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http://physics.stackexchange.com/questions/46277/general-question-on-aligning-a-quantization-axis?answertab=oldest
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# General question on aligning a quantization axis
I have a general question on how to work with quantization axis. Here is the setup:
I am looking at a single two-level atom placed at the origin $(0, 0, 0)$, which is unperturbed in the sense that no magnetic field is applied to it.
I now send in a coherent EM-field, resonant with the transition of the atom. With a right-handed coordinate system in mind, the field is linearly polarized along $y$ and propagagtes along $+x$.
Since there is no magnetic field applied, I am allowed to choose the quantization as I wish.
1. The easiest choice is to choose the $y$-axis directly such that the EM-field drives the $\pi$-transition of the atom.
2. If I had chosen the $z$-axis instead, then the EM-field would drive $\sigma$-transitions instead. Since no magnetic field is applied, they are equal to the $\pi$-transition, so we get exactly the same signal as we should.
3. Now say I had instead chosen to put my quantization axis along in the $(y, z)$-plane with an angle $A$ relative the $y$-axis. Is it valid to to decompose the quantization axis into a $\cos(A)$ part (driving the $\pi$-transition) and a $\sin(A)$ part (driving the $\sigma$-transitions)?
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http://physics.stackexchange.com/questions/26921/physical-interpretation-to-the-category-of-cfts?answertab=oldest
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# Physical interpretation to the category of CFTs
This question comes from reading Andre's question where I wandered whether that question even makes sense physically. In mathematics, VOAs form a category, does this category as a whole have a physical interpretation?
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1
I think a good question should be self-contained. This one is not. What's a VOA, what is the relation between your question and the question you're linking to? Please provide more background. – Marcin Kotowski Oct 15 '11 at 1:59
VOA=Vertex operator Algebra, which is like the chiral part of a CFT. The question I link to talks about embeddings of CFTs, which in particular would be morphisms between them. I wandered whether a "morphism" of CFT (as an inclusion would be) had a physical interpretation. – Reimundo Heluani Oct 15 '11 at 2:03
3
A naive guess would be that morphisms of VOAs would correspond physically to domain walls coupling different CFTs. – Andrew Neitzke Oct 20 '11 at 18:45
@Andy: thanks, that does sound like something a friend would say, but I suspect that in order to think this way you need to promote the category to a higher one and then these would be "1-morphisms" which in turn won't be the usual morphisms of VOA... and then again I'm getting further away from what I was hoping. – Reimundo Heluani Oct 20 '11 at 23:06
## 3 Answers
Any kind of objects, in physics or in math, is only genuinely defined when you also define what the homomorphisms between them are supposed to be, hence when you define the category that the objects form.
Notably the invertible homomorphisms (the isomorphism) encode which objects are equivalent. There may be objects that superficially look very different and neverteless have invertible morphisms between them. Without the morphisms, you would never be able to tell that these objects are in fact equivalent. In physics this important phenomenon is often referred to as "duality". See the Gannon-Höhn Database of Vertex Operator Algebras and Modular Categories for a list of nontrivial isomorphisms in the category of VOAs.
Next, the homomorphisms which behave like injections (the monomorphisms) are needed to even define what it means to have a subobject. Physically this means: a subsystem. You can't even say (precisely) "heterotic string" or "CY compactification" or the like without a notion of monomorphism of VOAs.
Then there are the monoidal structure morphisms. VOAs form a monoidal category under tensor product. So it makes sense to ask if the tensor product of a VOA with its dual has morphisms to and from the trivial VOA. If these exist in a certain way, this encodes self-duality of the VOA. The famous physical example here is the Moonshine VOA encoding a certain bosonic string compactifications. Mathematically this is a self-dual object of rank 24 and no grade 1-subspaces in the monoidal category of VOAs, and by a famous conjecture by Frenkel, Lepowsky and Meurman it is uniquely characterized by this universal property in the category of VOAs. This means that the Monster string background exists due to universal structure in the category of VOAs.
Finally, only with the category of VOAs in hand is it possible to check for equivalences to other categories and hence discover equivalences of VOAs as a whole to other structures. The famous example of such an equivalence of most fundamental importance to the physics descriped by VOAs is Huang's theorem which says that the category of VOAs is equivalent to that of holomorphic algebras over the operad $\mathcal{HS}$ of punctured holomorphic spheres. As opposed to the definition of VOAs, an $\mathcal{HS}$-algebra is manifestly a genus-0 holomorphic 2d CFT in that it is a rule that assigns a correlator to each punctured conformal sphere, such that these correlators obey the sewing law. Starting from this Huang's student Kong derived precisley the extra structure neccessary to promote a VOA to a full 2d CFT, hence to a full string theory background (see the references here). So it's the category theory of VOAs that fully informs us about their full physical meaning in the first place.
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I'm sorry Urs but I was hoping for a physical interpretation of the category structure and I what I can read of your answer is a physical interpretation of isomorphisms and to some extent of monomorphisms. I am very familiar with Huang's work, but calling that a physical interpretation is some leap, most people working on CFTs won't know what an operad is, and remember that CFTs are also used in condensed matter physics, it is not only a stringy thing. More than the monoidal structure (composite systems) I am interested in basic things like associativity for composition of morphisms and such. – Reimundo Heluani Oct 15 '11 at 21:07
Any kind of object only has an interpretation with a given definition of homomorphism. The very fact that the definition of a VOA has something to do with CFT rests in the morphisms. For instance, if we redefine the morphisms between VOAs to be linear maps of the underlying vector spaces, then VOAs would become equivalent to just vector spaces, and all the CFT structure were lost. Or, even more drastically, if you declared that there is precisely one morphism from any VOA to any other, then that would make the theory of VOAs become equivalent to the theory of the contractible space. – Urs Schreiber Oct 17 '11 at 6:08
– Urs Schreiber Oct 17 '11 at 6:13
Generally, if you pass from just the category of VOAs to that of full CFTs, you see more exmaples of the direct physical role of homomorphisms. For instance morphisms from a trivial (n+1)-dimensional theory to a nontrivial one define "twisted" n-dimensional theories, as in arxiv.org/abs/1108.0189 . In a similar fashion the very definition of rational 2d CFT is holographically given, as discussed in arxiv.org/abs/hep-th/0612306 . – Urs Schreiber Oct 17 '11 at 6:23
My first naive guess (in line with Andy's comment but not terribly well thought out) is that no, morphisms of VOAs are not physically terribly natural - rather the more natural thing to consider is an appropriate notion of bimodule for two VOAs, which would be domain walls between the corresponding chiral CFTs (or 3d TFTs, in the rational case).
To make an analogy one dimension down, let's think that we have an associative (or if you prefer $A_\infty$) algebra, and we use it to try to define a 2d TFT --- i.e. we can always integrate it over the circle to get a vector space (take Hochschild homology or center, depending on how you think of circles) and if it's fully dualizable (f.d. semisimple in the abelian case or "smooth proper" in the dg case) we can integrate it over 2-manifolds to get numbers. However from the TFT point of view what's important is the category of modules over the algebra rather than the algebra itself -- i.e., relations between two algebras are given by bimodules (ie Morita morphisms), not necessarily by morphisms of algebras. These are exactly domain walls between 2d field theories. (This is also natural from thinking of 2d TFTs as noncommutative varieties --- only in the commutative case does it really make sense to focus on maps of algebras, since in that case you can recover the algebra from the corresponding category or TFT).
[If you want honest 2d CFTs rather than modular functors then you want a modification of the story above..]
Likewise I think (following Costello and Lurie) of a VOA as what you attach to a point in a 2d chiral CFT or modular functor (ie we're attaching vector spaces to Riemann surfaces, obtained by integrating the VOA over the surface --- conformal blocks, aka chiral homology). The coarse topological analog is an E_2 algebra. In any case what seems physically meaningful is domain walls between these modular functors (or the corresponding 3d TFT if it makes sense), and these are "chiral bimodules" for two vertex algebras: something you can put on a wall, so that on either side you have bulk operators given by your two VOAs (and in particular there's also a "boundary OPE" structure on this chiral bimodule --- topologically this would be an associative algebra object in bimodules over two E_2 algebras).
Anyway to summarize usual morphisms of vertex algebras are special cases of something more natural, which are domain walls of chiral CFTs, which give monoidal functors between the monoidal categories of left modules ("boundary conditions") for the two VOAs..
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The category of CFT's (and related 3D TQFT) has been studied by Kapustin and Saulinas in their recent paper Topological boundary conditions in abelian Chern-Simons theory.
The quote "we obtain a classification of such theories up to isomorphism" from their abstract refers to a notion of isomorphism that is clearly higher-categorical (more precisely, it appears to be 3-categorical).
See also Surface operators in 3d Topological Field Theory and 2d Rational Conformal Field Theory by the same authors.
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Thanks André and David. Actually I was thinking on you by the "friend" in my comment above and was hoping that the answer would not have to involve higher categories. I'll accept BZ's answer just because it is an hour older and David may need the reputation to comment. – Reimundo Heluani Nov 3 '11 at 21:24
Thanks Reimundo - the inability to comment was incredibly frustrating! – David Ben-Zvi Nov 3 '11 at 23:21
If I understand correctly, Kapustin-Saulina are talking about a (higher) category of CFTs, not of VOAs -- in other words I think they're dealing implicitly with chiral bimodules for VOAs rather than morphisms. (In other words the natural notion of map of field theories doesn't correspond to a map of VOAs). – David Ben-Zvi Nov 3 '11 at 23:25
The CFT's in Kapustin-Saulina are not chiral (they are when the torus has definite signature), and the sentence "the natural notion of map of field theories doesn't correspond to a map of VOAs" is definitely correct. – André Nov 4 '11 at 8:30
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http://mathhelpforum.com/differential-geometry/112574-summation-print.html
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# Summation
Printable View
• November 5th 2009, 06:17 AM
Talamari
Summation
Thanks for taking the time to look at this, it's going to be one of those problems where I'm either missing something really obvious, or there's no simple way to approach it, so I thought I'd post it up on see if anyone has any ideas:
From this summation:
$\sum^{b}_{x=1} f(x)g(x)$
I want to isolate $\sum^{b}_{x=1} f(x)$, since I have a substitution for it.
My question is then, is there a simple way to do this? The only way I can see is to sum each part in turn, multiply the 2 together, and then manually subtract the unwanted terms this also spews forth.
Any help or thoughts would be appreciated, thanks again.
• November 5th 2009, 08:37 AM
HallsofIvy
Quote:
Originally Posted by Talamari
Thanks for taking the time to look at this, it's going to be one of those problems where I'm either missing something really obvious, or there's no simple way to approach it, so I thought I'd post it up on see if anyone has any ideas:
From this summation:
$\sum^{b}_{x=1} f(x)g(x)$
I want to isolate $\sum^{b}_{x=1} f(x)$, since I have a substitution for it.
My question is then, is there a simple way to do this? The only way I can see is to sum each part in turn, multiply the 2 together, and then manually subtract the unwanted terms this also spews forth.
Any help or thoughts would be appreciated, thanks again.
Excuse me, but what do you mean by $\sum^{b}_{x=1} f(x)g(x)$? Are f and g defined over some countable set? That's the only case in which a sum makes sense. In any case, I don't see any way to separate f and g.
All times are GMT -8. The time now is 04:51 AM.
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http://mathhelpforum.com/advanced-algebra/129452-defining-isomorphism.html
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# Thread:
1. ## Defining an isomorphism....
Prove that the field R of real numbers is isomorphic to the ring of all 2x2 matrics of the form (0 0; 0 a) with a a real number.
It says to let f(a)=(0 0; 0 a), but I don't know how to show that this is one to one and onto.
Thanks.
2. Originally Posted by zhupolongjoe
Prove that the field R of real numbers is isomorphic to the ring of all 2x2 matrics of the form (0 0; 0 a) with a a real number.
It says to let f(a)=(0 0; 0 a), but I don't know how to show that this is one to one and onto.
Thanks.
Are you sure you don't mean matrices of the form
$\left( \begin{array}{cc}<br /> a & 0 \\<br /> 0 & a \end{array}<br /> \right)$?
As the set you gave has no identity.
Think about it for a bit - what does surjective mean? take an arbitrary matrix of this form. Is this mapped onto by something? What if two matrices were mapped to by the same element, what would this mean with respect to the elements which were mapping to it?
Are you confident with the definitons of surjectivity and injectivity?
3. No, the book says matrices of the form (0 0; 0 a). Doesn't this have identity
(0 0; 0 1)? Identity doesn't have to be (1 0; 0 1)
4. Originally Posted by zhupolongjoe
No, the book says matrices of the form (0 0; 0 a). Doesn't this have identity
(0 0; 0 1)? Identity doesn't have to be (1 0; 0 1)
Yes, you're right. One day I will learn to think before I type.
5. When are two matrices equal? I guess answering this question will instantaneously make your map become one-to-one and onto.
6. Okay, I think I got it actually. Maybe I was just making this problem harder than it really is.
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http://mathoverflow.net/questions/21565?sort=oldest
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## Is it true that the only interesting topologies are metric topologies and weak topologies?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In "Infinite dimensional analysis, A hitchhikers guide" by Aliprantis and Border, they write that these 2 classes of topologies "by and large include everything of interest".
@Pete Clarke: I was asking if it holds for mathematics in general, and not just for functional analysis. From the answers I get the impression that their statement is a fairly good first approximation.
@Gerald Edgar: I thought I read in Mathoverflow that the Zariski topology can be regarded as a weak topology (at least in some cases).
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Probably MO is the wrong board for this question... posters here are fond of Zariski topology and such things. But probably Aliprantis & Border mean these are everything of interest for infinite-dimensional analysis. But even there they are wrong if you want to do Schwartz ditributions. – Gerald Edgar Apr 16 2010 at 13:20
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I'd just like to note that I voted for Gerald's comment because of the middle sentence, with the implication that the question would be better if refined a little to be clear what context it is in (for those of us who sadly lack any knowledge of said book). Having read the comment again, it could be read as "This looks like a functional analysis question and MO is full of algebraic geometers so you're probably in the wrong place.". I'm sure that that's not what Gerald meant and I'd like to be sure that everyone knows that MO is a great place for functional analysis! – Andrew Stacey Apr 16 2010 at 19:58
1
I think it is worth mentioning that the book by Aliprantis and Border is a book about the functional analysis and measure theory used in mathematical economics. – Michael Greinecker Jan 28 2011 at 7:16
## 5 Answers
Picking up on Gerald's interpretation of the question (namely, that it really focusses on infinite dimensional vector spaces) then I say: absolutely not!
For example, piecewise-smooth paths in some Euclidean space has a topology that is neither of these (it's an uncountable inductive limit of Frechet spaces). (Not that I particularly recommend this space!)
Close to Gerald's comment, "dual-Frechet" spaces (that is, the dual of a Frechet space with the strong topology) have very nice properties, almost as nice as Frechet spaces themselves. This class includes distributions with the "right" topology.
And that's the point, really. If you're only interested in, say, distributions for what they can say about compactly supported functions, then the weak topology is probably fine. However, if you are interested in distributions in their own right then the weak topology is very unlikely to be okay.
Here's an example from my research: I like infinite dimensional manifolds, and I quite like loop spaces. To construct the Dirac operator on a loop space, I needed to put an inner product on the cotangent bundle. So I needed, in effect, a continuous injective map $(L\mathbb{R}^n)^* \to H$ ($H$ being some standard Hilbert space). I can't do this with the weak topology any continuous map from $(L\mathbb{R}^n)^*$ with the weak topology to a normed vector space has to factor through a finite dimensional space. With the strong topology, though, it was no problem.
So, deal with metric and weak topologies if you like; but real analysts use the strong topology[1].
[1] Not sure what complex analysts use.
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5
+1 for the so-bad-its-good pun :D – Johannes Hahn Apr 16 2010 at 17:08
Thanks! Incidentally, if anyone can think of a better footnote, I'd be happy to edit it (with credits, but I don't think that I can split the reputation). – Andrew Stacey Apr 16 2010 at 20:00
### 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.
Every topological space X has the initial topology (or weak topology) with respect to the family of continuous functions from X to the Sierpiński space. (see http://en.wikipedia.org/wiki/Initial_topology.)
This is the two point space {a,b} with open sets: emptyset, {a} and {a,b} only. If U is any subset of X, then the function fU, mapping points in U to a and the rest to b, will be continuous if and only if U is open.
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It's better than that. The initial topology so generated is in fact isomorphic (as a lattice) to the set of continuous maps from $X$ to the Sierpinski space $\Sigma$, i.e., $O(X) \cong C(X,\Sigma)$. This is something every student of topology should be aware of. – Andrej Bauer Apr 18 2010 at 11:45
Andrej, yes, and thanks for the remark. Every function from X into Sigma is f_U for some U, and as I explained, U is open if and only if f_U is continuous. So the isomorphism is the association of U with f_U. But one thing I find incongruous in your suggestion is that although we may regard C(X,Sigma) as a lattice by imposing an order on Sigma (say, with a less than b), we have not put the order topology on Sigma. – Joel David Hamkins Apr 18 2010 at 20:13
Frechet spaces, limits of Frechet spaces were mentioned before. I'd like to emphasize a particularly important example (which was also mentioned before, but I want to extend it a bit): The space of test functions $\mathcal{D}(\Omega)$ is a strong inductive limit of Frechet spaces, neither metrizable nor do they carry the weak topology. These spaces are the foundation of distribution theory and therefore most important.
on the other hand the other two usual spaces of test function $\mathcal{E}(\Omega)$ and the schwartz space $\mathcal{S}(\mathbb{R}^n)$ are metrizable because they can be topologized by a countable family of (semi)norms.
The distribution space $\mathcal{D}'(\Omega), \mathcal{E}'(\Omega), \mathcal{S}'(\mathbb{R}^n)$ can be endowed with the weak topology (that is the pointwise convergence). But the strong topology is also common and this is again neither weak nor metrizable.
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Spaces $\mathcal{D}'(\Omega), \mathcal{E}'(\Omega), \mathcal{S}'(\mathbb{R}^n)$ the strong topologies are neigher weak nor metrizable. And yet textbooks commonly check things only with sequences in these spaces (rather than nets or filters). And commonly get it right. Interesting... – Gerald Edgar Apr 18 2010 at 11:52
Maybe the strong duals are sequential spaces? – Johannes Hahn Apr 18 2010 at 16:34
@ Gerald Edgar. The fact that using sequences works for the latter two spaces is not happenstance. It is because they are inductive limits of sequences of Banach spaces linked by compact mappings. This class of spaces was investigated by J. Sebastião e Silva (they are now called Silva spaces). A similar remark applies to the first space but this is more delicate---they are projective limits of sequences of Silva spaces, but of a special kind, namely with partitions of unity (in the functional analytic sense which is related, of course, to the classical version). – jbc Sep 18 at 5:28
So the meaning of "weak topology" is that which Joel defines. This was actually a new idea to me, but it seems very general (more general that what I would have called the "weak topology" in Functional Analysis). As Joel points out, interpreted in maximal generality, any topological space carries the weak topology in this sense.
The book actually very quickly specialises to weak topologies generated by continuous, real-valued functions. Then your space has to be completely regular. I think it would be fair to say that this does rule out some interesting examples.
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Of course, any question of the form "Is it the case that the things considered interesting by such-and-such small set of people are all the interesting things?" is going to have a negative answer. Worse, taking statements such as yours too literally and outside of appropriate context will just cause you to be closed-minded and unable to understand other people's ideas.
Even if you are interested only in functional analysis you will quickly hit against topologies which are not included among your topologies. For example, given an infinite-dimensional Hilbert space $H$, what is a good topology to put on its lattice of closed subspaces? (This topology won't even be Hausdorff.)
For a second example, consider how we reconstruct a compact Hausdorff space $X$ from its $C^{*}$-algebra $A = C(X,\mathbb{C})$ of continuous complex maps. As the points of the reconstructed space we take the maximal ideals in $A$, and declare that the closed sets are the ideals of $A$ (please correct me if I am getting closed/open and ideal/filter wrong here, I always do). Even though at the end of the day this topology turns out to be metrizable, that is entirely orthogonal to how the topology is defined and thought of.
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I certainly don't want to be closed minded. In contrast to metric spaces I find topological spaces extremely hard to understand, I can see the trees but I can't see the forest; I am seeing the definitions and the theorems but I am not seeing the ideas that are running underneath them. Perhaps the reason for this is that I don't have enough experience with non-metrizable topologies. I was looking for a heuristic that was more precise than "topological spaces generalize metric spaces". – teil Apr 18 2010 at 12:39
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http://mathhelpforum.com/differential-geometry/27506-1-manifolds.html
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# Thread:
1. ## 1-manifolds
Hi to all!
I'm a new member, my name is Riccardo.
I've a problem with 1-manifolds, i have to prove a proposition but i can't do it.
Here is the extract of the book with notation and the proposition 1 that i have to prove:
//
DEFINITION. A 1-manifold is a second countable Hausdorff topological
space X such that X can be covered by open sets each of which is
homeomorphic either to the open interval (0,1) or the half-open
interval [0, 1). Sets of the first type will be called O-sets, of the second
type H-sets, of either type I-sets, and the corresponding homeomorphsms
to these intervals will be called O-charts,H-charts, and I-charts, respectively.
