the remaining numbers for primality. You want to do as much sieving as possible, so fill maybe half the memory with primes and use those for sieving.
If S fits in memory and $n_2-n_1$ is not much smaller than S, use a sieve (sieve of Atkin or Eratosthenes). You may need to use a segmented version of the sieve, depending on how large $n_2-n_1$ is.
If $n_2-n_1$ is significantly smaller than S, make a bit array of (segments of?) the range, mark off small divisors, and test the remaining members for primality.
-
This means we should test for primality of each number in [n1, n2] which is not nice if the length of [n1 ,n2] contains for examples $2^{10}$ or $2^{50}$ numbers !!. Or we should generate all primes p < n2 which is not nice if n1 is a large number for examples $2^{50}$!! – Mohammed Marey Jul 26 2010 at 10:16
1
@Mohammed: The distance between primes is ~ ln(n), so even if you are only looking for the primes between [$2^{50}$, $2^{51}$) ($2^{50}$ numbers), you are still looking at many terabytes of space to store all those primes. – BlueRaja Jul 26 2010 at 16:29
1
Practically speaking, if you want to generate all primes in a large interval, you need to be able to store the primes up to the square root of the upper end of the interval in main memory. To sieve [2^50, 2^51] would require storing the 2,857,269 primes up to sqrt(2^51) which is pretty easy (11 MB using 32-bit integers). With 8 GB of main memory you could sieve an interval with upper bound of 2^68. – Charles Jul 26 2010 at 17:00
@Charles: Is there is a clear reference explaining in details your practical comment? – Mohammed Marey Jul 27 2010 at 11:46
1
@Mo: You'll have to if you're using the Sieve of Eratosthenes or Atkin. But if you're looking at $[2^{1000},2^{1000}+2^{20}]$ then you won't want to store the ~244 TB of primes up to the square root. Instead you should sieve out the multiples of the first thousand primes (removing 94% of the range), M-R the remaining numbers (removing ~98% of the candidates), and prove the primality of the remainder -- or test them to sufficient confidence, as desired. – Charles Jul 27 2010 at 16:39
show 3 more comments
The other answers have covered the right strategies, but here are a few additional comments.
On an idealized computer with an unbounded amount of memory, the cost of sieving $[n_1,n_2]$ for the prime $p$ is $C_1(n_2-n_1)/p$ for some constant $C_1$, the average cost of a memory read & write operation. Therefore, the cost of sieving $[n_1,n_2]$ for all primes up to some bound $B$ is $$C_1 (n_2 - n_1) \sum_{p\leq B} \frac{1}{p} = (n_2 - n_1) (C_1 \log \log B + O(1)),$$ by the 2nd Mertens Theorem. The numbers in $[n_1,n_2]$ that remain after sieving is approximately $$(n_2-n_1)\prod_{p \leq B} \left(1-\frac{1}{p}\right) = (n_2 - n_1) \frac{e^{-\gamma}}{\log B},$$ by the 3rd Mertens Theorem. Therefore, the time for testing the remaining numbers for primality is about $$C_2(n_2-n_1) \frac{e^{-\gamma}}{\log B},$$ where $C_2$ is the average cost of a single primality test. The optimal sieving bound $B$ is then approximately $\exp(e^{-\gamma} C_2/C_1)$.
The "constants" $C_1$ and $C_2$ are best determined empirically. In reality, the constant $C_1$ depends heavily on the length of the interval $[n_1,n_2]$ while the constant $C_2$ depends mostly on the size of $n_2$. The constant $C_2$ should vary smoothly with $n_2$, but there will be sharp increases in $C_1$ when the cache size is exceeded and a still larger increase if disk swaps are necessary. Here is a trick that will help you fit longer intervals into the cache, especially if you have multiple processors available. The cost of sieving is higher for small primes, to save some of this time, break the interval into arithmetic progressions $a + Mx$, where $\gcd(a,M) = 1$ and $M = 2\cdot 3 \cdots p_i$ is the product of the first $i$ primes. Each such progression can be sieved independently, possibly on different processors. Essentially, we're breaking the interval into $\phi(M) = (2-1)(3-1)\cdots(p_i-1)$ different progressions which each require $(n_2-n_1)/M$ bits to store in memory while the total sieving and testing times remains about the same. I have some C code that implements this strategy with $M = 2\cdot 3 \cdots 29$ if you're interested.
-
The fact that $\phi(30) = 8$ can also be exploited nicely to fit longer intervals into the cache. – jp Jul 27 2010 at 19:18
Good analysis. It's worth noting, though, that $C_2$ isn't really constant, in that it depends on the amount of sieving done! Suppose you use a Miller-Rabin test followed by ECPP if needed. The expected cost is cost(M-R) + cost(ECPP)/1000, since about 999/1000 are expected to fail the Miller-Rabin test. If 99% of the composites are sieved out, then the expected cost rises to cost(M-R) + cost(ECPP)/10. – Charles Jul 27 2010 at 20:09
@Dorais: Yes, I am interested to see your code. – Mohammed Marey Jul 28 2010 at 11:12
@Mohammed: Here is a download link - www-personal.umich.edu/~dorais/docs/… – François G. Dorais♦ Jul 28 2010 at 14:57
@Dorais: Thank you very much Dorais :) – Mohammed Marey Jul 28 2010 at 21:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 105, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9098785519599915, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/156/can-you-explicitly-write-r2-as-a-disjoint-union-of-two-totally-path-disconnected
|
## Can you explicitly write R^2 as a disjoint union of two totally path disconnected sets?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
An anonymous question from the 20-questions seminar:
Can you explicitly write R^2 as a disjoint union of two totally path disconnected sets?
-
1
I'd be happy with a non-explicit way to make such a partition. – Anton Geraschenko♦ Oct 6 2009 at 23:57
26
Here's a nonexplicit construction. Enumerate (well-order) all possible paths with order-type the continuum. By induction, put one point from each path in each of the sets of the partition. This is possible since each path contains continuum many points and at any stage of the induction, you've only chosen where less than continuum points go. – Eric Wofsey Oct 9 2009 at 2:32
@Eric: that's awesome! What else could you possibly do? – Anton Geraschenko♦ Oct 9 2009 at 20:17
3
What does "well-order with order-type the continuum"? Maybe "biject the set of paths with the first uncountable ordinal" so that all initial segments of the ordinal are numerable? – Mariano Suárez-Alvarez Nov 18 2009 at 6:52
6
Mariano: No, you are assuming CH there. It means "biject with the first ordinal in bijection with the reals." – Andres Caicedo Oct 31 2010 at 15:59
## 5 Answers
Let S be a subset of the reals such that S∩[a,b] and Sc∩[a,b] cannot be written as a countable union of closed sets for any a<b. This can be done (this explicit example of a non-Borel set achieves this). Let ℚ be the rationals. Then, A=(Sxℚ)U(Scxℚc) and B=(Sxℚc)U(Scxℚ) should do it.
The proof is as follows. Suppose that the curve t→(f(t),g(t)) lies in A, and consider a closed bounded interval I. As the curve lies in A, f(I)∩S = f(I∩g-1(ℚ))=∪x∈ℚf(I∩g-1(x)) is a union of countably many closed sets. By the choice of S, f(I) must be a single point. Hence, f is constant. Then, g is a continuous function mapping into either ℚ or ℚc, so is also constant. So A is totally path disconnected. The argument for B follows in the same way by exchanging S and Sc
-
1
+1 Wow, I really thought that this question was destined to never get a good answer, but that's a pretty explicit, and the proof looks good to me. – Anton Geraschenko♦ Oct 10 2009 at 3:00
1
@Anton - maybe you want to accept this answer? – Steven Gubkin Sep 16 at 15:16
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I we omit "path" in the formulation, then it cannot be done. I guess that is why it is formulated this way.
Suppose the plane could be written as the union of two totally disconnected sets. Intersect with a closed square to write the square as the union of two totally disconnected sets. But the square is compact, so these two sets are zero dimensional. Finally, the union of two zero dimensional sets has dimension at most 1. So that union cannot be the square.
So, in the construction, your two sets are totally path disconnected, but not totally disconnected!
-
@Gerald: The two sets in my construction are connected and locally connected, yet totally path disconnected. This came up in a recent MO question (mathoverflow.net/questions/46748/…). – George Lowther Nov 20 2010 at 21:14
@Gerald: What is your definition of "totally disconnected"? A subset of square which does not contain any connected subsets but the empty set or 1-element sets, can still have topological dimension = 1. Together with a 0-dimensional set they can cover the whole square. – Wlodzimierz Holsztynski Feb 24 at 20:42
Remedial question: what's a reference for the statement "the union of two zero dimensional sets has dimension at most 1"? – Greg Martin Mar 13 at 7:25
"the union of two zero dimensional sets has dimension at most 1" ... Decomposition Theorem $\dim X \le n$ if and only if $X$ can be written as a union of $n+1$ sets, each of dimension $\le 0$. Engleking, Dimension Theory (somewhere on pages 257 to 260). – Gerald Edgar Mar 13 at 13:22
Here's a stab at something explicit but without proofs. We may as well work with the square of the open interval ]0, 1[. Represent a number between 0 and 1 by its regular continued fraction, say x = [a_1, a_2, ..., a_n] if x is rational and x = [a_1, a_2, ...] if x is irrational. With this notation, define f(x) = [a_n, ..., a_2, a_1] if x is rational and f(x) = [a_2, a_1, a_4, a_3, ...] if x is irrational. So f is some brutally discontinuous involution. Then consider A = {(f(x), f(y)): x + y less than 1} and B the complement of A in the square of the interval. Gut feeling is that A and B are both totally disconnected.
Surely someone can improve on this.
-
1
I think it's probably not to hard to rule out any vertical or horizontal segments in A or B, but it's hard to imagine how you'd rule out arbitrary curves (even arbitrary line segments seems pretty hard). – Anton Geraschenko♦ Oct 9 2009 at 20:26
Let me try. QxQ and its complement in R^2. Does this work? Edit: this does not work, :(. Let me try something else then. Take S = (R\Q)x(R\Q). QxR U RxQ would be S complement. Let p be a prime number and let q_p be in Q. For each q_p, let J_p be the set of all rational numbers times the square root of p. Since Q is countable, there is a bijection between the set of primes and {q x R : q is rational} union {R x q : q is rational}. So let A be the union of all q_p X J_p and J_p X q_p. Take A union S and the complement of that in R^2 to be your disjoint union. Dunno if this works but at least it seems like a better attempt.
-
2
okay, I now see why this doesn't work, sorry. – Patrick Tam Oct 31 2010 at 12:17
@Patrick, see wonderful Hurewicz and Wallman, Dimension Theory :-) – Wlodzimierz Holsztynski Feb 24 at 21:03
I'll apply the following simple result:
THEOREM Let $f : I\rightarrow X$ be an arbitrary non-constant continuous function (a path) of interval $I:=[0;1]$ into an arbitrary topological space $X$. Then there exist continuous maps $\alpha:I\rightarrow I$ and $g:I\rightarrow X$ such that $f = g\circ \alpha$, and $g$ is not constant on any non-empty open subinterval of $I$.
Here is a simple positive solution for the question of this thread, and proof:
Let $\mathbb C := \mathbb R^2$ be the complex plane. Let $K \subseteq \mathbb C$ be a Knaster pseudo-arc. Let $$L := i\cdot K := \{i\cdot z : z \in K\}$$
where $i^2=-1$. Let $D$ be a dense countable subset of $\mathbb C$. Define $$A := \left(\bigcup_{d\in D}\ \left(d+K\right)\right)\cup\left(\bigcup_{d\in D}\ \left(d+L\right)\right)$$
where $d+X := \{d+x:x\in X\}$. Finally, let $$B := \mathbb C\setminus A$$
Then $\dim(B) = 0$, and $B$ does not contain any non-constant path.
Also, there does not exist any non-constant continuous map $f : I \rightarrow A$ --indeed, if there was one then we may assume that it is not constant on any open subinterval of $I$. Then the inverse images:
$(\bigcirc^{-1}f)(d+K)\quad$ and $\quad(\bigcirc^{-1}f)(d+L)$
would be 0-dimensional closed subsets of $I$, for every $d\in D$. Thus $I$ would be a countable union of 0-dimensional closed subsets, which is a contradiction. It means that $A$ does not contain any image of any non-constant path.
This completes a positive answer to the Question of this thread.
-
Images $f(I)$ of paths $f\rightarrow X$ in arbitrary metric spaces $X$ are the same as the connected and locally connected compact subsets of $X$ (Hahn-Mazurkiewicz Theorem). Each of them can be decomposed into a union of two smaller continua, if it contains more than one point. On the other hand the Knaster pseudo-arc $K$ is hereditarily indecomposable (all subcontinua of $K$ are homeomorphic to $K$, how nice!) hence it does not contain any image of a non-constant path. Probably I could use some of it to make the proof above a bit nicer. – Wlodzimierz Holsztynski Feb 25 at 4:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9278336763381958, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/55042/automorphism-group-of-a-scheme
|
## Automorphism group of a scheme
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi there,
I have a probably stupid question on schemes ... Let S be a scheme, and let A be its automorphism group. Does A carry a scheme structure itself, that is, can one see A as a group scheme ? Thanks !
-
## 5 Answers
The answer is yes when the scheme is flat and projective over the base. This follows from the existence of the Hom scheme, which in turn is proven via the existence of the Hilbert scheme.
A readable reference is Nitin Nitsure's part of the book Fundamental algebraic geometry. In particular Theorem 5.23 in his notes states that when S is noetherian, X projective and flat over S, and Y quasi-projective over S, then there exists a Hom-scheme parametrizing morphisms from X to Y over S. (The precise definition of the functor that this scheme represents is given in the text). The automorphism group scheme of X is then an open subscheme of $\mathrm{Hom}_S(X,X)$.
Addendum: The proof that Isom is open in Hom is very similar to the proof that Hom is open in Hilb. The map from the Hom scheme to $\mathrm{Hilb}_{X \times_S Y / S}$ is given by associating to a morphism $f : X \to Y$ its graph. Now the image consists of those $Z \subset X \times_S Y$ such that projection onto X induces an isomorphism $Z \cong X$, and the crucial part of the proof is showing that this is an open condition. (In Nitsure's notes this is Theorem 5.22.(b).) But then the condition that Z maps isomorphically onto Y is also open, and this is exactly the condition that defines the Isom scheme inside the Hom scheme.
-
Forgive the naive question, but is there a nice explanation of why being an automorphism is an open condition? – Daniel Loughran Feb 11 2011 at 12:02
Thanks ! Is there a "direct" way to do it for flat and projective schemes, that is, without using the Hilbert scheme ? (For instance when one works with base a field.) – THC Feb 15 2011 at 9:28
Not that I know of. Maybe you can show that any automorphism extends to the ambient projective space (by cooking up an equivariant ample line bundle?), and then embed the automorphism group scheme in $\operatorname{Aut} \mathbf{P}^n = \mathrm{PGL}(n+1)$. P.S. Perhaps you already know this and I am talking down to you, but when you work over a field, the adjective flat can be omitted -- every morphism to the spectrum of a field is flat. D.S. – Dan Petersen Feb 15 2011 at 11:59
Also, if you had asked Torsten or Sándor the same question, you would've gotten an authoritative answer. :p – Dan Petersen Feb 15 2011 at 12:04
Hi, Dan - yep, I knew (it's only "a for instance sentence") :-) – THC Feb 15 2011 at 15:23
show 2 more comments
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The automorphism group functor is not representable in general. Consider for instance the case of an algebraically closed field $k$ and $\mathbf A^2=\mathrm{Spec}\ k[x,y]$. Assume the automorphism group functor for this scheme is representable by an algebraic space $Aut$. For each $n$ we have an automorphism $(x,y)\mapsto(x+\sum_{1\leq i\leq n}t^iy^i,y)$ over $k[t]/(t^{n+1})$ giving a set of compatible morphisms $\mathrm{Spec}k[t]/(t^{n+1})\to Aut$. That system comes from a morphism $\mathrm{Spec}\ k[[t]]\to Aut$ (see Emerton's response to this question, which works also for maps into algebraic spaces at least as $k$ is algebraically closed). However, the corresponding automorphism would have to be $(x,y)\mapsto(x+\sum_{1\leq i}t^iy^i,y)$ which doesn't make sense.
-
Let $X\to B$ and $Y\to B$ be two flat, projective $B$-schemes ($S$ is already taken ;-). Then
Let $\mathscr Hom_B(X,Y)$ be the functor defined by `$$\mathscr Hom_B(X,Y)(Z)=\{Z{\rm -morphisms }\ X\times_B Z\to Y\times_B Z\}.$$` where $Z\to B$ is also a $B$-scheme. Then $\mathscr Hom_B(X,Y)$ is represented by an open $B$-subscheme $${\rm Hom}_B(X,Y)\subset {\rm Hilb}_B(X\times_BY).$$
The $\mathscr Hom$ functor has a subfunctor $\mathscr Isom$ consisting of those morphisms that define a relative isomorphism. This is represented by an open subscheme $${\rm Isom}_B(X,Y)\subset {\rm Hom}_B(X,Y).$$
Now if $B$ is a point, $X=Y$, then this $\rm Isom$ scheme can be identified with the automorphism group of $X$.
-
maybe you meant Z-morphisms $X\times_B Z \to Y\times_B Z$ in the definition of the functor? – Mattia Talpo Feb 10 2011 at 17:12
Yes, thanks. . – Sándor Kovács Feb 10 2011 at 17:46
For projective, non-uniruled varities, the birational automorphism group can also be given a scheme structure. The reference is "Hanamura: Structure of birational automorphism groups, I: non-uniruled varieties". As one expects, the automorphism group will be a subscheme of that.
-
This is a partial answer to THC's question about a "direct" way to do this.
Let $X$ be a smooth projective variety such that $\omega_X$ is ample. Then there is some $m\in\mathbb N$ such that $\mathscr L=\omega_X^{\otimes m}$ is very ample, so $X$ has an embedding into $\mathbb P^N=\mathbb P(H^0(X,\mathscr L))$. Obviously $\mathscr L$ is invariant under any automorphism, and hence so is $H^0(X,\mathscr L)$. This implies that the automorphisms of $X$ extend to automorphisms of the ambient $\mathbb P^N$. In other words ${\rm Aut}(X)$ can be identified with a quotient of a subgroup of ${\rm Aut}(\mathbb P^N)$ and hence it inherits a natural scheme structure.
On the other hand it actually also follows that in this case ${\rm Aut}(X)$ is finite (see below why) and in fact it seems to me that this argument could only possibly work if something close to that is true.
Assume that $X$ is embedded into a projective space and all automorphisms of $X$ are induced by automorphisms of the ambient projective space. Then ${\rm Aut}(X)$ has a scheme structure (as above) and let ${\rm Aut}^\circ (X)$ denote the connected component of ${\rm Aut}(X)$ containing the identity. Then the quotient ${\rm Aut}(X)/{\rm Aut}^\circ (X)$ is a discrete group which is the quotient group of a linear algebraic group and hence it is finite.
The action of ${\rm Aut}^\circ (X)$ (in this case) exhibits $X$ as a uniruled variety and hence if $X$ is not uniruled (e.g., when $\omega_X$ is ample) it implies that ${\rm Aut}(X)$ is finite.
In other words if the scheme structure on ${\rm Aut}(X)$ given by the Hilbert scheme argument has infinitely many components, then it cannot be the quotient of a subgroup of ${\rm Aut}(\mathbb P^N)$. This happens for instance for K3 surfaces with an infinite automorphism group, in particular for any supersingular K3.
-
Of course, Sándor's answer explains much more, but it's also worth pointing out that on an elliptic curve $E$, there is no $\mathrm{Aut} E$-invariant line bundle. Hence for every embedding of $E$ into projective space, almost all automorphisms do not extend to projective space. – ABayer Feb 20 2011 at 12:14
Dear Sandor, very informative ! Can you say a word about "and hence it inherits a natural scheme structure" ? (Why this is when dealing with a subgroup quotient ?) I think the finiteness outcome is not an obstruction for me, since eventually my automorphism schemes will look like Chevalley group schemes ... – THC Feb 21 2011 at 12:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 61, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9342503547668457, "perplexity_flag": "head"}
|
http://mathhelpforum.com/discrete-math/185589-question-about-recursion-theorem-print.html
|
# Question about recursion theorem
Printable View
• August 4th 2011, 04:20 AM
mrproper
I'm having difficulties understanding some material in a book I'm reading. (Hrbacek, K., Jech T., Introduction to set theory, 3rd Edt.). In the book there is a presentation of three variants of Recursion Theorem. The first variant states:
For any set $A$, any $a\in A$ and any function $g:A \times N \to A$ there exists unique infinite sequence $f:N \to A$ such that
a/ $f_{0}=a$
b/ $f_{n+1}=g(f_{n},n)$
I understand that theorem is meant to prove existence of functions, described recursively, like n! And I think I understand the proof. It introduces the term "n-step computation" which is finite sequence. The first element is the initial value and every next element is calculated from the previous, using g. Each separate computation is a function on some $n \in N$, hence it's a set of ordered pairs. Therefore each n-step computation is element of the power set of $N \times A$. Then the set of all computations may be defined:
$\{t \in P(N \times A) | \text{t is an n-step computation for some n}\}$
The proof procedes to show that all n-step computations for certain g are compatible functions, hence their union is a function. The proofs completes with establishing that the function has the desired properties.
So far so good, but then comes the second version - more general construct for functions like Fibonacci sequences in which each member of the sequence is calculated from several of the preceding ones:
For any set $S$ and any function $g:Seq(S) \rightarrow S$ there exists unique sequence $f:N \rightarrow S$ such that:
$f_{n}=g(<f_{0},f_{1},\cdots,f_{n-1}>)$ for all $n \in N$
( $Seq(S)$ means all finite sequences of elements of S, < and > are used to denote sequences and <> is used to denote the empty sequence).
The proof relies on the first version of recursion theorem. We set $A=Seq(S)$, $a=<>$ and $G:Seq(S) \times N \rightarrow Seq(S)$, defined as:
$G(t,n) = t \cup \{<n,g(t)>\}$ if t is a sequence of length n
$G(t,n) = <>$ otherwise
Here I'm loosing the idea behind the notation and any clue how the proof stated further on works. g must stand for any function that calculates value from a given sequence. It seems that application of the first version of the recursion theorem yields a sequence that can be described this way:
$F_{0}=<>$
$F_{n+1}=F_{n}\cup\{<n,g(t)>\}$
This is actually in the book. Then the book states by now it's easy to prove that every $F_{n}$ is actually member of the set of all functions on n into S $(S^n)$. But it's not obvious to me how it is so.
$F_{0}=<>$
$F_{1}=<>\cup\{<0,g(F_{0})>\}$ ( $g(F_{0})$ is some element of S)
$F_{2}=<>\cup\{<0,g(F_{0})>\}\cup\{<1,g(F_{1})>\}$
$\cdots$
The resulting sets of the type $\{<0,g(F_{0})>,<1,g(F_{1})>,\cdots\}$ do not look like function at all, if one takes a function to be set of ordered pairs. Am I reading the notation wrong here?
I am interested in other sources, where this theorem is proven this or other way, preferably more understandable for a newbie.
• August 4th 2011, 08:16 AM
MoeBlee
Quote:
Originally Posted by mrproper
The resulting sets of the type $\{<0,g(F_{0})>,<1,g(F_{1})>,\cdots\}$ do not look like function at all, if one takes a function to be set of ordered pairs.
Maybe, I'm missing something, but it looks like a set of ordered pairs to me. It has as members such ordered pairs as
<0 g(F_0)>
<1 g(F_1)>
etc.
and no other members.
All ordered pairs.
I don't see any difficulty with that.
• August 4th 2011, 11:17 AM
mrproper
You mean that I should consider that other definition of ordered pair which uses sequences? So the set becomes $\{(0,g(F_{0})), (1,g(F_{1})) ...}$
• August 4th 2011, 11:22 AM
MoeBlee
Definitions:
<j k> = {{j} {j k}}
x is an ordered pair <-> Ejk x=<j k>
f is a relation <-> Ax(x in f -> x is an ordered pair)
f is a function <-> (f is a relation & Axyz((<x y> in f & <x z> in f) -> y=z))
f is a sequence <-> (f is a function & dom(f) is an ordinal)
Perhaps what you're asking about is whether we use tuples or whether we use functions in certain contexts.
Let's define 'tuple' this way.
h is a tuple construction sequence
<->
h is a finite sequence &
An(n+1 in dom(h) -> Ej h(n+1)=<h(n) j>)
t is a tuple
<->
Eh(h is a tuple construction sequence & t in range(h))
I.e., a tuple is the result of using the ordered pairing operation n number of times for some natural number n.
Then, for any tuple there is the corresponding sequence, and vice versa:
The n-tuple <x y ... z> corresponds to the sequence {<0 x> <1 y> .... <n z>} and vice versa.
So, in most instances, it doesn't matter which we use, tuples or sequences.
In any case, the set you mentioned is clearly a function.
• August 4th 2011, 11:38 AM
mrproper
That's right. So I'm reading the proof correctly. But something worries me here... The recursion theorem is given early in the book as a tool for constructing isomorphisms. To the moment of introduction of the theorem the ordered pairs are defined only the way Kuratovski did. (a,b)={{a}, {a,b}}. And the author seems quite precise in such details. Why then he hasn't written $G(t,n)=t\cup\{(n,g(F_{n})\}$ instead of $G(t,n)=t\cup\{<n,g(F_{n})>\}$
Anyway, I'm trying to read this book from quite some time and sometimes I just "crash" when I try to understand what's encoded in the notation. Is there any source that can help me understand notation better?
• August 4th 2011, 11:49 AM
MoeBlee
I don't have the book. I'd have to see how the author specifies the usage of "( )", but is there any indication that "( )" is meant differently from "< >"? Sometimes authors use both "( )" and "< >" for the same thing:
(x y ... z) = <x y ... z> = <<<x y> ... > z>
And sometimes authors slip back and forth with the word 'sequence' where it may mean a tuple or it may the corresponding function on a natural number. In most contexts, you can see how it doesn't matter which of the very similar things is used, tuples or functions.
• August 4th 2011, 12:26 PM
mrproper
In this case it seems that it doesn't matter. The author states the same thing 10 or so pages further on.
I'm interested if the Recursion theorem in this book actually related to some of Kleene's recursion theorems. Kleene's recursion theorems are described and proved in Wikipedia, but it's formulation is not as newbie-firendly...
• August 4th 2011, 01:13 PM
MoeBlee
By the way, one difference is that the axiom schema of replacement is needed to prove:
EbAn((n is a natural number & t is an n-tuple of members of x) -> teb)
But the axiom schema of replacement is not needed to prove:
EbAn((n is a natural number & f is a sequence of length n of members of x) -> feb)
In other words, we need the axiom schema of replacement to prove there is the set of all tuples of members of x. But we don't need the axiom schema of replacement to prove there is the set of all finite sequences of members of x.
• August 5th 2011, 12:32 PM
bryangoodrich
I have the book, and I'll have to check if I'm recalling the correct one, but I believe they made a difference between ordered pairs and n-tuples (2-tuples). The difference, however, is slight and inconsequential to the results (more syntactical than semantical). That might explain the difference between the two uses. However, I could be thinking of another book.
• August 5th 2011, 09:26 PM
bryangoodrich
mrproper, you might want to have a look at section "5. Operations and Structures" in Chapter 3 (pp 56-7). Below is a snippet from that section.
_________________________________________________
We now proceed to give a general definition of a structure. In Chapter 2, we introduced a unary (1-ary), binary (2-ary),and ternary (3-ary) relations. The reason we did not talk about n-ary relations for arbitrary n is simply that we could not then handle arbitrary natural numbers. This obstacle is now removed, and the present section gives precise definitions of n-tuples, n-ary relations and operations, and n-fold cartesian products, for all natural numbers n.
We start with the definition of an ordered n-tuple. Recall that ordered pairs $(a_0, a_1)$ has been defined in Section 1 of Chapter 2 as a set that uniquely determines its two coordinates ... In analogy, an n-tuple should be a set that uniquely determines its n coordinates ...; that is, we want
$(*) \ (a_0, ..., a_{n-1}) = (b_0, ..., b_{n-1}) \text{ if and only if } a_i = b_i, \text{ for all } i = 0, ..., n-1$
But we already introduced a notion that satisfies (*): it is the sequence of length n ... [it] is just a reformulation of equality of functions, as in Lemma 3.2 in Chapter 2 (see page 24: Functions F and G are equal iff their domains are equal and for all points x in the domain F(x) = G(x))
_________________________________________________
Now, recall that under Section 3 "The Recursion Theorem" of Chapter 3, page 46, they define a sequence as "a function whose domain is either a natural number or N," (i.e., either the elements less than a given natural number or the set of all natural numbers itself). Thus, a sequence is defined as a function of a certain domain. It is obviously clear that if two sequences S and S' were both sequences of length n, then their domains would both be the natural number n, itself, and if they agreed at each coordinate (i.e., for every x in n, S(x) = S'(x)), then this satisfies Lemma 3.2 for equality of functions.