If X can be covered by O-sets it is a manifold without boundary, otherwise it
is a manifold with boundary.
From here on U and V will stand for I-sets in a 1-manifold and f and g will be
associated I-charts.
LEMMA. Suppose U∩V (set theoretic intersection) and U - V are
nonempty and let (x_n) be a sequence in U n V converging
to x in U - V (set theoretic difference). Then the sequence g(x_n)
has no limit point in g(V).
We say that U and V ouerlap if U∩V, U - V and V - U are nonempty.
DEFINITION. An open subinterval of (0,1) is lower if it is of the form (0, b) and
upper if it is of the form (a, 1). A subinterval which is either upper or lower is
called outer. It is easy to see that an open interval in (0,1) is outer if and only if it
contains a sequence with no limit point in (0,1). Similarly, in [0, 1), a subinterval is
called upper and outer if it is of the form (a, 1). (There are, by definition, no lower
open subintervals of [0, 1).) An open subinterval of [O,1) is outer if and only if it
contains a sequence with no limit point in [O,1).
PROPOSITION1. If U and V overlap and W is a component of U∩V, then f(W)
and g(W) are outer intervals.
Hint: Note that f(W) is a proper subinterval of f(U). Using the lemma show
that f(W) is an open interval. Then construct an appropriate sequence in f(W)
and use the lemma again.
//
Thank you in advance
2. Suppose $f|_{W}(x_n)\subset (0,1)$ has a limit point $0<b<1$. Use the lemma on the sets $U'=U\cap V-W, \ V'=W$ and the sequence $f^{-1}\circ (f|_{W})(x_n)$.
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http://mathoverflow.net/questions/32133?sort=oldest
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## Expressing adj(A) as a polynomial in A?
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Suppose $A\in R^{n\times n}$, where $R$ is a commutative ring. Let $p_i \in R$ be the coefficients of the characteristic polynomial of $A$: $\mathop{\mathrm{det}}(A-xI) = p_0 + p_1x + \dots + p_n x^n$.
I am looking for a proof that: $-\mathop{\mathrm{adj}}(A) = p_1 I + p_2 A + \dots + p_n A^{n-1}$.
In the case where $\mathop{\mathrm{det}}(A)$ is a unit, $A$ is invertible, and the proof follows from the Cayley-Hamilton theorem. But what about the case where $A$ is not invertible?
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This is a polynomial identity in n^2 variables A_{ij} with integer coefficients, so it holds over any commutative ring R if and only if it holds over a dense subset of C, so the invertible case is already enough. – Qiaochu Yuan Jul 16 2010 at 8:47
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In my humble opinion Qiaochu Yuan's answer is more convincing than Martin's. All you need to know (I think) is that $\mathbb Z[X_1,...,X_k]$ is a domain. I don't think things like the existence of algebraic closure, the density of diagonal matrices, Zariski's topology, ..., are necessary (or even helpful) in this context. I don't even think Qiaochu Yuan had to invoke complex numbers. I believe the statement is truly elementary, and can be (easily) proved by considering only countable sets. – Pierre-Yves Gaillard Jul 16 2010 at 11:40
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Pierre-Yves, I agree! See mathoverflow.net/questions/29271/…, where this was discussed $\textit{ad nausem}$ and Martin finally made a concession that it's not the most natural approach, but it's awesome (to him). – Victor Protsak Jul 17 2010 at 5:46
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@Qiaochu: How can your intuition be "model theoretic" (I assume it means "true over an arbitrary commutative ring") and be "about diagonalizable matrices and their eigenvalues" (which makes sense only when the ground ring is a field) $\textit{at the same time}?$ For matrix identities (and polynomial identities more generally), the right kind of intuition does $\textit{not}$ come from eigenvalues or complex numbers. See Bill's remark at the end of his answer and my comment. – Victor Protsak Jul 18 2010 at 9:04
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What I meant is that there are these results of the form "if a first-order sentence is true in X, it must also be true in Y, Z, W...." and my intuition about proving a polynomial identity (with coefficients in Z) by proving it for C^n is that it is a statement of this sort. Maybe this is not a good way to think about things; in any case I appreciate the clarification. – Qiaochu Yuan Jul 18 2010 at 16:24
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## 6 Answers
Here is a direct proof along the lines of the standard proof of the Cayley–Hamilton theorem. [This works universally, i.e. over the commutative ring $R=\mathbb{Z}[a_{ij}]$ generated by the entries of a generic matrix $A$.]
The following lemma combining Abel's summation and Bezout's polynomial remainder theorem is immediate.
Lemma Let $A(\lambda)$ and $B(\lambda)$ be matrix polynomials over a (noncommutative) ring $S.$ Then $A(\lambda)B(\lambda)-A(0)B(0)=\lambda q(\lambda)$ for a polynomial $q(\lambda)\in S[\lambda]$ that can be expressed as
$$q(\lambda)=A(\lambda)\frac{B(\lambda)-B(0)}{\lambda}+\frac{A(\lambda)-A(0)}{\lambda}B(0)=A(\lambda)b(\lambda)+a(\lambda)B(0) \qquad (*)$$
with $a(\lambda),b(\lambda)\in S[\lambda].$
Let `$A(\lambda)=A-\lambda I_n$ and $B(\lambda)=\operatorname{adj} A(\lambda)$` [viewed as elements of $S[\lambda]$ with $S=M_n(R)$], then
$$A(\lambda)B(\lambda)=\det A(\lambda)=p_A(\lambda)=p_0+p_1\lambda+\ldots+p_n\lambda^n$$ is the characteristic polynomial of $A$ and $$A(0)B(0)=p_0 \text{ and } q(\lambda)=p_1+\ldots+p_n\lambda^{n-1}$$
Applying $(*),$ we get
$$q(\lambda)=(A-\lambda I)b(\lambda)-\operatorname{adj} A \qquad (**)$$
for some matrix polynomial $b(\lambda)$ commuting with $A.$ Specializing $\lambda$ to $A$ in $(**),$ we conclude that
$$q(A)=-\operatorname{adj} A\qquad \square$$
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Looks like "hints" are more popular than full proofs. Why bother doing mathematics honestly if you can blog and handwave? – Victor Protsak Jul 18 2010 at 8:12
Dear Victor, one of your lines is too long for my browser, which makes your answer almost unreadable. I was wondering if you couldn't break this line. – Pierre-Yves Gaillard Jul 18 2010 at 8:22
Dear Pierre-Yves: Thank you for letting me know. I have shortened the display (*) and broken several other lines by displaying the formulas. If you continue to experience difficulties, please, state specifically which lines cause them. – Victor Protsak Jul 18 2010 at 8:56
Dear Victor: It's much better! But Line (*) is still a tiny bit too long for my browser. If I were you, I'd break it just before the second equal sign. (But it's a detail.) Thank you very much! – Pierre-Yves Gaillard Jul 18 2010 at 9:37
It's perfect now!!! – Pierre-Yves Gaillard Jul 18 2010 at 10:54
### 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.
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have polynomial form in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a domain, enjoys cancelation of elts $\ne 0$. Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint. As a result, many students cannot easily resist the obvious topological temptations and instead derive hairier proofs employing density arguments (e.g see elswhere in this thread).
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$
$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$
Now, for our pièce de résistance of topology, we derive the polynomial derivative purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$ and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition since it is linear and it takes the same value on the basis monomials $\rm x^n$. Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$$\rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$
$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg)$
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$, i.e. specialize the difference "quotient" from the product rule for differences. Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\;\rm det \:A\;$ in all of the examples given above.
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Yes, this is probably what Quiaochu had in mind. A couple of points: (1) There is some work left in showing that Cayley-Hamilton holds universally, which some books fail to emphasize (hence many people are unaware how to prove it, in line with your last comment); and (2) The usual proof of CH relies on factorization $\det(A-\lambda)=(A-\lambda)\operatorname{adj}(A-\lambda)$ and this is needed again in the "black box" proof you've presented in the form $\det A=A\operatorname{adj} A$ (hence after unwinding the argument, it ends up being applied twice). That's why I prefer a direct argument. – Victor Protsak Jul 18 2010 at 8:30
@Victor: but Quiaochu's hint employs a density argument, not polynomial universality - which is the essence of the matter in my approach. Yes, I presume Cayley-Hamilton for commutative rings but this is so widely known it is even in Wikipedia, besides Jacobson BA1, etc. (not to mention Nakayama inspired generalizations, e.g. Atiyah & Macdonald) – Bill Dubuque Jul 18 2010 at 15:16
"Density argument" is ambiguous: I interpreted "the invertible case is already enough" in the first comment as "$\{A:\det A\ne 0\}$ is a Zariski dense subset of the affine space of $n\times n$ matrices" (over $C$, as it were), which is exactly your "working generically allows us to cancel $d$". As you can see, I am not happy with handwaving in such matters: I often found that it masks incomplete understanding (cf the link I gave above to discussion of what constitutes a proof of CH itself). This is typically manifested in starting an answer with "Hint" :) – Victor Protsak Jul 18 2010 at 16:04
Here is a question that you might answer. As you say, CH is widely known, but what about the OP's identity? I have a feeling that this is something that once upon a time was equally standard (early 20th century, when Kronecker-Weierstrass-Frobenius methods were widely employed), but I can't remember seeing it stated explicitly in any recent texts. Do you know the history of this identity? When did this transition from "polynomial universality" to "eigenvalues and density" approach (also evident in the current treatment of Jordan normal form without developing elementary divisors) occur? – Victor Protsak Jul 18 2010 at 16:12
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@Victor: to ensure there is no confusion, I remark that my use of "generic" above is not intended to denote anything topological or geometrical. Rather, it is meant to be understood as exploiting the universality of a free objects. The proof I gave does not require any knowledge of topology or (algebraic) geometry. I'm not saying that imposing other such viewpoints isn't interesting or useful - just that such is not required for problems of this sort. Further doing so adds complexity to what is - at the heart - trivial (yet elegant) algebra. – Bill Dubuque Jul 18 2010 at 22:36
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I guess it is worth giving a fuller answer, and then Victor can tell me more precisely where I am missing some subtlety. As I said, the definition I know of the adjugate is that it is a matrix whose entries are polynomials in the entries $a_{ij}$ of $A$ and which satisfies $A \text{ adj}(A) = I \det A$ identically, e.g. over $\mathbb{Z}[a_{ij}]$. Assuming Cayley-Hamilton, we know that $p_0 I + p_1 A + ... + p_n A^n = 0$ identically and that $p_0 = \det A$, where $p_k \in \mathbb{Z}[a_{ij}]$ as well.
Specializing now to $a_{ij} \in \mathbb{C}$ and supposing that $A$ is invertible, we conclude that
$$A \text{ adj}(A) = - p_1 A - p_2 A^2 - ... - p_n A^n$$
implies
$$\text{adj}(A) = - p_1 I - p_2 A - ... - p_n A^{n-1},$$
as you say.
Lemma: The invertible $n \times n$ matrices are dense in the $n \times n$ matrices with the operator norm topology.
Proof. Let $A$ be a non-invertible $n \times n$ matrix, hence $\det A = 0$. The polynomial $\det(A - xI)$ has leading term $(-1)^n x^n$, hence cannot be identically zero, so in any neighborhood of $A$ there exists $x$ such that $A - xI$ is invertible.
But everything in sight is continuous in the operator norm topology, so the conclusion follows identically over $\mathbb{C}$ and hence identically.
(I should mention that this is not even my preferred method of proving matrix identities. Whenever possible, I try to prove them combinatorially by interpreting $A$ as the adjacency matrix of some graph. For example - confession time! - this is how I think about Cayley-Hamilton. This is far from the cleanest or the shortest way to do things, but my combinatorial intuition is better than my algebraic intuition and I think it's good to have as many different proofs of the basics as possible.)
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Sure, this is a nice proof. Depending upon taste, one might want to replace the "analytic" topology with the Zariski topology. The merit of this is that the Zariski density of the invertible matrices is even easier to establish: because $\mathbb{C}[x_1,\ldots,x_{n^2}]$ is a domain, its spectrum is an irreducible topological space, so every nonempty open subset -- like the set of all matrices with nonzero determinant -- is Z-dense. – Pete L. Clark Jul 18 2010 at 19:11
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Qiaochu: Multiply your first displayed line by $\text{adj}(A),$ get $$\det A\ \text{adj}(A) = \det A(- p_1 I - p_2 A - \ldots - p_n A^{n-1})$$ and universally cancel $\det A$ per Bill's answer. The rest (specialization and density) is crud. – Victor Protsak Jul 19 2010 at 0:14
As in it doesn't work or as in you don't like it? In any case, maybe I should present this style of argument not as "the proof," but as a way to check such results. Many matrix identities are easier to verify for A with a special form, and I think it's good to know that there exist theorems saying that this is enough. Whether one chooses to use them is a matter of taste, I think. – Qiaochu Yuan Jul 19 2010 at 0:39
@Q The equation in Victor's comment is indeed the equation that I started with, viz. $\rm d B = d C$ for $\rm d = det A$. – Bill Dubuque Jul 19 2010 at 1:38
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Also, I forgot to mention: there is a natural (i.e. not relying on tricks and special cases) enumerative combinatorics proof of CH in Stanton-White's book. – Victor Protsak Jul 19 2010 at 2:38
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As an arithmetic geometer, I have no choice but to use topological methods hand in hand with algebraic methods. Very likely necessity has been the mother of aesthetics here, but I find proofs of linear algebra facts using genericity arguments to be beautiful and insightful.
Qiaochu has shown how to answer the OP's question using these methods [he uses the "analytic" -- i.e., usual -- topology on $\mathbb{C}^n$, but close enough] assuming the Cayley-Hamilton theorem. Here I want to show that one can also prove the Cayley-Hamilton theorem quickly by these methods.
Step 1: To prove C-H as a polynomial identity, it is enough to prove that it holds for all $n \times n$ matrices over $\mathbb{C}$.
Proof: Indeed, to say C-H holds as a polynomial identity means that it holds for the generic matrix `$A = \{a_{ij}\}_{1 \leq i,j \leq n}$` whose entries are independent indeterminates over the ring $R = \mathbb{Z}[a_{ij}]$. But this ring embeds into $\mathbb{C}$ -- indeed into any field of characteristic zero and infinite absolute transcendence degree -- and two polynomials with coefficients in a domain $R$ are equal iff they are equal in some extension domain $S$.
Step 2: C-H is easy to prove for complex matrices $A$ with $n$ distinct eigenvalues $\lambda_1,\ldots,\lambda_n$.
Proof: The characteristic polynomial evaluated at $A$ is $P(A) = \prod_{i=1}^n(A-\lambda_i I_n)$. Let $e_1,\ldots,e_n$ be a basis of $\mathbb{C}^n$ such that each $e_i$ is an eigenvector for $A$ with eigenvalue $\lambda_i$. Then -- using the fact that the matrices $A - \lambda_i I_n$ all commute with each other -- we have that for all $e_i$,
$P(A)e_i = \left(\prod_{j \neq i} (A-\lambda_j I_n)\right) (A-\lambda_i I_n) e_i = 0.$
Since $P(A)$ kills each basis element, it is in fact identically zero.
Step 3: The set of complex matrices with $n$ distinct eigenvalues is a Zariski-open subset of $\mathbb{C}^n$: indeed this is the locus of nonvanishing of the discriminant of the characteristic polynomial. Since we can write down diagonal matrices with distinct entries, it is certainly nonempty. Therefore it is Zariski dense, and any polynomial identity which holds on a Zariski dense subset of $\mathbb{C}^{n^2}$ holds on all of $\mathbb{C}^{n^2}$.
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Right. To expand on my answer, let me tell you how I prove C-H when nobody's looking. First we consider the special case when A is the adjacency matrix of the graph on {1, 2, ... n} with one edge from n to n-1 to... to 1 and c_k edges from 1 to k. The entries in powers of A count walks on this graph, which count tilings of a line where there are c_k types of tiles of size k (subject to some conditions depending on where the walk starts and ends). There is a bijective proof that A^n = c_1 A^{n-1} + c_2 A^{n-2} ... : the bijection is "delete the first tile"... – Qiaochu Yuan Jul 18 2010 at 20:24
(cont.) which means, look at the first place the walk visits its starting vertex again, and delete that loop. On the other hand, it's not hard to show that x^n - c_1 x^{n-1} - ... is the characteristic polynomial of A. This proves C-H for all companion matrices (the combinatorial argument is "weighted," e.g. the c_k are variables) over C, hence for all diagonalizable matrices with distinct eigenvalues by conjugation, and hence for all matrices by whichever density argument you prefer. – Qiaochu Yuan Jul 18 2010 at 20:25
Pete, so if I understand you correctly, you prefer Weil-van der Waerden style "generic point" (over a universal field) to the Grothendieck generic point (i.e. zero ideal of $R$)? There are good reasons to prove identities such as CH directly, the comparative ease being one, but let me mention another: the usual proof $$p_A(\lambda)=(A-\lambda)\operatorname{adj}(A-\lambda), \lambda\to A\implies p_A(A)=0\ \square$$ immediately leads to a proof Nakayama's lemma, whereas I see only (technical) pain and no (conceptual) gain in "reducing" it to tricks with geometric points based on eigenvalues. – Victor Protsak Jul 18 2010 at 23:52
@VP, No, I'm not expressing any kind of systematic preference. This is just a proof that I like and find easy to remember -- no disparagement of other approaches is intended or felt. One point: in principle Zariski-density arguments are more general than genericity arguments. For instance, I believe that over any field $k$, the set of matrices which have distinct, $k$-rational eigenvalues is Zariski-dense, but the generic matrix does not have eigenvalues rational over its fraction field. However, proving this seems harder than reducing to the case of an alebraically closed field... – Pete L. Clark Jul 19 2010 at 0:16
...over any INFINITE field, I should have said. – Pete L. Clark Jul 19 2010 at 2:49
EDIT OF AUG. 31, 2010. The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on page 21 (proof of Proposition 2.4) of Introduction to Commutative Algebra by Atiyah and MacDonald. The argument can be phrased as follows.
Let $K$ be a commutative ring; let $n$ be a positive integer; let $A=(a_{ij})\in M_n(K)$ be an $n$ by $n$ matrix with entries in $K$; let $\chi$ be its characteristic polynomial; define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}\,A-a_{ij}$; let $(e_i)$ be the canonical basis of $K^n$; observe $$\sum_i\ \ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Applying (a trivial case of) Fubini's Theorem to the double sum $\sum_{i,j}\ c_{jk}\ b_{ij}\ e_i$, we get $\chi(A)=0$.
Thank you very much to darij grinberg! [I'm leaving the previous edits "for the record".] END OF EDIT OF AUG. 31, 2010.
EDIT OF DEC. 11, 2010. For a nice application of the Cayley-Hamilton Theorem, see this answer by Balazs Strenner.
PREVIOUS EDITS:
Here is a proof of the Cayley-Hamilton Theorem.
Let $K$ be a commutative ring, let $n$ be a positive integer, let $X$ be an indeterminate, let $A\in M_n(K)$ be an $n$ by $n$ matrix with coefficients in $K$, and let $\chi:=\det(X-A)$ be the characteristic polynomial. Equip $K^n$ with the $K[X]$-module structure induced by $A$. We must check $\chi K^n=0$. Form the right $M_n(K[X])$-module $$H:=\mathrm{Hom}_{K[X]}(K[X]^n,K^n).$$ Let $e\in H$ be the evaluation at $A$ (note $K[X]^n=K^n[X]$). As $e$ is surjective, it suffices to show $e\chi=0$. As $X-A$ divides $\chi$ on the left, it suffices to show $e(X-A)=0$. But this is obvious.
EDIT OF AUG. 1, 2010. Here is a diagrammatic rewriting of the argument.
EDIT OF AUG. 30, 2010. Here is a coordinate version of the above argument. [Compare with the proof of Propositon 3 page 81 of Weil's Basic Number Theory, and with the proof of Propositon 2.4 page 21 of Introduction to Commutative Algebra by Atiyah and MacDonald].
Weil's formulation. Put $$B(X)=(b_{ij}(X)):=X-A\in M_n(K[X]),$$ and let $C(X)=(c_{ij}(X))$ be the adjugate of $B(X)$. We have $$\sum_j\ c_{jk}(X)\ b_{ij}(X)=\delta_{ik}\ \chi(X)\in K[X].$$ Replacing $X$ with $A$, evaluating on $e_i$ (the $i$-th vector of the canonical basis of $K^n$), and summing over $i$ gives $$\sum_j\ c_{jk}(A)\ \sum_i\ b_{ij}(A)\ e_i=\chi(A)\ e_k\in K^n.$$ But the second sum is 0 by definition of $b_{ij}(X)$.
Atiyah-MacDonald's formulation. Put $A=(a_{ij})$ and define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}A-a_{ij}$; observe $$\sum_i\ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Computing $\sum_{i,j}\,c_{jk}\,b_{ij}\,e_i$ in two ways we get $\chi(A)=0$.