As you read through the Recursion Theorem here, do not think of them as merely functions that are themselves a set of ordered pairs. Think of them as members from $A^n, \text{ for all } a_i \in A, i = 0, 1, ..., n-1$. (Recall definition 3.13 on page 26: $B^A$ is the set of all functions on A into B.)
• August 6th 2011, 03:23 AM
bryangoodrich
I was reviewing Theorem 3.5 (p. 50), which states.
For any set S and any function $g:Seq(S) \rightarrow S$ there exists a unique sequence $f:N\rightarrow S$ such that
$f_n = g(f | n) = g(\langle f_0, ..., f_{n-1}\rangle), \text{ for all } n\in N.$
I think it is important to break this down so we're clear on what is being stated. First off, the Seq(S) is nothing but the union of all functions from n into S (p. 46):
$Seq(S) = \bigcup_{n\in N} S^n$
This is a complicated statement because the theorem is saying for any set S and for any function g that maps into S from, basically, the set of functions from n into S, then we can define another function f that maps from N into S (obviously, based on g). In particular, this is done by defining $f_n$ (f evaluated at n, see definition 3.1 on page 23). But what is $f_n$? It is nothing but a function defined at n, for all $n\in N$ where at each n it is equal to g evaluated at f restricted to n. But as the equality in the above statement goes, this is just g evaluated at a sequence of f for each numbered index before n.
The proof begins by defining the sequence (function):
$\langle F_n\ |\ n \in N\rangle = \langle f\ |\ n\in N\rangle$, with
$F_0 = \langle \rangle$
$F_{n+1} = F_n \cup \{ \langle n, g(F_n) \rangle\}, \text{ for all } n\in N$
For clarity on the solution, notice that $F_n$ consists of a sequence of the points into g as defined for $f_n$ in the theorem statement, and g maps into S. The proof will conclude by letting $f = \bigcup_{n\in N} F_n$, which is clearly a function from N into S.
Before we get there, however, we need to establish the existence of the defined sequence above. This follows from the Recursion Theorem, as pointed at on page 50. They show you the substitutions to make into the Theorem, which we could line up side-by-side, but I'll just state substituted below.
Recursion Theorem (substituted for this problem):
For any set Seq(S), any $f_n \in Seq(S)$, and any function $G:Seq(S)\times N \rightarrow Seq(S)$, there exists a unique infinite sequence $f:N \rightarrow Seq(S)$ such that
(a) $F_0 = \langle \rangle$
(b) $F_{n+1} = G(t, n)$, where
$G(t, n) = t \cup \{\langle n, g(t)\rangle\}, \text{ if t is a sequence of length n}, \langle \rangle, \text{ otherwise.}$
You should be able to see the clear correspondence with G(t, n) and <F>. Therefore, <F> is well defined and exists by the recursion theorem. The authors go on to state that it is easy to prove by induction that each $F_n$ belongs to $S^n$ and that $F_n \subseteq F_{n+1}$ for all $n\in N$. By previous results, this establishes that $\{F_n\ |\ n\in N\},$ is a compatible system of functions. Note, this is saying that the range of $F_n$ is a compatible system, and a compatible system just is (Definition 3.10 page 26) a set of functions for which any two functions in it are compatible; furthermore, any two functions are compatible if their points evaluated are equal for every point that belongs to the intersection of those function domains.
The domain of <F> is the natural numbers and its range is based on $f\ |\ n$ which maps to S. Thus, this establishes the concluding statement mentioned earlier:
Let $f = \bigcup_{n\in N} F_n$; then clearly $f:N\rightarrow S$ and $f\ |\ n = F_n, \text{ for all } n\in N$. Therefore, this concludes what was needed since,
$f_n = F_{n+1}(n) = g(F_n) = g(f\ |\ n)$
I think if you look at this final statement it should make things appear clearer because it tells you precisely what is going on. It is easy to get confused about the n's that are being thrown around. When F_i is being defined, you have to remember that it is a function of numbers, and as the statement above points out, any $F_{n+1}(n)$ just is g evaluated at $F_n$. Remember, further, why this follows: $F_n \subseteq F_{n+1}$. So what do we get by the definition of $F_{n+1}$? Remember, it was defined as a union.
• August 16th 2011, 06:32 AM
mrproper
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 72, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9445526599884033, "perplexity_flag": "head"}
|
http://math.stackexchange.com/questions/tagged/convergence+complex-analysis
|
# Tagged Questions
4answers
72 views
### Radius of Convergence of power series of Complex Analysis
I have come across the following few questions on past exams papers.. I know how to solve these type when it is of the form $a_nz^n$ but don't have a clue what to do with these. Any help would be ...
3answers
101 views
### showing $1 + z + z^2 + \dots$ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$
What test can I use to show that $1 + z + z^2 + \dots$ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$. I know $\displaystyle 1 + z + z^2 + \dots +z^n = \frac{1-z^{n+1}}{1-z}$ and as \$n \to ...
2answers
69 views
### Convergence of Roots for an analytic function
Show that the roots of $$f(z) = z^n+z^3+z+2 =0$$ converge to the circle $|z|=1$ as $n \to \infty$.
1answer
29 views
### Study the convergence of this sequence of functions
I have the following sequence of function: $$f_n(\lambda)=\bigg[\alpha-i\bigg(\lambda+\frac{1}{n}\bigg)\bigg]^{-1}-\bigg[\alpha-i\lambda\bigg]^{-1},\,\,\,\alpha\neq 0$$ and I have to study its ...
1answer
26 views
### how to show uniform convergence for sequence $u_n(z) = n z e^{-nz^2}$ such that $\Re[z^2] > 0$
How to show uniform convergence for $u_n(z) = n z e^{-nz^2}$ such that $\Re[z^2] > 0$ Here is my attempt letting $z = x + iy$: \begin{align*} |u_n - 0| &= |nz e^{-nz^2}| \\ &\le n ...
1answer
43 views
### help for convergence of an integral
Does this integral converges: $$\int_{-\infty}^{\infty}\frac{\mathrm e^{ -\mathrm{i}\omega t}}{\sqrt{(-1)^\frac{1}{3}(\mathrm{i}\omega)^\frac{4}{3}}}\,\mathrm{d}\omega$$ and why ?
1answer
29 views
### Zeros of the analytic limit of complex rational function
For $n\in\mathbb{N}$ let $r_n,\ s_n$ be two polynomials of $O(n)$ degrees with real positive coefficients and set $f_n=r_n/s_n$. Suppose there exists $c>0$ such that $\bullet$ if $z\in\mathbb{C}$ ...
1answer
38 views
### Zeros of the analytic limit of complex polynomials
For $n\in\mathbb{N}$ let $p_n$ be a polynomial of degree $n$. Suppose there exists $c>0$ such that $\bullet$ if $z\in\mathbb{C}$ is a zero of a $p_n$, then $|z^2+c|\leq c$ (note that in particular ...
1answer
105 views
### Does a convergent power series on a closed disk always converge uniformly?
If I have a power series $\displaystyle\sum_{i=0}^{+\infty} {a_iz^i} \in\mathbb{C}[[z]]$ with radius of convergence $r>0$ and I know that the series $\displaystyle\sum_{i=0}^{+\infty} {a_iz^i}$ ...
0answers
28 views
### Proof that a sequence of numbers $A_i$ is an infinite product of complex residues correct? [closed]
Is the above proof correct? Assume that theorem 2 is true without proof Thanks Edit: At the end it should be $2^n$, not $2n$
1answer
56 views
### Behavior at infinity.
Classify the behavior at $\infty$ for $$f(z)=\frac{\sin z}{z^2},\,g(z)=\frac{1}{\sin z},\,h(z)=\exp\left(\tan\frac{1}{z}\right).$$ So I just considered $f(1/z),g(1/z),h(1/z)$ at $z=0$. For $f$ I ...
1answer
70 views
### Radius of convergence of Maclaurin series for $\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$
What is the radius of convergence of the Taylor series about $z=0$ for $h(z)=\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$? Here's a plot ...
1answer
52 views
### Absolute convergence of $\sum\limits_{n=0}^{\infty} \frac {z} {(z+n)^2}$
I want to check the absolute convergence of $\displaystyle\sum_{n=0}^{\infty} \frac {z} {(z+n)^2}$ in the half plane $\Re(z)>0$, and to see if the convergence is uniform or locally uniform. How do ...
1answer
58 views
### Radius of convergence of $\sum_{-\infty}^{\infty}3^{-|n|}z^{2n}, z \in \mathcal{C}$
I want to find out the radius of the following power series of a complex variable: $\sum_{-\infty}^{\infty} 3^{-|n|} z^{2n}, z \in \mathbb{C}$ The ration test \$\lim_{n \to ...
2answers
113 views
### Confused by Laurent series
A typical problem related to Laurent series is this: For the function $\frac 1{(z-1)(z-2)}$, find the Laurent series expansion in the following regions: \$\\(a) |z|<1, \\ (b) 1<|z|<2, ...
1answer
50 views
### Confused over analytic functions, point convergence of power series
It is well-known that a power series sums to a function that is analytic at every point inside its circle of convergence and that conversely, if a function is analytic on an open disc then its Taylor ...
2answers
102 views
### How do I obtain the Laurent series for $f(z)=\frac 1{\cos(z^4)-1}$ about $0$?
I know that $$\cos(z^4)-1=-\frac{z^8}{2!}+\frac{z^{16}}{4!}+...$$ but how do I take the reciprocal of this series (please do not use little-o notation)? Or are there better methods to obtain the ...
1answer
57 views
### Convergence radius of power series
I am trying to solve an exercise, but i am not sure that the result i get at the end is correct...May i kindly ask you for a little help or a remark? Find the radius of convergence of the following ...
1answer
90 views
### Existence of Riemann-Liouville Integral
The Riemann Liouville integral is defined as: $\frac{1}{\Gamma\left(\nu\right)}\int\limits _{h}^{t}\left(t-\xi\right)^{\nu-1}f\left(\xi\right)d\xi$ It is supposed it does exist for all $\nu>0$ and ...
2answers
131 views
### Is the alternating sum of exp(-exp(n z)) analytic?
Define $f(z) = \frac{1}{\exp(\exp(z))} - \frac{1}{\exp(\exp(2z))} + \frac{1}{\exp(\exp(3z))} - \frac{1}{\exp(\exp(4z))} + ...$ $f(z) = \sum_{n = 1}^{\infty} (-1)^{n-1} \exp(-\exp(n z))$. Is $f(z)$ ...
1answer
252 views
### Is there a $k$ such that $a_n=\frac{n^k!}{(n^k!!)^2}$ converges?
Lately I have been playing around with the sequence $$a_n(k) := \frac{n^k!}{(n^k!!)^2}.$$ For $k=1$, it does not look much like it converges. I don't know $k=2$ it converges, but it doesn't really ...
2answers
265 views
### Is $\frac{1}{\exp(z)} - \frac{1}{\exp(\exp(z))} + \frac{1}{\exp(\exp(\exp(z)))} -\ldots$ entire?
Let $z$ be a complex number. Is the alternating infinite series $f(z) = \frac{1}{\exp(z)} - \frac{1}{\exp(\exp(z))} + \frac{1}{\exp(\exp(\exp(z)))} -\ldots$ an entire function ? Does it even converge ...
0answers
166 views
### Fibonacci Numbers - Complex Analysis [duplicate]
Possible Duplicate: Complex Analysis - Integral over a circle of radius R Hey guys~ Does anyone know where to find the solutions to this problem set on page 106 involving the fibonacci ...
1answer
256 views
### Radius of convergence for the exponential function
I'm studying physics and am currently following a course on complex analysis and in the section on analytic functions, the radius of convergence $R$ for power series was introduced. The Taylor ...
2answers
108 views
### Divergence of an infinite product
How can I prove that the infinite product $$\displaystyle\prod_{n=1}^{+\infty}(1+z^{2n})$$ diverges for $|z|>1$?
1answer
189 views
### Convergence of infinite product
This could be something which is already somewhere in the website, but I am unable to locate any. Prove $$\prod_{n=1}^{\infty} (1-z^n)$$ converges absolutely and uniformly on each compact subset of ...
3answers
395 views
### A question on convergence of series
Suppose $(z_i)$ is a sequence of complex numbers such that $|z_i|\to 0$ strictly decreasing. If $(a_i)$ is a sequence of complex numbers that has the property that for any $n\in\mathbb{N}$ ...
0answers
55 views
### Uniform convergence of complex exponent derivative
I'm trying to prove the following: Let $\Re z > 0$. Then $$\lim_{\varepsilon \to 0} \frac{t^{z + \varepsilon} - t^z}{\varepsilon} = t^z \log t$$ uniformly in $t \in [0,1]$. I've tried to ...
2answers
86 views
### Find the limit of $\frac{\bar{z}}{z}$ as $z$ goes to $0$.
I put it in exponential form to get $\dfrac{re^{-i \theta}}{re^{i \theta}}$ but I think I'll get $\frac{0}{0}$ which isn't defined and isn't a good enough proof to say it doesn't have a limit.
0answers
79 views
### Presentation Topic: Convergence rate of Power Series
I am required to do a 10-15 minutes presentation in a Complex Analysis class on the following idea: "Discuss convergence rates of power series, and numerical aspects of how to write good algorithms ...
2answers
80 views
### Where's the singular point of the series $\sum_{v\ge 1}v^{-2}z^v$?
The radius of convergence of the series $\sum_{v\ge 1}v^{-2}z^v$ is 1, so there must be a singular point on the boundary. But for every $|\zeta|=1$, \$|\sum_{v\ge 1}v^{-2}\zeta^v|\le\sum_{v\ge ...
1answer
98 views
### Infinite product of entire functions
Is it true that "an infinite product of entire functions $\{ f_{n}(z) \}_{n=1}^{\infty}$ is again an entire function"? $$\prod_{n=1}^{\infty}f_{n}(z)$$ (entire function is a function which is ...
1answer
229 views
### Error bound for Taylor approximation to $\tan(x)$
I'm looking to bound the error for the Taylor series of $\tan(x)$ so that I will know how many terms I need to go out to get a desired precision. I've already searched and came across this, but the ...
1answer
277 views
### Complex analysis: Radius of convergence of power series
$$\sum_{n=1}^\infty \frac{\cos n \theta}{(\sqrt{13})^{n+1}}x^n$$ Find the radius of convergence for the above series. I have learnt to use the root test and ratio test but neither of them seem to ...
0answers
108 views
### $L^{2}(\mathbb R)$- norm of entire function
Let $f(z)$ be an entire function defined by $$f(z)=\prod_{n=1}^{\infty}\bigg(1-\frac{z^{2}}{a_{n}^{2}}\bigg),\qquad z\in \mathbb C$$ where $\{a_{n}\}_{n=1}^{\infty}$ is a sequence of positive real ...
1answer
263 views
### Uniform convergence of infinite series
Suppose $f$ is a holomorphic function (not necessarily bounded) on $\mathbb{D}$ such that $f(0) = 0$. Prove the the infinite series $\sum_{n=1}^\infty f(z^n)$ converges uniformly on compact subsets ...
1answer
142 views
### Bounded power series without convergence on the boundary
Does there exist a holomorphic function $f\in H^\infty$ (i.e. holomorphic and bounded on the open unit disc) such that its power series expansion at $0$ doesn't converge anywhere in $\{|z|=1\}$?
1answer
91 views
### Convergence in the upper half-plane
I have a sequence $\{F_{n}(z)\}_{n=1}^{\infty}$ of analytic functions in the open upper half plane $\mathbb H$ and continuous on $\mathbb R$, such that $|F_{n}(z)|\leq 1$ for all $n\geq 1$, and all ...
0answers
222 views
### Uniform convergence of analytic functions
Let $f_{n}(z), g(z)$ be entire functions, for all $n\geq 1$. Suppose that $g(x)$ doesn't vanish on $\mathbb H\cup\mathbb R$ (so we have $\frac{f_{n}(z)}{g(z)}$ analytic on $\mathbb H\cup\mathbb R$). ...
0answers
250 views
### Convergence of sequence of analytic functions
Given $\{f_{n}(z)\}$, a sequence of analytic functions in the upper half plane $\mathbb C^{+}$, where each $f_{n}(z)$ has continuous extension to the real line, and $|f_{n}(z)|\leq 1$ for all \$z\in ...
1answer
123 views
### Zeros in the complex plane and convergence
I'm doing some number theory which requires some work in $\mathbb{C}$, but unfortunately my complex analysis is a little rusty. A text I am reading states the following: ...and given that ...
1answer
73 views
### Convergence to zero of a sequence
Let $c > 0$. I'm trying to show that the sequence $\displaystyle\sum\limits_{k=0}^n \left|\frac{n^k-\frac{n!}{(n-k)!}}{n^k}\right|\frac{c^k}{k!}$ converges to zero, as $n \to \infty$. I know that ...
2answers
99 views
### Absolute convergence for all values except the values $z=\left( 1+\frac {a} {m}\right) e^{\frac {2k\pi i} {m}}$
I am trying to show that the series \sum _{n=1}^{\infty }\dfrac {nz^{n-1}\left( \left( 1+\dfrac {1} {n}\right) ^{n}-1\right) } {\left( z^{n}-1\right) \left( z^{n}-\left( 1+\dfrac {1} {n}\right) ...
2answers
553 views
### Determining precisely where $\sum_{n=1}^\infty\frac{z^n}{n}$ converges?
Inspired by the exponential series, I'm curious about where exactly the series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges. I calculated ...
1answer
200 views
### Absolute Convergence of an Infinite Product based on Weierstrass's Factor Theorem
I am trying to show that \$\left\{ \left( 1-\dfrac {z} {\pi }\right) e^{\left( \dfrac {z} {\pi }\right) }\right\} \left\{ \left( 1+\dfrac {z} {\pi }\right) e^{\left( -\dfrac {z} {\pi }\right) }\right\} ...
1answer
378 views
### Every Cauchy sequence in $\mathbb{C}$ is bounded
Prove that every Cauchy sequence in $\mathbb{C}$ is bounded. In $\mathbb{R}$, this is a sketch of the proof that I recall: Let {${a_k}$} be Cauchy in $\mathbb{R}$, since $1\in\mathbb{R}$, ...
1answer
172 views
### Proof of Uniform Convergence
Prove that the sequence of functions $\{f_{n}(z) = (1+nz)^{-1}\mid n=1,2,...\}$ converges uniformly to $f(z)=0$ for $|z| \geq r > 0$. To answer the question, for a given choice of \$\epsilon > ...
0answers
145 views
### Complex Analysis: Power Series and Convergence
Let $f(z)=\sum_{n=0}^{\infty}c_nz^n$ have radius of convergence $R$. Problem Prove that $\sum_{n=0}^{\infty}\overline{c_n}z^n$ has radius of convergence $R$ and that ...
2answers
239 views
### Complex Analysis: Convergence of Power Series
Problem Let $p(z)=(z-a_1)(z-a_2)...(z-a_N)$, where $a_1, a_2, ..., a_N$ are distinct complex numbers. Let $M=\min_{1\le{k}\le{N}}|a_k|$. Prove that it is possible to express $\frac{1}{p(z)}$ as a ...
2answers
341 views
### Complex Analysis: Radius of convergence of Power Series
Let p be a polynomial of degree $k>0$. Prove that $\sum p(n)z^n$ has radius of convergence $1$ and that there exists a polynomial $q(z)$ of degree $k$ such that \sum_{n=0}^{\infty} p(n) ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 142, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.907460629940033, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/2100
|
Is there a coalgebraic characterisation of the hyperfinite II_1 factor?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Peter Freyd showed that the real interval [0, 1] is a final coalgebra for a functor on sets equipped with two points, which sends such a set to the 'wedge' of two copies of itself, identifying the second point of the first copy with the first point of the second copy.
The hyperfinite II_1 factor has trace values in [0, 1] and arises from a process of completing a union of finite subalgebras by a form of doubling discussed here.
Is there then a similar coalgebraic characterisation of the factor?
-
Perhaps this ought to be tagged "von neumann algebras" as well - and maybe even "functional analysis"? Not sure on the hierarchy of labels/tags used here on MO at the moment. Oh, and interesting question! I don't know the answer, but would just like to observe that since said factor is an injective von Neumann algebra, it should have "lots of morphisms going into it", which seems a first bit of encouragement. – Yemon Choi Oct 24 2009 at 1:24
1 Answer
I thought about this question some yesterday. As I was saying in the related post, von Neumann algebras are a non-commutative or quantum generalization of measurable spaces, `$C^*$` algebras are a non-commutative/quantum generalization of compact Hausdorff spaces, and both generalizations are contravariant. Your question has a covariant spirit, which is a bit misaligned but not an essential point for this particular question.
The more germane issue is that the interval is a topological space in Peter Freyd's construction, while the hyperfinite factor, like any von Neumann algebra, behaves as a measurable space. It is true that the hyperfinite factor is a non-commutative analogue of an interval, and you can construct it as the closure of a certain class of operators on $L^2([0,1])$. But Freyd's construction does not work for measurable maps, only continuously. In fact, as a measurable space, the unit interval doesn't have endpoints. It has points, sort-of, but not in any useful way. (To be precise, the measurable model of the interval here is the class of measures on it which are Lebesgue absolutely continuous, not all Borel measures.)
Maybe there is a `$C^*$`-algebra with a Freyd-type property, and which generates the hyperfinite factor. You would have to decide whether its "endpoints" are a classical bit or a qubit. You could try to work in the category of `$C^*$`-algebra homomorphisms, which tends to be short of morphisms from non-commutative objects. Or you could try to work in the category of completely positive maps on `$C^*$`-algebras, which has plenty of morphisms but also has other complications. Certainly you would want to reverse arrows in passing from topology to $C^*$-algebras. I don't know a whole lot about making `$C^*$`-algebras; I couldn't come up with anything.
-
You're right, Greg, there is something like a dualized version of Freyd's theorem. It characterizes L^1[0,1] among Banach spaces, by a simple universal property. One also gets the definition of integration on [0, 1] out of this. See maths.gla.ac.uk/~tl/glasgowpssl – Tom Leinster Nov 12 2009 at 23:55
Just to clarify: I'm not claiming this has anything to do with factors. – Tom Leinster Nov 12 2009 at 23:57
I can't claim credit for anticipating your example of a dualized Freyd's theorem. Indeed you found an initial object in a category of Banach spaces, decorated in vaguely the same way as Freyd's coalgebra structure. Of course, I took the question more literally than that. – Greg Kuperberg Nov 13 2009 at 0:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9538788199424744, "perplexity_flag": "middle"}
|
http://cms.math.ca/Events/winter12/abs/st
|
2012 CMS Winter Meeting
Fairmont Queen Elizabeth (Montreal), December 7 - 10, 2012
Partial Differential Equations and Spectral Theory
Org: Dmitry Jakobson (McGill) and Iosif Polterovich (Montréal)
[PDF]
ALMUT BURCHARD, University of Toronto
Perimeter under multiple Steiner symmetrizations [PDF]
Steiner symmetrization along n linearly independent directions in n-space transforms every compact set into a set of finite perimeter.
(Joint work with Gregory R. Chambers.)
YAIZA CANZANI, McGill University
Distribution of randomly propagated Schrödinger eigenfunctions [PDF]
This is joint work with Dmitry Jakobson and John Toth. Let $(M,g_0)$ be a compact Riemmanian manifold, and $V \in C^\infty(M)$. Let $P_0 (h) := -h^2\Delta_{g_0}+V$, be the semiclassical Schrödinger operator for $h \in (0,h_0]$. If $\varphi_h$ is an $L^2$-normalized eigenfunction of $P_0(h)$, then $\int_A |\varphi_h(x)|^2dv_{g_0}(x)$ is interpreted as the probability that a quantum particle of energy $\sim 1/h^2$ belongs to $A \subset M$. For a quantum particle with initial state $\varphi_h$, its evolution at time $t$ is described by the same probabilty density since $|e^{-\frac{it}{h} P_0(h) } \varphi_h| = |\varphi_h|$. However, since real life systems are usually affected by noise'', the time evolution is better decribed by the state $$\varphi^{(u)}_h(x)=e^{-\frac{it}{h} P_u(h) } \varphi_h$$ where $P_u(h)$ is some small perturbation of $P_0(h)$.
In this talk we consider a smooth family of perturbations $g_u$ of the reference metric $g_0$ for $u \in \mathcal B^k(\varepsilon) \subset \mathbb R^k$ of radius $\varepsilon>0$, and consider the perturbed Schr\"{o}dinger operators $P_{u}(h) := -h^2 \Delta_{g_u} +V$. For $t>0$ small, we study the moments of the real part of the perturbed eigenfunctions regarded as random variables $$Re \left(\varphi^{(\cdot)}_h(x)\right):\mathcal B^{k}(\varepsilon) \to \mathbb R \quad \quad \text{for } x\in M.$$
LAYAN EL HAJJ, McGill
Intersection bounds for planar Neumann eigenfunctions with interior analytic curves [PDF]
Let $\Omega \subset \mathbb{R}^2$ be a bounded piecewise smooth domain and $\phi_\lambda$ be a Neumann (or Dirichlet) eigenfunction with eigenvalue $\lambda^2$ and nodal set ${\mathcal N}_{\phi_{\lambda}} = \{ x \in \Omega; \phi_{\lambda}(x) = 0 \}.$ Let $H \subset \Omega$ be an interior $C^{\omega}$ curve. Consider the intersection number $$n(\lambda,H):= \# ( H \cap {\mathcal N}_{\phi_{\lambda}} ).$$ We first prove that for general piecewise-analytic domains, and under an appropriate goodness" condition on $H$ , $$n(\lambda,H) = {\mathcal O}_H(\lambda) \,\, (*)$$ as $\lambda \rightarrow \infty.$ We then prove that the bound in $(*)$ is satisfied in the case of quantum ergodic (QE) sequences of interior eigenfunctions, provided $\Omega$ is convex and $H$ has strictly positive geodesic curvature.
SURESH ESWARATHASAN, McGill University/CRM
Lower bounds for the Weyl Remainder [PDF]
This talk will address omega bounds for the remainder term in Weyl's Law for piecewise $C^{\infty}$ domains. This is joint work with Iosif Polterovich and John Toth.
RENJIE FENG, CRM and Mcgill Unviersity
Some recent results on Gaussian random fields over Kahler manifolds [PDF]
I will present some results on Gaussian Random homomorphic sections over Kahler manifolds which is a generalization of Kac's polynomials to Kahler manifolds. The distribution of zeros is studied by Bleher-Shiffman-Zelditch; critical points by Douglas-Shiffman-Zelditch. Recently, I studied the critical values on SU(2) random polynomials with Z. Wang; I also studied critical values of random holomorphic sections on compact Kahler manifolds under the both Gaussian and Spherical ensembles with S.Zelditch.
WENYING FENG, Trent University
Nonlinear spectra and related maps: generalizations and applications [PDF]
I will introduce the nonlinear spectrum defined by three European mathematicians (Furi, Martelli and Vignoli), and its further generalizations in different directions. Classes of nonlinear maps related to the nonlinear spectra such as stably solvable operators, 0-epi maps, (a, q)-stably solvable operators, L-stably solvable operators, their properties and applications will be discussed. I will also present some new results on nonlinear operators that can be decomposed into a linear operator and a nonlinear map. Applications to existence of solutions for some boundary value problems are obtained.
NASSIF GHOUSSOUB, University of British Columbia
Stability and Regularity in elliptic systems and 4th order equations [PDF]
We examine nonlinear fourth order eigenvalue problems on bounded domains of N-dimensional space. We show --among other things-- that for exponential nonlinearities, the extremal solution is smooth provided the dimension N is below 10.718. To do that, we isolate a new stability inequality satisfied by minimal solutions that is more amenable to estimates, as it allows a method of proof reminiscent of the second order case. This new approach leads to substantial improvements of various results on critical dimensions obtained recently by various authors. This is joint work with Craig Cowan.
NIKY KAMRAN, McGill University
Local energy decay for Dirac fields in the 5-dimensional Myers-Perry black hole geometry [PDF]
We consider massive Dirac fields evolving in the exterior region of a 5-dimensional Myers-Perry black hole. Our main result states that the local energy of such fields decays in a weak sense at late times. This is proved in two steps. First, using the separability of the Dirac equation, we prove the absence of a pure point spectrum for the corresponding Dirac operator. Second, using a new form of the equation adapted to the two rotations axes of the black hole, we show by a Mourre theory argument that the spectrum is absolutely continuous. The result then follows. This is joint work with Thierry Daude (Cergy-Pontoise).
ALEXEY KOKOTOV, Concordia
Krein formula and S-matrix for Euclidean Surfaces with Conical Singularities [PDF]
We use the Krein formula and the S-matrix formalism to give formulas for the zeta-regularized determinant of non-Friedrichs extensions of the Laplacian on Euclidean surfaces with conical singularities. This formula involves S(0) and we show that the latter can be expressed using the Bergman projective connection on the underlying Riemann surface. The talk is based on the joint work with Luc Hillairet (Nantes).
BENJAMIN LANDON, McGill University
The scattering length at positive temperature [PDF]
A positive temperature analogue of the scattering length of a potential $V$ can be defined via integrating the difference of the heat kernels of $-\Delta$ and $-\Delta + V$, with $-\Delta$ the Laplacian. An upper bound on this quantity is a crucial input in the derivation of a bound on the critical temperature of a dilute Bose gas, obtained in 2009 by R. Seiringer and D. Ueltschi. This bound on the critical temperature was given in the case of finite range potentials and sufficiently low temperature. In this paper, we improve the bound on the scattering length and extend it to potentials of infinite range. (Joint work with R. Seiringer)
ARIAN NOVRUZI, University of Ottawa
About the regularity of optimal convex domains [PDF]
We will discuss the issue of the regularity of a domain, minimizing a shape functional under the convexity constraint, in dimension two and higher. Joint work with J. Lamboley and M. Pierre.
FRÉDÉRIC ROCHON, UQAM
Compactness of relatively isospectral sets of surfaces with cusps [PDF]
We introduce a notion of relative isospectrality for surfaces with boundary having possibly non-compact ends either conformally compact or asymptotic to cusps. We obtain a compactness result for such families via a conformal surgery that allows us to reduce to the case of surfaces hyperbolic near infinity recently studied by Borthwick and Perry, or to the closed case by Osgood, Phillips and Sarnak if there are only cusps. This is a joint work with Pierre Albin and Clara Aldana.
ROBERT SEIRINGER, McGill University
A positive density analogue of the Lieb-Thirring inequality [PDF]
The Lieb-Thirring inequalities give a bound on the negative eigenvalues of a Schroedinger operator in terms of an $L^p$ norm of the potential. This is dual to a bound on the $H^1$-norms of a system of orthonormal functions. Here we extend these to analogous inequalities for perturbations of the Fermi sea of non-interacting particles, i.e., for perturbations of the continuous spectrum of the Laplacian by local potentials. (This is joint work with R. Frank, M. Lewin and E. Lieb.)
DAVID SHER, McGill University/CRM
The determinant on flat conic surfaces with excision of disks [PDF]
Let $(M,g)$ be a surface with a flat conical metric, and consider the surface $M_{\epsilon}$ obtained by removing disks of radius $\epsilon$ around any subset of the conical singularities. We investigate the asymptotic behavior of the determinant of the Laplacian on $M_{\epsilon}$, with Dirichlet boundary conditions, as $\epsilon$ approaches zero. Such families of surfaces were first studied by Khuri in the context of the Osgood-Phillips-Sarnak approach to compactness of isospectral sets of metrics. In particular, Khuri used these surfaces to prove that the determinant is not a proper map on the moduli space of flat metrics on a surface of genus $p$ with $n$ disks removed when $np\geq 1$. Our goal is to analyze the determinant of the Laplacian more closely as a function on this moduli space. As a first step, we sharpen and generalize Khuri's results; our main formula is an asymptotic expansion for the determinant of the Laplacian on $M_{\epsilon}$, with an error which vanishes as $\epsilon$ approaches zero. The proof uses determinant gluing formulas as well as an extension of an argument of Wentworth, which enables us to analyze the asymptotics of the Dirichlet-to-Neumann operator on $M_{\epsilon}$.
VITALI VOUGALTER, University of Cape Town
Existence and nonlinear stability of stationary states for the semi-relativistic Schroedinger-Poisson system [PDF]
We study the stationary states of the semi-relativistic Schroedinger-Poisson system in the repulsive (plasma physics) Coulomb case. In particular, we establish the existence and the nonlinear stability of a wide class of stationary states by means of the energy-Casimir method. Moreover, we establish global well-posedness results for the semi-relativistic Schroedinger-Poisson system in appropriate functional spaces.
## Sponsors
© Canadian Mathematical Society
© Canadian Mathematical Society : http://www.cms.math.ca/
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 4, "mathjax_asciimath": 2, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8764052391052246, "perplexity_flag": "head"}
|
http://polymathprojects.org/2009/07/26/mock-up-discussion-thread/?like=1&source=post_flair&_wpnonce=d64817105f
|
# The polymath blog
## July 26, 2009
### Mock-up discussion thread
Filed under: discussion,mock-up — Terence Tao @ 8:06 pm
This is a mockup of what a discussion thread for a polymath project would look like; it would run alongside the research thread, but is focused on strategy, exposition, and discussion by casual participants, as opposed to the more cutting edge research conducted by more active participants in the research thread.
Please feel free to make suggestions in this thread as to how the format and organisation of these projects (or this blog) could be improved.
## 23 Comments »
1. Some tweaks to the blog format here: I’ve changed the names for the “thumbs up” and “thumbs down” icons for comment rating from the generic “Rate Up” and “Rate Down” to the more specific “Useful” and “Needs work”, as this is what we really want the ratings to reflect for the polymath projects.
Also, this is not visible yet, but I’ve broken up the comments into chunks of 50 each; once the 51st comment is reached, the post will display the comments from 51-100 by default, with a link to go back to the previous comments. This may hopefully solve the problem of having to start a new thread every 50 or 100 comments or so. One can see an example of this at the xkcd blog, e.g. at http://blag.xkcd.com/2009/06/18/security-breach/#comments .
Please feel free, by the way, to suggest any design enhancements to this blog (suggestions from CSS experts would be particularly useful).
Comment by — July 27, 2009 @ 3:03 pm
2. In case anyone is interested, there were a number of reasons why we chose this particular wordpress theme (which is called Rubric, by the way). Firstly, it had automatic comment numbering; secondly, it was fairly uncluttered; but thirdly, it had a particularly wide column width for comments. I think part of the problem with reading 100+ comment threads, as is the case with polymath projects, is that the narrow columns of most blog themes make such threads extremely long. Presumably with more CSS customisation one can improve things even further, but I am an amateur only in these issues (but would welcome suggestions).
Comment by — July 27, 2009 @ 4:17 pm
3. Terry, I’d like to better understand the rationale for comment rating. Will the ratings be reflected somehow in the presentation of the comments? E.g., they might affect the order in which comments are shown (for an example, see http://news.ycombinator.com/item?id=726544 ), or might be used to suppress comments (like Slashdot). Or are they simply intended to help guide readers in some way, e.g., by helping highlight unusually insightful comments?
Comment by — July 28, 2009 @ 1:58 pm
• Well, as far as I know wordpress has no facility yet to order comments by rating, and this would interfere with comment numbering, so the former reason is not really a factor right now. The idea arose from the feedback from the minipolymath project, where it was noted that once the research thread exceeded 100 comments or so, it became difficult to catch up, because one did not know in advance what the key insights were, and also which arguments later proved to be incomplete or otherwise faulty. The hope is by supporting deep threading (currently we allow 10 nested levels of comments), together with ratings to identify particularly useful posts (as well as posts that ended up not working as intended), this may assist casual readers in keeping up. (Simultaneously with all this, we plan to have some volunteers summarise the research thread from time to time in the discussion thread. We will have to run another polymath experiment to see whether this all works, of course.)
Comment by — July 28, 2009 @ 2:37 pm
• So far I don’t think the ratings are working very well. I’ve read through the developing thread and so far the application of thumbs seems arbitrary; for example, why did this comment from Gil get a thumbs down?
Additionally, we’ve already discussed the importance of allowing mistakes and errors; the original screed from Dr. Gowers was to allow half-formed ideas, but I fear if people can start to get rated down that we will get less experimentation.
Finally if there’s some sort of filter code to only view comments over a certain threshold, I fear it could make things more rather than less confusing, because often key insights have only made sense in the context of the (more minor) comments that preceded them.
Comment by — July 30, 2009 @ 2:47 pm
• Jason – I’d like to wait until there’s a bit more data on the use of the comment ratings. It’s possible, for example, that they may turn out to be unexpectedly helpful when people summarize blog discussion on the wiki, as a way of quickly identifying crucial comments.
Comment by — July 31, 2009 @ 1:59 pm
4. My own feeling is that ratings should not be used to change the order of comments, or even to indicate whether you think that a comments was a good one, but rather to indicate to casual readers which comments they should read if they want to follow the essential points of a discussion without reading all the comments. For example, if somebody asks a subquestion that goes on to get discussed quite a lot, then the relevant comment should be related highly. And if someone else finds a counterexample to that question, then that too should be rated highly. But if someone proposes an approach that is not taken up, then that should not be rated highly (unless you think that it is in fact a very important comment and should not be ignored). I’m not sure I see any useful purpose for negative ratings, but maybe someone can suggest one.
Comment by — July 28, 2009 @ 2:41 pm
• I changed the labels for the thumbs up and thumbs down from “Rate Up” and “Rate Down” to “Useful” and “Needs Work” (you’ll see this if you hover the mouse over these labels). I agree that negative ratings should be used sparingly, but the one place where I see they could be useful is to flag a post which contained an argument which overclaimed a bit and was subsequently found to have some gaps.
Comment by — July 28, 2009 @ 3:41 pm
• If someone posted a flawed proof, I think that we should vote “Needs work” to indicate that the proof doesn’t work. On the other hand, the proof might contain some important ideas and be “Useful”. Right now, it is not possible to vote both, only your last vote counts. And the only way to cancel a vote is to vote the opposite. Is it possible to change this?
Comment by Sune Kristian Jakobsen — July 28, 2009 @ 5:00 pm
• Perhaps in this case many people would up-vote a “reply” comment explaining the error in the earlier argument. Whether the earlier argument got a lot of up-votes or a lot of down-votes or both, I think it would indicate to the reader that something important may be in that comment.
Comment by Ryan O'Donnell — July 29, 2009 @ 12:47 am
• By the way, I put a draft policy for ratings at
http://polymathprojects.org/general-polymath-rules/
but it, like all other polymath rules, are of course open to debate.
Comment by — July 28, 2009 @ 4:28 pm
5. There has been a request to move the comment metadata (author, timestamp, and “Reply” feature) to the top of the comment rather than at the bottom. Presumably this can be done by editing the CSS – does anyone know the precise string of CSS code that would do the trick?
Comment by — July 28, 2009 @ 3:40 pm
• Just to say that I made the request — the reason I did so was that I found that with the current layout, comments that reply to other comments look as though they have been written by the person who is named just above. For example, if you aren’t thinking too hard, this comment will look as though it is written by Terry. I have found this minor confusion slightly irritating, and I think my irritation would accumulate if I was in the middle of an active Polymath project.
Comment by — July 28, 2009 @ 3:57 pm
• an easier solution is to space out comments such that they’re more distinct – it also improves readability. just add to the stylesheet:
#commentlist li {
margin-top: 40px;
}
Comment by bliu — July 28, 2009 @ 6:19 pm
• Thanks for this. I added it to the CSS on a trial basis. It seems like an improvement to me.
(I’ve found that it doesn’t always seem to take effect, though. I assume that’s probably some type of caching issue, and will disappear.)
Edit: I see what’s going on. It’s displaying correctly in a kind of preview mode available to admins. But apparently we have to purchase the ability to modify the CSS, for general use. I’ll look into it.
Comment by — July 28, 2009 @ 6:37 pm
• I’ve enabled CSS modification. Looks like the spacing has taken effect.
Comment by — July 28, 2009 @ 6:57 pm
6. Omg I knew that! Rubric has automated commenting.
Comment by Anonymous — July 28, 2009 @ 4:04 pm
7. Another point. I think we should revisit a discussion that we had during Polymath1 about threading. In particular, the questions I would like to discuss are these: how much threading should we have, and what should it be used for?
Here are my views on the matter. I think we should have only very limited threading, and a reply to comment X should be given as a threaded reply only if it is a very local matter to comment X rather than a continuation of the general conversation.
To illustrate by an example, if comment X said,
I’ve just thought of a rather interesting conjecture. It seems to me that $x^n+y^n$ can never equal $z^n$ when $x$, $y$ and $z$ are positive integers.
then the comment
You obviously mean $n$ to be greater than $1$, and the conjecture is false for $n=2$. But for $n\geq 3$ it looks interesting.
would naturally be a threaded reply. However, the comment,
I’m assuming you require $n\geq 3$ here. If so, it occurs to me that if the conjecture is false, then one could create some elliptic curves with pretty strange properties. In fact, if all elliptic curves were modular, those properties might even be impossible. Could this be a fruitful line of attack?
then it would be a separate comment, even if there had been several comments in between. (If that was the case, then one would give the number of the comment one was reacting to.)
What I am describing is roughly the convention that we settled on with Polymath1, where I allowed depth-1 threading only. But there is room for disagreement here. Basically, what people think about this depends on how much they value they attach to the linearity of a blog, which makes it easy to see how the thoughts are developing and to keep track of which comments have been made recently, and how much to the ease of seeing later which comments were linked to which. I myself attach a high weight to the former, with the wiki providing a counterbalancing tree structure.
Comment by — July 28, 2009 @ 4:06 pm
• With regards to threading, I have a few comments. I was a casual participant in miniPolymath1, and I can say that it became too much of a headache to follow everything. Allowing depth-10 nesting of comments will certainly help, but I still think it will be intimidating for a casual reader to jump in simply based on the number of comments he has to sort through.
I am not sure that this is even feasible on these blogs, but I know how I would like to see comments (as a casual reader). I imagine a bulletin board style layout. At the top are a few links which look like
“Taking John’s idea from comment 10 further (59 comments)”
“Taking Dan’s idea from comment 19 further (30 comments)”
“Continuing on the ideas of comments 11-15 (100 comments)”
Below these links is the original discussion of the problem. People could continue to post observations which are disjoint from all the above links, such as fresh ideas, new approaches, and observations which apply across the board.
Clicking on “Taking John’s idea from comment 10 further” would lead to another page with the same layout. Again at the top we read “Taking X’s idea from comment n further” and below these links we see the discussion, but this time the discussion is only concerned with John’s idea. You get the idea.
To achieve such organization, perhaps those who are working full time on the project (or whoever has agreed to occasionally summarize the findings) could create these links and then be the first to post on the newly created page with their summary.
I am no expert on manipulating blogs so I understand that this may not be as easy as I’ve made it sound. But this sort of layout would be most pleasing to the eye of the casual reader who may just want to work on the most popular attempts (gauged by the number of comments next to each link). And it also allows more organization in the comments since we get to see the relative depth of the observation based on how many clicks into the thread we are.
Comment by Scott — July 30, 2009 @ 9:34 pm
8. Do you think it would be worthwhile to create a WordPress “page” for each project giving the problem statement and some background information and links? Or do you want to keep that sort of information just on the Wiki?
Comment by thlog — July 29, 2009 @ 2:47 am
• Well, ideally the wiki and the blog would be integrated, so that one could mirror a wiki page on the blog or vice versa (or even more ideally, there would be a single page object somewhere that both the wiki and the blog could access or edit, essentially simultaneously). [This, incidentally, is one of the capabilities of Google Wave.] But until such time, I guess the best option is to have a master page (on the wiki) for this sort of thing, and links to that page on the blog. So I’ve set up a wiki page now for all of the active projects, and on the blogroll on the right of the blog there are links to each of them.
Comment by — August 4, 2009 @ 3:04 pm
9. …
, ?
Comment by Mirtesky — May 21, 2011 @ 6:08 pm
10. Заброс удочки Одним из главных условий результативной ловли является грамотный заброс удочки, ведь неумелое обращение со снастью может распугать всю рыбу Удилище Kosadaka Cinergy Tele Bolognese Удочка на карпа Карп, который водится почти во всех водоемах со слабым течением, отлично ловится весной и летом в солнечную
Comment by — June 18, 2012 @ 12:54 am
RSS feed for comments on this post. TrackBack URI
Theme: Customized Rubric. Blog at WordPress.com.
%d bloggers like this:
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 10, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.959363579750061, "perplexity_flag": "middle"}
|
http://crypto.stackexchange.com/questions/671/how-is-the-mjh-double-block-length-hash-function-constructed
|
# How is the MJH double-block-length hash function constructed?
I'm looking for information on the MJH double-block-length hash function, but the best free source I could find is the diagram on page 18 of Efficient hashing using the AES Instruction Set (submitted to the ECRYPT II Hash Workshop 2011).
Is source code available for this function? What are the standard symbols used (M, V, Z) when describing hash functions?
-
I'll elaborate this to an answer later (if not someone else does it first), but here about the meanings of the symbols used in the diagram (as I understand them): n is the block size of the block cipher, k the key size. M and V are inputs to the compression function (M typically being one message block, V the state of the hash function), Z is the output (same size as V, and the new state of the hash function). – Paŭlo Ebermann♦ Sep 11 '11 at 1:01
## 1 Answer
I had a look at the paper introducing the MJH construction (MJH: A Faster Alternative to MDC-2).
It actually presents a different diagram than the one in the paper Efficient hashing using the AES Instruction Set your referenced – there $V_2$ and $M$ are swapped. I'll describe the original one here, and the variation below.
The core of MJH is a compression function, using a similar construction like the JH hash function (one of the SHA-3 finalists), hence the name.
### The JH construction
Given a (non-compressing) function $F : \{ 0,1\}^{2n} \to \{0,1\}^{2n}$, we define the compression function $JH[F] : \{0,1\}^{3n} \to \{0,1\}^{2n}$ as $JH[F](V_1, V_2, M) := (Z_1, Z_2)$, with $(X_1, X_2) = (V_1, V_2 \oplus M)$, $(Y_1, Y_2) = F(X_1, X_2)$ and $(Z_1, Z_2) := (Y_1 \oplus M, Y_2)$.
In the JH hash function, F is a specially-made (fixed) permutation.
### MJH's $F[\sigma, \theta]$
Given a $n$-bit-blockcipher $E$ with keysize $k = n$, $\sigma : \{0,1\}^{n} \to \{0,1\}^n$ an involution (i.e. $\sigma \circ \sigma = \mathrm{id}$) without fixed point (i.e. $\sigma(X) \neq X$) – an example would be an XOR with a non-zero constant –, and $\theta \in \mathbb{F}_{2^n} \setminus\{0,1\}$ a constant (so multiplication with $\theta$ is another non-trivial permutation).
We define $F[\sigma, \theta] : \{0,1\}^{2n} \to \{0,1\}^{2n}$ as $F(X, K) = (L, R)$, with $L = E_K(X) \oplus X$ and $R := \theta(E_K(\sigma(X)) \oplus \sigma(X) ) \oplus X$.
(The $E_K(X) \oplus X$ part is essentially the Davies-Meyer construction to make it one-way, even if $K$ is known. It is actually used twice here.)
### The final Hash function
Combining these ideas, we get our compression function $\tilde F[\sigma, \theta] := JH[F[\sigma, \theta]] : \{0,1\}^{3n} \to \{0,1\}^{2n}$.
We then apply the known Merkle–Damgård caining construction on this compression function, receiving our final hash function $H[\sigma, \theta] = MD[\tilde F[\sigma, \theta]]$.
(This image is from Wikimedia Commons, the others are made by me.)
In practice we now also have to choose some specific $\sigma$ and $\theta$, a specific block cipher, as well as an initialization vector and padding for MD - the security proof still applies for all of them (if the block cipher is good).
### Longer key variant
The paper also describes a variant of the compression function to be used when the key length $k$ is greater than the block size $n$. Then we use message blocks of size $k$, and split these blocks into two parts $z, z'$. $z$ (of size $n$)is used as before to be XOR'ed into the left half of the state before encryption and right half after, while $z'$ (of size $k-n$) is appended to the right half of the state to form the key for the block cipher.
By using a larger (hash) block size, this longer variant can be more efficient than the original one.
### The variant depicted in Efficient Hashing
For $n = k$, this is the same compression function as the original one, but it is used in a different way in the MD-construction: Here the message blocks are passed in the key position, instead of being XOR'ed into the parts of the state before and after encryption. (This XOR'ing instead uses one half of the original state.)
This gives a more obvious generalization into the $k \geq n$ case, as we don't have to split the message block and compose the key, but simply can pass the longer key to the block cipher.
I didn't check if the proofs given in the MJH paper also apply to this variant, though, and I don't know which version the authors of Efficient Hashing actually measured.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9059790968894958, "perplexity_flag": "middle"}
|
http://nrich.maths.org/36
|
### Number Round Up
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### Triangle Animals
How many different ways can you find to join three equilateral triangles together? Can you convince us that you have found them all?
### Button-up
My coat has three buttons. How many ways can you find to do up all the buttons?
# Plants
##### Stage: 1 and 2 Challenge Level:
Uncle Raj has three children. Next year, when they've had their birthdays, Naomi will be $5$, Alex will be $6$ and Chris will be $7$. The family has decided on something rather unusual for part of their presents.
All three children have their birthday in the late spring and since they are keen on gardening they are going to buy some plants for the garden, one for each year they have been alive.
Here is the plan of their house and garden:
You notice that there are three circular paths that cross over each other. Each child is to have a circle but there will be some bits that are shared, around the middle.
When the time comes, the four of them go off to the garden centre to choose the plants. They do not have a lot of money so they're looking for special offers. They find a very special offer which gives a good discount if you buy ten plants altogether. The three children say that that is no good because they need more than ten. But Uncle Raj realises they can manage with only ten.
They go to the cafe and have some cool drinks, and Uncle Raj draws a plan of the three paths and puts little marks to show the plants.
Here is his idea:
The children are fascinated to see that Naomi has $1$ and shares $4$, Alex has $2$ and shares $4$ and Chris has $3$ and shares $4$. They think that's rather cool and it saves them a lot of money. So they finish their drinks and off they go to buy their ten plants.
Well now it's your turn to have a go and find some different solutions.
REMEMBER:- You must use exactly ten plants (no more, no less)
REMEMBER:- The circles must contain $5$, $6$ and $7$ plants (no more, no less).
As you try, you may find that you are developing a system for getting the next one. If so, we'd love to hear about it. You might like to try to find them all, and write about all the things you notice about each solution.
You could print off Plants Recording Sheet.pdf to help you record them, if you like. However, you might find a different way of recording them altogether. Some people find it's easiest to do it quite large and have ten objects to move around in different places.
As with most of these challenges you can and should ask "I wonder what would happen if ...?" Well you might try a different number of objects (plants). You could try different numbers for each circle, as if the children were different ages.
Good luck and don't forget about sending in any results you have.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9705544114112854, "perplexity_flag": "middle"}
|
http://mathhelpforum.com/advanced-algebra/185556-show-set-not-subspace-r-4-a.html
|
# Thread:
1. ## Show a set is not a subspace of R^4
Hi all,
Having a problem trying to show that the set $<br /> \begin{bmatrix} 3a-b\\ ab \\ a+b\\ b+2a \end{bmatrix}<br />$
where a and b are real numbers, is not a subspace in R^4..
2. ## Re: Show a set is not a subspace of R^4
Let $a=1,b=2$ and consider the vector
$2\left[\begin{array}{c}3a-b\\ab\\a+b\\2a+b\end{array}\right]=2\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]=\left[\begin{array}{c}2\\4\\6\\8\end{array}\right]$.
Then $3a'-b'=2$, $a'b'=4$ and $a'+b'=6$. But this system has no solution:
$3a'-b'=2$ implies $b'=3a'-2$. Then $a'+b'=6$ gives us $a'+(3a'-2)=6$ and therefore $a'=2$ and $b'=4$. But then $a'b'=8$, contradicting the condition $a'b'=4$.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8900111317634583, "perplexity_flag": "head"}
|
http://physics.stackexchange.com/questions/53467/unified-field-theory-in-laymans-terms
|
# Unified field theory in layman's terms
I watched some videos on the unified field theory, specifically interviews with Michio Kaku and John Hagelin, and want to learn a bit more about it. I looked up the theory of everything, string theory and unified field theory but most of it goes over my head.
In one of the interviews with John Hagelin he describes the unified field as being the deepest level of the universe. He refers to it as intelligence or a field that has the potential for matter. So, what does this consists of if it isn't the smallest form of matter?
Also, is there a definite view in the scientific community that the unified field exists and what it is?
-
– grasshopper Feb 9 at 16:47
1
If you like this question, you may also enjoy reading this or this Phys.SE post. – Qmechanic♦ Feb 9 at 20:00
## 1 Answer
First we need to understand what the word "field" means in the context of physics. A field essentially is an object that assigns a value for each point in space and time. So if you have a set of properties that you are interested in, then the field would tell you what their exact values are for all of time and space. Basically, it is a function where you input your position in space and time, and out pops all the physical information you could possibly want to know about that position.
A unified field theory would then encompass some sort set of principles, axioms and equations that would tell you what the field is for all space and time and tell you how it behaves under changes in its parameters. In principle it would provide some predictive capability given some initial input.
The popular discussion of unified field theory centers around the desire to find a theory that can describe how the universe behaves under the action of the four known forces: Strong, Electromagnetic, Weak and Gravity. Historically, the archetypical unification model is Maxwell's success in unifying the electric and magnetic forces into the electromagnetic force.
One of the successes of the last 40 years was the success of unifying the electromagnetic and weak forces as part of electroweak theory. The discovery of the Higgs Boson has effectively proven that the electromagnetic and weak forces are in fact the same force, so in reality, there are only three forces that need to be unified: Strong, Electroweak and Gravity.
Grand Unified Theories, or GUTs, are an attempt to unify the Electroweak and Strong forces. Although the Standard Model effectively predicts behavior of those combined forces at low energies, it still falls short of true unification. It is expected that unification of the Strong and Electroweak forces will occur at energies higher than what we can probe currently, however it is possible that when LHC reaches full power we will see evidence some indication of how these forces might unify.
Quantum Gravity is generally used synonymously with Unified Field Theory, although this is not entirely accurate. Loop Quantum Gravity is a proposed theory of Quantum Gravity that does not include known particle physics, and can therefore not be considered a Unified Field Theory.
String Theory is a the most favored model for a Unified Field Theory. Through various dualities, most importantly the AdS/CFT duality, and through an understanding of asymptotic limits, String Theory can effectively morph itself to encompass most if not all of the current theories governing the four forces. The chief difficulty associated with the current theory is that the fundamental vacuum solution, or kernel, of the theory is unknown and there are $10^{500}$ possible solutions that would satisfy the constraints of the theory. This effectively means that we have a means of understanding how a field would be governed by String Theory, but no means to decide what that fundamental field actually is.
I would say that the view is that we have evidence that their is a fundamental field theory, and a path, but we simply do not have the resources to find out if we are right or wrong. It is unfortunately entirely possible that we will never know.
-
Thanks for the answer, I hope physicist will confirm one of theories soon, it will be interesting to see what impact it will have on the world. – grasshopper Feb 11 at 10:31
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9574301242828369, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/53673/why-is-it-ok-to-rely-on-the-fundamental-theorem-of-arithmetic-when-using-godel-nu/53740
|
Why is it OK to rely on the Fundamental Theorem of Arithmetic when using Gödel numbering?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Gödel's original proof of the First Incompleteness theorem relies on Gödel numbering. Now, the use of Gödel numbering relies on the fact that the Fundamental Theorem of Arithmetic is true and thus the prime factorization of a number is unique and thus we can encode and decode any expression in Peano Arithmetic using natural numbers.
My question is, how can we use a non-trivial result like the Fundamental Theorem of Arithmetic in the meta-language that describes Peano Arithmetic, when, the result actually requires a proof from within Peano Arithmetic itself, not like other trivial things we believe to be true (i.e. existence of natural numbers and the axioms for addition and multiplication which we want to interpret in the natural way - Platonism)?
I understand we could instead use a different way of enumeration, e.g. a pairing function and the Chinese Remainder Theorem or simply string concatenation, but, then the need for a proof of uniqueness when encoding and decoding remains and in general, I am interested in the structure of Gödel's original proof.
Basically, I have two ideas of how it might be possible to resolve this:
1. Prove the Fundamental Theorem of Arithmetic within a different sound (?) system.
2. Maybe there is nothing needed to be done to 'resolve' this, because I am just misinterpreting something and it is actually acceptable to use provable sentences of PA in the meta-language.
EDIT: I have realized how to make my question less confusing:
Say PA proves FTA. Then if we only assume PA is consistent, that does not rule out the possibility of FTA being false. Now, if FTA is false, then PA and the meta-language too includes a false statement and thus the whole proof is useless.
How is this resolved? Is it maybe related to the fact the originally we actually assume $\omega$-consistency and obviously, for each natural number $n$ separately its unique prime factorization can simply be found algorithmically?
-
2
The theorem states that your theory cannot be both consistent and complete, so it's a win-win. I don't see the problem. – Thierry Zell Jan 29 2011 at 4:28
It has been a while since I worked through that proof, but is the unique factorization really needed or just a canonical one (without the fact that as it happens it is unique)? What if you use a greedy factorization where you divide out all 2's you can then all 3's (from what is left) then all 4's etc. That gives a code. – Aaron Meyerowitz Jan 29 2011 at 20:51
6
Why are number theorists allowed to use numbers? Shouldn't they first establish that it is safe to use them? – Andrej Bauer Jan 30 2011 at 8:32
3
The point of my question is this: if all other mathematicians are allowed to take numbers for granted, why isn't Gödel? All he does is apply the mathematical method to the study of a certain system for manipulation of symbols known as "first order logic" and "Peano arithmetic". The subject of his study is no less mysterious, magical, or "fundamental" as all the other subjects of study in mathematics. – Andrej Bauer Jan 30 2011 at 18:49
4
And I am sure Gödel knew how to prove the fundamental theorem of arithmetic. Perhaps I should have been more precise: when Gödel does mathematics is allowed to use whatever mathematics is considered "standard" by his peers. My pet peeve is this: why do "ordinary" mathematicians think that "logicians" must secure the ground on which they stand, when those same mathematicians stand on the same ground together with logicians? – Andrej Bauer Jan 31 2011 at 13:40
show 2 more comments
5 Answers
Just like most mathematical theorems, you can formalize Godel's Theorems in some first order language (with some "standard" interpretation under which the formalization means what it's supposed to mean), turn the proof into a purely syntactic string of formulas, and figure out which formulas in that first order language are needed as axioms. I'm embarrassed to say I don't know exactly how strong the assumptions we need are to carry out the proof of Godel's Theorems, but there will be some weak fragment of ZFC probably not much stronger than PA which will do. So we would be using a theory slightly stronger than PA to establish the incompleteness of PA, but why should that be a problem?