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Why is your $e$ in $H$? – darij grinberg Jul 20 2010 at 9:58
Dear darij grinberg: Because it maps $K[X]^n$ to $K^n$ and is $K[X]$-linear. – Pierre-Yves Gaillard Jul 20 2010 at 10:09
Why does it map $K[X]^n$ to $K^n$ ? Shouldn't it map $K[X]^n$ to $\left(\mathrm M_n\left(K\right)\right)^n$ ? – darij grinberg Jul 20 2010 at 10:09
Dear darij grinberg: It maps $X^i v$ in $K[X]^n=K^n[X]$ to $A^i v$ in $K^n$. (Here $v$ is in $K^n$.) – Pierre-Yves Gaillard Jul 20 2010 at 10:15
Ah, so that' what you mean by evaluation at $A$! – darij grinberg Jul 20 2010 at 10:17
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This formula can be obtained during a proof of the Cayley-Hamilton theorem, as is indicated on its Wikipedia article. The essence of the argument is that Euclidean division by a monic polynomial (on the left, say), can be performed in the polynomial ring over any (unitary) ring, not necessarily commutative; this follows directly from consideration of what Euclidean division does, or by a simple inductive argument.
Since I care about polynomials being monic, I'l define the characteristic polynomial of a matrix $A$ to be $\chi_A=\det(I_nX-A)=\sum_{i=0}^nc_iX^i$ where $c_n=1$ (and $c_0=\det(-A)$), so the result to prove becomes $\mathrm{adj}(-A)=c_1I_n+c_2A+\cdots+c_{n-1}A^{n-2}+A^{n-1}=\sum_{i=1}^nc_iA^{i-1}$
Consider the noncommutative ring $M=\mathrm{Mat}_n(R)$, and using Euclidean division in $M[X]$ (in which $R[X]$ is embedded by mapping $r$ to $rI_n$) divide $\chi_A$ on the left by $X-A$; the quotient and remainder will be $\mathrm{adj}(X-A)$ and $0$ (by uniqueness). Writing the quotient $\mathrm{adj}(X-A)=\sum_{i=0}^{n-1}B_iX^i$, its coefficients $B_i\in M$ are determined in the division successively as $B_{n-1}=c_n=1$ and $B_{i-1}=c_i+AB_i$ for $i=n-1,\ldots,1$, which expands to $B_{i-1}=c_iA^0+c_{i+1}A^1+\cdots+c_nA^{n-i}$. In particular the constant coefficient of the quotient is $B_0=\sum_{i=1}^nc_iA^{i-1}$, but this is also $\mathrm{adj}(-A)$ (by substituting $X=0$ into $\mathrm{adj}(X-A)$).
To retrieve the Cayley-Hamilton theorem from the formula found, multiply on the left or right by $A$ and move the left hand side to the right.
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Having written half of that article, I may be biased, but I didn't write this part and I always liked that argument. – Ryan Reich Nov 7 2011 at 14:50
Thank you. Indeed I did contribute that part (and another half;-) to that article, in 2008. I should say that this argument is quite close to the proof by Victor Protsak, the main difference being mention of Euclidean division: to me the right hand side just begs to be interpreted as a coefficient of a Euclidean quotient by $X-A$ (more generally Horner-scheme sub-expressions do). By the way, the Cayley-Hamilton theorem can be obtained more directly here by interpreting the fact that the remainder is 0. – Marc van Leeuwen Nov 8 2011 at 10:35
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http://mathoverflow.net/questions/10957/a-hands-on-description-of-a-completion-of-the-free-commutative-monoid-on-counta
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## A hands-on description of a “completion” of the free commutative monoid on countably many generators
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This is basically an I'm-weak-at-algebraic-geometry question. I asked it as a warm-up question here, but Ilya N asked me to break that post up into several questions.
Consider the free commutative monoid $X$ on countably many generators. Let $A$ be the algebra of all functions $X \to \mathbb C$. Then $A$ is a commutative and cocommutative bialgebra, and hence a commutative monoid object in $(\text{CAlg})^{\rm op}$, where $\text{CAlg}$ is the category of commutative algebras. A global point of $A \in (\text{CAlg})^{\rm op}$ is an algebra homomorphism $A \to \mathbb C$. The set of global points of $A$ is naturally a commutative monoid that contains $X$ as a submonoid.
What are all the global points of $A$? I'm pretty sure that there are more than just the points of $X$. The form a monoid, because $A$ is a bialgebra: is there a reasonable description of the monoid structure?
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You probably want the monoid to be finite if you are thinking of the bialgebra of functions on it. – Mariano Suárez-Alvarez Jan 6 2010 at 20:20
It's fine as long as each element is the sum of two others in finitely many ways, as is the case here. – Reid Barton Jan 6 2010 at 20:27
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As far as I can tell, the question has nothing to do with the coalgebra structure. You're just asking if an infinite product of fields has exceptional augmentations. – S. Carnahan♦ Jan 6 2010 at 23:18
@Scott: that's totally correct. I meant to imply "what is their monoid structure?" I'll add that. (Man, I've been making lots of edits-to-questions, which annoy me when other people do it. Why can't people ask the questions they mean to begin with ?!? :) ) – Theo Johnson-Freyd Jan 7 2010 at 0:16
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I think that (editing question) kind of works as long as it's an extension of the question. That is, the rule is: when there's already a reply, it should continue to be a valid reply after editing the question. – Ilya Nikokoshev Jan 7 2010 at 0:24
## Preliminaries
First part of your question doesn't use the bialgebra structure. That is, you have a space of functions on countable many points which I'll denote $A = \mathbb C_1\times \mathbb C_2\times\cdots \times\mathbb C_n\times\cdots$ equipped with $+$ and $\times$ pointwise. You would like to classify all algebra maps $\varphi: A\to \mathbb C$.
To start, note that idempotents must go to idempotents, of which $\mathbb C$ has only two: $0$ and $1$. Moreover, $1\times1\times\cdots \mapsto 1$ (unless the whole map is trivial).
## Classification of points
Consider characteristic functions $\lambda_n$ which take value from point labeled $n$. As $\lambda_n^2 = \lambda_n$, we must have $\varphi(\lambda_n) = 0$ or $1$. There are three cases:
• there are two indices $i, j$ such that $\varphi(\lambda_i) = \varphi(\lambda_j) = 1$. This is impossible: $0 = \varphi(\lambda_i\lambda_j) = 1$.
• there exists exactly one index $i$ for which $\varphi(\lambda_i) = 1$. This implies $\varphi(f) = \varphi((1-\lambda_i)f + \lambda_if) = f(i)$.
• all functions $\lambda_i$, and, therefore, all finite sums from $A$, lie in the kernel of $\varphi$.
The maps of the latter type are indeed the "extra" points. Note, however, that they are "wild" and can be easily killed by some extra finitness assumptions, for example the following "limit of zeroes" property: if $\varphi$ restricted to all finite subsets is 0, then $\varphi$ is 0.
## Wild maps
You can construct examples of these wild maps using the axiom of choice. To do that, make the elements of the form $a\times a \times \cdots$ and all their finite modifications to map to $a$; denote those elements $A_0$. Next, select an arbitrary $T_1\in A$ which is transcendental over $A_0$ and map it to arbitrary $t_1\in \mathbb C$. This will fix all elements that lie in algebraic closure $A_1 = A_0(T_1)$. Do that again for $T_2$ and repeat until you have nothing left. At each step you're producing a correct map $A_n \to \mathbb C$, so you get the final map $A\to \mathbb C$ as the limit of those.
Conversely, any wild map can be constructed by the above process, assuming all necessary set-theoretic things. So, the answer is, the wild points are classified by maps from this terribly non-constructive sequence of $T_i$ (of cardinality the same as $A$, that is, continuum) to $\mathbb C$. Those are again in cardinality of continuum.
## Monoid structure
One now is reminded that $A$ came with a natural basis enumerated by numbers $(n_1, n_2, \dots, n_k, \dots) \in \mathbb Z^+\oplus\mathbb Z^+\oplus\cdots\oplus\mathbb Z^+\oplus\cdots$ (which is isomorphic to $\mathbb Z_{>0}^\times$). Therefore, it should be possible to add any two points (denoted $\oplus$). The following properties are clear:
• regular points add normally as $n_k = n_k' + n_k''$;
• adding wild point to either regular or wild point results in a wild point.
Now, although you could write explicitly some wild points, nearly all of them are too hard to describe. The best approximation to the resulting picture is probably this: imagine a set of wild points $W$; the whole space is $W \times \mathbb Z^+\times \mathbb Z^+\times\cdots$. The remaining question, therefore, is what monoid $W$ is equivalent to. For that, you need to settle these questions:
• is it true that you cannot have $w_1 \oplus w_2 \oplus \cdots\oplus w_n = r$ where $r$ is a regular point;
• whether you can subtract them;
• whether and how wild points are divisible by naturals.
## Operations on wild points
I think now is certainly time to post another question :)
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### 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.
When the field is $k=\mathbb F_2$ instead of $\mathbb C$, you can construct algebra morphisms from the $k$-algebra of functions on an infinite set to $k$ which are not evaluations, by using an ultrafilter.
Let $U$ be a non-principal ultrafilter on $X=\mathbb N$, and let $I$ be the ideal of $k(X)$, the algebra of $k$-valued functions on $X$, such that a function $f:X\to k$ is in $I$ iff $\{x\in X:f(x)=0\}\in U$. This is indeed a proper ideal, as one checks at once, and it is in fact maximal. Moreover, $k(X)/I$ has exactly two elements, so from $U$ we get a morphism $\epsilon:k(X)\to k$. As the ultrafilter $U$ is not principal, this map $\epsilon$ is not an evaluation.
(If you do the same with $\mathbb C$, one cannot prove that $k(X)/I$ is again $\mathbb C$; indeed, it is generaly not!) (On the other hand, the same is true for any finite field, I guess, because saying that the field has $q$ elements is a first order property of the field and Łoś's theorem applies... maybe someone with a firmer grasp on this can confirm this?)
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Yes, you can do the same thing on the algebra of all bounded functions (i.e. you get the Stone-Cech compactification of N) but I don't know what this says about the unbounded case. – Qiaochu Yuan Jan 7 2010 at 0:00
Apparently (mathoverflow.net/questions/3871), maxSpec of C(X) is \beta X, but the residue fields of the extra points can be wild. But they aren't always wild. – Theo Johnson-Freyd Jan 7 2010 at 0:14
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http://www.physicsforums.com/showthread.php?p=3999849
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Physics Forums
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Linearity of Maxwell's equations as a result of special relativity.
Quote by TrickyDicky The fact that in curved spacetime Maxwell eq. are also linear trivially reflects IMO the fact that the spatial part of curved spacetimes can be flat.
These two facts have nothing to do with each other, trivially or otherwise.
Maxwell's equations are
$$dF = 0, \qquad d \ast F = \ast J,$$
which are linear on ANY background whatsoever.
Blog Entries: 1 This thread has gone a bit beyond what I was expecting, and far beyond my knowledge of physics (I'm an incoming college freshman, so "Yang-Mills field theory" is a bit out of my grasp). I was hoping to create a simpler discussion based on the question "Does the linear way photons propogate in space-time have anything to do with how electric and magnetic fields add up?" It may be my fault for being unclear. Still, here's my go at a point that has been raised: Maxwell's equations are linear in curved spacetime. I believe that the generalization pf the electric Gauss' law for general relativity would be this: The electric flux through the surface of an infinitesimal enclosure around the charge $Q$ is equal to $4\pi Q\sqrt g$. Locally, I believe that this change cannot effect the linearity of the differential form of Maxwell's equations (in which you switch from ${\partial ^\mu }$ to ${\nabla ^\mu }$ in the covariant formulation; both operates being linear). However, personally, I feel that the covariant formulation of Maxwell's equation obfuscates some otherwise simple concepts so that the linearity --> superposition implication becomes less clear. So I suppose I ought to ask simply: Do electric fields and magnetic fields superimpose themselves linearly in curved spacetime? Or, given a system of moving point-charges in curved spacetime, is it possible to determine each contribution of electric and magnetic field at some arbitrary point?
Are you familiar with the Faraday field? I really wouldn't treat the electric and magnetic fields as separate. They're part of one object (Ben just alluded to it; it's usually denoted $F$). The Faraday field has six components, and so, just as you can find the component in a given direction for a vector field, the Faraday field always allows you to split it into electric and magnetic parts. You just have to choose a specific vector as the direction of time--time is what distinguishes electric from magnetic.
Blog Entries: 1
Quote by Muphrid Are you familiar with the Faraday field? I really wouldn't treat the electric and magnetic fields as separate. They're part of one object (Ben just alluded to it; it's usually denoted $F$). The Faraday field has six components, and so, just as you can find the component in a given direction for a vector field, the Faraday field always allows you to split it into electric and magnetic parts. You just have to choose a specific vector as the direction of time--time is what distinguishes electric from magnetic.
I know about that, but that is not really the issue (I assume the electromagnetic tensor ${F^{\mu \nu }}$ I alluded to is the Faraday field you are talking about). That is why I only mentioned the electric field in my OP, as you can view the magnetic field as a relativistic correction of the electrostatic field (equivalently, you can view the electric field as a relativistic correction of the magnetostatic field). They are indeed one and the same.
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Quote by Snicker I believe that the generalization pf the electric Gauss' law for general relativity would be this: The electric flux through the surface of an infinitesimal enclosure around the charge $Q$ is equal to $4\pi Q\sqrt g$.
In curved spacetime you should use the differential form; volume integrals do not always make sense
The Maxwell equations can be generalized to the Einstein-Maxwell equations:
http://en.wikipedia.org/wiki/Maxwell...rved_spacetime
Note that these are no longer linear due to the coupling of the el.-mag. field with the metric, which is a solution of the Einstein equation with the el.-mag. stress-energy tensor on the r.h.s. So formally the el.-mag. part seems to be linear (w.r.t. the el.-mag. fields) but the full equations aren't.
Quote by Snicker So I suppose I ought to ask simply: Do electric fields and magnetic fields superimpose themselves linearly in curved spacetime?
Yes, as long as you can neglect the back-reaction of el.-mag. fields to spacetime; so if you fix a space-time background you can still use the linear equations for the el.-mag. field.
Quote by Snicker Or, given a system of moving point-charges in curved spacetime, is it possible to determine each contribution of electric and magnetic field at some arbitrary point?
Yes, but again you have to fix spacetime and neglect back-reaction of el.-mag. fields; otherwise I don't think that using point-like charges will make sense.
Quote by Ben Niehoff These two facts have nothing to do with each other, trivially or otherwise. Maxwell's equations are $$dF = 0, \qquad d \ast F = \ast J,$$ which are linear on ANY background whatsoever.
You are right, of course.
I was thinking only in terms of the context in which certain equations could be derived, but obviously my statement is misleading, equation's linearity or nonlinearity per se is independent of any background.
Quote by Ben Niehoff No, QED includes non-linear effects (due to the fermion fields). However, the OP was about classical E&M, and we have been talking about classical gauge theories.
Ah, ok, so your statement about the link between linearity and abelian gauge symmetry holds only for classical theories?
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Quote by TrickyDicky Ah, ok, so your statement about the link between linearity and abelian gauge symmetry holds only for classical theories?
Yes and no. The operator equations look identical to the classical ones, but already classically you get a non-linearity when treating the current correctly.
Usually you have $j_\mu A^\mu$ which is linear in A; the current is fixed by hand. But in QED you have a fermionic current $j_\mu = \bar{\psi}\gamma_\mu\psi$ which introduces an interaction $\bar{\psi}\gamma_\mu\psi A^\mu$. w/o this term even QED (w/o fermions!) would be linear, no scattering, no bound states, superposition, ...; but this term adds an interaction between fermions and the electromagnetic field, which e.g. causes a (tiny) photon-photon scattering, i.e. linearity is lost. Two photons will scatter via exchange of virtual electron-positron pairs, whereas in classical electrodynamics nothing like that is present. Note that this is not related to quantization but to the different structure of the current - which in principle could also be studied in classical theories.
However, the theory remains gauge invariant and the fermionic current is still conserved via (the quantum analogue of) Noether's theorem. So non-linear interaction terms are compatible with gauge symmetry.
Quote by TrickyDicky When talking about Lorentz invariance most people don't include translations, that are included in the broader Poincare group, so certainly the equations of a theory that is Lorentz invariant in the usual sense, don't have to be linear, as Tom stoer's example of Yang-Mills theory shows. Maybe the OP is confused by the fact that Lorentz transformations are indeed linear. In any case Maxwell's equations linearity as previously commented is more related to the fact that it has an abelian gauge symmetry:U(1) in the Minkowskian formulation and also to the fact that it was originally formulated in pre-relativistic (pre-4spacetime) terms and the classical EM vector fields are based in 3D Euclidean geometry with absolute time as a parameter and this conditions the linearity of the EM field equations since the vector fields belong to the Euclidean vector space. The fact that in curved spacetime Maxwell eq. are also linear trivially reflects IMO the fact that the spatial part of curved spacetimes can be flat.
I don't see how U(1) could possibly support the Poincare group which includes the combination of rotations and boosts. Usually the Poincare group is associated with an SL algebra, isn't it? The SL group linearity is geometrically projective rather than Euclidean and that's what separates SL from GL groups, right? (From a Euclidean perspective, continuity relationships between series of projections are non-linear even if each projection is linear)
Quote by PhilDSP I don't see how U(1) could possibly support the Poincare group which includes the combination of rotations and boosts. Usually the Poincare group is associated with an SL algebra, isn't it? The SL group linearity is geometrically projective rather than Euclidean and that's what separates SL from GL groups, right? (From a Euclidean perspective, continuity relationships between series of projections are non-linear even if each projection is linear)
I 'm not sure how all these questions are related to my post, I only mentioned the Poincare group to stress that the Lorentz group is a subgroup of Poincare's that doesn't include translations.
Quote by TrickyDicky I 'm not sure how all these questions are related to my post, I only mentioned the Poincare group to stress that the Lorentz group is a subgroup of Poincare's that doesn't include translations.
And that's a good observation, especially when digging deeper into group representations (which I was attempting to do above). One complication with groups that can easily be overlooked is that it may not be sufficient to describe only one transformation at a time, but it may be necessary to account for all possible transformations concurrently and collectively.
Quote by tom.stoer However, the theory remains gauge invariant and the fermionic current is still conserved via (the quantum analogue of) Noether's theorem. So non-linear interaction terms are compatible with gauge symmetry.
Yes, again, as the gauge symmetry of non-abelian Yang-Mills gauge theories show. But then in those we must include other internal symmetries to form U(1)XSU(2) and the electroweak interaction.
However what I fail to see is how QED contains that non-linear term only with the U(1) abelian gauge symmetry. This might be getting offtopic and not related to relativity, so maybe I should open a new thread.
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Quote by TrickyDicky However what I fail to see is how QED contains that non-linear term only with the U(1) abelian gauge symmetry. This might be getting offtopic and not related to relativity, so maybe I should open a new thread.
This is a general construction principle for coupling gauge fields to spin 1/2 fields in the Lagrangian in a both Lorentz- and gauge invariant manner
$$\bar{\psi} \gamma^\mu \ldots (\partial_\mu - ig A_\mu^a T^a)\psi$$
where gamma and '...' mean a Clifford algebra, in () we have the covariant derivative (for the gauge group) and a=1..dim(G) labels the generators of the gauge group in the adjount rep. For U(1) we have a=1
This is the minimal coupling scheme and therefore gauge theory + fermions is always non-linear.
In gauge theories w/o fermions the Lagrangian reduces to
$$F_{\mu\nu}^a F_{\mu\nu}^a$$
which is non-linear when using a non-abelian gauge group b/c
$$F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a +gf^{a}_{bc}A_\mu^bA_\nu^c$$
For abelian gauge theories like U(1)*U(1)*... the structure constants f vanish identically, whereas for non-abelian groups already in the pure gauge sector the theory is necessarily non-linear.
Thanks Tom. I guess I interpreted the phrase "a gauge theory is linear if and only if its gauge group is Abelian" in a previous post as meaning "a gauge theory is non-linear if and only if its gauge group is non-Abelian" which are not exactly equivalent, the latter is wrong while the former is right.
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Quote by TrickyDicky However what I fail to see is how QED contains that non-linear term only with the U(1) abelian gauge symmetry. This might be getting offtopic and not related to relativity, so maybe I should open a new thread.
QED is not a pure gauge theory; it is coupled to fermions. The fermions provide the nonlinearity as Tom described above.
Generally, when we solve classical E&M problems, we consider the sources J to be non-dynamical. They are fixed, or are given by a predetermined forcing function (cf. antennas). In this case, the problem we are left to solve is actually pure E&M.
If you allow the sources to be dynamical--i.e., to allow the fields to move the charges in the way that they actually would in nature--then the theory as a whole is no longer linear. At least I think so. This fact may actually depend on the sources having a nonlinear coupling to the fields, as in the case with fermions.
In any case, we should be careful what we mean by "gauge theory". I've been referring to the pure gauge sector, possibly up to including external, non-dynamical sources. QED, QCD, etc. are "gauge theories coupled to matter".
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http://stats.stackexchange.com/questions/29117/regression-technique-for-data-comprised-of-categorical-explanatory-variables-a
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# Regression technique for data comprised of categorical explanatory variables & a continuous response variable
i suppose one way to characterize data is by a combination of the variable types that comprises it:
```` Continuous/Continuous | Continuous/Discrete
-----------------------|---------------------
Discrete/Discrete | Discrete/Continuous
````
Each of the four cells comprising the 2x2 table just above is comprised of two descriptors, one for the Explanatory Variables (EV), which we'll assume are all of the same type, and one for the Response Variable (RV):
````Explanatory Variable Type / Response Variable Type
````
The columns represent the EV type; the rows, the CV type.
Nearly all of the data i see can be placed in either of the two cells that comprise the first row.
So for instance, OLS Regression is a suitable model type for data in row1/col1; and for data in row1/col2, Logistic Regression is an appropriate model choice.
It's the second row, and in particular row2/col2, that my Question is directed to.
I'm aware of a few regression techniques like ordinal regression which handle a particular type of discrete variables (ordinary or rank, 1st, 2nd, 3rd,....) but i am interested in techniques for handling discrete data more generally, given that most categorical variables do not have an implicit ordinal relationship among the values that comprise them.
For instance:
````Sex | City_of_Residence | Car_Make&Model | Married? | DUI? | Prior_3P_Claims?
F | Cleveland | Chevy Camaro | No | No | Yes
````
And the Response Variable is continuous--e.g., building a model to predict the quotes offered by major auto insurers--a price.
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– whuber♦ May 24 '12 at 21:38
## 1 Answer
The type of regression is related to the dependent variable only. When the dependent variable is continuous, you can consider OLS regresssion, regardless of whether the independent variables are categorical or continuous or both. Ordinal independent variables are a bit tricky - sometimes they are treated as continuous sometimes as categorical.
If the independent variables are categorical, there are a variety of methods including effect coding and dummy coding to deal with them.
Of course, OLS makes assumptions beyond the idea that the dependent variable is continuous (or nearly so). e.g. it assumes that the residuals are independent and $\sim{N(0,1)}$
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I might be misreading the question, but it seems that in spirit it is asking for more than how to code a purely nominal explanatory variable. (The clue for that comes from the reference to ordinal regression.) Perhaps the best answer is "use some kind of dummy coding," but maybe, on the other hand, the question that's really lurking here concerns the choices one makes for modeling data that appear to be categorical but might have some structure: perhaps not an order, but something that would allow one to use many fewer parameters than categories. But maybe I'm being too imaginative ... – whuber♦ May 24 '12 at 22:00
Except that the last line says the response variable is continuous. I'm not sure what he wants. – Peter Flom May 24 '12 at 22:09
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Well, now we're making real progress! We have identified a possible point of misunderstanding in the question (which, as an expert consultant, you know is a key step towards developing a truly useful and correct answer). Let's ask the OP: @Doug, could you expand on this issue so we can create a common unambiguous understanding of what you are asking? – whuber♦ May 24 '12 at 22:13
Indeed! And @doug has been here a while, so an answer should be forthcoming. – Peter Flom May 24 '12 at 22:17
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@macro is right I mean $N(0,\sigma^2)$ and also about the errors, not the residuals, but we don't know the errors, only the residuals. – Peter Flom May 25 '12 at 14:42
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http://crypto.stackexchange.com/questions/3200/hash-function-from-narrower-block-cipher-operated-in-cbc-encryption-mode/4032
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# Hash function from narrower block cipher operated in CBC-encryption mode?
I am trying to build a public hash function (thus collision-resistant and preimage-resistant, and more generally behaving like a random oracle), with input a message $M$ of fixed size $|M|=m\cdot b$ bits, and output the hash $H(M)$ of fixed size $|H(M)|=h\cdot b$ bits, using as the single primitive a $b$-bit block cipher with key of $k\cdot b$ bits, operated in CBC-encryption mode.
One of the motivation of the construction is to maintain the confidentiality of the message under DPA attacks, assuming that the implementation of the cipher in CBC-encryption mode is DPA-protected. I thus require that all message-dependent data is manipulated using the $P$ and $K$ inputs of that trusted primitive: $$F(K,IV,s,P)\to C\text{ with }|K|=k\cdot b,|IV|=b,s>0,|P|=|C|=s\cdot b,$$ $$C[0]=\text{ENC}_K(P[0]\oplus IV), C[j]=\text{ENC}_K(P[j]\oplus C_{j-1})\mathsf{\text{ if }}1\le j<s.$$
I would like security beyond that of the underlying block cipher by a security parameter $q$, with an argument that any attack requires $2^{b\cdot q}$ queries to an encryption or decryption oracle implementing $\text{ENC}$ and the corresponding decryption, for some suitable definition of security/attack which, uh, is part of the question (in particular it seems necessary to restrict the amount of memory usable by the attacker, including but not limited to as a cache for the oracle). Performance is secondary (one envisioned application is a slow KDF), a security argument (or better proof) is a must, simplicity matters.
Assume there is enough memory for the whole message, and then some. Assume at least $\min(b,k\cdot\lceil b/2\rceil)$ bits of the block cipher's key are effective (which rules out simple-DES). A typical setup could be 3DES thus $b=64$, $k=3$, and $m=64$ ($4096$-bit input), $h=8$ ($512$-bit output), $q=4$ (security of $2^{256}$ block cipher queries or memory accesses). Assume $m\ge h>1$, and if that helps assume $m\gg h$ or/and $h\gg k$ or/and $k=1$ (e.g. AES-128, ignoring attacks marginally reducing its effective key size).
What appropriate definition(s) of security are applicable?
I apologize for late realization that this is an issue, and corresponding introduction of security parameter $q$. I now think I want a collision-resistant Universal One-Way Hash Function family per this terminology.
Is there some standard construction that fits?
My review of the existing found a few hashes made from block ciphers, but suitable for $h\le2$ only, not neatly constructible from my primitive when $h>1$, and aiming at efficiency rather than a strong security argument.
If nothing standard exists, my simplest candidate is the following CBC-HASH:
• assume an arbitrary public parameter $P$ of $p\ge k$ blocks which right $k$ blocks are suitable as an initial key for the cipher ($P$ parameterize the UOWHF family; in the simplest/original setup, $p=k$);
• append $P$ to message $M$ forming $X_0$ of $m+p$ blocks $X_0[j]$, that is $M[j]\to X_0[j]\mathsf{\text{ if }}0\le j<m, K[j]\to X_0[m+j]\mathsf{\text{ if }}0\le j<p$;
• repeat for $n$ rounds, numbered $r$ from $0$ to $n-1$:
• set $K_r$ as the right $k$ blocks of $X_r$, that is $X_r[m+p-k+j]\to K_r[j]\mathsf{\text{ if }}0\le j<k$;
• CBC-encipher $X_r$ using key $K_r$ and the $b$-bit big-endian representation of $r$ as $IV$, giving $X_{r+1}$, that is $F(K_r,(r)_b,m+p,X_r)\to X_{r+1}$;
• set output $H=H(M)$ as the right $h$ blocks of the left $m$ blocks of $X_n$, that is $X_r[m-h+j]\to H[j]\mathsf{\text{ if }}0\le j<h$.
Now I am wondering how the number of rounds $n$ and the size $p$ of the initial padding shall be chosen from $q$, $h$, $m$, $k$ (perhaps $b$), and the limits on parameters for a security argument.
Can you justify, prove, improve, break, or fix CBC-HASH?
The scheme now incorporates parameter $p$ controlling the size of the UOWHF family. I retract an earlier guess on $n$ made without even a clear security definition. I hereby vow to stop modifying this question except for fixing obvious typos; any addition will be an answer.
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Normally my advice is "never roll your own", but... Holy crap! – Polynomial Jul 11 '12 at 20:00
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@Polynomial well someone has to build those, they don't design themselves, right? :p – Thomas Jul 12 '12 at 1:21
I removed the community wiki status from this question, so answerers can get their reputation, too. – Paŭlo Ebermann♦ Jul 18 '12 at 12:38
## 1 Answer
If the motivation is to rely on a DPA protected implementation of a block cipher, hash function constructions based on a block cipher (think of Davies-Meyer for instance with say Rijndael at 256-bit plaintext) could be used. Alternatively, the issue of using small input size compression function to built secure hash functions with larger output size has been investigated in e.g. Nandi, INDOCRYPT 2005; Hirose, FSE 2006; Peyrin et al., Asiacrypt 2006 and following work, but is not entirely settled.
From an efficiency point of view, CBC-HASH is going to be at least $n$ times slower than any other hash function in the wild. (And practical security is always a balance between practicality, that is resources, and security.) In addition, the hash cannot be computed on the fly: one has to make $n$ passes over the message which is cumbersome in many scenarii (think embedded devices for instance).
Case $n=1$. Assuming the underlying block cipher is denoted by $E$, take whatever you want for the message blocks $M_i=N_i$, $l>i>1$ for some big $l$ and:
• $M_0=0$ and $M_1=E_K(R)$,
• $N_0=R$ and $N_1=E_K(0)$.
for some $R\neq 0$ if $IV=0$ and $R=IV$ otherwise. Then, note that the hash output is the same, so that you have a collision for $M$ and $N$. This is possible because $K$ and $IV$ are known values.
Case $n=2$. Choose some index $\xi$ and choose the message blocks $M_1$, ..., $M_\xi$ and $N_1$, ..., $N_\xi$ randomly and choose the remaining message blocks so that $M_i=N_i$, $i=\xi+1, ..., l$. Then, use the trick above to ensure the key for the next round is identical for both messages (i.e. create a collision at stage $\xi$ during the first pass instead of at the beginning). Then, wait for a collision to happen in the 2-nd round at block $\xi$. It happens with probability ${2^{block\_size/2}}$ for a primitive $E$ with block size $block\_size$. This appears to be possible as the first message blocks are randomly chosen.
Case $n>2$. This procedure can be followed up to any round $n$ with complexity $2^{block\_size*{n-1\over2}}$.
Example. With the 3DES as $E$ and an hash output size of 512 bits, this leads to a collision attack for CBC-HASH with 8 rounds that has complexity $2^{32\cdot 7}$ instead of the $2^{32*8}$ of the birthday paradox.
I am more generally concerned about the small "bandwidth" issue due to the use of an underlying CBC with a block cipher of small output size.
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The problem is that those constructions require a sufficient block size. In practice 256 or 512 bits. I think fgrieu's premise is a block size that's smaller than the desired hash size. | Additionally when you use such a construction with AES, it might expose AES's key schedule weaknesses. – CodesInChaos Oct 11 '12 at 21:38
CodeInChaos is right that my problem with Davies-Meyer and friends is the insufficient block size of available block ciphers; e.g. 128 bit at best in any hardware that I use, which is insufficient for a secure hash using DM. Also, I do not care so much for speed, as one of the application would be a purposely-slow hash, à la scrypt. – fgrieu Oct 12 '12 at 7:03
@CodesInChaos: if you assume AES message processing can be DPA protected, then so can AES key schedule, since they are based on the same primitives. In practice, key schedule is also protected. – bob Oct 12 '12 at 8:30
@bob: If Rijndael ever had 256-bit plaintexts, that was removed in AES. About my rationale: I am often using Smart Cards or HSM platforms where DES (resp. AES) is a fast hardware thing with 64-bit (resp. 128-bit) data input and 64/128/192 (resp. 128/192/256-bit) key input, hardware support for the XOR required by CBC, and extensive hardware DPA protection; it has a formally security-certified API including fast CBC encrypt (and slightly slower decrypt), anything else either is a subset, or is not formally security-certified. – fgrieu Oct 12 '12 at 9:11
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@fgrieu: you are correct that the case $n>1$ cannot be attacked since the key after first round was the same for both messages and due to the bijectivity of the block cipher, a collision could not occur. I updated the last paragraph accordingly. As I write, I'm concerned about the small "bandwidth", and there is probably a lot to do, but would require some more time. – bob Oct 22 '12 at 9:27
show 1 more comment
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http://mathoverflow.net/questions/102983?sort=oldest
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## Reference Request: de Rham vs. Dolbeault
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi everyone. I need the following statement:
For a Kahler manifold $X$, the natural map `$H^n(X,\mathbb{C})\to H^n(X,\mathcal{O})$` (from the sheaf extension) coincides with the Hodge projection $\Pr_{0,n}$, up to the de Rham isomorphism and the Dolbeault isomorphism.
Does anybody know a good reference?
P.S. Surely there must be a reference. I am much less interested in proofs: I think I know one.
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## 1 Answer
There are lots of references. Mainly every textbook which treats Hodge theory. Try to look at:
• Voisin: Hodge theory and complex algebraic geometry. I
• Huybrechts: Complex geometry
• Wells: Differential analysis on complex manifolds
• Griffiths, Harris: Principles of algebraic geometry
There, you will find mainly the proof in the case $n=2$, which is used to prove the Lefschez theorem on $(1,1)$-classes. The general case is a straightforward adaptation of that argument.
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1
Yes, but to write in a paper "The general case is a straightforward adaptation of that" is not a good manner, is it? And the proofs in Griffiths&Harris and Voisin are rather specific for $n=2$. (They may be generalized, to be sure, but I wouldn't say it is straightforward. Do not remember about other books). Do not take me wrong, but it doesn't look like a very good reference. – Alex Gavrilov Jul 24 at 13:29
The point is rather that you didn't say you needed this reference for a paper you are writing. In this case, sincerely, you can just state that fact as well-known. No referee would protest! – diverietti Jul 25 at 7:03
Yes, perhaps you are right. Anyway, it won't do any harm: if the referee insists, then I can write my own proof (with the reference to Griffiths&Harris for a special case). Of course I shoud have made my purpose clear from the beginning, so we could avoid this bit of confusion. However, I slill hope that someone may give me a reference for a complete proof, which is why I do not accept your answer. Do not mind this. By the way, I have Griffiths&Harris on my bookshelf. – Alex Gavrilov Jul 26 at 13:29
Of course I don't mind! If I find a complete reference I'll tell you! – diverietti Jul 26 at 16:24
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http://physics.stackexchange.com/questions/16179/poles-wavefunctions-transmission
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# Poles, wavefunctions, transmission
Why is it said that $\operatorname{sech}x$ (a transmission amplitude) has a simple pole on the imaginary axis? Thanks.
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1
Can you give some context? sech has no poles as far as I can tell, as it is a continuous, strictly positive, bound function. I would vote down but I can't. – adavid Oct 25 '11 at 19:10
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@adavid, the imaginary axis! The poles are on the imaginary axis. – user1631 Oct 25 '11 at 20:00
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Thanks @user1631. When I see $x$, I think $\mathbb{R}$. I guess $z$ would have got me on the right track. And next time, please do give some more context... – adavid Oct 25 '11 at 20:56
– Qmechanic♦ Nov 19 '12 at 19:15
## 1 Answer
To have a simple pole at $a$ means that $f(z) \sim 1/(z-a)^{n}$, with $n=1$. I.e., the function does not diverge with $1/z^2$ or a larger power.
For more details on the poles of $\operatorname{sech}(z)$, check out this answer. As simple as the poles of $\operatorname{sech}(z)$ may be, there is an infinite amount of them (as many as there are zeroes to $\operatorname{cosh}(z)$).
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Thanks, and sorry about the misleading variable choice. – simpleton Oct 25 '11 at 23:25
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http://physics.stackexchange.com/questions/43510/bound-states-in-a-double-delta-function-potential
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# Bound States in a Double Delta Function Potential [closed]
Let $V(x) = −u \delta(x) - v \delta(x − a)$ where $u, v > 0$ correspond to a potential with two $\delta$ wells. Let $v > u$. If $a$ is very large, there is certainly a bound state: the particle sits in the $\delta$-well. As $a$ decreases to a certain critical value, the bound state disappears. I need help finding that value.
My idea was: Before the bound state disappears, its energy approaches $0$. I'm trying to assume that the energy $E$ is a very small negative number, solve the Schrodinger equation, and find the suitable value of $a$, but I'm having trouble doing this.
Would someone be able to help me with this problem?
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2
I would suggest at least finding the bound state and its energy before you try to do anything sexy with limits and such. – DJBunk Nov 5 '12 at 23:43
Do know how to find the eigenvectors and eigenvalues in a case like this? – DJBunk Nov 6 '12 at 3:09
I got something for you, give me a minute – Dylan Sabulsky Nov 6 '12 at 3:14
## closed as too localized by dmckee♦Nov 7 '12 at 14:16
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.
## 1 Answer
I'm not going to answer your exact question, but this is a good example (from an old copy of Griffith's that my loser chem bro uses [real women and men of physics use Shankar and Sakurai]
Consider the double delta-function potential $$V(x)=-\alpha[\delta(x+a)+\delta(x-a)]$$ where $a$ and $\alpha$ are positive constants. Hope this helps! -Dylan
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Hi Dylan - you shouldn't directly put in scans from a textbook, instead type out the relevant parts of the material, using block quote syntax if necessary. Could you edit this accordingly? (Perhaps someone else will be willing to do it, but don't count on that.) – David Zaslavsky♦ Nov 6 '12 at 18:29
Hey David, I'll try to later on today. Sorry about that! – Dylan Sabulsky Nov 6 '12 at 22:08
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http://math.stackexchange.com/questions/250998/diffeomorphism-on-the-torus
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# Diffeomorphism on the torus
Let $S : \mathbb{R}^n → \mathbb{R}^n$ be linear invertible map, then $S$ projects to $\mathbb{T}^n$ diffeomorphism if and only if $S ∈ GL_n(\mathbb{Z})$.
I can't prove the right to left implication.
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That $S$ projects to a diffeo of $T^n$ means that $S$ is $\mathbb{Z}^n$-equivariant. – Neal Dec 4 '12 at 21:50
## 1 Answer
I'm assuming that for you $\mathbb T^n = \mathbb R^n / \Gamma$, where $\Gamma$ is a full lattice in $\mathbb R^n$. In this case, the left to right implication is just the fact that if $S \in \text{GL}_n(\mathbb Z)$ then $S(\Gamma) \subset \Gamma$, hence universal property of the quotient concludes.
In fact, being the action of $\Gamma$ over $\mathbb R^n$ properly discontinuous, you can see $\mathbb T^n$ as a quotient in the category of differentiable manifolds. Hence your $S$ factors through $\mathbb T^n$ producing a map $\mathbb T^n \to \mathbb T^n$. In the same way you can factors its inverse, hence you get a diffeomorphism of the torus (you check that the two induced morphisms are inverse each other using again the universal property).
Edit: To do the converse, you don't even need the universal property. Assume in fact that $S$ is a linear invertible map inducing a diffeomorphism of the torus. Apparently I'm not allowed to draw commutative diagrams... So let $p \colon \mathbb R^n \to \mathbb T^n$ be the canonical projection and let $\overline{S} \colon \mathbb T^n \to \mathbb T^n$ the map induced by $S$. Then $p \circ S = \overline{S} \circ p$. This means that $p(S(\Gamma)) = \overline{S}(p(\Gamma)) = \overline{S}([\mathbf 0]) = [\mathbf 0]$, hence $S(\Gamma) \subset \Gamma$. Since $\Gamma = \mathbb Z^n$, and since the vectors $\mathbf e_i = (0,\ldots,1,\ldots,0)$ form a basis both for $\Gamma$ and for $\mathbb R^n$, you can write down the matrix of $S$ explicitly: in the $i$-th column you will have the coefficients of $S(\mathbf e_i)$ with respect to this basis; since $S(\Gamma) \subset \Gamma$ you see that $S(\mathbf e_i) \in \mathbb Z^n$, hence the coefficients must be integers! This implies $S \in M_n(\mathbb Z)$.
Now, your hypothesis is that $\overline{S}$ is a diffeomorphism of $\mathbb T^n$. Let $f \colon \mathbb T^n \to \mathbb T^n$ be its inverse. Since the action of $\mathbb Z^n$ over $\mathbb R^n$ is properly discontinous, then the canonical projection $p$ is a covering; since $\mathbb R^n$ is simply connected, this is the universal covering. Thus $f$ lifts (uniquely!) to a continuous map $T \colon \mathbb R^n \to \mathbb R^n$. Universal property of universal covering implies that $T$ is the inverse of $S$, hence $T \in \text{GL}_n(\mathbb R)$. The same reasoning of above shows $T \in M_n(\mathbb Z)$. Therefore $S, T \in \text{GL}_n(\mathbb Z)$.