The axioms needed for the proofs of Godel's Theorems are probably pretty natural, probably pretty close to PA, and probably have a natural interpretation. If you believe these axioms have this interpretation, then you would have no problem with Godel's proofs or the interpretation of the theorems. If not, then you're probably pretty close to believing PA is inconsistent, in which case you would probably:
1. Accept that the formalized versions of Godel's Theorems follow from whatever axioms are used, but only because you believe those axioms are inconsistent.
2. Deny that the formalized versions of Godel's Theorems mean what they're supposed to mean, and just regard what's happening in point 1 as a valid string of symbolic manipulations.
3. Accept the natural language meaning of Godel's Theorems, in spite of point 2, for trivial reasons, since they say, "if PA is consistent, ..."
EDIT: (in response to your edit) So we're assuming PA proves FTA, PA is consistent, and FTA might be false? What do you mean by "false," you mean false in the standard intepretation? In that case, PA would be false in the standard interpretation. Now if we take Godel's first theorem to say, "If PA is consistent, then there is a true formula in the standard interpretation which is not provable from PA," then this meta-theorem is certainly true.
EDIT: Ignas requested an explanation of some of the basics to make sense of my claim, "If PA proves FTA, and FTA fails in the standard model, then so does PA." It's too big to fit in a comment so I'm adding it to my response:
Let $\mathcal{L}$ denote the first order language of number theory, we'll have lower case Greek letters vary over sentences of $\mathcal{L}$, upper case Greek letters vary over sets of sentences of $\mathcal{L}$, and upper case Roman letters vary over $\mathcal{L}$ structures. We write $M \vDash \varphi$ to denote that $\varphi$ is true in the model $M$, i.e. when its symbols are interpreted according to $M$. Tarski's definition of truth for a sentence in a given model is by recursion on the complexity of the sentence. We write $M \vDash \Sigma$ if every member of $\Sigma$ is true in $M$. We write $\Sigma \vDash \varphi$ if for every $M$, $M \vDash \Sigma$ implies $M \vDash \varphi$, i.e. if every model of $\Sigma$ is also a model of $\varphi$.
For provability, we write $\Sigma \vdash \varphi$ to say that there is a (finite) proof of $\varphi$ using (finitely many) sentences from $\Sigma$ as axioms.
The Soundness Theorem states that for all $\Sigma ,\ \varphi$, if $\Sigma \vdash \varphi$ then $\Sigma \vDash \varphi$. It's this theorem, with PA in place of $\Sigma$ and FTA in place of $\varphi$, that I'm using to establish the claim you're asking about. The converse of this theorem is also true; it's Godel's Completeness Theorem. Putting these two theorems together, they say that the relations $\vdash$ and $\vDash$ are the same relation between sets of sentences and sentences. One (perhaps not immediately obvious) way to rephrase this is, "being true in every model is the same as being provable from no axioms." Contrast this Godel's Incompletness Theorem, which says that "being true in the standard model of number theory is not the same as being provable from PA."
-
1
I am pretty sure that the incompleteness theorems can be proved in PA. I.e., you can prove something like "if ZFC is consistent then so is ZFC+$\neg$Con(ZFC)" in PA (the same for ZFC replaced by weaker theories such as PA itself). – Stefan Geschke Jan 29 2011 at 5:58
I guess it depends on how you word Godel's Theorem. Take the first one for instance. If it says something like, "if basic arithmetic is consistent, then there's a formula that's true in $\mathbb{N}$ but not provable from basic arithmetic" then to express this you need some theory capable of expressing "true in $\mathbb{N}$." PA can't even talk about $\mathbb{N}$. I suppose there may be equivalent formluations of Godel's Theorem which sidestep the need to talk about $\mathbb{N}$, but then the question becomes: What axioms do you need to prove that equivalence? – Amit Kumar Gupta Jan 29 2011 at 6:09
I see. This is not a problem if you formulate the first incompleteness theorem as something like "if T is r.e. axiomatizable, consistent, and allows representations of recursive functions and relations, then T is incomplete". This gets you that PA is incomplete (if consistent, which we all believe to be the case) and once you are in a theory that allows it to talk about $\mathbb N$ in any reasonable way, then you get the theorem in the form that you quote. – Stefan Geschke Jan 29 2011 at 6:23
@Amit Kumar Gupta: whatever metatheory you formalize the proof in can talk about $\mathbb{N}$, though. – Carl Mummert Jan 30 2011 at 23:14
Regarding your edit: Yes, I meant false under standard interpretation. What do you mean by "PA would be false"? Inconsistent? Well, it does not have to be as long as PA does not prove $\neg$FTA. What am I misunderstanding here? – Ignas Jan 31 2011 at 0:32
show 4 more comments
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Stefan has explained the essential points well, but let me add some details and a reference. For the first incompleteness theorem, one can work with a very weak theory of arithmetic. One needs to be able to define the relevant recursive predicates and functions (in particular those that describe the coding process) and prove that these represent, in the theory under consideration, the intended functions and relations. Here "represent" is a rather weak requirement. A formula $\phi(x,y)$ represents a function $f$ in a theory $T$ if, for each natural number $n$, $T$ proves the formula `$(\forall y)(\phi([n],y)\iff y=[f(n)])$`, where the square brackets mean the numeral for the number inside. (Note that $T$ need not prove `$\forall x\exists y\phi(x,y)$`.) A weak theory like Robinson's Q suffices to represent all recursive functions and predicates, so the first incompleteness theorem applies to all theories with at least the strength of Q. For the second incompleteness theorem, on the other hand, one needs to prove, in the theory under consideration, some basic properties of the coding, for example that concatenations can always be coded. PA suffices for this (and so do some weaker theories; `$\Sigma^0_1$` induction seems to suffice). So the second incompleteness theorem is usually stated for theories that have at least the strewngth of PA. The details of what needs to be proved, as well as most of the details of the proofs, can be found in Shoenfield's book "Mathematical Logic"; I believe they're also in Goedel's original paper. (I'm away from home and writing this on an unfamiliar computer, which I can't persuade to preview any of my TeX code, so I apologize if it contains errors.)
-
@Andreas , do you know of a reference that pursues the question of how little/much of PA is actually required? I seem to recall "Metamathematics of first order arithmetic" treats some of this, but don't recall the explicit suggestion that $\Sigma^0_1$ induction seems enough (I do not have the book with me, though, so I cannot currently check). – Andres Caicedo Jan 29 2011 at 20:59
2
I learned only recently that Pudlak proved in 1985, by model-theoretic techniques, that Q does not prove Con(Q), along with finer results. This is in the paper "Cuts, consistency statements and interpretations" from the JSL volume 50. I don't think that anyone has proved a sort of reverse-mathematics result that characterizes the amount of induction required to verify the Hilbert-Bernays conditions. $Sigma^0_1$ induction is certainly enough, but I don't remember about weaker theories at the moment. – Carl Mummert Jan 30 2011 at 23:12
@Carl, many thanks! I'll check the reference. – Andres Caicedo Jan 31 2011 at 1:03
Feferman proves a version of the second incompleteness theorem in his system $FS_0$: citeseerx.ist.psu.edu/viewdoc/… I don't think $FS_0$ is weaker than $\Sigma_1$ induction though. – Timothy Chow Jan 31 2011 at 15:58
I am not absolutely sure I understand your problems with the meta-language and such.
You take your favorite base theory, let's say PA or something stronger like ZFC. But I am quite sure that PA is enough. You prove the fundamental theorem of arithmetic in PA (or rather, a la Goedel, you prove that the $\beta$-function does what it is supposed to do). This tells you that you can code finite sequences and hence that you can talk about strings, languages, and proofs.
Now you continue and define a formula Con(PA) that expresses that PA is consistent. In the same way you can define a formula Con(PA+$\neg$Con(PA)), that expresses the obvious thing. Note that the PA that appears in these formulas is not the meta-mathematical PA but the internal PA of the model that you are in. Anyhow. This is just the usual confusion that arises with the incompleteness theorems. Now you prove, still only using PA, the statement "Con(PA) implies Con(PA+$\neg$Con(PA))" and you are done.
This is a purely number theoretic proof.
-
Here is a way to think about it. When you use something like Fund Theorem of Arithmetic to prove Godel numbering, you are referring to the abstract set $\mathbf N$, which "exists". The Fundamental Theorem of Arithmetic is simply true about this set, irrespective of any logical system which one is constructing. If you like, you can prove the Fundamental Theorem of Arithmetic, more or less rigorously, in a metalanguage. But the proof in this metalanguage is completely orthogonal to any proofs one might construct in the logical system, such as Peano axioms.
-
1
This may well be true, but it is not the issue at hand. The numbering of formulas and its reliance or not on the FTA happens inside the theory (as explained by the answers below). – Andres Caicedo Jan 29 2011 at 22:14
1
Oh no no no, that would be pretty much the worst case, since claiming that FTA is "simply true" would render the whole proof rather vague and non-rigorous. And on the contrary, as Andres pointed out, the proofs in the meta-language are actually established within PA itself, since PA is the meta language too. – Ignas Jan 30 2011 at 1:39
1
The Godel incompleteness theorem shows that $Th(\cal N)$ is undecidable. Any first-order statement, such as FTA, which is true on $\mathbf N$ is part of $Th(\cal N)$. Proving that it is part of $Th(\cal N)$ can be regarded as simply a fact (which one prove if one wishes). – David Harris Jan 31 2011 at 23:24
It's a trick of encoding. You encode the formulas of the theory into a fixed collection of integers which have a certain form.
Now, you have defined somewhere in the background a natural notion of multiplication, addition, and equality of integers. So you are free to ask questions about whether or not two numbers are equal, and whether or not for some number $x$, there is a formula in the language of PA, for which $x$ is the encoding of, and this allows you to go backwards, in the metatheory. (that is take a number and see if it encodes a sentence)
That being said, you are never starting with a number and going backwards in the formal theory (that is to say turning it into the sentence it encodes, as this would require FTA), you are always going forward, taking the Gödel encoding of a given sentence.
It's this careful dance and intermingling of the syntax and semantics that makes the proof of this theorem so much fun to go over.
Hope this clears it up a little bit. I know it probably doesn't completely answer your question.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 58, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.946240246295929, "perplexity_flag": "head"}
|
http://planetmath.org/proofoffermatstheoremstationarypoints
|
# proof of Fermat’s Theorem (stationary points)
Suppose that $x_{0}$ is a local maximum (a similar proof applies if $x_{0}$ is a local minimum). Then there exists $\delta>0$ such that $(x_{0}-\delta,x_{0}+\delta)\subset(a,b)$ and such that we have $f(x_{0})\geq f(x)$ for all $x$ with $|x-x_{0}|<\delta$. Hence for $h\in(0,\delta)$ we notice that it holds
$\frac{f(x_{0}+h)-f(x_{0})}{h}\leq 0.$
Since the limit of this ratio as $h\to 0^{+}$ exists and is equal to $f^{{\prime}}(x_{0})$ we conclude that $f^{{\prime}}(x_{0})\leq 0$. On the other hand for $h\in(-\delta,0)$ we notice that
$\frac{f(x_{0}+h)-f(x_{0})}{h}\geq 0$
but again the limit as $h\to 0^{+}$ exists and is equal to $f^{{\prime}}(x_{0})$ so we also have $f^{{\prime}}(x_{0})\geq 0$.
Hence we conclude that $f^{{\prime}}(x_{0})=0$.
To prove the second part of the statement (when $x_{0}$ is equal to $a$ or $b$), just notice that in such points we have only one of the two estimates written above.
Major Section:
Reference
Type of Math Object:
Proof
## Mathematics Subject Classification
26A06 One-variable calculus
## Recent Activity
May 17
new image: sinx_approx.png by jeremyboden
new image: approximation_to_sinx by jeremyboden
new image: approximation_to_sinx by jeremyboden
new question: Solving the word problem for isomorphic groups by mairiwalker
new image: LineDiagrams.jpg by m759
new image: ProjPoints.jpg by m759
new image: AbstrExample3.jpg by m759
new image: four-diamond_figure.jpg by m759
May 16
new problem: Curve fitting using the Exchange Algorithm. by jeremyboden
new question: Undirected graphs and their Chromatic Number by Serchinnho
## Info
Owner: paolini
Added: 2003-07-15 - 12:58
Author(s): paolini
## Versions
(v5) by paolini 2013-03-22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.879562258720398, "perplexity_flag": "middle"}
|
http://stats.stackexchange.com/questions/17320/mixing-and-dividing-point-processes
|
# Mixing and dividing point processes
At the following figure at left side two realizations of point processes with different density (intensity) $\lambda_1$ and $\lambda_2$ is being mixed matching the center of the belonging areas to build a point process in the middle with intensity $\lambda$. Then randomly selected points as two sets extracted from it as shown at the right side.
Questions:
Is $\lambda=\lambda_1+\lambda_2$? and Is $\lambda=\lambda_3+\lambda_4$?
If two at left side were Poisson PP, Is the middle one a Poisson PP?
How about the two at right hand side?
-
4
The key words you are looking for are superposition and thinning of the Poisson process. The answer, with some qualifications, is yes. But, an affirmative answer depends intimately on (i) the independence of the two processes in the first case and (ii) how the splitting is done in the second case. :) – cardinal Oct 20 '11 at 13:55
Thanks for keywords. I would appreciate it if you would give a complete explanation as an answer. For (i) since both are Poisson PP they are independent (I think). For (ii) say a Poisson random selector can be sued. – Developer Oct 20 '11 at 14:15
2
As cardinal said, the independence of the point processes is important. You can easily define two dependent poisson processes whose superposition would not be a poisson process; for example: say points in process #2 are exactly the same as those in process #1, just shifted to the right by 1 unit. – Karl Oct 20 '11 at 15:49
@Karl: I like the essence of your example, though the second process is not quite a Poisson process since the probability of an arrival in $[0,1)$ is zero in the second case. :) – cardinal Oct 20 '11 at 17:40
2
@cardinal - I was thinking of Point processes on the full plane. – Karl Oct 20 '11 at 18:14
show 1 more comment
## 1 Answer
To answer this question we need a little background and notation. In the general terminology let $N$ denote a of point process in the plane, which means that for any Borel set, $A$, in the plane, $N(A)$ is an integer valued (including $+\infty$) random variable, which counts the number of points in $A$. Moreover, $A \mapsto N(A)$ is a measure for each realization of the point process $N$.
Associated with the point process is the expectation measure $$A \mapsto \mu(A) := E(N(A))$$ where the expectation is always well defined, since $N(A) \geq 0$, but may be $+\infty$. It is left as an exercise to verify that $\mu$ is again a measure. To avoid technical issues lets assume that $\mu(\mathbf{R}^2) < \infty$, which is also reasonable if the process only really lives on a bounded set such as the box in the figure that the OP posted. It implies that $N(A) < \infty$ a.s. for all $A$.
The following definitions and observations follow.
• We say that $N$ has intensity $\lambda$ if $\mu$ has density $\lambda$ w.r.t. the Lebesgue measure, that is, if $$\mu(A) = \int_A \lambda(x) \mathrm{d}x.$$
• If $N_1$ and $N_2$ are two point processes we define the superposition as the sum $N_1 + N_2$. This is equivalent to superimposing one point pattern on top of the other.
• If $N_1$ and $N_2$ are two point processes (independent or not) with intensities $\lambda_1$ and $\lambda_2$ then the superposition has intensity $\lambda_1 + \lambda_2$.
• If $N_1$ and $N_2$ are independent Poisson processes then the superposition is a Poisson process. To show this we first observe that $N_1(A) + N_2(A)$ is Poisson from the convolution properties of the Poisson distribution, and then that if $A_1, \ldots, A_n$ are disjoint then $N_1(A_1) + N_2(A_1), \ldots, N_1(A_n) + N_2(A_n)$ are independent because $N_1$ and $N_2$ are independent and Poisson processes themselves. These two properties characterize a Poisson process.
Summary I: We have shown that whenever a point process is a sum, or superposition, of two point processes with intensities then the superposition has as intensity the sum of the intensities. If, moreover, the processes are independent Poisson the superposition is Poisson.
For the remaining part of the question we assume that $N(\{x\}) \leq 1$ a.s. for all singleton sets $\{x\}$. Then the point process is called simple. Poisson processes with intensities are simple. For a simple point process there is a representation of $N$ as $$N = \sum_i \delta_{X_i},$$ that is, as a sum of Dirac measures at the random points. If $Z_i \in \{0,1\}$ are Bernoulli random variables, a random thinning is the simple point process $$N_1 = \sum_i Z_i \delta_{X_i}.$$ It is quite clear that with $$N_2 = \sum_i (1-Z_i) \delta_{X_i}$$ it holds that $N = N_1 + N_2$. If we do i.i.d. random thinning, meaning that the $Z_i$'s are all independent and identically distributed with success probability $p$, say, then
$$N_1(A) \mid N(A) = n \sim \text{Bin}(n, p).$$ From this, $$E(N_1(A)) = E \big(E(N_1(A) \mid N(A))\big) = E(N(A)p) = p \mu(A).$$
If $N$ is a Poisson process it should be clear that for disjoint $A_1, \ldots, A_n$ then $N_1(A_1), \ldots, N_1(A_n)$ are again independent, and $$\begin{array}{rcl} P(N_1(A) = k) & = & \sum_{n=k}^{\infty} P(N_1(A) = k \mid N(A) = n)P(N(A) = n) \\ & =& e^{-\mu(A)} \sum_{n=k}^{\infty} {n \choose k} p^k(1-p)^{n-k} \frac{\mu(A)^n}{n!} \\ & = & \frac{(p\mu)^k}{k!}e^{-\mu(A)} \sum_{n=k}^{\infty} \frac{((1-p)\mu(A))^{n-k}}{(n-k)!} \\ & = & \frac{(p\mu(A))^k}{k!}e^{-\mu(A) + (1-p)\mu(A)} = e^{-p\mu(A)}\frac{(p\mu(A))^k}{k!}. \end{array}$$ This shows that $N_1$ is a Poisson process. Similarly, $N_2$ is a Poisson process (with mean measure $(1-p)\mu$). What is left is to show that $N_1$ and $N_2$ are, in fact, independent. We cut a corner here and say that it is actually sufficient to show that $N_1(A)$ and $N_2(A)$ are independent for arbitrary $A$, and this follows from $$\begin{array}{rcl} P(N_1(A) = k, N_2(A) = r) & = & P(N_1(A) = k, N(A) = k + r) \\ & = & P(N_1(A) = k \mid N(A) = k + r) P(N(A) = k + r) \\ & = & e^{-\mu(A)} {k+r \choose k} p^k(1-p)^{r} \frac{\mu(A)^{k+r}}{(k+r)!} \\ & = & e^{-p\mu(A)}\frac{(p\mu(A))^k}{k!} e^{-(1-p)\mu(A)}\frac{((1-p)\mu(A))^r}{r!} \\ & = & P(N_1(A) = k)P(N_2(A) = r). \end{array}$$
Summary II: We conclude that i.i.d. random thinning with success probability $p$ of a simple point process, $N$, with intensity $\lambda$ results in two simple point processes, $N_1$ and $N_2$, with intensities $p\lambda$ and $(1-p)\lambda$, respectively, and $N$ is the superposition of $N_1$ and $N_2$. If, moreover, $N$ is a Poisson process then $N_1$ and $N_2$ are independent Poisson processes.
It is natural to ask if we could thin independently without assuming that the $Z_i$'s are identically distributed and obtain similar results. This is possible, but a little more complicated to formulate, because the distribution of $Z_i$ then has to be linked to the $X_i$ somehow. For instance, $P(Z_i = 1 \mid N) = p(x_i)$ for a given function $p$. It is then possible to show the same result as above but with the intensity $p\lambda$ meaning the function $p(x)\lambda(x)$. We skip the proof. The best general mathematical reference covering spatial point processes is Daley and Vere-Jones. A close second covering statistics and simulation algorithms, in particular, is Møller and Waagepetersen.
-
+1, great answer! – mpiktas Oct 21 '11 at 2:38
+1 Reading this answer is really amazing and helpful. I personally learned lots of thing. It is one of the most complete answers that I have ever received. I really appreciate it. – Developer Oct 21 '11 at 12:11
@Developer, thanks. Glad I could be of assistance. – NRH Oct 23 '11 at 19:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 76, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9390327334403992, "perplexity_flag": "head"}
|
http://www.physicsforums.com/showthread.php?p=4008315
|
Physics Forums
## Measurement of spin due to the force on a magnet
Can we measure the spin of a particle by measuring the force on a Magnet?
I wish to consider a modified stern gerlach like apparatus where a Magnet of mass M is attached to a spring. Can we measure Spin of the particle by studying the deflection of the magnet?
Even if it is possible, I can see No way to calculate the force on the magnetic.
I think the consideration that the spring is composed of several degrees of excitation becomes important to analyze this problem.
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Mentor Blog Entries: 27 How tiny of a magnet do you have to be able to detect that small force? Whenever you want to figure out something like this, you have to do a back-of-the-envelope calculation on the quantitative aspect of it. Figure out what strength of force that something like this would produce, and then see if there is a realistic setup that will be able to detect it! Zz.
My guess would be that the Force of the particle = Force on the Magnet = μ∇B. The change in momentum = μ∇B*δT = μ∇B L/P where δT = L/P Where L is the extension P is the momentum.
Mentor
Blog Entries: 27
## Measurement of spin due to the force on a magnet
And what "numbers" do you get? The magnitude will determine what you will need to detect such a thing. This is what I mean by being quantitative! It is how physicists design experiments.
Zz.
Numbers and whether such an experiment is performable is not my focus. My main interest is to understand how such a setup must be treated in theory. On one hand We have the measurement apparatus(The Spring+Magnet System) that is described by classical laws. On the Other Hand We have a Spin, whose state is described by a wave function. I wish to undertake a an analysis of what measurement apparatus is using a simple system. I think It becomes very important to take into consideration that the spring is composed of many degrees of freedom.
Mentor
Blog Entries: 27
Quote by Prathyush Numbers and whether such an experiment is performable is not my focus. My main interest is to understand how such a setup must be treated in theory. On one hand We have the measurement apparatus(The Spring+Magnet System) that is described by classical laws. On the Other Hand We have a Spin, whose state is described by a wave function. I wish to undertake a an analysis of what measurement apparatus is using a simple system. I think It becomes very important to take into consideration that the spring is composed of many degrees of freedom.
Ah, but the devil is in the DETAILS!
What will kill your spring-magnet system is that you will NOT be able to find a magnet sensitive enough to measure such a thing. That's why a quantitative value is crucial! I'm trying to teach you how to think through something like this when you are trying to figure out what can and cannot be done. Your dismissal of any quantitative evaluation will turn your idea into something useless if you are not aware of what can be realistically done!
Why do you think we have SQUIDs, and why do you think these are the most common devices to measure magnetic flux?
Zz.
Recognitions: Homework Help Science Advisor Torsion balances can easily measure forces of ~10^(-8) N, which is equivalent to the gravitational force between two 1kg-masses at 10cm separation. This number is 200 years old, so you can probably add a lot of sensitivity now. Stern and Gerlach used a magnetic field gradient of 10T/cm, multiplied by the magnetic moment of an electron this gives a force of 20^(-20)N per atom. This is 100 years old, so maybe you get better gradients now (or the same gradients with less mass). You should have multiple atoms in the field at the same, of course. Can close the gap between both numbers?
Mentor
Blog Entries: 27
Quote by mfb Torsion balances can easily measure forces of ~10^(-8) N, which is equivalent to the gravitational force between two 1kg-masses at 10cm separation. This number is 200 years old, so you can probably add a lot of sensitivity now. Stern and Gerlach used a magnetic field gradient of 10T/cm, multiplied by the magnetic moment of an electron this gives a force of 20^(-20)N per atom. This is 100 years old, so maybe you get better gradients now (or the same gradients with less mass). You should have multiple atoms in the field at the same, of course. Can close the gap between both numbers?
But this doesn't measure the "spin of a particle", per the OP.
Putting a bunch of atoms together and measuring the collective magnetic moment isn't that straightforward. Anyone who has done NMR or EPR experiment (as in electron paramagnetic resonance) can tell you, the atom-atom interaction and temperature WILL affect the orientation of neighboring atoms. One can already see this in solids where the nature of the Heisenburg coupling, for example, can determine if something is ferromagnetic or antiferromagnetic.
Zz.
Recognitions: Science Advisor Basically this is the classical Einstein de Haas experiment from 1915.
Recognitions:
Homework Help
Science Advisor
Well, it would measure the spin of a lot of particles, preferably a polarized beam.
Anyone who has done NMR
Did that. But not with a beam of particles.
Quote by mfb Torsion balances can easily measure forces of ~10^(-8) N, which is equivalent to the gravitational force between two 1kg-masses at 10cm separation. This number is 200 years old, so you can probably add a lot of sensitivity now. Stern and Gerlach used a magnetic field gradient of 10T/cm, multiplied by the magnetic moment of an electron this gives a force of 20^(-20)N per atom. This is 100 years old, so maybe you get better gradients now (or the same gradients with less mass). You should have multiple atoms in the field at the same, of course. Can close the gap between both numbers?
So It appears that sort of some classical version of this experiment can be performed where A Continuous stream of atoms are sent through the apparatus and we can measure the average force on the magnet. That should be directly related to average spin of the particles that enter the setup.
Though measurements for single spin would be difficult to perform.
Is it meaningful to ask the question what is the Impulse the magnet feels when a single atom passes through. How does a record of that Impulse correspond to measurement of the electron.
Quote by DrDu Basically this is the classical Einstein de Haas experiment from 1915.
The Einstein de Haas experiment involves several electrons, we are interested in the effect due to one spin.
Recognitions:
Homework Help
Science Advisor
Quote by Prathyush Is it meaningful to ask the question what is the Impulse the magnet feels when a single atom passes through. How does a record of that Impulse correspond to measurement of the electron.
It depends on the speed of the atom and the length of the setup. A thermal atom with ~1km/s and 10cm length of the setup would give 100µs, multiplied with the force this corresponds to 10^(-24) kg m/s.
Quote by mfb 20^(-20)N per atom
Typo, should be 10^(-20)
Is there any chance that this type of experiment could be adapted to provide a similar measurement, at least indirectly? http://arstechnica.com/science/2012/...ust-two-atoms/
Quote by PhilDSP Is there any chance that this type of experiment could be adapted to provide a similar measurement, at least indirectly? http://arstechnica.com/science/2012/...ust-two-atoms/
I am not sure, how do you intend to set it up?
Quote by mfb It depends on the speed of the atom and the length of the setup. A thermal atom with ~1km/s and 10cm length of the setup would give 100µs, multiplied with the force this corresponds to 10^(-24) kg m/s.
It appears Such an impulse would be rather Difficult/Impossible to measure.
But say one can somehow construct an apparatus that is sensitive enough to measure The force applied by the spring. I am very to understand How to apply A formalism That deals with springs that are to be treated Classically and Spins that must be treated Quantum Mechanically. In specific How does One see that Impulse must be quantized.
Recognitions: Homework Help Science Advisor Interesting aspect with a spring. The energy corresponding to a momentum is given by $E=\frac{p^2}{2m}$. Using m=1kg and p from above, this corresponds to 5*10-49J or ~5*10-15Hz. The single atom could not excite the whole spring/magnet system for any reasonable setup.
Thank you ZapperZ and mfb, I see this is a Highly impractical situation. I will try to look else. Maybe research up on SQUIDS. Can some one please refer me to a lucid introduction to the subject.
Thread Tools
| | | |
|-----------------------------------------------------------------------|-------------------------------|---------|
| Similar Threads for: Measurement of spin due to the force on a magnet | | |
| Thread | Forum | Replies |
| | General Physics | 3 |
| | Quantum Physics | 3 |
| | General Physics | 15 |
| | Introductory Physics Homework | 0 |
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.933171808719635, "perplexity_flag": "middle"}
|
http://physics.stackexchange.com/questions/tagged/angular-momentum?sort=votes&pagesize=30
|
# Tagged Questions
The conserved quantity arising from a rotational invariance. Combine with rotational-dynamics for the classical mechanics approach and quantum-mechanics for the QM interpretation
9answers
3k views
### Is Angular Momentum truly fundamental?
This may seem like a slightly trite question, but it is one that has long intrigued me. Since I formally learned classical (Newtonian) mechanics, it has often struck me that angular momentum (and ...