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Thanks for the quick reply. I'm reading on wiki about the universal property of the quotient. However I really have trouble proving the converse statement. I already changed the question to correct this. – user47709 Dec 4 '12 at 22:14
Well, assume that $f \colon \mathbb T^n \to \mathbb T^n$ is a diffeomorphism induced by some linear invertible map $S$. Then, since $S$ factors through $\mathbb T^n$ it follows that $S(\Gamma) \subset \Gamma$. I'm assuming that for you $\Gamma = \mathbb Z^n$. In this case, the image of $\mathbf e_i$ must be of the form $\sum_{j = 1}^n a_j \mathbf e_j$ with $a_J \in \mathbb Z$, hence $S \in M_n(\mathbb Z)$, hence $S \in \text{GL}_n(\mathbb Z)$. – Mauro Porta Dec 4 '12 at 22:26
I don't understand: for example, why does a mapping $\left(\begin{matrix}2 & 0 \\ 0 & 1\end{matrix}\right)$ produce a diffeomorphism instead of a covering? – Alexei Averchenko Dec 4 '12 at 22:26
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It's me to not understand. Your matrix is not in $\text{GL}_2(\mathbb Z)$, hence it doesn't produce a diffeomorphism of the torus. – Mauro Porta Dec 4 '12 at 22:28
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Then I don't know. I'm a little bit sceptical because I don't see why a map $\mathbb T^n \to \mathbb T^n$ should be induced necessarily from a linear map in $\mathbb R^n$. This is true for a complex torus of dimension $1$ (i.e. endowed with complex structure), but I'm not sure about the general case. – Mauro Porta Dec 4 '12 at 22:49
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http://mathhelpforum.com/calculus/34731-value-sin-x-different-intervals.html
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# Thread:
1. ## Value of sin x for different intervals
I have been asked what the average value of the sin x for the interval greater or equal to 0 and less than or equal to 2 pi. I just plugged in the two end values and since both are = 0 the answer is, of course, 0.
Then I was asked what the average value of this is between 0 and pi
Obviously I am doing something wrong because the answer is 2/pi!!! Please advise on how to do these correctly. Thank you very much
2. You are right actually!
The average is 2/pi. Why do you think it is wrong?
3. Average value of a function f(x) in the interval between a and b is
$\frac{\int^{b}_{a}f(x)~dx}{b-a}$
4. ## value of sin x
So, if I have this right it goes like this:
it would be the value of -cos x from 0 to pi divided by pi - 0
-cos pi + -cos 0 /pi = 1+1 / pi or 2/pi Is this correct figuring? Thanks so much for helping out quickly!
5. Originally Posted by Frostking
So, if I have this right it goes like this:
it would be the value of -cos x from 0 to pi divided by pi - 0
-cos pi + -cos 0 /pi = 1+1 / pi or 2/pi Is this correct figuring? Thanks so much for helping out quickly!
Yes
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http://unapologetic.wordpress.com/2009/05/21/arml-scrimmage-power-question-2/?like=1&source=post_flair&_wpnonce=3fdfac1636
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# The Unapologetic Mathematician
## ARML Scrimmage Power Question
I helped the Howard County and Baltimore County ARML teams practice tonight by joining the group of local citizens and team alumni to field a scrimmage team. As usual, my favorite part is the power question. It follows, as printed, but less the (unnecessary) diagrams:
Consider the function
$\displaystyle\phi(t)=\left(\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right)=(x,y)$
which maps the real number $t$ to the a coordinate in the $x$-$y$ plane. Assume throughout that $q$, $r$, $s$, $t$, and $u$ are real numbers.
(1) Compute $\phi(1)$, $\phi(1/2)$, $\phi(2)$, $\phi(-1)$, $\phi(-1/2)$, and $\phi(-2)$. Sketch a plot of these points, superimposed on the unit circle.
(2) Show that $\phi$ is one-to-one. That is, show that if $\phi(s)=\phi(t)$, then $s=t$.
(3) Let $(x_\phi,y_\phi)$ be the intersection point between the unit circle and the line connecting $(0,1)$ and $(t,0)$. Prove that $\phi(t)=(x_\phi,y_\phi)$.
(4) Show that $(x,y)$ is an ordered pair of rational numbers on the unit circle different from $(0,1)$ if and only if there is a rational number $t$ such that $\phi(t)=(x,y)$. (This result allows us to deduce that there are infinitely (countably) many rational points on the unit circle.)
According to problem 3, $\phi(t)$ is a particular geometric mapping of a single point on the real line to the unit circle. Now, we will be concerned with the relationship between the pairs of points, which will lead to a way of doing arithmetic by geometry. Use these definitions:
• Let $\left\{\phi(s),\phi(t)\right\}$ be a “vertical pair” if either $s=t=1$, or $s=t=-1$, or $st\neq0$ and $\phi(s) and$latex \phi(t)\$ are two different points on the same vertical line.
• Let $\left\{\phi(s),\phi(t)\right\}$ be a “horizontal pair” if either $s=t=0$, or $\phi(s)$ and $\phi(t)$ are two different points on the same horizontal line.
• Let $\left\{\phi(s),\phi(t)\right\}$ be a “diametric pair” if $\phi(s)$ and $\phi(t)$ are two different end points of the same diameter of the circle.
(5) (a) Prove that for all $s$ and $t$, $\left\{\phi(s),\phi(t)\right\}$ is a vertical pair if and only if $st=1$.
(b) Prove that for all $s$ and $t$, $\left\{\phi(s),\phi(t)\right\}$ is a horizontal pair if and only if $s=-t$.
(c) Determine and prove a relationship between $s$ and $t$ that is a necessary and sufficient condition for $\left\{\phi(s),\phi(t)\right\}$ to be a diametric pair.
(6) (a) Suppose that $\left\{\phi(s),\phi(t)\right\}$ is not a vertical pair. Then, the straight line through them (if $\phi(s)=\phi(t)$, use the tangent line to the circle at that point) intersects the $y$-axis at the point $(0,b)$. Find $b$ in terms of $s$ and $t$, and simplify and prove your answer.
(b) Draw the straight line through the point $(1,0)$ and $(0,b)$, where $(0,b)$ is the point described in problem (5a). Let $\phi(u)$ denote the point of intersection of this line and the circle. Prove that $u=st$.
(7) (a) Suppose that $\left\{\phi(s),\phi(t)\right\}$ is not a horizontal pair. Then, the straight line through them (if $\phi(s)=\phi(t)$, use the tangent line to the circle at that point) intersects the horizontal line $y=1$ at the point $(a,1)$. Find $a$ in terms of $s$ and $t$, and simplify and prove your answer.
(b) Draw the straight line through the point $(0,-1)$ and $(a,1)$, where $(a,1)$ is the point described in problem (6a). Let $\phi(u)$ denote the point of intersection of this line and the circle. Prove that $u=s+t$.
(8) Suppose $q$, $r$, $s$, and $t$ are distinct real numbers such that $qr=st$ and such that the line containing $\phi(q)$ and $\phi(s)$ intersects the line containing $\phi(r)$ and $\phi(t)$. Find the $y$-coordinate of the intersection point in terms of $s$ and $t$ only.
(9) Let $s$ and $t$ be distinct real numbers such that $st>0$. Given only the unit circle, the $x$- and $y$- axes, the points $\phi(s)$ and $\phi(t)$, and a straitedge (but no compass), determine a method to construct the point $\phi(\sqrt{st})$ that uses no more than $5$ line segments. Prove why the construction works and provide a sketch.
(10) Given only the unit circle, the $x$-and $y$- axes, the point $(1,1)$, and a straightedge (but no compass), describe a method to construct the point $\left(-\frac{2\sqrt{3}}{3},0\right)$.
### Like this:
Posted by John Armstrong | Uncategorized
## 2 Comments »
1. Oh, I didn’t realize that was you! I don’t know if you remember me but I was on the alumni team as well.
Comment by Adeel Khan | May 21, 2009 | Reply
2. Sure. You caught me misparsing that “evil” number question.
Comment by | May 21, 2009 | 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.
|
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http://mathhelpforum.com/differential-equations/101294-basic-diff-eq-question-finding-function-given-properties.html
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Thread:
1. Basic Diff EQ question: finding a function with given properties
Find a function f(y) with the property that: for each solution y(x) of dy/dx = f(y), the limiting value lim as x-> +inf of y(x) equals 3 if y(0) > 0.
Any help?
2. A possible solution of the problem is a function $y(x)$ with the following derivative $y^{'} (x)$...
$y^{'} < 0$ with $y>3$ or $y<0$
$y^{'} =0$ with $y=3$ or $y=0$
$y^{'} >0$ with $0<y<3$
A DE the solution of which has these properties is, among others, the following ...
$y^{'} = 3 y - y^{2}, y(0)=y_{0}>0$
... as you can easily verify...
Kind regards
$\chi$ $\sigma$
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http://mathhelpforum.com/advanced-algebra/82116-fields-rings-homomorphisms.html
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# Thread:
1. ## Fields, Rings and Homomorphisms
Let F be a field and R a ring and suppose phi: F --> R is a homomorphism.
Prove that either phi is one-to-one or phi is the trivial homomorphism (that is, phi(a) = 0 for all a in F).
2. Originally Posted by Coda202
Let F be a field and R a ring and suppose phi: F --> R is a homomorphism.
Prove that either phi is one-to-one or phi is the trivial homomorphism (that is, phi(a) = 0 for all a in F).
Hint1: The kernel of a ring homorphism gives rise to an ideal.
Hint2: What are the ideals of a field?
3. Mind you, some authors – e.g. D.J.H. Garling in A Course in Galois Theory (1986, Cambridge University Press) – insist that a ring homomorphism should map the multiplicative identity of one ring to that of the other. If you adopt such a definition, the second possibility would not be possible: all homomorphisms from a field to a ring would then have to be injective.
4. Originally Posted by JaneBennet
Mind you, some authors – e.g. D.J.H. Garling in A Course in Galois Theory (1986, Cambridge University Press) – insist that a ring homomorphism should map the multiplicative identity of one ring to that of the other. If you adopt such a definition, the second possibility would not be possible: all homomorphisms from a field to a ring would then have to be injective.
I like to distinguish between "homomorphism between rings" and "homomorphism between commutative unitary rings". So when I see "ring homomorphism" all I think of is $\phi(ab) = \phi(a)\phi(b),\phi(a+b)=\phi(a)+\phi(b)$. However, if I see "commutative ring homomorphism" (it is rather common to called commutative unitary rings simply by commutative rings) then I think of the additional condition $\phi(1) =1$.
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http://crypto.stackexchange.com/questions/3654/malleability-attacks-against-encryption-without-authentication?answertab=votes
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# Malleability attacks against encryption without authentication
Suppose there is a message that is encrypted with AES-128-CBC. The message is as follows, new lines are used to delimit the 16 byte boundary for each block:
````Wire funds from:
Alice to Bob in
the amount of $
1
````
Because this message is encrypted using CBC mode, any modification of the first block of cipher text would propagate throughout the message. However, the amount of the fund transfer falls in the last block. I am under the assumption that an attacker can use the property of Malleability to alter the last block and produce a valid amount that is greater than \$1 in far fewer than$2^{128}\$ operations.
Is this a correct assumption? Is there a name for this attack or do you know of a real world example?
My instinct is that if you just need one byte of the 16 byte block to be of a specific value then it would only require $2^8$ operations because the other 120 bits of the plain text could be any value. For example lets say we wanted the last block of cipher text to decrypt to `9\0` (where `\0` denotes a single null byte, and the message would be interpreted as \$9), then it would take only$2^{16}\$ operations to find this cipher text block.
Is there something I am missing?
-
## 3 Answers
First of all, you stated:
Because this message is encrypted using CBC mode, any modification of the first block of cipher text would propagate throughout the message.
Actually, that's not true. Here's the CBC mode operation in the decryption direction:
(Public domain image from Wikimedia Commons.)
If you examine the process closely, you will see that a specific plaintext block depends only on two ciphertext blocks (or a ciphertext block and the IV for the very first plaintext block).
So, if you modify a ciphertext block $N$ (say, flip one of the bits), then the resulting plaintext will have block $N$ garbled (because the modified block goes through the decryption, and unless we picked a ciphertext block which we already knew the decryption of, the result will be unpredictable). In addition, block $N+1$ will have the corresponding bits modified (in our example, the same bit will be flipped). And, there will be no other changes; all other plaintext blocks will be exactly what they were in the original message.
So, how can we use this observation? Well, if block $N$ is the one we want to affect, and we don't mind if block $N-1$ decrypts to something unpredictable, the obvious approach is to modify block $N-1$ of the ciphertext. If we guess that the real plaintext is $A$ (which corresponds to \$1), and we want it to read$B$(which corresponds to (\$1000000), then we exclusive or $A \oplus B$ into ciphertext block $N-1$; this will cause the last ciphertext block to read exactly as we want it to.
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Are you suggesting that if I replace ciphertext block N-1 with A⊕B⊕(ciphertext block N-1) I will have the message B for the plain text block N? – Rook Aug 26 '12 at 21:48
@Rook: yes, that is one approach I am suggesting. It works best if it's the first block you're trying to modify; this approach then leads you to alter the IV (which, if it is sent with the packet, you can modify without causing any other blocks to decrypt unexpectedly). – poncho Aug 26 '12 at 23:01
+1 creative attack – Rook Aug 27 '12 at 0:46
Yes, in theory, but this doesn't usually work in practice. You are right in that the last-block ciphertext in this example is malleable to some extent, but not as much as you seem to suggest. It is true that here, if you were to change the last ciphertext block so that the first byte of its corresponding plaintext corresponds to some value you chose, it would only take you $2^7$ trials on average.
However, as you identified, it would make the rest of the block garbage. As this is the last block, it often happens to contain the padding information, and it will therefore be impossible to correctly decrypt the message with overwhelming probability (as the padding will be malformed), which will defeat the point and reveal your tampering as well.
With regard to padding, it depends on the padding mode used - some modes will make this attack rather successful, but others will make it more difficult. For instance, from this page, bit padding will make it hard on you, because you will probably need a lot of consecutive zero bits in the plaintext block for decryption to succeed, whereas with byte padding, you only need the last byte to have the right value in it for decryption to work, which is easier to achieve (well, technically, you also need a bunch of zero bytes before, but some implementations skip that step - which is a vulnerability, by the way).
Now, if the rest of the block is not padding data (for instance, if padding is disabled as a result of the plaintext being guaranteed to be a multiple of 16 bytes in length, or if padding is stored somewhere else), and is completely ignored by the receiver for some reason (this could work with a null byte if it is interpreted as a null-terminated string, as you suggested), then this works and you have a malleable ciphertext weakness that can be exploited if the attacker knows the rough format of the plaintext.
This is a good example of why encryption without authentication is worthless, and you should always attach a MAC to any ciphertext to ensure it arrives intact to its destination.
-
How is padding checked for consistency? Do you know of a good paper that explains this? – Rook Aug 26 '12 at 21:17
@Rook See my edit. – Thomas Aug 26 '12 at 21:27
+1 good stuff. PKCS7 would severally undermine this attack. – Rook Aug 26 '12 at 21:36
To learn more about this sort of attack, see my answer Don't use encryption without message authentication, where I detail many examples of systems that were broken because they used encryption without authentication.
You will see that there is a wide variety of attacks that may be applicable, depending upon specific details of how the system works. You haven't provided enough details to assess exactly which of these might be applicable, and frankly it's probably not worth your time to analyze a hypothetical system in great detail, but there are many: reaction attacks, ciphertext modification, and more.
So, read about the failures of deployed systems. I think you'll learn a lot about how this can fail and why cryptographers recommend you should always use authenticated encryption, any time you think you need to encrypt.
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|
http://unapologetic.wordpress.com/2008/12/19/the-symmetrizer-and-antisymmetrizer/?like=1&_wpnonce=34114a6dfa
|
# The Unapologetic Mathematician
## The Symmetrizer and Antisymmetrizer
Today we’ll introduce two elements of the group algebra $\mathbb{F}[S_n]$ of the symmetric group $S_n$ which have some interesting properties. Each one, given a representation of the symmetric group, will give us a subrepresentation of that representation.
The symmetrizer gets its name from the way that it takes an arbitrary vector and turns it into one that the symmetric group will act trivially on. Specifically, we use the element
$\displaystyle S=\frac{1}{n!}\sum\limits_{\pi\in S_n}\pi$
That is, we take all $n!$ permutations in the symmetric group and — in a sense — average them out. If we compose this with any permutation $\hat{\pi}$ we find
$\displaystyle\begin{aligned}\hat{\pi}S=\hat{\pi}\frac{1}{n!}\sum\limits_{\pi\in S_n}\pi\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}\hat{\pi}\pi\end{aligned}$
But as $\pi$ runs over all the permutations in the group, $\hat{\pi}\pi$ does as well. So this is just the symmetrizer $S$ back again. The upshot is that if we have a representation $\rho:S_n\rightarrow\mathrm{GL}(V)$ we find that
$\displaystyle\begin{aligned}\left[\rho(\hat{\pi})\right]\left(\left[\rho(S)\right](v)\right)=\left[\rho(\hat{\pi}S)\right](v)\\=\left[\rho(S)\right](v)\end{aligned}$
Thus every vector in the image of $\rho(S)$ is left unchanged by the action of any permutation. That is, $\mathrm{Im}\left(\rho(S)\right)\subseteq V$ is a subrepresentation on which $S_n$ acts trivially.
The antisymmetrizer, on the other hand, on the other hand, will give us vectors on which the symmetric group acts by the signum representation. We use the group algebra element
$\displaystyle A=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\pi$
Now if we compose this with any permutation $\hat{\pi}$ we find
$\displaystyle\begin{aligned}\hat{\pi}A=\hat{\pi}\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\pi\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\hat{\pi}\pi\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\hat{\pi})\mathrm{sgn}(\hat{\pi}\pi)\hat{\pi}\pi\\=\mathrm{sgn}(\hat{\pi})\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\hat{\pi}\pi)\hat{\pi}\pi\\=\mathrm{sgn}(\hat{\pi})\hat{\pi}A\end{aligned}$
Now given a representation $\rho:S_n\rightarrow\mathrm{GL}(V)$ we find that
$\displaystyle\begin{aligned}\left[\rho(\hat{\pi})\right]\left(\left[\rho(A)\right](v)\right)=\left[\rho(\hat{\pi}A)\right](v)\\=\left[\rho(\mathrm{sgn}(\hat{\pi})A)\right](v)\\=\mathrm{sgn}(\hat{\pi})\left[\rho(A)\right](v)\end{aligned}$
Thus every vector in the image of $\rho(A)$ is multiplied by the signum of any permutation. That is, $\mathrm{Im}\left(\rho(A)\right)\subseteq V$ is a subrepresentation on which $S_n$ acts by the signum representation.
Now, one thing to be careful about: I haven’t said that the subrepresentations are nontrivial. That is, when we (anti)symmetrize a representation, the subrepresentation we get may be zero — maybe no vectors in the representation transform trivially or by the signum representation. In fact, let’s check what happens when we multiply the symmetrizer and antisymmetrizer:
$\displaystyle\begin{aligned}SA=\left(\frac{1}{n!}\sum\limits_{\pi\in S_n}\pi\right)\left(\frac{1}{n!}\sum\limits_{\hat{\pi}\in S_n}\mathrm{sgn}(\hat{\pi})\hat{\pi}\right)\\=\frac{1}{n!}\frac{1}{n!}\sum\limits_{\substack{\pi\in S_n\\\hat{\pi}\in S_n}}\mathrm{sgn}(\hat{\pi})\pi\hat{\pi}\\=\frac{1}{n!}\frac{1}{n!}\sum\limits_{\substack{\pi\in S_n\\\hat{\pi}\in S_n}}\mathrm{sgn}(\pi)\mathrm{sgn}(\pi\hat{\pi})\pi\hat{\pi}\\=\frac{1}{n!}\frac{1}{n!}\left(\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\right)\left(\sum\limits_{\pi\hat{\pi}\in S_n}\mathrm{sgn}(\pi\hat{\pi})\pi\hat{\pi}\right)\\=\frac{1}{n!}\frac{1}{n!}\left(\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\right)A\\=0\end{aligned}$
Where the sum comes to zero because we’re just adding up $\frac{n!}{2}$ terms where $\mathrm{sgn}(\pi)=1$ and $\frac{n!}{2}$ where $\mathrm{sgn}(\pi)=-1$, and so everything cancels out. That is, the symmetric part of an antisymmetrized representation is trivial. Similarly, the antisymmetric part of a symmetrized representation is trivial.
### Like this:
Posted by John Armstrong | Algebra, Representation Theory
## 5 Comments »
1. [...] and tomorrow I want to take last Friday’s symmetrizer and antisymmetrizer and apply them to the tensor representations of , which we know also carry symmetric group [...]
Pingback by | December 22, 2008 | Reply
2. [...] of , which actions commute with each other. Our antisymmetric tensors are the image of a certain action from the symmetric group, which is an intertwiner of the action. Thus we have a one-dimensional representation of , which [...]
Pingback by | December 31, 2008 | Reply
3. [...] if its value is unchanged when we swap two of its inputs. Equivalently, it commutes with the symmetrizer. That is, it must kill everything that the symmetrizer kills, and so must really define a linear [...]
Pingback by | October 22, 2009 | Reply
4. [...] may well not be symmetric itself. Instead, we will take the tensor product of and , and then symmetrize it, to give . This will be bilinear, and it will work with our choice of grading, but will it be [...]
Pingback by | October 26, 2009 | Reply
5. [...] may not be antisymmetric itself. Instead, we will take the tensor product of and , and then antisymmetrize it, to give . This will be bilinear, but will it be [...]
Pingback by | October 27, 2009 | 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.
|
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http://physics.stackexchange.com/questions/20212/expansion-of-helmholtz-energy/20214
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# Expansion of Helmholtz energy
To get an expansion of Helmholtz energy of
a) an ideal gas b) a Van der waals gas
we must integrate
$\left ( \frac{\delta A }{\delta V} \right )_{T}=-P$
I saw the solution is :
Can you explain why is that?
-
## 1 Answer
Consider just part (a). we use the ideal gas law for $P$, that is $$-P=-\frac{nRT}{V}$$ and substituting this in for $P$ we get $$\delta A=-\frac{nRT}{V}\delta V\implies\Delta A=-nRT\int_{V_1}^{V_{2}}\frac{1}{V}dV$$ which gives $$\Delta A=-nRT(\ln(V_2)-\ln(V_1))=-nRT\ln\left(\frac{V_2}{V_2}\right)$$ I'm not sure where the $n$ went... It's key we held $T$ constant. Hope this helps.
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http://stats.stackexchange.com/questions/2125/whats-the-difference-between-correlation-and-simple-linear-regression/30760
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# What's the difference between correlation and simple linear regression?
Is it cheating if I know the answer, but ask it anyways?
I figure it's good for people to know :)
In particular, I am referring to the Pearson product-moment correlation coefficient
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3
I have no idea what this question is about. – Robby McKilliam Aug 26 '10 at 1:35
11
It sounds like it is about the difference between correlation and simple linear regression. – Brett Magill Aug 26 '10 at 2:19
– gung Sep 20 '12 at 19:39
## 6 Answers
What's the difference between the correlation between $X$ and $Y$ and a linear regression predicting $Y$ from $X$?
First, some similarities:
• the standardised regression coefficient is the same as Pearson's correlation coefficient
• The square of Pearson's correlation coefficient is the same as the $R^2$ in simple linear regression
• Neither simple linear regression nor correlation answer questions of causality directly. This point is important, because I've met people that think that simple regression can magically allow an inference that X causes Y.
Second, some differences:
• The regression equation (i.e., $a + bX$) can be used to make predictions on $Y$ based on values of $X$
• While correlation typically refers to the linear relationship, it can refer to other forms of dependence, such as polynomial or truly nonlinear relationships
• While correlation typically refers to Pearson's correlation coefficient, there are other types of correlation, such as Spearman's.
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Here is an answer I posted on the graphpad.com website:
Correlation and linear regression are not the same. Consider these differences:
• Correlation quantifies the degree to which two variables are related. Correlation does not fit a line through the data.
• With correlation you don't have to think about cause and effect. You simply quantify how well two variables relate to each other. With regression, you do have to think about cause and effect as the regression line is determined as the best way to predict Y from X.
• With correlation, it doesn't matter which of the two variables you call "X" and which you call "Y". You'll get the same correlation coefficient if you swap the two. With linear regression, the decision of which variable you call "X" and which you call "Y" matters a lot, as you'll get a different best-fit line if you swap the two. The line that best predicts Y from X is not the same as the line that predicts X from Y (unless you have perfect data with no scatter.)
• Correlation is almost always used when you measure both variables. It rarely is appropriate when one variable is something you experimentally manipulate. With linear regression, the X variable is usually something you experimentally manipulate (time, concentration...) and the Y variable is something you measure.
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In the single predictor case of linear regression, the standardized slope has the same value as the correlation coefficient. The advantage of the linear regression is that the relationship can be described in such a way that you can predict (based on the relationship between the two variables) the score on the predicted variable given any particular value of the predictor variable. In particular one piece of information a linear regression gives you that a correlation does not is the intercept, the value on the predicted variable when the predictor is 0.
In short - they produce identical results computationally, but there are more elements which are capable of interpretation in the simple linear regression. If you are interested in simply characterizing the magnitude of the relationship between two variables, use correlation - if you are interested in predicting or explaining your results in terms of particular values you probably want regression.
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Correlation analysis only quantifies the relation between two variables ignoring which is dependent variable and which is independent. But before appliyng regression you have to calrify that impact of which variable you want to check on the other variable.
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From correlation we can only get an index describing the linear relationship between two variables; in regression we can predict the relationship between more than two variables and can use it to identify which variables x can predict the outcome variable y.
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– gung Sep 20 '12 at 19:32
Correlation is an index (just one number) of the strength of a relationship. Regression is an analysis (estimation of parameters of a model and statistical test of their significance) of the adequacy of a particular functional relationship. The size of the correlation is related to how accurate the predictions of the regression will be.
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http://physics.stackexchange.com/questions/tagged/gravity?sort=votes&pagesize=30
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# Tagged Questions
Gravity is an attractive force that affects and is effected by all mass and - in general relativity - energy, pressure and stress. Prefer newtonian-gravity or general-relativity if sensible.
13answers
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### How does gravity escape a black hole?
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### Why are our planets in the solar system all on the same disc/plane/layer?
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6answers
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### Can I survive a free fall using a ramp and a rope?
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### Why can't we feel the Earth turning?
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### Vassiliev Higher Spin Theory and Supersymmetry
Recently there is renewed interest in the ideas of Vassiliev, Fradkin and others on generalizing gravity theories on deSitter or Anti-deSitter spaces to include higher spin fields (utilizing known ...
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### experimental bounds on spacetime torsion
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### Kerr Geometry, Separability and Twistors
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### Could gravity be an emergent property of nature?
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3answers
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### Why is gravity such a unique force?
My knowledge on this particular field of physics is very sketchy, but I frequently hear of a theoretical "graviton", the quantum of the gravitational field. So I guess most physicists' assumption is ...
4answers
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### Would you be weightless at the center of the Earth?
If you could travel to the center of the Earth (or any planet), would you be weightless there?
6answers
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### The speed of gravity?
Sorry for the layman question, but it's not my field. Suppose this thought experiment is performed. Light takes 8 minutes to go from the surface of the Sun to Earth. Imagine the Sun is suddenly ...
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### How can a black hole emit X-rays?
Considering that a black hole's gravity prevents light from escaping, how can a black hole emit X-rays? Visible light and X-rays are both electromagnetic radiation, shouldn't the black hole's gravity ...
8answers
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### Why are L4 and L5 lagrangian points stable?
This diagram from wikipedia shows the gravitational potential energy of the sun-earth two body system, and demonstrates clearly the semi-stability of the L1, L2, and L3 lagrangian points. The blue ...
3answers
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### Are modified theories of gravity credible?
I'm a statistician with a little training in physics and would just like to know the general consensus on a few things. I'm reading a book by John Moffat which basically tries to state how GR makes ...
2answers
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### Does gravity travel at the speed of light?
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8answers
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### How exactly does curved space-time describe the force of gravity?
I understand that people explain (in layman's terms at least) that the presence of mass "warps" space-time geometry, and this causes gravity. I have also of course heard the analogy of a blanket or ...
1answer
251 views
### Flames with no gravity?
I was watching "Solaris" (Tarkovsky) today, and noticed this: in some moment the space station changed orbit and the people inside experienced zero-gravity. At that moment, a candlestick passed ...
8answers
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### How can anything ever fall into a black hole as seen from an outside observer?
The event horizon of a black hole is where gravity is such that not even light can escape. This is also the point I understand that according to Einstein time dilation will be infinite for a ...
3answers
600 views
### Why doesn't a fly fall off the wall?
Pretty simple question, but not an obvious answer at least not to me. I mean you can't just place a dead fly on the wall and expect it to stay there, he will fall off due to gravity. At first I ...
7answers
591 views
### Is the quantization of gravity necessary for a quantum theory of gravity?
The other day in my string theory class, I asked the professor why we wanted to quantize gravity, in the sense that we want to treat the metric on space-time as a quantum field, as opposed to, for ...
1answer
61 views
### Instantons and Non Perturbative Amplitudes in Gravity
In perturbative QFT in flat spacetime the perturbation expansion typically does not converge, and estimates of the large order behaviour of perturbative amplitudes reveals ambiguity of the ...
7answers
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### What does it mean to say “Gravity is the weakest of the forces”?
I can understand that on small scales (within an atom/molecule), the other forces are much stronger, but on larger scales, it seems that gravity is a far stronger force; e.g. planets are held to the ...
8answers
958 views
### What are the reasons to expect that gravity should be quantized?
What I am interested to see are specific examples/reasons why gravity should be quantized. Something more than "well, everything else is, so why not gravity too". For example, isn't it possible that a ...
7answers
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### Is it theoretically possible to shield gravitational waves?
Electromagnetic waves can be shielded by a perfect conductor. What about gravitational waves?
3answers
323 views
### Why don't I feel pressure on my body when swimming under water?
If I put a couple of lead bricks on my foot, there would be a definite sensation of a heavy, perhaps even painful, force. Calculating the pressure for $20 kg$ of lead over a $100 cm^2$ area of my ...
2answers
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### Why aren't gas planets and stars fuzzy?
The edge of Jupiter looks very sharp. Even more bothersome, the edge of the sun looks sharp, aside from kind of a soup of particles floating above it. The sun's surface has an incredibly low ...
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533 views
### How do horseshoe orbits work?
An asteroid was recently discovered that is in a horseshoe orbit with respect to the earth. Is there an intuitive explanation for these orbits? It seems that the earth acts as a repulsive force where ...
5answers
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### Is Gravity an entropic force after all?
Recently, there was a rapid communication published in Phys.Rev.D (PRD 83, 021502), titled "Gravity is not an entropic force", that claimed that an experiment performed in 2002 with ultra cold ...
3answers
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### Why does each celestial object spin on its own axis?
AFAIK all the celestial objects have a spin motion around its axis. What is the reason for this? If it must rotate by some theory, what decides it's direction and speed of rotation? Is there any ...
5answers
2k views
### Why do rockets have multiple stages?
What is the advantage for rockets to have multiple stages? Wouldn't a single stage with the same amount of fuel weigh less? Note I would like a quantitative answer, if possible :-)
6answers
608 views
### Particle wavefunction and gravity
Suppose a particle has 50% probability of being at location $A$, and 50% probability being at location $B$ (see double slit experiment). According to QM the particle is at both $A$ and $B$ at the same ...
3answers
3k views
### How do we determine the mass of a black hole?
Since by definition we cannot observe black holes directly, how do astronomers determine the mass of a black hole? What observational techniques are there that would allow us to determine a black ...
3answers
803 views
### Significance of the second focus in elliptical orbits
1.In classical mechanics, using Newton's laws, the ellipticity of orbits is derived. It is also said that the center of mass is at one of the foci. 2.Each body will orbit the center of the mass of ...
5answers
707 views
### What is a good mathematical description of the Non-renormalizability of gravity?
By now everybody knows that gravity is non-renormalizable, what is often lacking is a simplified mathematical description of what that means. Can anybody provide such a description?
1answer
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### Does gravitation of a sphere equal gravitation of a point?
Under Newtonian model of gravity, a perfect sphere creates the same gravitation field as a point mass in its center. General Relativity describes gravitation differently. How much this difference ...
1answer
45 views
### Relationship between Weak Cosmic Censorship and Topological Censorship
The weak cosmic censorship states that any singularity cannot be in the causual past of null infinity (reference). The topological censorship hypothesis states that in a globally hyperbolic, ...
2answers
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### What is the terminal velocity for a mobile phone
You may have seen the story of the iPhone which was dropped from perhaps 13,500 feet by a skydiver - it survived. This made me wonder how to work out the terminal velocity for something like that. ...
1answer
129 views
+50
### Is period of rotation relative?
My question is inspired by the following answer by voix to another problem: "There is a real object with relativistic speed of surface - millisecond pulsar. The swiftest spinning pulsar currently ...
5answers
937 views
### The square in the Newton's law of universal gravitation is really a square?
When I was in the university (in the late 90s, circa 1995) I was told there had been research investigating the $2$ (the square of distance) in the Newton's law of universal gravitation. ...
1answer
940 views
### Does a guitar sound different in zero (or micro) gravity?
Seeing a video of astronaut Chris Hadfield playing a guitar on the International Space Station made me wonder if a guitar or other stringed instrument played in zero-G would sound any different than ...
2answers
84 views
### Why is Jupiter so sharply defined?
In photographs of Jupiter, the limb seems extremely definite. Being a gas giant, my naive self thinks that the atmosphere should have a more gradual cut off, creating a hazy effect similar to that on ...
3answers
857 views
### “Speed” of Gravity and Speed of Light
Some threads here touching speed of gravity made me think about that. This lead to some questions. The speed of gravity was not measured until today (at least there are no undebated papers to that ...
2answers
247 views
### Does relativistic mass have weight?
If an object was sliding on an infinitely long friction-less floor on Earth with relativistic speeds (ignoring air resistance), would it exert more vertical weight force on the floor than when it's at ...
7answers
3k views
### Tesla's theory of gravity
I was reading up on Tesla's Wikipedia page last night, and I came across this: When he was 81, Tesla stated he had completed a "dynamic theory of gravity". He stated that it was "worked out ...
4answers
1k views
### How does gravitational lensing account for Einstein's Cross?
Einstein's Cross has been attributed to gravitational lensing. However, most examples of gravitational lensing are crescents known as Einstein's rings. I can easily understand the rings and crescents, ...
5answers
7k views
### jumping into water
Two questions: Assuming you dive head first or fall straight with your legs first, what is the maximal height you can jump into water from and not get hurt? In other words, an H meter fall into ...
8answers
3k views
### Is Gravity Energy?
This might be stupid, but is gravity a form of energy? And, if so, couldn't we use it for power?
5answers
6k views
### How exactly does time slow down near a black hole?
How exactly does time slow down near a black hole? I have heard this as a possible way of time traveling, and I do understand that it is due in some way to the massive gravity around a black hole, but ...
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http://mathoverflow.net/questions/25509/effectively-closed-computable-functions
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## Effectively closed computable functions
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I've recently been interested in the following type of functions. A total computable function f:N→N is effectively closed if there is a computable function p such that f[N \ We] = N \ Wp(e), where We is the e-th c.e. set.
Have effectively closed functions been studied? If so, what are they normally called?
I would also appreciate pointers to some uses and/or alternative characterizations of effectively closed functions.
Motivation. It is well-known that there is a near-perfect analogy between the adjectives computable and continuous. For example, a total function f:N→N is computable if and only if it is effectively continuous, i.e. there is a computable function p such that f-1[We] = Wp(e). [For the backward implication, let q be a computable function such that Wq(n) = {n} and use the composite p∘q to enumerate the graph of f.] A similar trick shows that a total function is effectively open if and only if it is computable. However, a total computable function is not necessarily effectively closed since that entails that the range of f is computable and, indeed, that f maps every computable set onto a computable set. Also, the notion is nontrivial since non-constant polynomials and increasing functions are effectively closed.
Update. Joel David Hamkins gave the following characterization of effectively closed computable functions: they are the computable functions f:N→N for which there is a computable b:N→N such that f-1(n) ⊆ {0,1,...,b(n)} for every n ∈ N. Although I accepted Joel's answer, the main question is still open.
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François, it's a great problem! – Joel David Hamkins May 22 2010 at 2:40
## 2 Answers
I like your concept a lot, and have been able to find a characterization.
Suppose that $f:N\to N$ is effectively closed in your sense.
First, as you mentioned, it is easy to see that $\text{ran}(f)$ is computable, since by taking $W_e$ to be empty your equation shows that $\text{ran}(f)$ is both c.e. and co-c.e.
Second, I claim that $f$ is finite-to-one. To see this, suppose that $f^{-1}(k)$ is infinite for some $k$. Define a c.e. set $W_e$ as follows: At stage $s$, if we see that $k$ is still not in $W_{\rho(e),s}$, the state-$s$ approximation to $W_{\rho(e)}$, then enumerate the next element of $f^{-1}(k)$ into $W_e$. (Although this definition may look circular, since I am defining $W_e$ by reference to $W_{\rho(e)}$, the definition is legitimate by an application of the Recursion Theorem. That is, I really define $W_{r(e)}$, and then find $e$ such that $W_e=W_{r(e)}$.) Note that if $k$ is never enumerated into $W_{\rho(e)}$, then I will eventually put all of $f^{-1}(k)$ into $W_e$, which will result in $k\notin f[N-W_e]$, but $k\in N-W_{\rho(e)}$, a contradiction. Alternatively, if $k\in W_{\rho(e),s}$, then $f^{-1}(k)\cap W_e$ has at most $s$ members, and so there are $a\in N-W_e$ with $f(a)=k$, placing $k$ into $f[N-W_e]$ but not in $N-W_{\rho(e)}$, again a contradiction.
A similar argument shows actually that the function $k\mapsto f^{-1}(k)$ is computable. Namely, define the set $W_e$ by the following procedure. At stage $s$, look at every $k\leq s$, and if $k\notin W_{\rho(e),s}$, then enumerate all of $f^{-1}(k)\cap s$ into $W_e$. (Again, appeal to Recursion Theorem to get such an $e$.) In other words, as long as $k$ is not in $W_{\rho(e),s}$, then we put all elements of $f^{-1}(k)$ below $s$ into $W_e$.
If $k\notin W_{\rho(e)}$, then $f^{-1}(k)\subset W_e$, and so $k\notin f[N-W_e]$, contradicting $k\in N-W_{\rho(e)}$. Thus, $W_{\rho(e)}=N$. From this, it follows that $W_e=N$. Now, note that $k\in W_{\rho(e)}$ implies $k\in W_{\rho(e),s_k}$ for some stage $s_k$, and so $f^{-1}(k)$ is a subset of $s_k$. By applying $f$ to each value below $s_k$, we see that the map $k\mapsto f^{-1}(k)$ is a computable function.
This means that $f$ has a particularly simple form. Namely, there is a computable partition $N=\bigsqcup_k B_k$, with each $B_k$ finite, such that $f$ maps elements of $B_k$ to $k$. (Note that some $B_k$ may be empty.)
Conversely, every function with such a form is computably closed in your sense. Suppose that $f$ arises from such a computable partition of $N$ into finite sets $B_k$. Given any program $e$, enumerate $k$ into $W_{\rho(e)}$ when all of $B_k$ gets enumerated into $W_e$. It follows that $f[N-W_e]=N-W_{\rho(e)}$, as desired.
This provides a characterization of the effectively closed computable functions:
Theorem. A computable function $f:N\to N$ is effectively closed if and only if $f$ is finite-to-one and the map $k\mapsto f^{-1}(k)$ is computable.
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This is great! I had gotten the finite-to-one part using a completely different argument, but not the computable bound on $f^{-1}(k)$. – François G. Dorais♦ May 23 2010 at 4:04
This doesn't answer the main question, but I'm accepting it anyway since the argument is so beautiful! – François G. Dorais♦ May 23 2010 at 4:10
Thanks, François, you are very kind. – Joel David Hamkins May 23 2010 at 11:59
I think there's a minor error in the 6th paragraph: "that $W_e=N$" should be "that $W_e = ran(f)$." – François G. Dorais♦ May 23 2010 at 15:30
I think $W_e=N$ is correct. The point is that once you know $W_{\rho(e)}=N$, then $N-W_{\rho(e)}$ is empty, so it had better be that $W_e=N$ or else $f[N-W_e]$ won't be empty. Or have I misunderstood? – Joel David Hamkins May 23 2010 at 16:47
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.
I think that this is probably studied in Russian/Markov School of Constructivism. A good starting point might be the chapters on Russian Constructivism in Michael Beeson's book:
Michael Beeson, "Foundations of Constructive Mathematics: Metamathematical Studies", Springer, 1985
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Is there a particular reason you are citing this reference? Most of us don't have this book, so perhaps you could supply a relevant sentence or two -- otherwise such a comment is not very useful. – András Salamon Aug 14 2010 at 15:26
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http://mathoverflow.net/questions/38204/morphisms-of-frobenius-manifolds-definitions-and-examples/38257
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## Morphisms of Frobenius manifolds: Definitions and examples?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Frobenius manifolds arise in the study of quantum cohomology and mirror symmetry -- roughly they are manifolds (or varieties or whatever) such that the tangent spaces are Frobenius algebras (there are more conditions to be satisfied -- flat metric, flat identity, etc). A basic reference is Manin's book "Frobenius manifolds, quantum cohomology, and moduli spaces".
I have studied Frobenius manifolds for a while now, and I have so far never seen any definitions nor any examples of morphisms (other than, perhaps, isomorphisms).
Are there any reasonable definitions of morphism of Frobenius manifolds? Are there any interesting examples?
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## 1 Answer
This is going to be more a non-answer than an answer:
I don't think there is a useful notion of morphism between Frobenius manifolds, besides the notions of (local) isomorphisms, and of a sub-Frobenius manifold (which is not the same thing as a submanifold of a Frobenius manifolds).
To argue completely in intrinsic terms: if you require that a morphism preserves the metric on the tangent spaces, then it is necessarily a local embedding. Locally, the image is then a sub-Frobenius manifold, i.e. a flat submanifold $M \subset N$ such that the multiplication on the tangent bundle $T_N \otimes T_N \to T_N$ maps $T_M \otimes T_M$ to $T_M$.
Examples of sub-Frobenius manifolds arise naturally, e.g. in quantum cohomology, $\oplus_p H^{p, p}(X, \mathbb{C})$ is a sub-Frobenius manifold of $H^*(X, \mathbb{C})$.