1answer
429 views
### Why are our planets in the solar system all on the same disc/plane/layer?
I always see pictures of the solar system where our sun is in the middle and the planets surround the sun. All these planets move on orbits on the same layer. Why?
4answers
2k views
### What is spin as it relates to subatomic particles?
I often hear about subatomic particles having a property called "spin" but also that it doesn't actually relate to spinning about an axis like you would think. Which particles have spin? What does ...
6answers
1k views
### Rotation angle of a giant lily when a child crawls on its rim
Below is a picture of Giant Water Lily. Scientific Name: Victoria Amazonica. Leaves of some of these could be as big as 3 m diameter and carry a weight of 45kg spread evenly and can support a child. ...
5answers
2k views
### What if the universe is rotating as a whole?
Suppose in the milliseconds after the big bang the cosmic egg had aquired some large angular momentum. As it expanded, keeping the momentum constant (not external forces) the rate of rotation would ...
4answers
614 views
### Why does every thing spin?
The origin of spin is some what a puzzle to me, every thing spin from galaxies to planets to weather to electrons. Where has all the angular momentum come from? Why is it so natural? I was also ...
8answers
4k views
### Why don't spinning tops fall over?
One topic which was covered in university, but which I never understood, is how a spinning top "magically" resists the force of gravity. The conservation of energy explanations make sense, but I don't ...
3answers
319 views
### Do particles have different spins in different frames of reference?
Let's say we have two photons, whose momentum vectors point to opposite directions. Also spin angular momentum vectors of the photons point to opposite directions. (Sum of spins is zero) Now we ...
3answers
119 views
### Why are spacecraft made to “spin” after launch?
At some point after launch, usually just before or after separation from the last booster stage, spacecraft are often made to "spin" (about the axis of their trajectory)? See e.g this You Tube video. ...
6answers
2k views
### How do you explain spinning tops to a nine year old?
Why don't spinning tops fall over? (The young scientist version) My nine year old son asked me this very question when playing with his "Battle Strikers" set. Having studied Physics myself, I am very ...
2answers
2k views
### How to define angular momentum in other than three dimensions?
In classical mechanics with 3 space dimension the angular momentum is defined as $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ In relativistic mechanics we have the 4-vectors $x^{\mu}$ and $p^{\mu}$, ...
2answers
81 views
### When are there enough Casimirs?
I know that a Casimir for a Lie algebra $\mathfrak{g}$ is a central element of the universal enveloping algebra. For example in $\mathfrak{so}(3)$ the generators are the angular momentum operators ...
3answers
817 views
### Adding 3 electron spins
I've learned how to add two 1/2-spins, which you can do with C-G-coefficients. There are 4 states (one singlet, three triplet states). States are symmetric or antisymmetric and the quantum numbers ...
3answers
691 views
### Why does a curling rock curl?
In the game of curling, players "curl" a granite "rock" (of precise size and roughly a flattened cylinder) down a "sheet" of ice towards a target; the "rock" will curve in its path in the direction of ...
2answers
376 views
### Where the angular momentum has gone?
I came across an interesting problem when I prepared for the preliminary exam on electromagnetism. Below is the problem in its original words: A metallic sphere of mass, $m$, and radius, $a$, ...
3answers
488 views
### Questions about the Solar System
Most images you see of the solar system are 2D and all planets orbit in the same plane. In a 3D view, are really all planets orbiting in similar planes? Is there a reason for this? I'd expect that ...
2answers
345 views
### Lie bracket for Lie algebra of $SO(n,m)$
How does one show that the bracket of elements in the Lie algebra of $SO(n,m)$ is given by $$[J_{ab},J_{cd}] ~=~ i(\eta_{ad} J_{bc} + \eta_{bc} J_{ad} - \eta_{ac} J_{bd} - \eta_{bd}J_{ac}),$$ ...
1answer
2k views
### Why exactly does a boomerang return back to the thrower?
I was always intrigued by the phenomena that govern the returning back of a boomerang to the thrower. Even if it is dependent on various factors such as velocity, aerodynamics, wind resistance and ...
1answer
484 views
### How does one experimentally determine chirality, helicity, spin and angular momentum?
If I've got an instance of a fundamental particle, how can I separate out the measurements of these three concepts? (I think) I understand the theory behind them, and why the particles in the ...
5answers
650 views
### Will freely rotatable polarizers align?
Will two freely rotatable linear polarizers (placed in sequence and at some angular offset less than, say, 45 degrees) eventually align if you shine (plenty of) unpolarized light at the first one? If ...
2answers
352 views
### Addition of Angular Momementa in deeply bound situations, proton spin crisis
In Landau and Lipshitz's introductory book on Quanum Mechanics, "Quantum Mechanics Non-Relativistic Theory, Third Edition: Volume 3", chapter XIV (page 433 in the edition on Amazon) is "Addition of ...
2answers
589 views
### Intrinsic angular momentum in classical mechanics
Please note, I am only interested in classical mechanics discussion on this. Please do not involve quantum mechanics. Inspired by this question: Is Angular Momentum truly fundamental? My question ...
4answers
467 views
### What causes a soccer ball to follow a curved path?
Soccer players kick the ball in a linear kick, though you find it to turn sideways, not even in one direction. Just mid air it changes that curve's direction. Any physical explanation? Maybe this ...
5answers
2k views
### Why are some galaxies flat?
What is the explanation for the flatness of some galaxies? (If it's due to their rotation then why they are rotating, why some other galaxies are not flat etc., I would like to hear a nice and ...
3answers
279 views
### Can one make a ball rotate around a vertical axis using only a combination of horizontal axis rotations?
This is a nice problem that I would like to share. Problem: In a public garden, there a statue consisting of a spherical stone and a stone cup. The ball is 1 meter in diameter and weighs at least a ...
3answers
89 views
### Rotationally invariant body and principal axis
Suppose a rigid body is invariant under a rotation around an axis $\mathsf{A}$ by a given angle $0 \leq \alpha_0 < 2\pi$ (and also every multiple of $\alpha_0$). Is it true that in this case the ...
1answer
728 views
### Why do galaxies and water going down a plug hole spin?
We all experience things spinning, whether it's water down a drain, the earth on its axis, planets round the sun, or stars in a galaxy - even electrons round an atom. But why is spin so common in ...
1answer
266 views
### Classical vs. Quantum use of the spin 4-vector
I have a few basic questions about the Pauli-Lubanski spin 4-vector S. I've used it in quantum mechanical calculations as an operator, that is to say each of the components of S is a matrix operator ...
3answers
849 views
### Conservation of angular momentum for a rigid body rotating about a fixed point
Picture a rigid body such as a sledge hammer. Imagine that the base of the handle is attached to a fixed point such that it can rotate but not translate. I give the hammer a good push to get it ...
1answer
138 views
### Why doesn't my particle simulation end in a flat disc?
I've made a 3d particle simulator where particles are attracted to each other by the inverse of the square radius. The purpose of my experiment is to see if this alone would create a flat disk (like ...
2answers
910 views
### maximum rotational speed
I am wondering if there is a limit to rotational speed of an object just like there is one for translation speed ? what are the implications of general relativity for rotating objects ?
4answers
274 views
### Why do 3d spheres and gravity tend to rotating discs on one plane?
Whether is it our solar system or a whole galaxy, there is usually a massive object (star or black hole) at the centre with gas and objects rotating around it. The gravitational effect of the ...
3answers
393 views
### Why does spin have a discrete spectrum?
Why is it that unlike other quantum properties such as momentum and velocity, which usually are given through (probabilistic) continuous values, spin has a (probabilistic) discrete spectrum?
6answers
545 views
### Why does the earth rotate? [duplicate]
Possible Duplicate: Why does every thing spin? So why would the earth, or any planet for that matter, rotate along an axis? I know of no force which could come into play here, so i assume ...
2answers
730 views
### How is angular momentum conserved when a spinning top finally stops spinning?
Where does the top's angular momentum get transferred to? Does it very slightly change the angular momentum of the table, and then the angular momentum of the Earth?
3answers
397 views
### confused about ball rolling and rotational energy
I was solving a practice test problem and it was just a conservation of energy problem where a spherical ball is falls from a height h to the ground such that mgh = \frac{1}{2}mv^2 + ...
2answers
1k views
### Relation between total orbital angular momentum and symmetry of the wavefunction
My question essentially revolves around multi-electron atoms and spectroscopic terms. I understand the idea that the total wavefunction for Fermions should be antisymmetric. Consider as an example, ...
2answers
215 views
### Do we have any effect on the rotation of he earth?
We walk or ride on our vehicles to our destinations daily. Does our movement have any effect on the rotation of the earth according to Newton's law? What will be the effect if we move all the peoples ...
2answers
96 views
### What maintains quark spin alignments in baryons?
What maintains quark spin alignments in baryons? The $uud$ proton and $udd$ neutron are both spin 1/2, implying that two of their spin 1/2 quarks are always parallel and the other is always opposed. ...
3answers
384 views
### Can the spin of a photon change during its “life”?
Or is the spin set in one of two possible states at its moment of creation and does not change for the rest of the duration of its "life"?
1answer
192 views
### Effect of the tail of the cat in the falling cat problem
To explain why a falling cat can turn by 180 degree without external torque and without violation of the conservation of angular momentum, one usually models the cat as two cylinders as in ...
1answer
338 views
### Angular momentum and EM wave
Is there any sense in saying that circularly polarized EM waves have angular momentum?
1answer
343 views
### Simultaneously commuting set
How does one determine the members of an simultaneously commuting set (of operators)? For example, I have read that for orbital angular momentum, the set is {$H,L^2,L_z$}. How does one know that these ...
2answers
63 views
### How do objects change their axis of rotation?
If I hold a pencil at its end and spin it, throwing it upwards, it will spin about its end, but will soon start spinning around its center. How is this? I would draw the following torque diagram for ...
3answers
336 views
### How does Hubble's constant affect the Earth's orbit
If Hubble's constant is $2.33 \times 10^{-18} \text{ s}^{-1}$ and the earth orbits the sun with average distance of 150 million kilometers; Does that mean the earth's orbital radius increases ...
0answers
186 views
### Coupling Coefficients in SO(4)
I have two equations (from two distinct authors) for the decomposition of a coupling coefficient of SO(4) (i.e. Wigner 3j-symbol for SO(4)). In the first: ...
1answer
109 views
### Sum of angular momentum of all electrons in a magnet
Can the sum of angular momentum of all rotating electrons in all the aligned atoms in a permanent magnet have a significant contribution to the macro angular momentum of the magnet? If yes, why does ...
4answers
3k views
### Why does a ballerina speed up when she pulls in her arms?
My friend thinks it's because she has less air resistance but I'm not sure.
2answers
93 views
### Where does a star's angular momentum go as its spin slows down?
So we know that stars slow down as they age. But total angular momentum must be conserved. Where does that angular momentum go? The dissipation of Earth's tides somehow transfers Earth's angular ...
2answers
482 views
### Half wave plate and angular momentum
Given: A half wave plate freely floating in space. Circularly polarized light, falling perpendicularly to it. The plate changes polarisation of the beam to the opposite one. Therefore it ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9282777905464172, "perplexity_flag": "middle"}
|
http://en.wikipedia.org/wiki/Dynamic_programming
|
# Dynamic programming
For the programming paradigm, see Dynamic programming language.
In mathematics, computer science, and economics, dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems. It is applicable to problems exhibiting the properties of overlapping subproblems[1] and optimal substructure (described below). When applicable, the method takes far less time than naive methods that don't take advantage of the subproblem overlap (like depth-first search).
The idea behind dynamic programming is quite simple. In general, to solve a given problem, we need to solve different parts of the problem (subproblems), then combine the solutions of the subproblems to reach an overall solution. Often when using a more naive method, many of the subproblems are generated and solved many times. The dynamic programming approach seeks to solve each subproblem only once, thus reducing the number of computations: once the solution to a given subproblem has been computed, it is stored or "memo-ized": the next time the same solution is needed, it is simply looked up. This approach is especially useful when the number of repeating subproblems grows exponentially as a function of the size of the input.
Dynamic programming algorithms are used for optimization (for example, finding the shortest path between two points, or the fastest way to multiply many matrices). A dynamic programming algorithm will examine all possible ways to solve the problem and will pick the best solution. Therefore, we can roughly think of dynamic programming as an intelligent, brute-force method that enables us to go through all possible solutions to pick the best one. If the scope of the problem is such that going through all possible solutions is possible and fast enough, dynamic programming guarantees finding the optimal solution. The alternatives are many, such as using a greedy algorithm, which picks the best possible choice "at any possible branch in the road". While a greedy algorithm does not guarantee the optimal solution, it is faster. Fortunately, some greedy algorithms (such as minimum spanning trees) are proven to lead to the optimal solution.
For example, let's say that you have to get from point A to point B as fast as possible, in a given city, during rush hour. A dynamic programming algorithm will look into the entire traffic report, looking into all possible combinations of roads you might take, and will only then tell you which way is the fastest. Of course, you might have to wait for a while until the algorithm finishes, and only then can you start driving. The path you will take will be the fastest one (assuming that nothing changed in the external environment). On the other hand, a greedy algorithm will start you driving immediately and will pick the road that looks the fastest at every intersection. As you can imagine, this strategy might not lead to the fastest arrival time, since you might take some "easy" streets and then find yourself hopelessly stuck in a traffic jam.
Sometimes, applying memoization to a naive basic recursive solution already results in an optimal dynamic programming solution, however many problems require more sophisticated dynamic programming algorithms. Some of these may be recursive as well but parametrized differently from the naive solution. Others can be more complicated and cannot be implemented as a recursive function with memoization. Examples of these are the two solutions to the Egg Dropping puzzle below.
## History
The term dynamic programming was originally used in the 1940s by Richard Bellman to describe the process of solving problems where one needs to find the best decisions one after another. By 1953, he refined this to the modern meaning, referring specifically to nesting smaller decision problems inside larger decisions,[2] and the field was thereafter recognized by the IEEE as a systems analysis and engineering topic. Bellman's contribution is remembered in the name of the Bellman equation, a central result of dynamic programming which restates an optimization problem in recursive form.
Bellman explains the reasoning behind the term dynamic programming in his autobiography, Eye of the Hurricane: An Autobiography (1984). He explains:
"I spent the Fall quarter (of 1950) at RAND. My first task was to find a name for multistage decision processes. An interesting question is, Where did the name, dynamic programming, come from? The 1950s were not good years for mathematical research. We had a very interesting gentleman in Washington named Wilson. He was Secretary of Defense, and he actually had a pathological fear and hatred of the word, research. I’m not using the term lightly; I’m using it precisely. His face would suffuse, he would turn red, and he would get violent if people used the term, research, in his presence. You can imagine how he felt, then, about the term, mathematical. The RAND Corporation was employed by the Air Force, and the Air Force had Wilson as its boss, essentially. Hence, I felt I had to do something to shield Wilson and the Air Force from the fact that I was really doing mathematics inside the RAND Corporation. What title, what name, could I choose? In the first place I was interested in planning, in decision making, in thinking. But planning, is not a good word for various reasons. I decided therefore to use the word, “programming” I wanted to get across the idea that this was dynamic, this was multistage, this was time-varying I thought, lets kill two birds with one stone. Lets take a word that has an absolutely precise meaning, namely dynamic, in the classical physical sense. It also has a very interesting property as an adjective, and that is its impossible to use the word, dynamic, in a pejorative sense. Try thinking of some combination that will possibly give it a pejorative meaning. It's impossible. Thus, I thought dynamic programming was a good name. It was something not even a Congressman could object to. So I used it as an umbrella for my activities."
The word dynamic was chosen by Bellman to capture the time-varying aspect of the problems, and because it sounded impressive.[3] The word programming referred to the use of the method to find an optimal program, in the sense of a military schedule for training or logistics. This usage is the same as that in the phrases linear programming and mathematical programming, a synonym for mathematical optimization.[4]
## Overview
Figure 1. Finding the shortest path in a graph using optimal substructure; a straight line indicates a single edge; a wavy line indicates a shortest path between the two vertices it connects (other nodes on these paths are not shown); the bold line is the overall shortest path from start to goal.
Dynamic programming is both a mathematical optimization method and a computer programming method. In both contexts it refers to simplifying a complicated problem by breaking it down into simpler subproblems in a recursive manner. While some decision problems cannot be taken apart this way, decisions that span several points in time do often break apart recursively; Bellman called this the "Principle of Optimality". Likewise, in computer science, a problem that can be broken down recursively is said to have optimal substructure.
If subproblems can be nested recursively inside larger problems, so that dynamic programming methods are applicable, then there is a relation between the value of the larger problem and the values of the subproblems.[5] In the optimization literature this relationship is called the Bellman equation.
### Dynamic programming in mathematical optimization
In terms of mathematical optimization, dynamic programming usually refers to simplifying a decision by breaking it down into a sequence of decision steps over time. This is done by defining a sequence of value functions V1, V2, ..., Vn, with an argument y representing the state of the system at times i from 1 to n. The definition of Vn(y) is the value obtained in state y at the last time n. The values Vi at earlier times i = n −1, n − 2, ..., 2, 1 can be found by working backwards, using a recursive relationship called the Bellman equation. For i = 2, ..., n, Vi−1 at any state y is calculated from Vi by maximizing a simple function (usually the sum) of the gain from decision i − 1 and the function Vi at the new state of the system if this decision is made. Since Vi has already been calculated for the needed states, the above operation yields Vi−1 for those states. Finally, V1 at the initial state of the system is the value of the optimal solution. The optimal values of the decision variables can be recovered, one by one, by tracking back the calculations already performed.
### Dynamic programming in computer programming
There are two key attributes that a problem must have in order for dynamic programming to be applicable: optimal substructure and overlapping subproblems. If a problem can be solved by combining optimal solutions to non-overlapping subproblems, the strategy is called "divide and conquer" instead. This is why mergesort and quicksort are not classified as dynamic programming problems.
Optimal substructure means that the solution to a given optimization problem can be obtained by the combination of optimal solutions to its subproblems. Consequently, the first step towards devising a dynamic programming solution is to check whether the problem exhibits such optimal substructure. Such optimal substructures are usually described by means of recursion. For example, given a graph G=(V,E), the shortest path p from a vertex u to a vertex v exhibits optimal substructure: take any intermediate vertex w on this shortest path p. If p is truly the shortest path, then it can be split into subpaths p1 from u to w and p2 from w to v such that these, in turn, are indeed the shortest paths between the corresponding vertices (by the simple cut-and-paste argument described in Introduction to Algorithms). Hence, one can easily formulate the solution for finding shortest paths in a recursive manner, which is what the Bellman-Ford algorithm or the Floyd-Warshall algorithm does.
Overlapping subproblems means that the space of subproblems must be small, that is, any recursive algorithm solving the problem should solve the same subproblems over and over, rather than generating new subproblems. For example, consider the recursive formulation for generating the Fibonacci series: Fi = Fi−1 + Fi−2, with base case F1 = F2 = 1. Then F43 = F42 + F41, and F42 = F41 + F40. Now F41 is being solved in the recursive subtrees of both F43 as well as F42. Even though the total number of subproblems is actually small (only 43 of them), we end up solving the same problems over and over if we adopt a naive recursive solution such as this. Dynamic programming takes account of this fact and solves each subproblem only once.
Figure 2. The subproblem graph for the Fibonacci sequence. The fact that it is not a tree indicates overlapping subproblems.
This can be achieved in either of two ways:[citation needed]
• Top-down approach: This is the direct fall-out of the recursive formulation of any problem. If the solution to any problem can be formulated recursively using the solution to its subproblems, and if its subproblems are overlapping, then one can easily memoize or store the solutions to the subproblems in a table. Whenever we attempt to solve a new subproblem, we first check the table to see if it is already solved. If a solution has been recorded, we can use it directly, otherwise we solve the subproblem and add its solution to the table.
• Bottom-up approach: Once we formulate the solution to a problem recursively as in terms of its subproblems, we can try reformulating the problem in a bottom-up fashion: try solving the subproblems first and use their solutions to build-on and arrive at solutions to bigger subproblems. This is also usually done in a tabular form by iteratively generating solutions to bigger and bigger subproblems by using the solutions to small subproblems. For example, if we already know the values of F41 and F40, we can directly calculate the value of F42.
Some programming languages can automatically memoize the result of a function call with a particular set of arguments, in order to speed up call-by-name evaluation (this mechanism is referred to as call-by-need). Some languages make it possible portably (e.g. Scheme, Common Lisp or Perl), some need special extensions (e.g. C++, see[6]). Some languages have automatic memoization built in, such as tabled Prolog and J, which supports memoization with the M. adverb.[7] In any case, this is only possible for a referentially transparent function.
## Example: Mathematical optimization
### Optimal consumption and saving
A mathematical optimization problem that is often used in teaching dynamic programming to economists (because it can be solved by hand[8]) concerns a consumer who lives over the periods $t=0,1,2,\ldots,T$ and must decide how much to consume and how much to save in each period.
Let $c_t$ be consumption in period $t$, and assume consumption yields utility $u(c_t)=\ln(c_t)$ as long as the consumer lives. Assume the consumer is impatient, so that he discounts future utility by a factor $b$ each period, where $0<b<1$. Let $k_t$ be capital in period $t$. Assume initial capital is a given amount $k_0>0$, and suppose that this period's capital and consumption determine next period's capital as $k_{t+1}=Ak^a_t - c_t$, where $A$ is a positive constant and $0<a<1$. Assume capital cannot be negative. Then the consumer's decision problem can be written as follows:
$\max \sum_{t=0}^T b^t \ln(c_t)$ subject to $k_{t+1}=Ak^a_t - c_t \geq 0$ for all $t=0,1,2,\ldots,T$
Written this way, the problem looks complicated, because it involves solving for all the choice variables $c_0, c_1, c_2, \ldots , c_T$. (Note that $k_0$ is not a choice variable—the consumer's initial capital is taken as given.)
The dynamic programming approach to solving this problem involves breaking it apart into a sequence of smaller decisions. To do so, we define a sequence of value functions $V_t(k)$, for $t=0,1,2,\ldots,T,T+1$ which represent the value of having any amount of capital $k$ at each time $t$. Note that $V_{T+1}(k)=0$, that is, there is (by assumption) no utility from having capital after death.
The value of any quantity of capital at any previous time can be calculated by backward induction using the Bellman equation. In this problem, for each $t=0,1,2,\ldots,T$, the Bellman equation is
$V_t(k_t) \, = \, \max \left( \ln(c_t) + b V_{t+1}(k_{t+1}) \right)\text{ subject to }k_{t+1}=Ak^a_t - c_t \geq 0$
This problem is much simpler than the one we wrote down before, because it involves only two decision variables, $c_t$ and $k_{t+1}$. Intuitively, instead of choosing his whole lifetime plan at birth, the consumer can take things one step at a time. At time $t$, his current capital $k_t$ is given, and he only needs to choose current consumption $c_t$ and saving $k_{t+1}$.
To actually solve this problem, we work backwards. For simplicity, the current level of capital is denoted as $k$. $V_{T+1}(k)$ is already known, so using the Bellman equation once we can calculate $V_T(k)$, and so on until we get to $V_0(k)$, which is the value of the initial decision problem for the whole lifetime. In other words, once we know $V_{T-j+1}(k)$, we can calculate $V_{T-j}(k)$, which is the maximum of $\ln(c_{T-j}) + b V_{T-j+1}(Ak^a-c_{T-j})$, where $c_{T-j}$ is the choice variable and $Ak^a-c_{T-j} \ge 0$.
Working backwards, it can be shown that the value function at time $t=T-j$ is
$V_{T-j}(k) \, = \, a \sum_{i=0}^j a^ib^i \ln k + v_{T-j}$
where each $v_{T-j}$ is a constant, and the optimal amount to consume at time $t=T-j$ is
$c_{T-j}(k) \, = \, \frac{1}{\sum_{i=0}^j a^ib^i} Ak^a$
which can be simplified to
$c_{T}(k) \, = \, Ak^a$, and $c_{T-1}(k) \, = \, \frac{Ak^a}{1+ab}$, and $c_{T-2}(k) \, = \, \frac{Ak^a}{1+ab+a^2b^2}$, etc.
We see that it is optimal to consume a larger fraction of current wealth as one gets older, finally consuming all remaining wealth in period $T$, the last period of life.
## Examples: Computer algorithms
### Dijkstra's algorithm for the shortest path problem
From a dynamic programming point of view, Dijkstra's algorithm for the shortest path problem is a successive approximation scheme that solves the dynamic programming functional equation for the shortest path problem by the Reaching method.[9][10][11]
In fact, Dijkstra's explanation of the logic behind the algorithm,[12] namely
Problem 2. Find the path of minimum total length between two given nodes $P$ and $Q$.
We use the fact that, if $R$ is a node on the minimal path from $P$ to $Q$, knowledge of the latter implies the knowledge of the minimal path from $P$ to $R$.
is a paraphrasing of Bellman's famous Principle of Optimality in the context of the shortest path problem.
### Fibonacci sequence
Here is a naïve implementation of a function finding the nth member of the Fibonacci sequence, based directly on the mathematical definition:
``` function fib(n)
if n = 0 return 0
if n = 1 return 1
return fib(n − 1) + fib(n − 2)
```
Notice that if we call, say, `fib(5)`, we produce a call tree that calls the function on the same value many different times:
1. `fib(5)`
2. `fib(4) + fib(3)`
3. `(fib(3) + fib(2)) + (fib(2) + fib(1))`
4. `((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))`
5. `(((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))`
In particular, `fib(2)` was calculated three times from scratch. In larger examples, many more values of `fib`, or subproblems, are recalculated, leading to an exponential time algorithm.
Now, suppose we have a simple map object, m, which maps each value of `fib` that has already been calculated to its result, and we modify our function to use it and update it. The resulting function requires only O(n) time instead of exponential time (but requires O(n) space):
``` var m := map(0 → 0, 1 → 1)
function fib(n)
if map m does not contain key n
m[n] := fib(n − 1) + fib(n − 2)
return m[n]
```
This technique of saving values that have already been calculated is called memoization; this is the top-down approach, since we first break the problem into subproblems and then calculate and store values.
In the bottom-up approach, we calculate the smaller values of `fib` first, then build larger values from them. This method also uses O(n) time since it contains a loop that repeats n − 1 times, but it only takes constant (O(1)) space, in contrast to the top-down approach which requires O(n) space to store the map.
``` function fib(n)
if n = 0
return 0
var previousFib := 0, currentFib := 1
else repeat n − 1 times // loop is skipped if n=1
var newFib := previousFib + currentFib
previousFib := currentFib
currentFib := newFib
return currentFib
```
In both examples, we only calculate `fib(2)` one time, and then use it to calculate both `fib(4)` and `fib(3)`, instead of computing it every time either of them is evaluated.
Note that the above method actually takes $\Omega(n^2)$ time for large n because addition of two integers with $\Omega(n)$ bits each takes $\Omega(n)$ time. (The nth fibonacci number has $\Omega(n)$ bits.) Also, there is a closed form for the Fibonacci sequence, known as Binet's formula, from which the $n$-th term can be computed in approximately $O(n(\log n)^2)$ time, which is more efficient than the above dynamic programming technique. However, the simple recurrence directly gives the matrix form that leads to an approximately $O(n\log n)$ algorithm by fast matrix exponentiation.
### A type of balanced 0–1 matrix
Consider the problem of assigning values, either zero or one, to the positions of an n × n matrix, with n even, so that each row and each column contains exactly n / 2 zeros and n / 2 ones. We ask how many different assignments there are for a given $n$. For example, when n = 4, four possible solutions are
$\begin{bmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{bmatrix} \text{ and } \begin{bmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{bmatrix} \text{ and } \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix} \text{ and } \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}.$
There are at least three possible approaches: brute force, backtracking, and dynamic programming.
Brute force consists of checking all assignments of zeros and ones and counting those that have balanced rows and columns ($n/2$ zeros and $n/2$ ones). As there are $\tbinom{n}{n/2}^n$ possible assignments, this strategy is not practical except maybe up to $n=6$.
Backtracking for this problem consists of choosing some order of the matrix elements and recursively placing ones or zeros, while checking that in every row and column the number of elements that have not been assigned plus the number of ones or zeros are both at least n / 2. While more sophisticated than brute force, this approach will visit every solution once, making it impractical for n larger than six, since the number of solutions is already 116,963,796,250 for n = 8, as we shall see.
Dynamic programming makes it possible to count the number of solutions without visiting them all. Imagine backtracking values for the first row – what information would we require about the remaining rows, in order to be able to accurately count the solutions obtained for each first row value? We consider k × n boards, where 1 ≤ k ≤ n, whose $k$ rows contain $n/2$ zeros and $n/2$ ones. The function f to which memoization is applied maps vectors of n pairs of integers to the number of admissible boards (solutions). There is one pair for each column, and its two components indicate respectively the number of ones and zeros that have yet to be placed in that column. We seek the value of $f((n/2, n/2), (n/2, n/2), \ldots (n/2, n/2))$ ($n$ arguments or one vector of $n$ elements). The process of subproblem creation involves iterating over every one of $\tbinom{n}{n/2}$ possible assignments for the top row of the board, and going through every column, subtracting one from the appropriate element of the pair for that column, depending on whether the assignment for the top row contained a zero or a one at that position. If any one of the results is negative, then the assignment is invalid and does not contribute to the set of solutions (recursion stops). Otherwise, we have an assignment for the top row of the k × n board and recursively compute the number of solutions to the remaining (k − 1) × n board, adding the numbers of solutions for every admissible assignment of the top row and returning the sum, which is being memoized. The base case is the trivial subproblem, which occurs for a 1 × n board. The number of solutions for this board is either zero or one, depending on whether the vector is a permutation of n / 2 $(0, 1)$ and n / 2 $(1, 0)$ pairs or not.
For example, in the two boards shown above the sequences of vectors would be
```((2, 2) (2, 2) (2, 2) (2, 2)) ((2, 2) (2, 2) (2, 2) (2, 2)) k = 4
0 1 0 1 0 0 1 1
((1, 2) (2, 1) (1, 2) (2, 1)) ((1, 2) (1, 2) (2, 1) (2, 1)) k = 3
1 0 1 0 0 0 1 1
((1, 1) (1, 1) (1, 1) (1, 1)) ((0, 2) (0, 2) (2, 0) (2, 0)) k = 2
0 1 0 1 1 1 0 0
((0, 1) (1, 0) (0, 1) (1, 0)) ((0, 1) (0, 1) (1, 0) (1, 0)) k = 1
1 0 1 0 1 1 0 0
((0, 0) (0, 0) (0, 0) (0, 0)) ((0, 0) (0, 0), (0, 0) (0, 0))
```
The number of solutions (sequence in OEIS) is
$1,\, 2,\, 90,\, 297200,\, 116963796250,\, 6736218287430460752, \ldots$
Links to the MAPLE implementation of the dynamic programming approach may be found among the external links.
### Checkerboard
Consider a checkerboard with n × n squares and a cost-function c(i, j) which returns a cost associated with square i, j (i being the row, j being the column). For instance (on a 5 × 5 checkerboard),
| 5 | 4 | 3 | 2 | 1 |
|-----|-----|-----|-----|-----|
| 6 | 7 | 4 | 7 | 8 |
| 7 | 6 | 1 | 1 | 4 |
| 3 | 5 | 7 | 8 | 2 |
| – | 6 | 7 | 0 | – |
| – | – | *5* | – | – |
| | | | | |
Thus c(1, 3) = 5
Let us say you had a checker that could start at any square on the first rank (i.e., row) and you wanted to know the shortest path (sum of the costs of the visited squares are at a minimum) to get to the last rank, assuming the checker could move only diagonally left forward, diagonally right forward, or straight forward. That is, a checker on (1,3) can move to (2,2), (2,3) or (2,4).
| 5 | 4 | 3 |
|-----|-----|-----|
| | | |
| | | |
| | | |
| x | x | x |
| | o | |
| | | |
This problem exhibits optimal substructure. That is, the solution to the entire problem relies on solutions to subproblems. Let us define a function q(i, j) as
q(i, j) = the minimum cost to reach square (i, j).
If we can find the values of this function for all the squares at rank n, we pick the minimum and follow that path backwards to get the shortest path.
Note that q(i, j) is equal to the minimum cost to get to any of the three squares below it (since those are the only squares that can reach it) plus c(i, j). For instance:
| 5 | 4 | 3 |
|-----|-----|-----|
| | | |
| | A | |
| B | C | D |
| | | |
| | | |
| | | |
$q(A) = \min(q(B),q(C),q(D))+c(A) \,$
Now, let us define q(i, j) in somewhat more general terms:
$q(i,j)=\begin{cases} \infty & j < 1 \text{ or }j > n \\ c(i, j) & i = 1 \\ \min(q(i-1, j-1), q(i-1, j), q(i-1, j+1)) + c(i,j) & \text{otherwise.}\end{cases}$
The first line of this equation is there to make the recursive property simpler (when dealing with the edges, so we need only one recursion). The second line says what happens in the last rank, to provide a base case. The third line, the recursion, is the important part. It is similar to the A,B,C,D example. From this definition we can make a straightforward recursive code for q(i, j). In the following pseudocode, n is the size of the board, `c(i, j)` is the cost-function, and `min()` returns the minimum of a number of values:
```function minCost(i, j)
if j < 1 or j > n
return infinity
else if i = 1
return c(i, j)
else
return min( minCost(i-1, j-1), minCost(i-1, j), minCost(i-1, j+1) ) + c(i, j)
```
It should be noted that this function only computes the path-cost, not the actual path. We will get to the path soon. This, like the Fibonacci-numbers example, is horribly slow since it wastes time recomputing the same shortest paths over and over. However, we can compute it much faster in a bottom-up fashion if we store path-costs in a two-dimensional array `q[i, j]` rather than using a function. This avoids recomputation; before computing the cost of a path, we check the array `q[i, j]` to see if the path cost is already there.
We also need to know what the actual shortest path is. To do this, we use another array `p[i, j]`, a predecessor array. This array implicitly stores the path to any square s by storing the previous node on the shortest path to s, i.e. the predecessor. To reconstruct the path, we lookup the predecessor of s, then the predecessor of that square, then the predecessor of that square, and so on, until we reach the starting square. Consider the following code:
``` function computeShortestPathArrays()
for x from 1 to n
q[1, x] := c(1, x)
for y from 1 to n
q[y, 0] := infinity
q[y, n + 1] := infinity
for y from 2 to n
for x from 1 to n
m := min(q[y-1, x-1], q[y-1, x], q[y-1, x+1])
q[y, x] := m + c(y, x)
if m = q[y-1, x-1]
p[y, x] := -1
else if m = q[y-1, x]
p[y, x] := 0
else
p[y, x] := 1
```
Now the rest is a simple matter of finding the minimum and printing it.
``` function computeShortestPath()
computeShortestPathArrays()
minIndex := 1
min := q[n, 1]
for i from 2 to n
if q[n, i] < min
minIndex := i
min := q[n, i]
printPath(n, minIndex)
```
``` function printPath(y, x)
print(x)
print("<-")
if y = 2
print(x + p[y, x])
else
printPath(y-1, x + p[y, x])
```
### Sequence alignment
In genetics, sequence alignment is an important application where dynamic programming is essential.[3] Typically, the problem consists of transforming one sequence into another using edit operations that replace, insert, or remove an element. Each operation has an associated cost, and the goal is to find the sequence of edits with the lowest total cost.
The problem can be stated naturally as a recursion, a sequence A is optimally edited into a sequence B by either:
1. inserting the first character of B, and performing an optimal alignment of A and the tail of B
2. deleting the first character of A, and performing the optimal alignment of the tail of A and B
3. replacing the first character of A with the first character of B, and performing optimal alignments of the tails of A and B.
The partial alignments can be tabulated in a matrix, where cell (i,j) contains the cost of the optimal alignment of A[1..i] to B[1..j]. The cost in cell (i,j) can be calculated by adding the cost of the relevant operations to the cost of its neighboring cells, and selecting the optimum.
Different variants exist, see Smith–Waterman algorithm and Needleman–Wunsch algorithm.
### Tower of Hanoi puzzle
A model set of the Towers of Hanoi (with 8 disks)
An animated solution of the Tower of Hanoi puzzle for T(4,3).
The Tower of Hanoi or Towers of Hanoi is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.
The objective of the puzzle is to move the entire stack to another rod, obeying the following rules:
• Only one disk may be moved at a time.
• Each move consists of taking the upper disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.
• No disk may be placed on top of a smaller disk.
The dynamic programming solution consists of solving the functional equation
S(n,h,t) = S(n-1,h, not(h,t)) ; S(1,h,t) ; S(n-1,not(h,t),t)
where n denotes the number of disks to be moved, h denotes the home rod, t denotes the target rod, not(h,t) denotes the third rod (neither h nor t), ";" denotes concatenation, and
S(n, h, t) := solution to a problem consisting of n disks that are to be moved from rod h to rod t.
Note that for n=1 the problem is trivial, namely S(1,h,t) = "move a disk from rod h to rod t" (there is only one disk left).
The number of moves required by this solution is 2n − 1. If the objective is to maximize the number of moves (without cycling) then the dynamic programming functional equation is slightly more complicated and 3n − 1 moves are required.[13]
### Egg dropping puzzle
The following is a description of the instance of this famous puzzle involving n=2 eggs and a building with H=36 floors:[14]
Suppose that we wish to know which storeys in a 36-storey building are safe to drop eggs from, and which will cause the eggs to break on landing. We make a few assumptions:
• An egg that survives a fall can be used again.
• A broken egg must be discarded.
• The effect of a fall is the same for all eggs.
• If an egg breaks when dropped, then it would break if dropped from a higher window.
• If an egg survives a fall, then it would survive a shorter fall.
• It is not ruled out that the first-floor windows break eggs, nor is it ruled out that the 36th-floor windows do not cause an egg to break.
If only one egg is available and we wish to be sure of obtaining the right result, the experiment can be carried out in only one way. Drop the egg from the first-floor window; if it survives, drop it from the second-floor window. Continue upward until it breaks. In the worst case, this method may require 36 droppings. Suppose 2 eggs are available. What is the least number of egg-droppings that is guaranteed to work in all cases?
To derive a dynamic programming functional equation for this puzzle, let the state of the dynamic programming model be a pair s = (n,k), where
n = number of test eggs available, n = 0, 1, 2, 3, ..., N − 1.
k = number of (consecutive) floors yet to be tested, k = 0, 1, 2, ..., H − 1.
For instance, s = (2,6) indicates that two test eggs are available and 6 (consecutive) floors are yet to be tested. The initial state of the process is s = (N,H) where N denotes the number of test eggs available at the commencement of the experiment. The process terminates either when there are no more test eggs (n = 0) or when k = 0, whichever occurs first. If termination occurs at state s = (0,k) and k > 0, then the test failed.
Now, let
W(n,k) = minimum number of trials required to identify the value of the critical floor under the worst case scenario given that the process is in state s = (n,k).
Then it can be shown that[15]
W(n,k) = 1 + min{max(W(n − 1, x − 1), W(n,k − x)): x = 1, 2, ..., k; n = 2, ..., N; k = 2, 3, 4, ..., H
with W(n,1) = 1 for all n > 0 and W(1,k) = k for all k. It is easy to solve this equation iteratively by systematically increasing the values of n and k.
An interactive online facility is available for experimentation with this model as well as with other versions of this puzzle (e.g. when the objective is to minimize the expected value of the number of trials.)[15]
Faster DP solution using a different parametrization
Notice that the above solution takes $O( n k^2 )$ time. This can be improved to $O( n k \log k )$ time by binary searching on the optimal $x$ in the above recurrence, since $W(n-1,x-1)$ is increasing in $x$ while $W(n,k-x)$ is decreasing in $x$, thus a local minimum of $\max(W(n-1,x-1),W(n,k-x))$ is a global minimum. However, there is a much faster solution that involves a different parametrization of the problem:
Let $k$ be the total number of floors such that the eggs break when dropped from the $k$th floor (The example above is equivalent to taking $k=37$).
Let $m$ be the minimum floor from which the egg must be dropped to be broken.
Let $f(t,n)$ be the maximum number of values of $m$ that are distinguishable using $t$ tries and $n$ eggs.
Then $f(t,0) = f(0,n) = 1$ for all $t,n \geq 0$.
Let $x$ be the floor from which the first egg is dropped in the optimal strategy.
If the first egg broke, $m$ is from $1$ to $x$ and distinguishable using at most $t-1$ tries and $n-1$ eggs.
If the first egg did not break, $m$ is from $x+1$ to $k$ and distinguishable using $t-1$ tries and $n$ eggs.
Therefore $f(t,n) = f(t-1,n-1) + f(t-1,n)$.
Then the problem is equivalent to finding the minimum $x$ such that $f(x,n) \geq k$.
To do so, we could compute ${ f(t,i) : 0 \leq i \leq n }$ in order of increasing $t$, which would take $O( n x )$ time.
Thus, if we separately handle the case of $n=1$, the algorithm would take $O( n \sqrt{k} )$ time.
But the recurrence relation can in fact be solved, giving $f(t,n) = \sum_{i=0}^{n}{ \binom{t}{i} }$, which can be computed in $O(n)$ time using the identity $\binom{t}{i+1} = \binom{t}{i} \frac{t-i}{i+1}$ for all $i \geq 0$.
Since $f(t,n) \leq f(t+1,n)$ for all $t \geq 0$, we can binary search on $t$ to find $x$, giving an $O( n \log k )$ algorithm.
### Matrix chain multiplication
Matrix chain multiplication is a well known example that demonstrates utility of dynamic programming. For example, engineering applications often have to multiply a chain of matrices. It is not surprising to find matrices of large dimensions, for example 100×100. Therefore, our task is to multiply matrices A1, A2, .... An. As we know from basic linear algebra, matrix multiplication is not commutative, but is associative; and we can multiply only two matrices at a time. So, we can multiply this chain of matrices in many different ways, for example:
((A1 × A2) × A3) × ... An
A1×(((A2×A3)× ... ) × An)
(A1 × A2) × (A3 × ... An)
and so on. There are numerous ways to multiply this chain of matrices. They will all produce the same final result, however they will take more or less time to compute, based on which particular matrices are multiplied. If matrix A has dimensions m×n and matrix B has dimensions n×q, then matrix C=A×B will have dimensions m×q, and will require m*n*q scalar multiplications.
For example, let us multiply matrices A, B and C. Let us assume that their dimensions are m×n, n×p, and p×s, respectively. Matrix A×B×C will be of size m×s and can be calculated in two ways shown below:
1. Ax(B×C) This order of matrix multiplication will require nps + mns scalar multiplications.
2. (A×B)×C This order of matrix multiplication will require mnp + mps scalar calculations.
Let us assume that m = 10, n = 100, p = 10 and s = 1000. So, the first way to multiply the chain will require 1,000,000 + 1,000,000 calculations. The second way will require only 10,000+10,0000 calculations. Obviously, the second way is faster, and we should multiply the matrices using that arrangement of parenthesis.
Therefore, our conclusion is that the order of parenthesis matters, and that our task is to find the optimal order of parenthesis.
At this point, we have several choices, one of which is to design a dynamic programming algorithm that will split the problem into overlapping problems and calculate the optimal arrangement of parenthesis. The dynamic programming solution is presented below.
Let's call m[i,j] the minimum number of scalar multiplications needed to multiply a chain of matrices from matrix i to matrix j (i.e. Ai × .... × Aj, i.e. i<=j). We split the chain at some matrix k, such that i<=k <j, and try to find out which combination produces minimum m[i,j].
The formula is:
``` if i = j, m[i,j]= 0
if i < j, m[i,j]= min over all possible values of k {{m[i,k]+m[k+1],j] + p_(i-1)*p_k*p_j}}
```
where k is changed from i to j - 1.
p_(i-1) is the row dimension of matrix i, p_k is the column dimension of matrix k, p_j is the column dimension of matrix j.
This formula can be coded as shown below, where input parameter "chain" is the chain of matrices, i.e. A1, A2, ... An:
``` function OptimalMatrixChainParenthesis(chain)
n = length(chain)
for i = 1, n
m[i,i] = 0 //since it takes no calculations to multiply one matrix
for len = 2, n
for i = 1, n - len + 1
for j = i + len -1
m[i,j] = infinity //so that the first calculation updates
for k = i, j-1
q = m[i, k] + m[k+1, j] + p_(i-1)*p_k*p_j
if q < m[i, j] // the new order of parenthesis is better than what we had
m[i, j] = q //update
s[i, j] = k //record which k to split on, i.e. where to place the parenthesis
```
So far, we have calculated values for all possible m[i, j], the minimum number of calculations to multiply a chain from matrix i to matrix j, and we have recorded the corresponding "split point" s[i, j]. For example, if we are multiplying chain A1×A2×A3×A4, and it turns out that m[1, 3] = 100 and s[1, 3] = 2, that means that the optimal placement of parenthesis for matrices 1 to 3 is (A1×A2)×A3 and to multiply those matrices will require 100 scalar calculation.
This algorithm will produce "tables" m[, ] and s[, ] that will have entries for all possible values of i and j. The final solution for the entire chain is m[1, n], with corresponding split at s[1, n]. Unraveling the solution will be recursive, starting from the top and continuing until we reach the base case, i.e. multiplication of single matrices.
Therefore, the next step is to actually split the chain, i.e. to place the parenthesis where they (optimally) belong. For this purpose we could use the following algorithm:
``` function PrintOptimalParenthesis(s, i, j)
if i = j
print "A"i
else
print "(" PrintOptimalParenthesis(s, i, s[i, j]) PrintOptimalParenthesis(s, s[i, j] + 1, j) ")"
```
Of course, this algorithm is not useful for actual multiplication. This algorithm is just a user-friendly way to see what the result looks like.
To actually multiply the matrices using the proper splits, we need the following algorithm:
``` function MatrixChainMultiply(chain from 1 to n) //returns the final matrix, i.e. A1×A2×... ×An
OptimalMatrixChainParenthesis(chain from 1 to n) //this will produce s[ . ] and m[ . ] "tables"
OptimalMatrixMultiplication(s, chain from 1 to n) //actually multiply
```
``` function OptimalMatrixMultiplication(s, i, j) //returns the result of multiplying a chain of matrices from Ai to Aj in optimal way
if i < j
//keep on splitting the chain and multiplying the matrices in left and right sides
LeftSide = OptimalMatrixMultiplication(s, i, s[i, j])
RightSide = OptimalMatrixMultiplication(s, s[i, j] + 1, j)
return MatrixMultiply(LeftSide, RightSide)
else if i = j
return Ai //matrix at position i
else
print "error, i <= j must hold"
```
``` function MatrixMultiply(A, B) //function that multiplies two matrices
if columns(A) = rows(B)
for i = 1, rows(A)
for j = 1, columns(B)
C[i, j] = 0
for k = 1, columns(A)
C[i, j] = C[i, j] + A[i, k]*B[k, j]
return C
else
print "error, incompatible dimensions."
```
## Algorithms that use dynamic programming
• Recurrent solutions to lattice models for protein-DNA binding
• Backward induction as a solution method for finite-horizon discrete-time dynamic optimization problems
• Method of undetermined coefficients can be used to solve the Bellman equation in infinite-horizon, discrete-time, discounted, time-invariant dynamic optimization problems
• Many string algorithms including longest common subsequence, longest increasing subsequence, longest common substring, Levenshtein distance (edit distance)
• Many algorithmic problems on graphs can be solved efficiently for graphs of bounded treewidth or bounded clique-width by using dynamic programming on a tree decomposition of the graph.
• The Cocke–Younger–Kasami (CYK) algorithm which determines whether and how a given string can be generated by a given context-free grammar
• Knuth's word wrapping algorithm that minimizes raggedness when word wrapping text
• The use of transposition tables and refutation tables in computer chess
• The Viterbi algorithm (used for hidden Markov models)
• The Earley algorithm (a type of chart parser)
• The Needleman–Wunsch and other algorithms used in bioinformatics, including sequence alignment, structural alignment, RNA structure prediction
• Floyd's all-pairs shortest path algorithm
• Optimizing the order for chain matrix multiplication
• Pseudo-polynomial time algorithms for the subset sum and knapsack and partition problems
• The dynamic time warping algorithm for computing the global distance between two time series
• The Selinger (a.k.a. System R) algorithm for relational database query optimization
• De Boor algorithm for evaluating B-spline curves
• Duckworth–Lewis method for resolving the problem when games of cricket are interrupted
• The value iteration method for solving Markov decision processes
• Some graphic image edge following selection methods such as the "magnet" selection tool in Photoshop
• Some methods for solving interval scheduling problems
• Some methods for solving word wrap problems
• Some methods for solving the travelling salesman problem, either exactly (in exponential time) or approximately (e.g. via the bitonic tour)
• Recursive least squares method
• Beat tracking in music information retrieval
• Adaptive-critic training strategy for artificial neural networks
• Stereo algorithms for solving the correspondence problem used in stereo vision
• Seam carving (content aware image resizing)
• The Bellman–Ford algorithm for finding the shortest distance in a graph
• Some approximate solution methods for the linear search problem
• Kadane's algorithm for the maximum subarray problem
## References
1. S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani, 'Algorithms', p 173, available at http://www.cs.berkeley.edu/~vazirani/algorithms.html
2. ^ a b Eddy, S. R., What is dynamic programming?, Nature Biotechnology, 22, 909–910 (2004).
3. Nocedal, J.; Wright, S. J.: Numerical Optimization, page 9, Springer, 2006..
4. "M. Memo". J Vocabulary. J Software. Retrieved 28 October 2011.
5. Stokey et al., 1989, Chap. 1
6. Sniedovich, M. (2006), "Dijkstra’s algorithm revisited: the dynamic programming connexion" (PDF), Journal of Control and Cybernetics 35 (3): 599–620. Online version of the paper with interactive computational modules.
7. Denardo, E.V. (2003), Dynamic Programming: Models and Applications, Mineola, NY: Dover Publications, ISBN 978-0-486-42810-9
8. Sniedovich, M. (2010), Dynamic Programming: Foundations and Principles, Taylor & Francis, ISBN 978-0-8247-4099-3
9. Moshe Sniedovich (2002), "OR/MS Games: 2. The Towers of Hanoi Problem,", INFORMS Transactions on Education 3 (1): 34–51.
10. Konhauser J.D.E., Velleman, D., and Wagon, S. (1996). Which way did the Bicycle Go? Dolciani Mathematical Expositions – No 18. The Mathematical Association of America.
11. ^ a b
## Further reading
• . An accessible introduction to dynamic programming in economics. The link contains sample programs.
• Bellman, Richard (1954), "The theory of dynamic programming", 60: 503–516, doi:10.1090/S0002-9904-1954-09848-8, MR 0067459 . Includes an extensive bibliography of the literature in the area, up to the year 1954.
• Bellman, Richard (1957), Dynamic Programming, Princeton University Press . Dover paperback edition (2003), ISBN 0-486-42809-5.
• Bertsekas, D. P. (2000), Dynamic Programming and Optimal Control (2nd ed.), Athena Scientific, ISBN 1-886529-09-4 . In two volumes.
• Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L.; Stein, Clifford (2001), (2nd ed.), MIT Press & McGraw-Hill, ISBN 0-262-03293-7 . Especially pp. 323–69.
• Dreyfus, Stuart E.; Law, Averill M. (1977), The art and theory of dynamic programming, Academic Press, ISBN 978-0-12-221860-6 .
• Giegerich, R.; Meyer, C.; Steffen, P. (2004), "A Discipline of Dynamic Programming over Sequence Data", Science of Computer Programming 51 (3): 215–263, doi:10.1016/j.scico.2003.12.005 .
• Meyn, Sean (2007), Control Techniques for Complex Networks, Cambridge University Press, ISBN 978-0-521-88441-9 .
• S. S. Sritharan, "Dynamic Programming of the Navier-Stokes Equations," in Systems and Control Letters, Vol. 16, No. 4, 1991, pp. 299–307.
• Stokey, Nancy; Lucas, Robert E.; Prescott, Edward (1989), Recursive Methods in Economic Dynamics, Harvard Univ. Press, ISBN 978-0-674-75096-8 .
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 126, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9073872566223145, "perplexity_flag": "middle"}
|
http://mathoverflow.net/questions/86019/software-package-to-manipulate-representations
|
## Software package to manipulate representations
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi,
I am looking for a software package that will allow me to experiment with the irreducible representations of lie groups (for example, $SL(2,p)$) over the complex field and over finite fields. That is, I would like to get the corresponding matrices for group elements. Thanks, Shachar
-
The CHEVIE package for GAP3 or MAPLE can be useful. math.rwth-aachen.de/~CHEVIE Unfortunately, the GAP4 version is still pending. – Leandro Vendramin Jan 18 2012 at 19:29
## 1 Answer
For representations over the complex field, I know that GAP does a good job. (I'm not sure if it can do modular representations as well, but I wouldn't be surprised.)
Here is some example code to get you started:
````G:=SL(2,3);;
reps:=IrreducibleRepresentations(G);;
Elements(G);
List(G,g->g^reps[5]);
````
This prints the elements of the group $SL(2,3)$:
````[ [ [ 0*Z(3), Z(3)^0 ], [ Z(3), 0*Z(3) ] ],
[ [ 0*Z(3), Z(3)^0 ], [ Z(3), Z(3)^0 ] ],
[ [ 0*Z(3), Z(3)^0 ], [ Z(3), Z(3) ] ],
[ [ 0*Z(3), Z(3) ], [ Z(3)^0, 0*Z(3) ] ],
[ [ 0*Z(3), Z(3) ], [ Z(3)^0, Z(3)^0 ] ],
[ [ 0*Z(3), Z(3) ], [ Z(3)^0, Z(3) ] ],
[ [ Z(3)^0, 0*Z(3) ], [ 0*Z(3), Z(3)^0 ] ],
[ [ Z(3)^0, 0*Z(3) ], [ Z(3)^0, Z(3)^0 ] ],
[ [ Z(3)^0, 0*Z(3) ], [ Z(3), Z(3)^0 ] ],
[ [ Z(3)^0, Z(3)^0 ], [ 0*Z(3), Z(3)^0 ] ],
[ [ Z(3)^0, Z(3)^0 ], [ Z(3)^0, Z(3) ] ],
[ [ Z(3)^0, Z(3)^0 ], [ Z(3), 0*Z(3) ] ],
[ [ Z(3)^0, Z(3) ], [ 0*Z(3), Z(3)^0 ] ],
[ [ Z(3)^0, Z(3) ], [ Z(3)^0, 0*Z(3) ] ],
[ [ Z(3)^0, Z(3) ], [ Z(3), Z(3) ] ],
[ [ Z(3), 0*Z(3) ], [ 0*Z(3), Z(3) ] ],
[ [ Z(3), 0*Z(3) ], [ Z(3)^0, Z(3) ] ],
[ [ Z(3), 0*Z(3) ], [ Z(3), Z(3) ] ],
[ [ Z(3), Z(3)^0 ], [ 0*Z(3), Z(3) ] ],
[ [ Z(3), Z(3)^0 ], [ Z(3)^0, Z(3)^0 ] ],
[ [ Z(3), Z(3)^0 ], [ Z(3), 0*Z(3) ] ],
[ [ Z(3), Z(3) ], [ 0*Z(3), Z(3) ] ],
[ [ Z(3), Z(3) ], [ Z(3)^0, 0*Z(3) ] ],
[ [ Z(3), Z(3) ], [ Z(3), Z(3)^0 ] ]
]
````
followed by complex matrices representing each element:
````[ [ [ 0, -1 ], [ 1, 0 ] ],
[ [ E(3)^2, -E(3) ], [ 1, 0 ] ],
[ [ -E(3), -E(3)^2 ], [ 1, 0 ] ],
[ [ 0, 1 ], [ -1, 0 ] ],
[ [ E(3), E(3)^2 ], [ -1, 0 ] ],
[ [ -E(3)^2, E(3) ], [ -1, 0 ] ],
[ [ 1, 0 ], [ 0, 1 ] ],
[ [ E(3), E(3)^2 ], [ 0, 1 ] ],
[ [ E(3)^2, -E(3) ], [ 0, 1 ] ],
[ [ 1, 0 ], [ E(3), E(3)^2 ] ],
[ [ -E(3)^2, E(3) ], [ E(3), E(3)^2 ] ],
[ [ 0, -1 ], [ E(3), E(3)^2 ] ],
[ [ 1, 0 ], [ -E(3)^2, E(3) ] ],
[ [ 0, 1 ], [ -E(3)^2, E(3) ] ],
[ [ -E(3), -E(3)^2 ], [ -E(3)^2, E(3) ] ],
[ [ -1, 0 ], [ 0, -1 ] ],
[ [ -E(3)^2, E(3) ], [ 0, -1 ] ],
[ [ -E(3), -E(3)^2 ], [ 0, -1 ] ],
[ [ -1, 0 ], [ E(3)^2, -E(3) ] ],
[ [ E(3), E(3)^2 ], [ E(3)^2, -E(3) ] ],
[ [ 0, -1 ], [ E(3)^2, -E(3) ] ],
[ [ -1, 0 ], [ -E(3), -E(3)^2 ] ],
[ [ 0, 1 ], [ -E(3), -E(3)^2 ] ],
[ [ E(3)^2, -E(3) ], [ -E(3), -E(3)^2 ] ]
]
````
Change "5" to other numbers to see different representations. Also, note that the GAP symbol
````Z(p)
````
denotes a generator of the multiplicative group of the finite field $\mathbb{F}_p$. Similarly,
````E(k)
````
denotes a primitive $k^{\mbox{th}}$ root of unity.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.6758967638015747, "perplexity_flag": "middle"}
|
http://stats.stackexchange.com/questions/31830/how-to-estimate-storage-needs-using-the-pert-distribution-for-filesizes-how-to/31872
|
# How to estimate storage needs using the PERT distribution for filesizes? How to aggregate them without falling into the flaw of extremes?
Lets say I know that I am going to store the information of 10,000 people each year for 4 years, that is 40,000 files. Now If I estimate that on the best case scenario the information from each person is going to weight .25MBytes, I believe there is a 50% chance that it will weight 1Mbyte, and I believe that the worst case scenario is that each person will require 5Mbytes...