To argue with a little more context: a more flexible notion of morphisms of Frobenius manifolds wouldn't be useful unless it arises naturally. However, even for the most simple situations you could think of, say if you compare the quantum cohomology of a product $X \times Y$ with the quantum cohomology of $X$ and $Y$, there does not seem to be a useful morphism between the Frobenius manifolds involved. Instead, the (Frob. mfd of) QC of $X \times Y$ is a tensor product of the QCs of $X$ and $Y$. Other nice statements are known for Grassmannians $G(k, n)$, whose QC is a kind of mixture of anti-symmetric and symmetric $k$-fold tensor product of the QC of $P^{n-1}$ (see papers by subsets of Bertram/Ciocan-Fontanine/Kim/Sabbah on abelian/non-abelian quotients). Again, a very nice statement, but no morphisms of Frob. manifolds anywhere.
[If I may sneak in a little advertising: If $\tilde X$ is the blow-up of $X$ at a point, then the QC of then the QC of $\tilde X$ has a partial compactification, and the boundary is a Frob. submanifold isomorphic to QC of $X$. Well, unfortunately that's a little bit of a lie, as the multiplication has a pole on the added boundary divisor, and so arXiv:math.AG/0403260 doesn't read quite as nicely as the statement above. But again, while there is almost an inclusion of QC of $X$ into QC of $\tilde X$, there is no morphism in the other direction.]
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http://physics.stackexchange.com/questions/24680/a-conceptual-problem-with-euler-bernoulli-beam-theory-and-euler-buckling
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A conceptual problem with Euler-Bernoulli beam theory and Euler buckling
Euler-Bernoulli beam theory states that in static conditions the deflection $w(x)$ of a beam relative to its axis $x$ satisfies
$$EI\frac{\partial^4}{\partial x^4}w(x)=q(x)\ \ \ \ (1)$$
where $E$ is Young's modulus, $I$ is the second moment of area and $q(x)$ is an external loading. This equation can be rather easily derived. The standard interpretation is that $EIw''(x)$ is the internal bending moment that an element of the rod exerts on its neighbour.
The standard analysis of Euler buckling is that if in addition one has an axial compression of force $F$, it also exerts an internal moment, equal to $Fw$. Torque balance then means
$$EI w''(x)+Fw=0\ ,\ \ \ \ (2)$$
the solution of which is clearly $$w=A \sin\left(\sqrt{\frac{EI}{F}x}+\phi\right)\ .$$ and from this one immediately obtains Euler buckling criterion $F_c=\pi^2\frac{EI}{L^2}$.
All this is standard textbook stuff. BUT, if there is no axial force, then Eq. (2) reads
$$EIw''(x)=0$$
which is much stronger with (1). That is, a general solution to (1) (the ones given in wikipedia, for example) does not satisfy (2). How can one resolve this?
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But wouldn't the last equation correspond to the first one with $q(x)\equiv 0$? – Raskolnikov May 1 '12 at 8:40
1
@Raskolnikov No, it has 2 derivatives less. – yohBS May 1 '12 at 14:03
Yes, but a solution of the equation with two derivatives is still a solution of the one with four derivatives. The converse is not true, but what is lost in appearance must reside somewhere in boundary or other conditions. – Raskolnikov May 1 '12 at 19:42
1 Answer
The buckling problem with $F=0$ is $EIw''=0$, with boundary values $w(0)=w(L)=0$. The corresponding beam problem with $q=0$ is $EIw''''=0$, with boundary values $w(0)=w(L)=0$ and $w''(0)=w''(L)=0$, since there is no applied moment at the ends. Both problems have the same solution: $w(x)=0$. Raskolnikov is right: it is in the boundary conditions. You don't want the general solution.
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You have to specify the first derivatives too on both ends in the fourth derivative problem. – Ron Maimon May 4 '12 at 1:00
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http://gowers.wordpress.com/2007/10/04/when-are-two-proofs-essentially-the-same/?like=1&_wpnonce=9e687b4e3a
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# Gowers's Weblog
Mathematics related discussions
## When are two proofs essentially the same?
A couple of years ago I spoke at a conference about mathematics that brought together philosophers, psychologists and mathematicians. The proceedings of the conference will appear fairly soon—I will give details when they do. My own article ended up rather too long, because I found myself considering the question of “essential equality” of proofs. Eventually, I cut that section, which was part of a more general discussion of what we mean when we attribute properties to proofs, using informal (but somehow quite precise) words and phrases like “neat”, “genuinely explanatory”, “the correct” (as opposed to merely “a correct”), and so on. It is an interesting challenge to try to be as precise as possible about these words, but I found that even the seemingly more basic question, “When are two proofs the same?” was pretty hard to answer satisfactorily. Since it is also a question on which we all have views (since we all have experience of the phenomenon), it seems ideal for a post. You may have general comments to make, but I’d also be very interested to hear of your favourite examples of different-seeming proofs that turn out, on closer examination, to be based on the same underlying idea (whatever that means).
A general remark that becomes clear quickly when one thinks about this is that there are fairly standard methods for converting one proof into another, and when we apply such a method then we tend to regard the two proofs as not interestingly different. For example, it is often possible to convert a standard inductive proof into a proof by contradiction that starts with the assumption that $X$ is a minimal counterexample. In fact, to set the ball rolling, let me give an example of this kind: the proof that every number can be factorized into primes.
The usual approach is the minimal-counterexample one: if there is a positive integer that cannot be factorized, let $n$ be a minimal one; $n$ is not prime, so it can be written as $ab$ with both $a$ and $b$ greater than 1; by minimality $a$ and $b$ can be factorized; easy contradiction.
Now let’s give the inductive version. We need the “strong” principle of induction. Suppose, then, that we have proved that every integer up to $n-1$ can be factorized and are trying to factorize $n$. If $n$ is already prime, then we are done. Otherwise, we can write $n=ab$ with both $a$ and $b$ greater than 1. But then $a$ and $b$ are less than $n$, so by induction they can be factorized. Putting together those factorizations gives a factorization for $n$ and the inductive step is complete.
(I missed out the proof that the induction starts, just as I did not check in the first proof that there exists a number that can be factorized.)
That’s a rather boring example of sameness of proofs—boring because they are so obviously the same, and one can even point to a mechanical procedure that converted one into the other (which can be applied to many other proofs). More interesting are examples where the sameness becomes apparent only after a more complicated process of transformation. At this point, I’d like to mention the theorem that I discussed in great detail in the bit that I removed from my article: the irrationality of $\sqrt{2}$.
Very briefly, here’s the standard proof. If $\sqrt{2}$ is rational, then we can write it as $p/q$. Let us do so in such a way that $p$ and $q$ are not both even. We know that $p^2=2q^2$, so $p^2$ is even, so $p$ is even, so $p=2r$ for some integer $r$. This gives us $4r^2=2q^2$, so $q^2=2r^2$, so $q^2$ is even, so $q$ is even, which is a contradiction.
Now, equally briefly, here is another proof. Suppose again that $\sqrt{2}=p/q$ and let $p/q$ be written in its lowest terms. Now $\sqrt{2}=\frac{1}{\sqrt{2}-1}-1=\frac{2-\sqrt{2}}{\sqrt{2}-1}$. Substituting $p/q$ for $\sqrt{2}$ and tidying up, this gives us that $\frac{p}{q}=\frac{2q-p}{p-q}$. But $p<2q$, so the denominator of the right-hand side is less than $q$, which contradicts the minimality of $q$ (and hence $p$ too, since their ratio is determined).
Now there are some similarities between those two arguments: both assume that $\sqrt{2}=p/q$ and aim for a contradiction, assuming that $p/q$ is in its lowest terms. But there are also definite differences. For example, the first proof doesn’t actually care whether $p$ and $q$ are minimized: it just wants them not both to be even. The second proof doesn’t care at all about the factorizations of $p$ and $q$ but does care about their sizes. So I’d maintain that they are different proofs. Having said that, I once put that to Terence Tao in a conversation and he immediately adopted a more general perspective from which one could regard the two arguments as different ways of carrying out the same essential programme. It had something to do with $SL_2(\mathbb{Z})$ if I remember correctly. Terry, if you felt like reminding me of exactly what you said, that would be a perfect illustration of “non-obvious equivalence” between proofs.
Here, though, is a third argument for the irrationality of $\sqrt{2}$. You just work out the continued-fraction expansion. It comes out to be $1+{\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}}}$. Since the continued-fraction expansion of any rational number terminates, $\sqrt{2}$ is irrational.
Here’s a fourth. Let $p$ and $q$ be coprime and suppose that $p^2$ does not equal $2q^2$. Then $p+2q$ and $p+q$ are also coprime. We also find that $(p+2q)^2-2(p+q)^2=-(p^2-2q^2)$. From this observation we can build a succession of fractions $p_n/q_n$, each in their lowest terms, with $q_n$ tending to infinity and with $|p_n^2-2q_n^2|$ all equal. From that we find that $|2-p_n^2/q_n^2|$ has order of magnitude $1/q_n^2$, and from that it is easy to verify that $|\sqrt{2}-p_n/q_n|$ has order of magnitude $1/q_n^2$ as well. But it is impossible to find a sequence of rationals with denominators tending to infinity that approach a rational this quickly. Indeed, $\frac{p}{q}-\frac{r}{s}=\frac{ps+qr}{qs}$, which for fixed $s$ has order of magnitude $1/q$. (For large and coprime $p$ and $q$, the difference cannot be zero.)
I won’t demonstrate it here, but it’s not too hard an exercise to see that the second, third and fourth proofs are all essentially the same. (At some point, perhaps I’ll put a link to a more detailed account of exactly why, at least for the second and third. The fourth has only just occurred to me.) For instance, the construction of the sequence $(p_n,q_n)$ in the fourth proof is the same as the construction of the continued-fraction expansion of $\sqrt{2}$ (as lovers of Pell’s equation will know). Also, the way that we produced $(p_n,q_n)$ from $(p_{n-1},q_{n-1})$ is just the reverse of the way that we produced a smaller fraction from $p/q$ in the second proof. The fourth proof is perhaps very slightly different, in that it involved inequalities, but that was not a fundamental difference. (It would be interesting to be precise about why not—this is what I haven’t got round to thinking about yet.)
To repeat: for the time being I am interested in responses of two kinds. First, I’d like to see lots of examples of proofs that turn out to be essentially the same when they might not initially seem that way. Secondly, I’m keen to see examples of “conversion techniques”—that is, methods for transforming a proof into another that is not interestingly different. See this comment for some interesting links, though here I am not so much looking for a formal theory right down at the level of logical formulae. Rather, I would like as good a picture as possible of high-level equivalence of proofs.
Some general questions are quite interesting. For instance, if two proofs are essentially the same, must there always be some more general perspective from which one can see that the only differences between them consist in arbitrary choices that do not affect the argument in an important way? (A simple example of what I mean is something like replacing “Let $\epsilon=\delta/5$” by “Let $\epsilon=\delta/3$” in an argument in real analysis. But there are much more interesting examples of this.) Also, is “essential equivalence” really an equivalence relation, or could one morph in several stages from one proof to another that was “genuinely different”? (My own feeling, by the way, is that the morph would itself be a demonstration of essential equivalence, but perhaps a really good example might change my mind on that.) Is it ever possible to give a completely compelling argument that two proofs are genuinely different? How would one go about such a task? Could one attach an invariant to a proof, perhaps?
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### 68 Responses to “When are two proofs essentially the same?”
1. John Armstrong Says:
October 4, 2007 at 3:22 pm | Reply
I think it should end up being analogous to a more concrete situation I can sketch.
Consider a formal system, and construct from it a category. The objects are the well-formed formulæ, and the morphisms between them are proofs. This is particularly well-trodden ground. Better mathematicians than I have pointed out that in the predicate calculus, conjunction behaves like a product, implication behaves like exponentiation, and so on.
Anyhow, from this, let’s add another layer of formal structure: rewrite rules. A proof says that we can replace part of a formula by another chunk of formula, but a rewrite rule will say we can replace part of a proof by another chunk of proof. These will play the role of 2-morphisms in our category.
John Baez has spoken in his seminar about the particular case of typed lambda calculi, which are cartesian closed categories. When we change the tools like beta-reductions from equalities to arrows, we start talking about cartesian closed 2-categories, which hopefully will give insight into the use of lambda-calculi as models of computation, since the 2-morphisms now have the sense of a process.
Anyhow, at the higher level you’re talking about, we still have a notion of statements as objects and proofs as morphisms from hypotheses to comclusions. The equivalences of proofs you’re looking for should be 2-isomorphisms.
But there are 2-morphisms that are not invertible. In particular, you have 2-isomorphisms between the second, third, and fourth proof of the irrationality of $\sqrt{2}$. Is it possible that the first proof is connected by only a 2-morphism? That is, can its proof be “rewritten” into one of the other forms by using a non-invertible rule, or vice versa?
2. Theo Says:
October 4, 2007 at 4:10 pm | Reply
Here are two worked examples that I wrote a while back. First, I discussed the infinitude of primes: I argue that Euclid’s proof and Fürstenberg’s proof are essentially the same. Then, I talked about the Brouwer fixed-point theorem in two dimensions, in which I slowly perturb the traditional category-theoretic and Sperner’s-lemma-based proofs towards each other, but I concluded that those proofs are merely “homotopic”, not “essentially the same”. In both entries, my method of analysis was simply a careful unpacking of the proofs into their most basic statements.
3. Terence Tao Says:
October 4, 2007 at 5:12 pm | Reply
Dear Tim,
I can think of four general ways in which one might try to capture the notion of equivalence between proof A and proof B: semantic, syntactic, algorithmic, and conceptual.
In the semantic (or model-theoretic) approach, one would try to describe the largest set of models to which proof A “naturally” generalises, and compare that set to the corresponding set for proof B. For instance, proof A of a linear algebra fact about real vector spaces might easily extend to complex vector spaces, but not to vector spaces with torsion or to modules of commutative rings, PIDs, or whatever, whereas proof B might have a different range of generalisation. There are at least two difficulties though with this approach; firstly, a proof sometimes has to be rewritten or otherwise “deconstructed” until it becomes obvious how to extend it properly, e.g. a proof involving real vector spaces, which relies crucially at one point on (say) the fact that the reals are totally ordered, may need to be modified a bit so that one no longer relies on that fact, so that extension becomes possible. Of course, one could argue that one has a genuinely different proof at this point. The second problem is if the proof itself uses model-theoretic techniques; for instance, in the previous post we discussed proofs which naturally worked for vector spaces over finite fields, but then could be extended to other vector spaces by model-theoretic considerations. One may need some sort of “second order model theory” (ugh) to properly analyse such proofs semantically.
Examples of syntactic approaches would include the ones that John described above, or use ideas from “proof mining” (which are useful, for instance, in converting “infinitary” proofs to “finitary” ones or vice versa). One can also work in the spirit of “reverse mathematics”: declare a certain “base theory” to be “obvious”, and then isolate the few remaining non-obvious steps in a proof which are external to that theory. For instance, consider my post on Pythagoras’ theorem. This theorem is of course a result in Euclidean geometry. As a base theory, one could use the strictly smaller theory of affine geometry, which can handle concepts such as linear algebra and area, but not lengths, angles, and rotations. Relative to affine geometry, one can deconstruct a proof of Pythagoras, and what one eventually observes is that at some point in that proof one must (implicitly or explicitly) use the fact that rotations preserve area and/or length. By isolating the one or two non-affine steps in the proof one can get a handle on the extent to which two proofs of Pythagoras are “equivalent”. In PDE, I like to compare arguments by assuming that every harmonic analysis estimate one needs, or every algebraic identity (e.g. conservation law or monotonicity formula) one needs, is “obvious”, leaving only the higher-level logic of the proof (e.g. iteration arguments, or expressing exact solutions as limits of approximate solutions, etc.).
The algorithmic approach is fairly similar to the syntactic one, but it only works well if the proof itself can be expressed as an algorithm. So instead of comparing proofs, let us compare two constructions, Construction A and Construction B, of some type of object (e.g. an expander graph). One crude way to detect differences between these constructions is to look at their complexity: for instance, Construction A might be polynomial time and Construction B be exponential time, which is fairly convincing evidence that the two constructions are “different”. But suppose now they are both exponential time. One could still argue that Construction B is equivalent to Construction A if one could run Construction B in (say) polynomial time assuming that every step in Construction A could be called as an “oracle” in O(1) time, and vice versa. This would say that, modulo polynomial errors, Construction A and Construction B use the “same” non-trivial ingredients, though possibly in a different order. So for instance, if one had a Gaussian elimination oracle which ran in time O(1) (once the relevant matrix was loaded into memory, of course), could one obtain a constructive Steinitz exchange lemma in time, say, linear in the size of the data (i.e. quadratic in the dimension n?) That would be a convincing way to conclude that Steinitz is “essentially a special case of” Gaussian elimination.
The last approach – conceptual – looks harder to formalise; this is the type of thing I had discussed in my “good mathematics” article. There is definitely a sense in which two different proofs of the same result (or of analogous results in different fields) are somehow “facing the same difficulty”, and “resolving it the same way” at some high level, even if at a low level the two proofs and results are quite different (e.g. one result concerns arithmetic progressions in subsets of Z_N and uses Fourier analysis and additive combinatorics, the other concerns multiple recurrence in a measure-preserving system and uses ergodic theory and spectral theory). One could imagine in those cases that there is some formal Grothendieckian abstraction in which the two proofs could be viewed as concrete realisations of a single abstract proof, but in practice (especially when “messy” analysis is involved) I think one has to instead proceed non-rigorously, by deconstructing each proof into vague, high-level “strategies” and “heuristics” and then comparing them to each other.
4. Emmanuel Kowalski Says:
October 4, 2007 at 5:59 pm | Reply
One result for which one typically sees many different proofs in undergraduate/beginning graduate studies is the Weierstrass approximation theorem of continuous functions on a segment by polynomials. There are two “schools of thought” which seem to emerge from those I know:
(1) the Stone-Weierstrass version
(2) the convolution/regularization version; among these one can include for instance the Bernstein-polynomials proof, though at first sight it may look quite different if phrased in probabilistic terms.
Are (1) and (2) really “different”? I think yes (in part because it seems they naturally generalize to different things), but maybe others can show some links…
5. Terence Tao Says:
October 4, 2007 at 6:20 pm | Reply
Hmm. I do remember that conversation about $\sqrt{2}$ several years ago, but can’t quite reconstruct what I was thinking of at the time. Right now I can see that all four proofs touch in some way or another on $SL(2,Z)$, but in different ways.
For the first proof, you are (implicitly) using the identity $\frac{\sqrt{2}}{1} = \frac{2}{\sqrt{2}}$. Equivalently, the linear transformation $T: (x,y) \mapsto (y,2x)$ has $(1,\sqrt{2})$ as an eigenvector, and so it suffices to show that T has no eigenvector in ${\Bbb Z}^2$ in the sector $\{ x < y < 2x \}$. One then observes from parity considerations that (x,y) cannot be an eigenvector if y is odd. But then (x,y) lies in the range of $T({\Bbb Z})^2$, and $T^{-1}(x,y) = (y/2,x)$ has a smaller x coordinate, and so we can set up an infinite descent $(x,y) \mapsto T^{-1}(x,y)$ of eigenvectors, which is not possible in ${\Bbb Z}^2$.
The second proof is based on the identity $\frac{\sqrt{2}}{1} = \frac{2-\sqrt{2}}{\sqrt{2}-1}$, or equivalently that vector $(1,\sqrt{2})$ is an eigenvector for the linear transformation $T: (x,y) \mapsto (y-x,2x-y)$. So the irrationality of $\sqrt{2}$ is equivalent to asserting that this particular transformation T does not have an eigenvector in ${\Bbb Z}^2$ in $\{ x < y < 2x \}$. We then observe that Tv has a smaller x coordinate than v; since T preserves eigenvectors, we again have an infinite descent. But this proof avoids the use of parity; it’s something to do with the fact that this transformation has determinant -1 (and thus can be inverted within the integers) whereas the previous one had determinant -2 and so needed a parity condition to invert. (This is where $SL(2,{\Bbb Z})$ is supposed to come in and explain everything, but I can’t recall exactly how.)
The third argument is based on the fact that $1+\sqrt{2}$ is a fixed point of $y \mapsto 2 + \frac{1}{y} = \frac{2y+1}{y}$, and thus $(1, 1+\sqrt{2})$ is an eigenvector of $T: (x,y) \mapsto (y, 2y+x)$ in the sector $\{ x < y < 2x\}$. This time, the determinant is 1, so we are genuinely in $SL(2,{\Bbb Z})$. The continued fraction algorithm can be viewed projectively as a dynamical system on the first quadrant of the plane which maps $(x,y)$ to $(x,y-x)$ when $y \geq x$ and $(x,y)$ to $(y,x)$ otherwise. This dynamical system terminates for any element of ${\Bbb Z}^2$ by an infinite descent; but when applied to an eigenvector $(x,y)$ of T, it maps to $T^{-1}(x,y)$ after three steps, giving a contradiction again.