Would it be ok for me to use the PERT formula (O + 4M + P)/6 and conclude ((.25 + 4(1) +5)/6 = 1.54) that my storage needs will be 40,000x1.5 = 60,000 Mbytes = 60 Gigabytes ?
I am trying to reach the answers "How likely would that be? ¿How can I know if that is X% likely?" numbers using Mathematica, so far it seems I can answer: How likely would that be? by writing this:
$Probability[x < 1.54, x \[Distributed] PERTDistribution[{0.25, 5}, 1]]$
I get a 0.55 probability of X being smaller than 1.54. ¿Correct?
I am asking this because I need to estimate the hardware requirements for a software system, and all the information I am given is that "I am going to store the information of 10,000 people each year for 4 years, that is 40,000 files". Once the system works for some months I will be able to use its behavior to predict its needs in a more accurately way, but in the meanwhile, I need to be able to give an initial estimate... and I am not sure on how to do that...
So. would it be safe to say that I have a 0.55 probability of my storage needs being smaller than 60 Gigabytes?
I am having some trouble beliving this is right, so I made the follwing experiment:
$Table[Fold[Plus, 0, RandomVariate[PERTDistribution[{.25, 5}, 1], 40000]], {10000}]$
To simulate the total size that I would need and, the thing is that this giving me a different answer than I would think:
If I do this:
$Quantile[Table[Fold[Plus, 0, RandomVariate[PERTDistribution[{.25, 5}, 1], 40000]], {10000}], 95/100]$
Answer: $61930.7$
It should give me the same answer as this (Should'nt it?) why it doesn't?
$(Quantile[PERTDistribution[{.25, 5}, 1], 95/100]) (40000)$
Answer: $121,938$
But it does not, it seems like estimating a single files size with 95% confidence and then multiplying it for 40,000 is very different...
Looking at my statistics books, this looks like the mistake in chapter 17 in "The Flaw of Averages" by by Dr. Sam Savage "The Flaw of Extremes" "In bottom - up budgeting, reporting the 90th percentile of cash needs leads to ever thicker layers of unnecessary cash as the figures are rolled up to higher levels. Even more harmful things result from focusing on above - or below - average results, such as test scores or health - related statistics."
I think that is why there is a difference, in this case, it is not bottom - up budgeting, but it is bottom - up file storage estimation, and the same rules should apply ¿no?
¿Is there a way to deal with this "symbolically" that is, without having to wait for slow simulated generation of sizes?
-
## 2 Answers
Let's say your estimate of a distribution is pretty much right. It looks like a log normal distribution with parameters 0 and 0.7:
The mean of a variable with this distribution is 1.278 and the median is 1.00 (fitting with your estimate of the 50% likelihood value); the variance is 1.032 (more than you would estimate from the PERT method if I understand it correctly). If we took 40,000 independent instances of this variable and added them together we would get a new variable that is normally distributed (because of the central limit theorem), with mean $40000\times 1.278=51100$ and a standard distribution of only $\sqrt{40000*1.032}=203$. In other words, you are virtually certain the needed amount is within say 600 of 51100. This is the "flaw of averages" issue - if the random variables are independent then you really can say with a lot of certainty what is going to happen when you add enough of them up together; you don't need to add the worst likely scenario n times.
But lets take another possibility - there's no individual level randomness at all, all the users will end up using exactly the same amount of storage, perhaps because storage is dictated by environmental factors eg what software they have to use. This isn't as odd as it may sound, because technological change is surely a big source of the uncertainty in how much storage is needed, and its something that impacts on each individual in a common way. In this case, effectively your initial estimates just get multiplied by 40,000; if your worst case scenario of 5MB turns out to be true, it's true for everyone, every year and you need 200,000 MB of storage.
These two scenarios look like the following:
Of course, the reality probably lies somewhere between these two. There's a chunk of uncertainty that impacts on everyone equally (including your own estimation skills); and then some individual randomness.
I appreciate this hasn't exactly answered your question, and I haven't engaged with the Mathematica or PERT issues at all. From what I can understand of your question and the other answer, it looks like the relevant Mathematica function is more in line with my cautious "everyone makes the same decision" approach. This seems to (sensibly) put a lot of weight on the uncertainty about the whole distribution and external environment.
I suppose my answer's theme, such as it is is that there is a lot of uncertainty about the uncertainty here - and that is just when we deal with the known unknowns. If we add into this any uncertainty about your initial parameter estimates, things get even wilder. So I would be very cautious about relying on the maths other than as a ballpark
-
I am trying to reach the answers "How likely would that be? ¿How can I know if that is X% likely?" numbers using Mathematica, so far it seems I can answer: How likely would that be? by writing this:
Probability[x<1.54,x[Distributed]PERTDistribution[0.25,5,1]] I get a 0.55 probability of X being smaller than 1.54. ¿Correct?
Yes your assumptions are correct and you have used the Mathematica function ( $Probability[x < 1.54, x \[Distributed] PERTDistribution[{0.25, 5}, 1]]$ ) in the right way.
You also have a 0.45 chance of your storage needs being larger than 60GB as well as 0.55 chance of it being smaller.
If you specify 128GB of storage you will have a 0.964 chance of your storage not being exceeded.
If you specify 256GB of storage you will have a probability of almost 1 that your storage will not be exceeded.
Unless you are using RAM or SSD why not specify 1 TB and be done with it :)
But it does not, it seems like estimating a single files size with 95% confidence and then multiplying it for 40,000 is very different... ¿where is the mistake? also ¿is there a way to deal with this mistake "symbolically" that is, without having to wait for slow generation of sizes?
Yes there is. Use the Mathematica function `Probability` and your chosen PERTDistribution, which will integrate the probability distribution correctly for you.
For your report perhaps you could consider including a graph such as this:
Produced using the following Mathematica command:
````With[{uLim = 5},
Plot[CDF[PERTDistribution[{0.25, 5}, 1], x], {x, 0, uLim},
FrameTicks -> {With[{ts = Range[0, uLim, 0.25]}, {ts,
40 ts}\[Transpose]], Automatic},
GridLines -> {Range[0, uLim, 0.25], Range[0, 1, 0.05]},
Frame -> True, LabelStyle -> Directive[Bold, Larger],
FrameLabel -> {"Space (GB) ",
"Probability that Space is Sufficient"}]]
````
which uses the Cumulative Distribution Function for the PERT distribution, `CDF` in Mathematica, which is effectively an aggregate of the probability that is below some value ( think of it as a plot of instantaneous values of the `Probability` function ).
You can then pose the question "What probability of storage exhaustion is acceptable to the organisation ?".
I'd be a little wary of the values of storage size for probabilities approaching unity, as the real distribution of file sizes is likely to have heavier tails than accounted for by the PERT distribution.
You may also want to consider your estimate of user numbers. You have a nice distribution estimate for file size, accounting for the potential variability in that quantity. However,you have a single fixed estimate for user numbers, 40,000, with no equivalent allowance for the potentially significant variation in this value. It might be wise to account for this uncertainty on your model.
-
¿why not specify 1 TB and be done with it? Server resources upgrade cycle, they are offering me a server with 140Gbytes, or, a server with 1TB, if I choose wrong, I have to live with my wrong choice for 18 months before being able to apply for an upgrade (internal multi-national fortune 100 company with this silly impossible-to-change policy). Now the easy thing is to ask for the 1TB but they want a document justifying it BEFORE I even start coding the sytem that is going to run there (I also need to do the same with processor power, RAM usage and network traffic) – Luxspes Jul 8 '12 at 3:26
I am being told (by some books I am reading) that one shouldn't just aggregate estimates... but in this case I am estimating the size of one file and then "aggregating" it 40,000 times to guess the storage capacity... ¿am I not violating this "do not just aggregate estimates" rule? ¿shouldn't I be using some special formula to estimate the full storage capacity instead of just making a multiplication? – Luxspes Jul 8 '12 at 3:32
I agree that I should err on the side of caution in my spec, but, regardless of the possible error in my input parameters.. would my estimation be a lot more precise if, for example i were to generate a number of samples of data, and then re-compute thestatistics of those samples? – Luxspes Jul 8 '12 at 9:18
For example, I could generate the 40,000 file sizes, 1000 times, and analize the resulting 1000 total file sizes (using something like $Table[RandomVariate[PERTDistribution[{0, 1}, 0.5], sampleSize], {i, numOfSamples}];$ But... ¿should I? – Luxspes Jul 8 '12 at 9:22
Sorry @image_doctor but it appears that you were wrong, theres is a big difference between generate the 40,000 file sizes, 10,000 times and the current answer...(see the changes in the question) – Luxspes Jul 9 '12 at 3:06
show 4 more comments
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9428341388702393, "perplexity_flag": "middle"}
|
http://mathhelpforum.com/algebra/113526-linear-quadratic-exponential-equations.html
|
# Thread:
1. ## Linear,quadratic,exponential equations?
How do you know if an equation is linear, quadratic or exponential?
Y=3X^2+4X-4
Y=2(4)^X
y=3X+2
Those equations are examples. If someone could tell me the rule for knowing if it's linear, quadratic or exponential, i would appreciate it.
2. Is linear if the highest order exponent is 1. This is given no exponents are fractional or negative. Your 3rd example is linear.
$y = 3x+2 = 3x^1+2$
Is quadratic if the highest order exponent is 2. This is given no exponents are fractional or negative. Your 1st example is linear.
the 2nd is exponential as the unknown is in the exponent.
3. Thanks....
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8883260488510132, "perplexity_flag": "middle"}
|
http://mathematica.stackexchange.com/questions/tagged/computational-geometry?sort=votes&pagesize=30
|
# Tagged Questions
Questions on constructing graphical objects using relatively complex computations relating to the mathematical structures defining those objects. Examples include convex hulls, mathematical constructions, symmetries, genuses of curves and graphs and programmatic constructions of polyhedra.
3answers
8k views
### How to create word clouds?
Word clouds are rather useless fancy and visually appealing plots, where words are plotted with different sizes according to their frequency in a corpus. Many applications exist out there (Wordle, ...
4answers
4k views
### How to peel the labels from marmalade jars using Mathematica?
How can I detect and peel the label from the jar below (POV, cylinder radius, jar contents are all unknown) to get something like this, which is the original label before it was stuck on the jar? ...
2answers
4k views
### How can I calculate a jigsaw puzzle cut path?
I want to generate a path to cut an arbitrary shape into a set of jigsaw puzzle pieces. All pieces must be unique to preclude placing a piece in the wrong spot. Pieces must be interlocking such that ...
8answers
3k views
### How to check if a 2D point is in a polygon?
Background: I use code from An Efficient Test For A Point To Be In A Convex Polygon Wolfram Demonstration to check if a point ( mouse pointer ) is in a ( convex ) polygon. Clearly this code fails for ...
2answers
838 views
### Animating a Voronoi Diagram
edit: Excellent answers have been provided and I made an animation which is suitable for my use, however, all the examples rely on bitmap/rasterized data; is there a vector based approach? I would ...
6answers
2k views
### Circuit drawing in Mathematica
This past semester I taught an introductory electromagnetism course and had quite a nice time using Mathematica to draw all sorts of figures and diagrams (mostly for problems and etc.). However, I was ...
3answers
455 views
### Build a refined grid based on intersecting line
I honestly have no idea where to begin with this problem. In summary, I have a 2D coarse grid with an intersecting line. For an easy example, let's assume it's a 4x4 grid. I wish to pass through ...
1answer
574 views
### How to create regular (planar) graphs?
How to programmatically create and plot regular planar graphs with $k = 3, 4$ or $6$ (not hypercubes) and regular nonplanar graphs of $k = 8$ (see figure)? Note that what matters is the average ...
5answers
586 views
### Voronoi diagrams for generators other than points
Any suggestions how to determine Voronoi diagram for sites other than points, as e.g. in the picture below? Input is a raster image.
3answers
418 views
### Efficiently determining if 3D points are within a surface composed of polygons
This is the 2nd part of a previous question which I edited to make into 2 separate questions: Extracting polygons from 3D contour plot surface As an extension of my earlier question involving simple ...
3answers
643 views
### Creating a 2D meshing algorithm in Mathematica
As what is proving to be a difficult, but entertaining task, I am attempting to adapt a 2D meshing algorithm created for MATLAB and port it to Mathematica. I understand meshing functions already exist ...
2answers
407 views
### Movable text on a curve
Having an arbitrary curve defined as InterpolatingFunction, what is the best way to place a text on this curve? The text generally has two rows, for example: ...
1answer
528 views
### How can the {x,y,z} points that fall on the outer boundary of a set of values be selected and smoothly surfaced?
For a given set of x,y,z values, that may, or may not form a uniform shape, how can the center of the data cloud be found, and the surface points be located and a solid smooth surface created from ...
3answers
499 views
### Procedure to find direction of triangle
I'm trying to write a procedure or function to find the direction in which a 2D triangle points. The triangle is assumed to be isosceles. While I can see the basic outline of what I want to do, making ...
2answers
491 views
### Efficient way of finding shortest distance between two sets of points in mathematica
I have two sets of 3D points, say a = RandomReal[1, {10, 3}]; b = RandomReal[1, {10, 3}]; I wanna find the first N pairs that have shortest distance between the ...
2answers
906 views
### How to calculate volume of convex hull and volume of a 3D object
I have a random 3D data points. How to calculate volume of the convex hull and volume of the object.
5answers
446 views
### How can I define a 3D version of the built-in VoronoiDiagram function?
Can anybody point me in a direction that will guide me to extend the VoronoiDiagram function in Mathematica to handle 3D (three dimensional) situations (i.e. points in 3D)? Any help will be greatly ...
2answers
202 views
### Equidistant points on a polyline
Set of 2D-points connected by a polyline B-spline function: ...
2answers
418 views
### Generating convex polyhedron from face planes?
Suppose I have lists of normals and points for planes. There's a convex polyhedron whose faces lie on these planes and are bounded by plane intersections. What would be the easiest way to produce an ...
3answers
443 views
### How to determine the convex hull of some text?
How can I get the co-ordinates of the convex hull of a piece of Text?
3answers
576 views
### How to speed up the function DelaunayTriangulation?
First define a function meshGrid to generate some points: ...
4answers
285 views
### Arranging connector lines
Heike gave an absolutely wonderful answer to my question about arranging subplots around a main plot and including connector lines. This is the result: Starting from Heike's answer, what is the ...
2answers
652 views
### How to get actual triangles from DelaunayTriangulation[]?
The ComputationalGeometry package has a DelaunayTriangulation[] function. It returns a list of points connected to each point, ...
2answers
595 views
### Finding a Concave Hull
I have a 3d clustered data: Is there any other way to get concavehull of 3D data points?
1answer
369 views
### Implementation of Balaban's Line intersection algorithm in Mathematica
I'm trying to implement a Brillouin Zone algorithm within Mathematica, including the generation of Brillouin zones of higher order in 2D and 3D. There is a nice implementation of generating these ...
3answers
655 views
### Computing the genus of an algebraic curve
How-to compute the genus of an algebraic curve in Mathematica ? In my case the algebraic curve is explicitly defined by a polynomial.
2answers
302 views
### Graphics3D: Finding intersection of 3d objects and lines
I found these two nice links 1) intersecting graphics 2) Implementation of Balaban's Line intersection algorithm in Mathematica which works for 2d. However, I need to find whether a ray(line) ...
0answers
100 views
### Has anyone implemented cohomology for complex manifolds?
I'm investigating the cohomology of complex manifolds, and need to construct chain complexes, Mayer-Vietoris sequences etc. I use Mathematica for numerically tracing manifolds with geodesics, but ...
2answers
328 views
### Checking if a point is in a convex 3D polyhedron
Extending from these questions How to check if a 3D point is in a planar polygon? and How to check if a 2D point is in a polygon?. I'm trying to do this to render specific shapes made up of spheres. ...
1answer
188 views
### Finding the perimeter, area and number of sides of a Voronoi cell
Does anyone have any suggestions how to determine the perimeter, area and number of sides of each Voronoi cell in Voronoi diagram?
0answers
322 views
### Proving inequalities with Mathematica
Question summary: I would like to learn some tips and tricks on how to prove inequalities with Mathematica. I'm studying various inequalities in triangle that have the form $R+ar + bs\geq 0$, where ...
3answers
253 views
### Distances between points in periodic cube
How can one implement more efficiently/elegantly/memory savvily the following function which returns a matrix of all Euclidian distances between points in 3D within a cube of width ...
2answers
277 views
### Create triangular mesh from random list of points
I have a list of points. I would like to take these points and create a mesh of triangles from them, making sure triangles don't overlap. So here's a list of points: ...
1answer
173 views
...
1answer
446 views
### Construct a simple mesh or tetrahedral mesh from 3D image surface
I have a 3D Y-Shape hollow tube, not so good surface. Import["http://dl.dropbox.com/u/68983831/tube02.vtk", "Graphics3D"] I tried to use following vertex data ...
6answers
283 views
### How do I draw a hemisphere?
I want to draw a solid or partially transparent hemisphere above a partially transparent cuboid object in Graphics3D. However, I do not know how to do this s.t. only half the sphere is drawn. Here's ...
3answers
342 views
### Generating a non-convex polyhedron from a list of vertex coordinates
I want to include a figure in a paper I am writing on Combinatorial Geometry which features a non-convex polyhedron given by the following vertices, EDIT: I was unaware that Mathematica could convert ...
2answers
252 views
### Convex hull of a 3D object?
I am trying to find a convex hull command for a Graphics3D object. Does it exist in Mathematica? ...
2answers
335 views
### Finding concave hull for separated small clusters
Data : data3D = Import[file, "VertexData"]; Graphics3D[Point[data3D]] How to find concave polygon for separated small clusters.
3answers
675 views
### Randomly packing spheres of fixed radius within a cube
How can I have Mathematica randomly place spheres in a cube so they won't overlap? The cube is $20 \times 20 \times 20$, and the spheres have a radius of $0.7$.
1answer
598 views
### Delaunay Triangulation for 3D Surface Data
I want to do a Delaunay triangulation on scattered 3D surface data. Mathematica itself does it only for 2D through the command DelaunayTriangulation[], which gives ...
2answers
245 views
### Calculating a minimum bounding box for a set of 3-space coordinates / spheres
I have a set of 3-space coordinates for the atoms of a molecule (I could also transform them into spheres with radii corresponding to the atoms they represent). I would like to place this molecule ...
3answers
213 views
### Point belonging to a Disk in a Cartesian Coordinate System
The image below represents a human subject fixations while observing this abstract pattern for 3 seconds. I would like to know how much time they spent looking at the actual disk. That is a red point ...
3answers
1k views
### Mathematica function intersection points with 3D grid
I need to produce a 3-dimensional equispaced grid over a given function in a way, that I can calculate intersection points of the function with the grids edges. So my first question is how to produce ...
3answers
275 views
### How to ensure that Polygon[list] plots a simple polygon?
Consider the following code which plots a triangle. p = {{0, 0}, {.2, 0}, {0, .2}}; {Cyan, Polygon[Dynamic[p]]} // Graphics Then adding (for example) ...
2answers
670 views
### How to process .vtk file
I want to extract the cluster of points inside 3D box, is there any way one can do this in Mathematica?
2answers
436 views
### How to check if a 3D point is in a planar polygon?
Following up on ndroock1's question, I naively tried to apply the solutions to a 3D point and polygon and they didn't work. For example, functions involving ArcTan ...
2answers
167 views
### Vectors in a spherical shell
I have written code that randomly generates a 3D vector of random magnitude. I now want to create a histogram of how many vectors lie in the concentric spherical shells (n*delta r, (n+1)delta r) ...
0answers
89 views
### Computing Ehrhart's polynomial for a convex polytope
Is there a Mathematica implementation for computing the Ehrhart polynomial of a convex polytope which is specified either by its vertices or by a set of inequalities? I am interested in knowing this ...
1answer
170 views
### A GraphicsMeshConvexHull[] peculiarity
I have been unable to explain the behavior of GraphicsMeshConvexHull[] on the following (highly simplified) example: ...
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 2, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9017321467399597, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/211787/uniform-distribution-on-the-unit-circle-in-the-complex-plane
|
# Uniform distribution on the unit circle (in the complex plane)
I was trying to prove that for a standard complex Gaussian variable $Z$ it holds that $|Z|^2$ is exponentially distributed with parameter 1, $\frac{Z}{|Z|}$ is uniformly distributed on the unit circle $S^1:=\{z\in\mathbb{C} | |z|=1\}$ and that the two are independent.
At some point I began asking myself:
How does one describe the uniform distribution on the unit circle $S^1$?
I resolved to say that it is the complex r.v. $e^{i\theta}$ where $\theta$ is uniformly distributed on $[0,2\pi]$. This seemed to work out fine (c.f. Byron's answer to this question).
However, if this is correct then this small argument will go through:
Let $f:S^1 \rightarrow \mathbb{R}$ be bounded. Then $$E[f(Z)]=\int_{0}^{2\pi}{f(e^{i\theta})\frac{1}{2\pi}}d\theta=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz,$$
where for the last equation $z=e^{i\theta}$ and thus $\frac{dz}{d\theta}=ie^{i\theta}$ i.e. $\frac{dz}{iz}=\frac{dz}{ie^{i\theta}}={d\theta}$. So:
Is $\frac{1}{2\pi i z}$ some kind of density for a uniformly distributed random variable on $S^1$?
(I write "some kind" as it cannot be one because the unit circle has Lebesque-measure 0 and hence the induced probability measure cannot be absolutely continuous to it.)
Thanks for clearing my lack of clarity.
-
The unit circle only has measure $0$ as a subset of $\mathbb{C}$. But you're looking only at functions defined on the unit circle, so it becomes the base set of your measure space, and as such can have measure $\geq 0$. Regarding $\frac{1}{2\pi iz}$ - how can the density of a probability distribution be complex? You'd have to define what that means first... – fgp Oct 12 '12 at 18:42
By "density" I mean that $\mathbb{P}(Z \in B)=\int_{B}{\frac{1}{2\pi i z}}$ for any arc $B$ on the unit circle. I believe (but am not sure) that this always gives a real number. The unit circle has measure zero so: $\mathbb{P}(Z\in S^1)=1$ but $\lambda_{\mathbb{C}}(S^1)=0$. So "$\mathbb{P}(Z\in \bullet) << \lambda_{\mathbb{C}}$" doesn't hold, does it? – AndreasS Oct 12 '12 at 19:01
But wait this "density" would only make sense for connected arcs, wouldn't it? – AndreasS Oct 12 '12 at 19:04
It always gives a real number because you shows that it's actually just a funny way to write an integral over the unit circle for a function with domain $\mathbb{R}$. I still don't understand what the lesbegue measure on $\mathbb{C}$ has to do with it - you're only looking at the unit circle, and your "density" is defined only on the unit circle... – fgp Oct 12 '12 at 19:19
Ok, I realize now that the "$dz$" indicates that the integral is something different to the usual Lebesque-measure idea I had in mind. It is a line integral. That it was a fancy way of writing the integral came also to my mind but I wondered if there is something more in this presentation... However, can I now go on and say that - looking only on the unit circle - this gives me some sort of "density"? Or do I just mix up the relatively simple idea that a r.v. uniformly distributed on $S^1$ is just of the form $e^{i\theta}$? – AndreasS Oct 12 '12 at 20:37
show 1 more comment
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9498378038406372, "perplexity_flag": "head"}
|
http://scicomp.stackexchange.com/questions/5036/levenberg-optimizer-halts-quickly-when-given-more-variables-or-fewer-constraint
|
# Levenberg optimizer halts quickly when given more variables, or fewer constraints
I'm using the g2o C++ optimization library to refine a GPS trajectory using accelerometer data.
The program uses a Levenberg-Marquardt optimizer over data points representing the position and orientation (vector part of quaternion) of the accelerometer. The only constraints are the accelerations in the local reference frame.
Error is measured as the difference between the accelerometer reading, and the local second derivative of the path traced by the data points — the change in velocity from (a to b) to (b to c), disregarding change in orientation, plus the upward acceleration counteracting gravity.
The initial path is a polygon drawn between GPS readings.
The optimizer runs for five iterations of gradient descent, and seems to make good adjustments. The polygon corners get rounded off a bit. But then it seems to get stuck, and Levenberg's lambda goes up for five iterations before the program terminates.
# BUT
It does a lot more refinement if I further constrain it using data from the gyroscope. However, not at as much as I'd expect, and the gyro data needs have artificially high Fisher information.
Working on segments of the whole problem doesn't terminate so quickly. It does more refinement using acceleration alone.
What could be going wrong? The inability to refine the initial estimate by acceleration alone seems to indicate a serious problem.
-
## 2 Answers
It could be that your optimization problem is very badly conditioned when using only the accelerometer data. In other words, the accelerometer data might not sufficiently constrain the parameters so that many different paths adequately fit the data. In terms of the minimization problem this means that you'd have a large "flat spot" at the minimum of different paths with nearly the same objective value.
To test for this, look at the condition number of $J^{T}J$ at the optimal trajectory. If this matrix is ill-conditioned, then you do indeed have a badly conditioned problem.
This would explain the increase in the LM's $\lambda$. If $J^{T}J$ is approaching singularity, LM will increase $\lambda$ to overcome that singularity. This results in shorter and shorter steps, and eventually the LM code just gives up.
-
1
Is this consistent with the behavior that any partition of the problem does not stall? If only some variables are at a broad plateau, which may be a local minimum with wiggle room, will it affect the convergence of the whole problem this way? Since some variables are bound to converge before others, I'm surprised if this is the case… – Potatoswatter Jan 21 at 1:36
The last vertex (data point) in the initial estimate was not getting initialized.
I'm not sure why this choked the solver. Something to do with the error from one outlier point propagating to the other data points, and pulling everything out of the minimum. It still doesn't make sense that everything would be pulled out after so few iterations, though, and the effect isn't obvious on inspection of the premature solution.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9260411858558655, "perplexity_flag": "middle"}
|
http://mathforum.org/mathimages/index.php?title=The_Golden_Ratio&oldid=31546
|
# The Golden Ratio
### From Math Images
Revision as of 09:47, 7 June 2012 by Azavez1 (Talk | contribs)
The Golden Ratio
Fields: Algebra and Geometry
Image Created By: azavez1, Massachusetts Institute of Technology
Website: The Math Forum
The Golden Ratio
The golden number, often denoted by lowercase Greek letter "phi", is ${\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...$. The term golden ratio refers to the ratio $\varphi$ : 1. The image to the right is a warped representation of dividing and subdividing a rectangle into the golden ratio. The result is fractal-like. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number.
# Basic Description
The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa uses the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids. However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.
### Misconceptions about the Golden Ratio
In his paper, Misconceptions about the Golden Ratio, George Markowsky investigates many claims about the golden ratio appearing in man-made objects and in nature. Specifically, he claims that the golden ratio does not appear in the Parthenon or the Great Pyramids, two of the more common beliefs. He also disputes the belief that the human body exhibits the golden ratio. To read more, click here!
[1]
## A Geometric Representation
### The Golden Ratio in a Line Segment
The golden number can be defined using a line segment divided into two sections of lengths a and b. If a and b are appropriately chosen, the ratio of a to b is the same as the ratio of a + b to a and both ratios are equal to φ. The line segment above (left) exhibits the golden proportion. The line segments above (right) are also examples of the golden ratio. In each case,
$\frac{{\color{Red}\mathrm{red}}+\color{Blue}\mathrm{blue}}{{\color{Blue}\mathrm{blue}} }= \frac{{\color{Blue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} }= \varphi .$
The golden rectangle is made up of line segments exhibiting the golden proportion. Remarkably, when a square is cut off of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle above.
### Triangles
The golden number, φ, is used to construct the golden triangle, an isoceles triangle that has legs of length $\varphi \times r$ and base length of $1 \times r$ where $r$ can be any constant. It is above and to the left. Similarly, the golden gnomon has base $\varphi \times r$ and legs of length $1 \times r$. It is shown above and to the right. These triangles can be used to form regular pentagons (pictured above) and pentagrams.
The pentgram below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio.
$\frac{{\color{SkyBlue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} } = \frac{{\color{Red}\mathrm{red}} }{{\color{Green}\mathrm{green}} } = \frac{{\color{Green}\mathrm{green}} }{{\color{Magenta}\mathrm{pink}} } = \varphi .$
These triangles can be used to form fractals and are one of the only ways to tile a plane using pentagonal symmetry. Two fractal examples are shown below.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Algebra, Geometry
# Mathematical Representations of the Golden Ratio
## An Algebraic Derivation of Phi
[Click to expand]
How can we derive the value of φ from its characteristics as a ratio? We may algebraically solve for the ratio (φ) by observing that ratio satisfies the following property by definition:
$\frac{b}{a} = \frac{a+b}{b} = \varphi$
[Click to hide]
Let $r$ denote the ratio :
$r=\frac{a}{b}=\frac{a+b}{a}$.
So
$r=\frac{a+b}{a}=1+\frac{b}{a}$ which can be rewritten as
$1+\cfrac{1}{a/b}=1+\frac{1}{r}$ thus,
$r=1+\frac{1}{r}$
Multiplying both sides by $r$, we get
${r}^2=r+1$
which can be written as:
$r^2 - r - 1 = 0$.
Applying the quadratic formula An equation, $\frac{-b \pm \sqrt {b^2-4ac}}{2a}$, which produces the solutions for equations of form $ax^2+bx+c=0$ , we get $r = \frac{1 \pm \sqrt{5}} {2}$.
The ratio must be positive because we can not have negative line segments or side lengths. Because the ratio has to be a positive value,
$r=\frac{1 + \sqrt{5}}{2} = 1.61803399... =\varphi$.
The golden ratio can also be written as what is called a continued fraction,a fraction of infinite length whose denominator is a quantity plus a fraction, which latter fraction has a similar denominator, and so on. This is done by using recursion.
[Click to expand]
[Click to hide]
We have already solved for φ using the following equation:
${\varphi}^2-{\varphi}-1=0$.
We can add one to both sides of the equation to get
${\varphi}^2-{\varphi}=1$.
Factoring this gives
$\varphi(\varphi-1)=1$.
Dividing by $\varphi$ gives us
$\varphi -1= \cfrac{1}{\varphi }$.
Adding 1 to both sides gives
$\varphi =1+ \cfrac{1}{\varphi }$.
Substitute in the entire right side of the equation for $\varphi$ in the bottom of the fraction.
$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }$
Substituting in again,
$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}}$
$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$
This last infinite form is a continued fraction
If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction (the finite displays above it), replacing $\varphi$ by 1, we produce the ratios between consecutive terms in the Fibonacci sequence.
$\varphi \approx 1 + \cfrac{1}{1} = 2$
$\varphi \approx 1 + \cfrac{1}{1+\cfrac{1}{1}} = 3/2$
$\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1} } } = 5/3$
$\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5$
Thus we discover that the golden ratio is approximated in the Fibonacci sequence.
$1,1,2,3,5,8,13,21,34,55,89,144...\,$
| | | |
|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|
| $1/1$ | $=$ | $1$ |
| $2/1$ | $=$ | $2$ |
| $3/2$ | $=$ | $1.5$ |
| $8/5$ | $=$ | $1.6$ |
| $13/8$ | $=$ | $1.625$ |
| $21/13$ | $=$ | $1.61538462...$ |
| $34/21$ | $=$ | $1.61904762...$ |
| $55/34$ | $=$ | $1.61764706...$ |
| $89/55$ | $=$ | $1.61818182...$ |
$\varphi = 1.61803399...\,$
As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio. Many real world applications of the golden ratio are related to the Fibonacci sequence. For more real-world applications of the golden ratio click here!
In fact, we can prove this relationship using mathematical Induction.
[Click to show proof]
[Click to hide proof]
Since we have already shown that
$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$,
we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above.
First, let $x_1=1$, $x_2=1+\frac{1}{1}=1+\frac{1}{x_1}$, $x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2}$ and so on so that $x_n=1+\frac{1}{x_{n-1}}$.
These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as $f_n=f_{n-1}+f_{n-2}$ where $f_1=1$,$f_2=1$, and so $f_3=1+1=2$, $f_4=1+2=3$ and so on.
We want to show that $x_n=\frac{f_{n+1}}{f_n}$ for all n.
First, we establish our base case. We see that $x_1=1=\frac{1}{1}=\frac{f_2}{f_1}$, and so the relationship holds for the base case.
Now we assume that $x_k=\frac{f_{k+1}}{f_{k}}$ for some $1 \leq k < n$ (This step is the inductive hypothesis). We will show that this implies that $x_{k+1}=\frac{f_{(k+1)+1}}{f_{k+1}}=\frac{f_{k+2}}{f_{k+1}}$.
By our definition of $x_n$, we have
$x_{k+1}=1+\frac{1}{x_k}$.
By our inductive hypothesis, this is equivalent to
$x_{k+1}=1+\frac{1}{\frac{f_{k+1}}{f_{k}}}$.
Now we only need to complete some simple algebra to see
$x_{k+1}=1+\frac{f_k}{f_{k+1}}$
$x_{k+1}=\frac{f_{k+1}+f_k}{f_{k+1}}$
Noting the definition of $f_n=f_{n-1}+f_{n-2}$, we see that we have
$x_{k+1}=\frac{f_{k+2}}{f_{k+1}}$
Since that was what we wanted to show, we see that the terms in our continued fraction are represented by ratios of Fibonacci numbers.
The exact continued fraction is $x_{\infty} = \lim_{n\rightarrow \infty}\frac{f_{n+1}}{f_n} =\varphi$.
## Proof of the Golden Ratio's Irrationality
[Click to expand]
[Click to hide]
Remarkably, the Golden Ratio is irrational, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers. We will use the method of contradiction to prove that the golden ratio is irrational.
Suppose $\varphi$ is rational. Then it can be written as fraction in lowest terms $\varphi = b/a$, where a and b are integers.
Our goal is to find a different fraction that is equal to $\varphi$ and is in lower terms. This will be our contradiction that will show that $\varphi$ is irrational.
First note that the definition of $\varphi = \frac{b}{a}=\frac{a+b}{b}$ implies that $b > a$ since clearly $b+a>b$ and the two fractions must be equal.
Now, since we know
$\frac{b}{a}=\frac{a+b}{b}$
we see that $b^2=a(a+b)$ by cross multiplication. Writing this all the way out gives us $b^2=a^2+ab$.
Rearranging this gives us $b^2-ab=a^2$, which is the same as $b(b-a)=a^2$.
Dividing both sides of the equation by $(b-a)$ and $a$ gives us that
$\frac{b}{a}=\frac{a}{b-a}$.
Since $\varphi=\frac{b}{a}$, we can see that $\varphi=\frac{a}{b-a}$.
Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since $a<b$, we know that $\frac{a}{b-a}$ must be in lower terms than $\frac{b}{a}$.
Since we have found a fraction of integers that is equal to $\varphi$, but is in lower terms than $\frac{b}{a}$, we have a contradiction: $\frac{b}{a}$ cannot be a fraction of integers in lowest terms. Therefore $\varphi$ cannot be expressed as a fraction of integers and is irrational.
# For More Information
• Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 2-19.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# References
1. ↑ "Parthenon", Retrieved on 16 May 2012.
# Future Directions for this Page
-animation?
http://www.metaphorical.net/note/on/golden_ratio http://www.mathopenref.com/rectanglegolden.html
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 113, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.932096540927887, "perplexity_flag": "head"}
|
http://en.wikipedia.org/wiki/Split_short_exact_sequence
|
# Exact sequence
(Redirected from Split short exact sequence)
An exact sequence is a concept in mathematics, especially in ring and module theory, homological algebra, as well as in differential geometry and group theory. An exact sequence is a sequence, either finite or infinite, of objects and morphisms between them such that the image of one morphism equals the kernel of the next.
## Definition
In the context of group theory, a sequence
$G_0 \;\xrightarrow{f_1}\; G_1 \;\xrightarrow{f_2}\; G_2 \;\xrightarrow{f_3}\; \cdots \;\xrightarrow{f_n}\; G_n$
of groups and group homomorphisms is called exact if the image (or range) of each homomorphism is equal to the kernel of the next:
$\mathrm{im}(f_k) = \mathrm{ker}(f_{k+1}).\!$
Note that the sequence of groups and homomorphisms may be either finite or infinite.
A similar definition can be made for certain other algebraic structures. For example, one could have an exact sequence of vector spaces and linear maps, or of modules and module homomorphisms. More generally, the notion of an exact sequence makes sense in any category with kernels and cokernels.
### Short exact sequence
The most common type of exact sequence is the short exact sequence. This is an exact sequence of the form
$A \;\overset{f}{\hookrightarrow}\; B \;\overset{g}{\twoheadrightarrow}\; C$
where ƒ is a monomorphism and g is an epimorphism. In this case, A is a subobject of B, and the corresponding quotient is isomorphic to C:
$C \cong B/f(A).$
(where f(A) = im(f)).
A short exact sequence of abelian groups may also be written as an exact sequence with five terms:
$0 \;\xrightarrow{}\; A \;\xrightarrow{f}\; B \;\xrightarrow{g}\; C \;\xrightarrow{}\; 0$
where 0 represents the zero object, such as the trivial group or a zero-dimensional vector space. The placement of the 0's forces ƒ to be a monomorphism and g to be an epimorphism (see below).
If instead the objects are groups not known to be abelian, then multiplicative rather than additive notation is traditional, and the identity element -- as well as the trivial group -- is often written as "1" instead of "0". So in that case a short exact sequence would be written as follows:
$1 \;\xrightarrow{}\; A \;\xrightarrow{f}\; B \;\xrightarrow{g}\; C \;\xrightarrow{}\; 1$
## Example
Consider the following sequence of abelian groups:
$\Bbb{Z} \;\overset{2\cdot}{\hookrightarrow}\; \Bbb{Z} \twoheadrightarrow \Bbb{Z}/2\Bbb{Z}$
The first operation forms an element in the set of integers, Z, using multiplication by 2 on an element from Z j = 2i. The second operation forms an element in the quotient space, j = i mod 2. Here the hook arrow $\hookrightarrow$ indicates that the map 2⋅ from Z to Z is a monomorphism, and the two-headed arrow $\twoheadrightarrow$ indicates an epimorphism (the map mod 2). This is an exact sequence because the image 2Z of the monomorphism is the kernel of the epimorphism.
This sequence may also be written without using special symbols for monomorphism and epimorphism:
$0\to \Bbb{Z} \;\xrightarrow{2\cdot}\; \Bbb{Z} \to \Bbb{Z}/2\Bbb{Z}\to 0$
Here 0 denotes the trivial abelian group with a single element, the map from Z to Z is multiplication by 2, and the map from Z to the factor group Z/2Z is given by reducing integers modulo 2. This is indeed an exact sequence:
• the image of the map 0→Z is {0}, and the kernel of multiplication by 2 is also {0}, so the sequence is exact at the first Z.
• the image of multiplication by 2 is 2Z, and the kernel of reducing modulo 2 is also 2Z, so the sequence is exact at the second Z.
• the image of reducing modulo 2 is all of Z/2Z, and the kernel of the zero map is also all of Z/2Z, so the sequence is exact at the position Z/2Z
Another example, from differential geometry, especially relevant for work on the Maxwell equations:
$\Bbb{H}_1\ \xrightarrow{\mbox{grad}}\ \Bbb{H}_\mbox{curl}\ \xrightarrow{\mbox{curl}}\ \Bbb{H}_\mbox{div}\ \xrightarrow{\mbox{div}}\ \Bbb{L}_2$,
based on the fact that on properly defined Hilbert spaces,
$\mbox{curl}\,(\mbox{grad}\,f ) = \nabla \times (\nabla f) = 0$
$\mbox{div}\,(\mbox{curl}\,\vec v ) = \nabla \cdot \nabla \times \vec{v} = 0$
in addition, curl-free vector fields can always be written as a gradient of a scalar function (as soon as the space is assumed to be simply connected, see Note 1 below), and that a divergenceless field can be written as a curl of another field.[1]
Note 1: this example makes use of the fact that 3-dimensional space is topologically trivial.
Note 2: $\Bbb{H}_\mbox{curl}\$ and $\Bbb{H}_\mbox{div}\$ are the domains for the curl and div operators respectively.
## Special cases
To make sense of the definition, it is helpful to consider what it means in relatively simple cases where the sequence is finite and begins or ends with 0.
• The sequence 0 → A → B is exact at A if and only if the map from A to B has kernel {0}, i.e. if and only if that map is a monomorphism (one-to-one).
• Dually, the sequence B → C → 0 is exact at C if and only if the image of the map from B to C is all of C, i.e. if and only if that map is an epimorphism (onto).
• A consequence of these last two facts is that the sequence 0 → X → Y → 0 is exact if and only if the map from X to Y is an isomorphism.
Important are short exact sequences, which are exact sequences of the form
$0 \rightarrow A~\overset{f}{\rightarrow}~B~\overset{g}{\rightarrow}~C \rightarrow 0$
By the above, we know that for any such short exact sequence, f is a monomorphism and g is an epimorphism. Furthermore, the image of f is equal to the kernel of g. It is helpful to think of A as a subobject of B with f being the embedding of A into B, and of C as the corresponding factor object B/A, with the map g being the natural projection from B to B/A (whose kernel is exactly A).
## Facts
The splitting lemma states that if the above short exact sequence admits a morphism t: B → A such that t o f is the identity on A or a morphism u: C → B such that g o u is the identity on C, then B is a twisted direct sum of A and C. (For groups, a twisted direct sum is a semidirect product; in an abelian category, every twisted direct sum is an ordinary direct sum.) In this case, we say that the short exact sequence splits.
The snake lemma shows how a commutative diagram with two exact rows gives rise to a longer exact sequence. The nine lemma is a special case.
The five lemma gives conditions under which the middle map in a commutative diagram with exact rows of length 5 is an isomorphism; the short five lemma is a special case thereof applying to short exact sequences.
The importance of short exact sequences is underlined by the fact that every exact sequence results from "weaving together" several overlapping short exact sequences. Consider for instance the exact sequence
$A_1\to A_2\to A_3\to A_4\to A_5\to A_6 \,\!$,
which implies that there exist objects Ck in the category such that
$C_k \cong \ker (A_k\to A_{k+1}) \cong \operatorname{im} (A_{k-1}\to A_k)$.
Suppose in addition that the cokernel of each morphism exists, and is isomorphic to the image of the next morphism in the sequence:
$C_k \cong \operatorname{coker} (A_{k-2}\to A_{k-1})$
(This is true for a number of interesting categories, including any abelian category such as the abelian groups; but it is not true for all categories that allow exact sequences, and in particular is not true for the category of groups, in which coker(f): G → H is not H/im(f) but $H / {\left\langle \operatorname{im} f \right\rangle}^H$, the quotient of H by the conjugate closure of im(f).) Then we obtain a commutative diagram in which all the diagonals are short exact sequences:
Note that the only portion of this diagram that depends on the cokernel condition is the object C7 and the final pair of morphisms A6 → C7 → 0. If there exists any object $A_{k+1}$ and morphism $A_k \rightarrow A_{k+1}$ such that $A_{k-1} \rightarrow A_k \rightarrow A_{k+1}$ is exact, then the exactness of $0 \rightarrow C_k \rightarrow A_k \rightarrow C_{k+1} \rightarrow 0$ is ensured. Again taking the example of the category of groups, the fact that im(f) is the kernel of some homomorphism on H implies that it is a normal subgroup, which coincides with its conjugate closure; thus coker(f) is isomorphic to the image H/im(f) of the next morphism.
Conversely, given any list of overlapping short exact sequences, their middle terms form an exact sequence in the same manner.
## Applications of exact sequences
In the theory of abelian categories, short exact sequences are often used as a convenient language to talk about sub- and factor objects.
The extension problem is essentially the question "Given the end terms A and C of a short exact sequence, what possibilities exist for the middle term B?" In the category of groups, this is equivalent to the question, what groups B have A as a normal subgroup and C as the corresponding factor group? This problem is important in the classification of groups. See also Outer automorphism group.
Notice that in an exact sequence, the composition fi+1 o fi maps Ai to 0 in Ai+2, so every exact sequence is a chain complex. Furthermore, only fi-images of elements of Ai are mapped to 0 by fi+1, so the homology of this chain complex is trivial. More succinctly:
Exact sequences are precisely those chain complexes which are acyclic.
Given any chain complex, its homology can therefore be thought of as a measure of the degree to which it fails to be exact.
If we take a series of short exact sequences linked by chain complexes (that is, a short exact sequence of chain complexes, or from another point of view, a chain complex of short exact sequences), then we can derive from this a long exact sequence (i.e. an exact sequence indexed by the natural numbers) on homology by application of the zig-zag lemma. It comes up in algebraic topology in the study of relative homology; the Mayer–Vietoris sequence is another example. Long exact sequences induced by short exact sequences are also characteristic of derived functors.
Exact functors are functors that transform exact sequences into exact sequences.
## References
• Spanier, Edwin Henry (1995). Algebraic Topology. Berlin: Springer. p. 179. ISBN 0-387-94426-5.
• Adhikari, M.R.; Adhikari, Avishek (2003). Groups, Rings and Modules with Applications. India: Universities Press. p. 216. ISBN 81-7371-429-0.
1.
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9326445460319519, "perplexity_flag": "head"}
|
http://mathhelpforum.com/calculus/35917-power-series.html
|
# Thread:
1. ## Power Series
I need help for the following problem.
Find the radius of convergence for the power series:
$\sum_{j=1}^{\infty} (x/2)^j$
2. Originally Posted by larson
I need help for the following problem.
Find the radius of convergence for the power series:
$\sum_{j=1}^{\infty} (x/2)^j$
By the root test, the series converges for all $x$ such that $\lim_{j \to \infty} |(x/2)^j|^{1/j} < 1$
don't forget to check the end points
3. Originally Posted by Jhevon
By the root test, the series converges for all $x$ such that $\lim_{j \to \infty} |(x/2)^j|^{1/j} < 1$
don't forget to check the end points
how do i know what my endpoints are?
4. Hello,
Find the radius of convergence R, then look at the convergence/divergence of the series if x=R and x=-R
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8737251162528992, "perplexity_flag": "middle"}
|
http://math.stackexchange.com/questions/54661/polynomial-equations-pa-b-0-for-matrices-that-ensure-ab-ba/122273
|
# Polynomial equations $p(A, B) = 0$ for matrices that ensure $AB = BA$
Let $k$ be a field with characteristic different from $2$, and $A$ and $B$ be $2 \times 2$ matrices with entries in $k$. Then we can prove, with a bit art, that $A^2 - 2AB + B^2 = O$ implies $AB = BA$, hence $(A - B)^2 = O$. It came to a surprise for me when I first succeeded in proving this, for this seemed quite nontrivial to me.
I am curious if there is a similar or more general result for the polynomial equations of matrices that ensures commutativity. (Of course, we do not consider trivial cases such as the polynomial $p(X, Y) = XY - YX$ corresponding to commutator)
p.s. This question is purely out of curiosity. I do not know even this kind of problem is worth considering, so you may regard this question as a recreational one.
-
2
Perhaps it would be a good idea to post (at least an idea of) your proof for the $A^2 - 2AB + B^2 = 0$ case that you could solve? – Srivatsan Jul 30 '11 at 20:58
13
My argument critically depends on the structure of $2 \times 2$ matrix, as you can see: Let $C = A - B$. Then $C^2 = AB - BA = CA - AC$. Our aim is to prove that $AB - BA = C^2$ vanishes. Suppose $C$ is invertible. Then $I = C^{-1}C^2 C^{-1} = AC^{-1} - C^{-1}A$, but this is impossible for $\mathrm{char} k \neq 2$ by taking trace. In particular, we have $\det C = 0$. Then by Cayley-Hamilton theorem, $C^2 = (\mathrm{tr}C) C$. Then $0 = \mathrm{tr}(CA - AC) = \mathrm{tr}(C^2) = (\mathrm{tr} C)^2$, hence $\mathrm{tr}C = 0$ and $C^2 = 0$. – sos440 Jul 30 '11 at 21:05
2
I agree that the proof does need a bit of art :-) . – Srivatsan Jul 30 '11 at 21:12
2
– Peter Sheldrick Jul 30 '11 at 21:30
1
– Peter Sheldrick Jul 30 '11 at 22:58
show 4 more comments
## 2 Answers
I'm neither an expert on this field nor on the unrelated field of the facts I'm about to cite, so this is more a shot in the dark. But: Given a set of matrices, the problem whether there is some combination in which to multiply them resulting in zero is undecidable, even for relatively small cases (such as two matrices of sufficient size or a low fixed number of $3\times3$ matrices).
A solution to one side of this problem (the "is there a polynomial such that..." side) looks harder (though I have no idea beyond intuition whether it really is!) than the mortality problem mentioned above. If that is actually true, then it would at least suggest that $AB = BA$ does not guarantee the existance of a solution (though it might still happen).
In any case, the fact that the mortality problem is decidable for $2 \times 2$ matrices at least shows that the complexity of such problems increases rapidly with dimension, which could explain why your result for $2$ does not easily extend to higher dimensions.
Apologies for the vagueness of all this, I just figured it might give someone with more experience in the field a different direction to think about the problem. If someone does want to look that way, this paper has the mentioned results as well as references to related literature.
-
Thanks for an answer. I'm also not an expert on this, and I doubt if I can understand any of the comments above or your linked material. Nonetheless, your sincere answer will help me when I'm prepared. – sos440 Mar 25 '12 at 1:45
Your question is very interesting, unfortunately that's not a complete answer, and in fact not an answer at all, or rather a negative answer.
You might think, as generalization of $A^2+B^2=2AB$, of the following matrix equation in $\mathcal M_n\Bbb C$ : $$(E) :\ \sum_{l=0}^k (-1)^k \binom kl A^{k-l}B^l = 0.$$
This equation implies the commutativity if and only if $n=1$ of $(n,k)=(2,2)$, which is the case you studied.However, the equation (E) has a remarkable property : if $A$ and $B$ satisfy (E) then their characteristic polynomials are equal. Isn't it amazing ? You can have a look at this paper for a proof.
-
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.951545238494873, "perplexity_flag": "head"}
|
http://mathoverflow.net/questions/14054?sort=newest
|
## Where can I find a comprehensive list of equations for small genus modular curves?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Does there exist anywhere a comprehensive list of small genus modular curves $X_G$, for G a subgroup of GL(2,Z/(n))\$? (say genus <= 2), together with equations? I'm particularly interested in genus one cases, and moreso in split/non-split cartan, with or without normalizers.
Ken Mcmurdy has a list here for $X_0(N)$, and Burcu Baran writes down equations for all $X_{ns}^+(p)$ of genus <=2 in this preprint.
-
## 5 Answers
No, there does not exist a comprehensive list of equations: the known equations are spread out over several papers, and some people (e.g. Noam Elkies, John Voight; and even me) know equations which have not been published anywhere.
When I have more time, I will give bibliographic data for some of the papers which give lists of some of these equations. Some names of the relevant authors: Ogg, Elkies, Gonzalez, Reichert.
In my opinion, it would be a very worthy service to the number theory community to create an electronic source for information on modular curves (including Shimura curves) of low genus, including genus formulas, gonality, automorphism groups, explicit defining equations...In my absolutely expert opinion (that is, I make and use such computations in my own work, but am not an especially good computational number theorist: i.e., even I can do these calculations, so I know they're not so hard), this is a doable and even rather modest project compared to some related things that are already out there, e.g. William Stein's modular forms databases and John Voight's quaternion algebra packages.
It is possible that it is a little too easy for our own good, i.e., there is the sense that you should just do it yourself. But I think that by current standards of what should be communal mathematical knowledge, this is a big waste of a lot of people's time. E.g., by coincidence I just spoke to one of my students, J. Stankewicz, who has spent some time implementing software to enumerate all full Atkin-Lehner quotients of semistable Shimura curves (over Q) with bounded genus. I assigned him this little project on the grounds that it would be nice to have such information, and I think he's learned something from it, but the truth is that there are people who probably already have code to do exactly this and I sort of regret that he's spent so much time reinventing this particular wheel. (Yes, he reads MO, and yes, this is sort of an apology on my behalf.)
Maybe this is a good topic for the coming SAGE days at MSRI?
Addendum: Some references:
Kurihara, Akira On some examples of equations defining Shimura curves and the Mumford uniformization. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 25 (1979), no. 3, 277--300.
$\$
Reichert, Markus A. Explicit determination of nontrivial torsion structures of elliptic curves over quadratic number fields. Math. Comp. 46 (1986), no. 174, 637--658.
http://www.math.uga.edu/~pete/Reichert86.pdf
$\$
Gonzàlez Rovira, Josep Equations of hyperelliptic modular curves. Ann. Inst. Fourier (Grenoble) 41 (1991), no. 4, 779--795.
http://www.math.uga.edu/~pete/Gonzalez.pdf
$\$
Noam Elkies, equations for some hyperelliptic modular curves, early 1990's. [So far as I know, these have never been made publicly available, but if you want to know an equation of a modular curve, try emailing Noam Elkies!]
$\$
Elkies, Noam D. Shimura curve computations. Algorithmic number theory (Portland, OR, 1998), 1--47, Lecture Notes in Comput. Sci., 1423, Springer, Berlin, 1998.
http://arxiv.org/abs/math/0005160
$\$
An algorithm which was used to find explicit defining equations for $X_1(N)$, $N$ prime, can be found in
Pete L. Clark, Patrick K. Corn and the UGA VIGRE Number Theory Group, Computation On Elliptic Curves With Complex Multiplication, preprint.
http://math.uga.edu/~pete/TorsCompv6.pdf
This is just a first pass. I probably have encountered something like 10 more papers on this subject, and I wasn't familiar with some of the papers that others have mentioned.
-
Really nice answer. One question: Do any of the algorithmic methods you mention for computing defining equations work integrally, or only over Q? – Tyler Lawson Feb 4 2010 at 4:44
@TL: good question! (Sounds familiar, in fact.) Off the top of my head, I would say that the key issue is which of these algorithms work over a field of positive characteristic, and for which characteristics? E.g. "my" algorithm (a.k.a.: the most immediately obvious algorithm) for $X_1(N)$ will work verbatim over a field of characteristic not dividing $N$, hence (I'm pretty sure) over $\mathbb{Z}[\frac{1}{N}]$. Doing things correctly at primes of bad reduction will be much harder in general, I fear, although in some very special cases (e.g. genus 0!) you can work these things out. – Pete L. Clark Feb 4 2010 at 6:07
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Explicit equations for X_1(N) that have been optimized to reduce both the degree and coefficient sizes are available for N <= 50 at http://math.mit.edu/~drew/X1_curves.txt. These were obtained using the algorithm described in http://arxiv.org/abs/0811.0296.
EDIT: Tables of defining equations for X_1(N) for N <= 189 are now available at http://www-math.mit.edu/~drew/X1_altcurves.html
-
There is also the paper by Broker, Lauter and Sutherland "Modular polynomials via isogeny volcanoes" http://arxiv.org/abs/1001.0402 which gives a fast (in practice) algorithm to calculate modular polynomials $\Phi_l(X,Y)$ (this is the polynomial satisfying $\Phi_l(j(z),j(lz)) = 0$ where $j$ is the usual $j$-invariant) for $l$ prime which is a highly-singular model for $X_0(l)$, and other analogous polynomials associated, for example, with the modular function $f$ which generates a function field associated with a congruence subgroup of degree 72 over $\Gamma_0(1)$. Sutherland just spoke here yesterday on this. For example he can calculate $\Phi_l(X,Y)$ for $l$ about 20000. The interesting feature in this algorithm is that he can calculate $\Phi_l$ modulo a small prime without actually calculating it over $\mathbb{Z}$ and reducing.
In the papers by Cummins and Pauli and Yang they essentially do their calculations by using "modular units" (cf. Kubert and Lang) which are explicit functions on $X(N)$ (sometimes with character) for which we know the divisor, and then combining them in various ways and using Riemann-Roch type calculations. The method by Broker, Lauter and Sutherland uses the modular interpretation of $\Phi_l$ in terms of isogenies, in a rather clever way. I feel that eventually this will be the way to go.
-
1
+1: there is lots more good work being done in this area than I knew about. MO strikes again! – Pete L. Clark Feb 4 2010 at 19:39
Cummins and Pauli have calculated generators for the function fields of all congruence subgroups of $\text{PSL}_2(\mathbb{Z})$ of genus $\le 24$ in:
http://www.mathstat.concordia.ca/faculty/cummins/congruence/
I haven't looked at this for a few months but I believe that the companion paper http://www.emis.de/journals/EM/expmath/volumes/12/12.2/pp243_255.pdf discusses the generators. In the meantime there is a paper by Yifan Yang "Defining equations of modular curves" Advances in Mathematics Volume 204, Issue 2, 20 August 2006, Pages 481-508
which gives tables of equations for many modular curves, and discusses a methodology for finding "good" equations (i.e. those with small coefficients and a small number of terms in the defining polynomials)
-
These are very nice tables for what they contain, but I didn't see any data about defining equations. Am I missing something? (If you will permit a pedantic remark: I can tell you what the generators are for the function field of any integral algebraic curve over $\mathbb{Q}$: $x$ and $y$. It's the relation that's not so easy...) – Pete L. Clark Feb 4 2010 at 6:10
@VM: In the paper of Cummins and Pauli I don't see any data about equations or function fields (and again, please let me know if I'm missing it). On the other hand, the paper of Yang seems like a must-read for those interested in the problem. – Pete L. Clark Feb 4 2010 at 14:43
Galbraith's thesis has a bunch:
http://www.isg.rhul.ac.uk/~sdg/thesis.html
-
1
There is code in Magma packages to do ModularCurveQuotient which is $X_0(N)$ mod Atkin-Lehners, via Galbraith's thesis. The looking at it seems that you can just change ModularForms(N,2) to ModularForms(Gamma1(N),2) in the function internals and hope to work with no Atkin-Lehners. This gives a canonical embedding to $C^{g-1}$ if so. Why you want this for $g=48$ with $X_1(50)$ as 1035 quadrtics is unclear but it ran in 2 minutes. – Junkie May 1 2010 at 1:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9219403862953186, "perplexity_flag": "middle"}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.