The fourth proof has a slight typo: $(p+q)^2$ should be $2(p+q)^2$. The identity $(p+2q)^2 - 2(p+q)^2 = -(p^2-2q^2)$ is equivalent to the assertion that $\begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} p & q \\ 2q & p \end{pmatrix} = \begin{pmatrix} p+2q & p+q \\ 2(p+q) & p+2q\end{pmatrix}$ has a determinant which is the negative of the determinant of $\begin{pmatrix} p & q \\ q & p \end{pmatrix}$. So the sequences $p_n, q_n$ really have to do with powers of the matrix $T := \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$, or equivalently the transform $T: (q,p) \mapsto (p+q,p+2q)$. I think your argument here is basically equivalent to the observation that $T(q,p) \wedge (1,\sqrt{2}) = (1-\sqrt{2}) (q,p) \wedge (1,\sqrt{2})$, setting up an infinite descent for $(q,p) \wedge (1,\sqrt{2})$ which is incompatible with $\sqrt{2}$ being rational. (Actually, your proof is a little different, it seems to essentially use the fact that $(q,p) \wedge (1,\sqrt{2})$ stays bounded under iteration, while (q,p) goes to infinity, but it is much the same thing.
6. Terence Tao Says:
October 4, 2007 at 6:32 pm | Reply
Dear Emmanuel,
I think the key to both proofs of Weierstrass’s theorem is that at some point one needs to obtain an approximation to the identity near every point of the domain, since by taking linear combinations of such things one gets a dense subset of C(X). To get the approximation to the identity, it suffices (by closure under multiplication) to get a function which is 1 at a specified point x_0 and less than 1 elsewhere, since one can then raise this function to a large power to improve the quality of the approximation.
It is at this point that proofs (1) and (2) diverge. Proof (2) simply writes down a function with the desired property, e.g. $1 - (x-x_0)^2$. Proof (1) proceeds by observing that, since polynomials separate points, one can find a function which is 1 at x_0 and less than 1 at some arbitrarily specified other point x_1. One then takes the “min” of several of these functions (exploiting compactness) to get what one needs (modulo controllable errors). There is then an auxiliary step to observe that the min operation can itself be approximated by polynomials, which can be done either by appeal to Weierstrass (which is a little circular, admittedly, unless one then uses proof (2)) or by explicitly writing down an approximant.
So I guess proof (2) is a shortcut version of proof (1), in which one uses the explicit structure of the domain X=[0,1] and of the ring of polynomials to write down polynomial approximations to the identity.
7. Emmanuel Kowalski Says:
October 4, 2007 at 6:56 pm | Reply
Here’s another example of result where there seem to be quite a few proofs (at least in the sense that books may include two or three): the existence of Brownian motion as a mathematical object.
Here again, I know two apparently different concepts at play:
(1) some kind of abstract existence theorem for stochastic processes, plus a general continuity criterion (Kolmogorov);
(2) writing down an “explicit” (in some sense) sequence of processes which converge (in law) to the required one (Lévy, the invariance principle, Bernstein polynomials, …)
Here I can see a fairly strong link between the two, in that the proof of convergence in law is often (explicitly or not) quite similar to the general continuity criterion. But does it qualify has making the proofs “the same”?
8. Kenny Says:
October 4, 2007 at 7:14 pm | Reply
In response to John Armstrong, I think the problem with that sort of approach to proof equivalence is that it tends to get bogged down in technicalities very fast, because it focuses too much on the specific syntactic moves in a proof. All sorts of tricky things come up with cut-eliminations (which might apparently change the number of times a rule is invoked in the proof) or even just doing operations in different orders.
This is a very difficult project that people have thought about for a while, but I don’t know if people have really gotten anywhere on it yet. We probably need more good examples, as Gowers is suggesting, to get our intuitions really going, so we can even figure out what criteria a good notion of proof equivalence should satisfy. (Like whether or not it should even be an equivalence relation!)
Or maybe it would help to approach this in some sort of opposite direction? If we start by thinking about what makes “the correct” proof of a theorem, then perhaps that will shed light on what sorts of modifications to a proof are inessential?
9. John Armstrong Says:
October 4, 2007 at 7:39 pm | Reply
Kenny, you’re right that that specific approach gets bogged down in the specific moves you choose. But, so does the formalist approach to mathematics in general.
We generally “do mathematics” at a much higher level than strict formalism, whether or not we believe in the formalist view. Still, there’s an unspoken analogy between how a high-level mathematical proof works and how a low-level formal deduction works. All I’m saying is that there’s a way of talking about “proof-morphisms” in the formalist approach, and so there should be an analogue of such things in the actual way we do proofs in real mathematics.
10. Emmanuel Kowalski Says:
October 4, 2007 at 7:42 pm | Reply
Yet another example… (I fiind this very interesting, as people can see…)
There are quite a few proofs of the fact that
$\displaystyle{L(1,\chi)=\sum_{n\geq 1}{\chi(n)n^{-1}}}$
is non-zero for a non-trivial Dirichlet character $\chi$; the crucial case is when $\chi$ is a real-valued character (like the Legendre symbol). (And this fact is itself the crucial final ingredient in proving Dirichlet’s theorem on primes in arithmetic progressions)
Here the two most distinct approaches seem in stark contrast: one is the algebraic proof of Dirichlet, which uses the exact class number formula to express $L(1,\chi)$ as an obviously non-zero number (e.g., $L(1,\chi)$ is, in the case $\chi(-1)=-1$, essentially equal to $h/\sqrt{|d|}$ where $h$ is the class number of an associated imaginary quadratic field), and the other is analytic, and there are many variants, e.g., compare upper/lower estimates for
$\displaystyle{\sum_{n<x}{n^{-1/2}\sum_{ab=n}{\chi(a)}}.}$
Despite the differences, there are very strong links; in particular the inner sum above is the number of ideals of norm $n$ in the quadratic field occuring in the first proof, and this “explains” its properties, in particular that it is $\geq 1$ for $n$ a square (though one does not need the quadratic field to prove this).
However, I wouldn’t say the proofs are “the same”; the first requires as an extra step some form of quadratic reciprocity; moreover, this proof more or less remains stuck there, whereas the other is really the basis for many arguments towards lower bounds for $L(1,\chi)$, and then for lower bounds for class numbers.
11. gowers Says:
October 4, 2007 at 11:27 pm | Reply
I’ll see if I can think of some less simple examples, but a fairly elementary situation where two proofs can be regarded as essentially the same is when a proof relies on compactness and you can choose whether to use open covers or convergent subsequences. For example, to prove that every continuous real-valued function on $f$ on $[0,1]$ is bounded we can either assume not and find a sequence $(x_n)$ such that $f(x_n)$ tends to infinity, find a limit point $x$, and derive an easy contradiction, or we can use continuity to surround each point with an open interval where $f$ is bounded and pick a finite cover of such intervals.
Why do we want to call these two proofs essentially the same? After all, the equivalence between Heine-Borel and Bolzano-Weierstrass, though simple, is not a complete triviality. However,the basic fact one uses in both proofs is the same: that every $x$ is contained in an interval in which $f$ is bounded. The rest of the proof is somehow standard compactness window-dressing: if you want to prove it by contradiction, you use Bolzano-Weierstrass, and if you want to prove it “forwards” then you use Heine-Borel. In general, I think I could justify an assertion along the following lines: if you use Heine-Borel in a proof then you can convert it into a proof using Bolzano-Weierstrass by zeroing in on the fact that you plug into Heine-Borel and plug it into Bolzano-Weierstrass instead.
Of course, this can’t be completely correct, because there are compact topological spaces that fail to be sequentially compact. But I think it’s safe as long as we stick to subsets of $\mathbb{R}^n$, say.
Just to check, let’s prove that the distance between a closed set $F$ and a disjoint closed bounded set $K$ is positive. The Heine-Borel proof tells us to include each point of $K$ in an open ball that’s disjoint from $F$. And now I remember something that once worried me, which is that we have to apply a small trick: we cover $K$ by finitely many open balls $B(x_i,\delta_i)$ in such a way that $B(x_i,2\delta_i)$ is disjoint from $F$. Then if $\delta$ is the minimal $\delta_i$, we find that the distance from $K$ to $F$ is at least $\delta$, since each point $x$ is distance at most $\delta_i$ from some $x_i$ and hence distance at least $\delta_i$ from $F$.
The Bolzano-Weierstrass proof works out more simply. If the result is false, then pick a sequence of points $x_1,x_2,\dots$ in $K$ with $d(x_i,F)$ tending to 0. This has a convergent subsequence, with a limit point $x$, and we also get a sequence in $F$ that converges to $x$, so $x\in F$, by the fact that $F$ is closed, contradicting the disjointness.
The thing that bothered me is that there was no analogue in that proof of the “replace $\delta_i$ by $2\delta_i$” trick, whereas usually one likes to feel that it’s impossible to avoid the work in a proof — all you can do is shift it around. It’s doubly strange because in a certain sense compactness is stronger than sequential compactness: it implies sequential compactness more easily than sequential compactness implies it. And yet the proof using sequential compactness was easier.
So perhaps I won’t after all manage to give some precise “translation scheme” for interchanging sequential compactness proofs and compactness proofs.
This reminds me of a very irritating mathematical difficulty I once had and never resolved. A very nice result in Banach spaces was proved (by Nicole Tomczak-Jaegermann) with the help of induction on countable ordinals. All my experience had taught me that such proofs could be converted into equivalent proofs using a double induction. But I was unable to be precise about how, and I never managed to convert that one.
12. Top Posts « WordPress.com Says:
October 5, 2007 at 2:50 am | Reply
[...] When are two proofs essentially the same? A couple of years ago I spoke at a conference about mathematics that brought together philosophers, psychologists and […] [...]
13. gowers Says:
October 5, 2007 at 9:13 am | Reply
(Just) after a good night’s sleep I think I sorted out my problem in my previous comment. One can prove the theorem about distances between sets by noting that the distance $d(x,F)$ is a continuous function of $x$, which therefore attains its minimum on $K$. The latter fact can easily be proved by Heine-Borel. The “key fact” that is then needed is that every point $x$ is contained in a ball, every point of which has positive distance from $F$. This is what was achieved more constructively with the $2\delta_i$ above. Precisely the same fact is also needed in the sequences proof, but this time it is hidden inside the simple statement that $x\in F$. How do we prove this? Well, for each $x_i$ in the convergent subsequence we take some $y_i\in F$ such that $d(x_i,y_i)\leq\delta_i$. We then need to know that the $y_i$ converge to the same limit, and this is the point where addition of distances and the triangle inequality are involved.
So this result does, after all, illustrate the general idea that if two proofs are essentially the same then they should involve the same amount of work. I should, however, qualify that, since it is possible to insert unnecessary work into a proof without changing its essential character. So it might be better to say that they involve the same “irreducible core” of work, or something like that.
A clear message that comes out of Terry’s first comment is that a general way of distinguishing between two proofs is to show that one proof “yields more” than the other. This could be an algorithm, or an efficient algorithm, or an explicit construction, or a generalization (either through a weaker hypothesis or through a stronger conclusion).
Here’s a rather extreme example: the usual combinatorial proof of van der Waerden’s theorem could not be regarded as the same as proving it by first proving Szemeredi’s theorem and then deducing van der Waerden’s theorem, since the latter also gives you a theorem that was an open problem long after the former was first proved. Of course, that counts as putting in unnecessary work if you use one of the proofs of Szemeredi’s theorem that needs van der Waerden’s theorem (or something very similar) as a step along the way. But not all of them do.
This raises another issue, very closely related to sameness of proofs. It’s the idea that one statement can be “stronger” than another. If two theorems A and B are both proved, then in a stupid sense they are equivalent, since each one implies the other. (To prove B just junk A and prove B, and vice versa.) The non-stupid sense in which we use the word “implies” requires a “genuine use” of a statement used to imply another statement. But that’s another fairly hard idea to make precise. (Maybe a category theorist would like to say that A genuinely implies B if there is a proof of B from A that is not homotopic to the proof A implies “0=0″ implies B. So now we’re back to sameness of proofs.)
Returning to the word “stronger,” it’s also interesting that we often describe one statement as stronger than another when we are about to prove that the two statements are equivalent. We say that A is stronger than B if “A implies B” is the “trivial direction” of the equivalence. For example, the statement that there is a perfect matching in a bipartite graph is “stronger” than the statement that the graph satisfies Hall’s condition. Or the statement that a group $G$ “belongs to one of these families or else is one of these 26 groups” is stronger than the statement that $G$ is finite and simple.
I mention this just to make the point that, in the light of Terry’s observations, perhaps I should refine my earlier questions. Are there good examples of genuinely different proofs of the same statement that are not different because of their other consequences? If so, then how might one go about showing that they were genuinely different? Here, I’m trying to rule out simple examples such as proving B directly versus proving a much stronger statement A and deducing B. Incidentally, the previous paragraph gives, I think, another way in which two proofs can be different: one proof may use axioms while the other uses a classification theorem and then verifies the statement for each example that comes out of the classification.
14. James Says:
October 5, 2007 at 10:46 am | Reply
Does anyone know if anyone has created toy logics in which it’s possible to say something interesting about homotopies between proofs?
15. Emmanuel Kowalski Says:
October 5, 2007 at 4:37 pm | Reply
Coming back to the example of non-vanishing of $L(1,\chi)$, it comes to me that at a certain (very high) level, the two approaches are “the same”: the algebraic proof and analytic proof can be said to approach the result by deriving it from the fact that the product $\zeta(s)L(s,\chi)$, where $\zeta(s)$ is the Riemann zeta function, has a pole at $s=1$. However, the approaches diverge then in how to approach this goal, and I think that at a lower level the difference is genuine.
This suggests that it might be useful to look at similarity of proofs by trying to see at what level — if any — they start to diverge. This makes sense because two proofs of the same statement are identical at the highest level (where the statement itself is seen as an axiom, or a black box for further arguments), so the divergence level is (in a very rough sense) always well-defined.
16. Danny Calegari Says:
October 5, 2007 at 5:47 pm | Reply
What about the following proof: a rational number is the same thing
as an assignment of an integer to every prime, which is zero for all
but finitely many primes, i.e. a divisor (by uniqueness of prime factorization).
Squaring acts on rational numbers (in this representation) by multiplying all
these integers by 2. So a number is the square of a rational if and only
if all these numbers are even. 2 assigns 0 to every prime except 2, to which
it assigns 1 (which is not even).
This is a “proof by canonical form”: find a canonical form for a certain class
of objects in which the object in question has a form in which the desired
property is obvious. In this respect it’s a little bit like proof 3 above (except
that the canonical form is for 2 rather than for sqrt(2)). Maybe this should
be called the “A=B” method (after Wilf-Zeilberger . .)
17. gowers Says:
October 5, 2007 at 7:06 pm | Reply
Danny, I like that point. The proof you mention could also be thought of as a proof that uses a classification theorem — in this case classifying the rationals by their multiplicative structure — so it illustrates well what I was trying to say in an earlier comment about another way in which proofs can be different. (But in this case one could perhaps argue that you only need to know how 2 appears in the factorization, and even just the parity, so it’s not quite clear whether one really does want to call it different, or just the same with some extra unneeded stuff thrown in. I’m not sure what my opinion is there.)
While musing on Terry Tao’s remarks about $SL_2(\mathbb{Z})$ I came up with another proof. I was thinking about whether there was a nice infinite set of proofs, of which the first two I gave were just two particularly simple examples. Terry’s suggestion is to think about $2\times 2$ integer matrices with $\begin{pmatrix} \sqrt 2\\ 1 \end{pmatrix}$ as an eigenvector. The general form of such a matrix is $\begin{pmatrix} a&2b\\b&a\\ \end{pmatrix}$. It was convenient if the determinant was $\pm 1$, but that gives you the Pell equation $a^2-2b^2=\pm 1$ and means that you’ll end up with powers of the matrix $\begin{pmatrix} 1&2\\1&1 \end{pmatrix}$ already considered. However, here’s a proof based on a matrix with determinant $-2$.
First, observe that $\frac{6-4\sqrt{2}}{3\sqrt{2}-4}$ is equal to $\sqrt{2}$. Observe also that $6-4\sqrt{2}$ is strictly between $\null 0$ and $1$. It follows that if $p/q$ is a fraction that equals $\sqrt{2}$, then $\frac{6q-4p}{3p-4q}$ is a smaller fraction that also equals $\sqrt{2}$. I haven’t yet worked out the most general proof along these lines.
A rather silly argument that combines the first two proofs I gave earlier is to use a matrix with an entry that is half an odd integer. It’s possible to use a suitable matrix of this kind to convert $p/q$ into a smaller fraction, but you need the fact that $q$ is even to be sure that that the numerator and denominator of the smaller fraction are integers. I won’t embarrass myself by putting up the details.
18. gowers Says:
October 5, 2007 at 9:51 pm | Reply
A thought that occurred to me a while after reading Emmanuel’s last comment, and also connected with the different proofs of the irrationality of $\sqrt{2}$, and indeed Terry’s idea of thinking about constructions, is this. Often a mathematical statement begins with a universal quantifier, but to prove it one ends up needing to construct something. To take an example, we might decide to prove the irrationality of $\sqrt{2}$ by infinite descent, and then decide that what we needed was a $2\times 2$ integer matrix with $\begin{pmatrix}\sqrt{2}\\ 1\\ \end{pmatrix}$ as an eigenvector with eigenvalue less than 1. This approach to the problem then turns into the problem of constructing such a matrix, which can be done in many different ways. Given two such constructions, one might say that the proof was the same at the top level (since the idea for proving the universally quantified statement was the same) but different at a lower level (since the idea for constructing a matrix was different).
Actually, in this example you don’t really need much of an idea to find the matrix. You just note that it has the form $\begin{pmatrix} a&2b\\ b&a\\ \end{pmatrix}$ and find that all you need to make the proof work is to satisfy the inequality $0<a+b\sqrt{2}<1$. Above, I took $a=-4$ and $b=3$, but obviously one can find many many other examples: indeed, for any non-zero integer $b$ there is an $a$ that does the job (though, amusingly, I’m using the irrationality of $\sqrt{2}$ when I say that).
19. manuel Says:
October 7, 2007 at 11:37 pm | Reply
Can we prove that there is no bijection between the integers and the real numbers without using a “diagonal argument”? Could this be a result with essentially one proof?
20. Jonah Sinick Says:
October 8, 2007 at 8:44 am | Reply
Manuel,
Cantor’s original proof that the real numbers are uncountable is (essentially) as follows:
We claim that no sequence f(n): N —> [a, b] is onto. Let f(n) be any sequence and let a_1 < b_1 be the first two distinct members of the sequence f(n). Then f(n) must attain at least two distinct values in [a_1, b_1], call the first ones a_2 < b_2.
By completeness of the reals the above sequence of nested closed intervals has a nonempty intersection. Say c is in the intersection. If f(N) = c then since c is in [a_N, b_N] we have N > 2N which is impossible. So c must not be in the image of f(n).
Should this proof be regarded as essentially the same as the diagonal argument? I’m not sure. My own judgment is that it is because at bottom both proofs are applications of the completeness of the reals (as they must be!) and don’t depend on any other properties of the reals. But this judgment seems highly subjective to me.
How about the usual approaches to the fundamental theorem of algebra?
http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Proofs
I consider the proofs listed in the wikipedia article under complex analysis and topology to be “essentially the same” as one another while the proofs listed under “algebraic proofs” seem essentially different from the proofs by complex analysis and topology. I would be interested in others’ responses to this example.
21. Jonah Sinick Says:
October 8, 2007 at 8:47 am | Reply
… f(n) must attain at least two distinct values in [a_1, b_1], call the first ones a_2 2i….
I’m genuinely perplexed at why my comment typeset so poorly. I apologize for the boldface letters.
22. gowers Says:
October 8, 2007 at 10:43 am | Reply
Jonah, I’ve corrected your first comment so that it looks OK — you’d put < instead of “ampersand lt semicolon” and because the next letter was a b you ended up with a lot of bold face. In general, I’ve got a policy of quietly editing people’s comments when they don’t compile properly. I don’t change what anyone says though.
I’ve thought about the two proofs of the uncountability of the reals, and in my view they really are the same. Here’s a slight variant of Cantor’s argument as presented by you. Suppose that $[0,1]$ is countable, and enumerate it as $a_1,a_2,\dots$. Then inductively construct nested closed intervals $\null [ p_{n}, q_{n} ]$ of positive length such that $a_n\notin [p_n,q_n]$. This is obviously possible, and can be done in many different ways. (Your argument is different only in that you chose a specific way of constructing these intervals, which happened to depend on the sequence $(a_n)$.) Then an element of the intersection of the intervals $\null [p_n,q_n]$ cannot be any of the $a_n$.
To turn that into the usual diagonal argument, you simply choose $\null [p_n,q_n]$ to be an interval of all points with a given first $n$ digits in their decimal expansion (except that it’s slightly more convenient to take the right end-point of this interval as well). You choose the $(n+1)$st digit to be 3 or 6 according to whether the $(n+1)$st digit of $a_n$ is or is not one of 5 or 6, say.
So this fits nicely into the scheme of what I was mentioning earlier. We actually prove a more precise statement: for every countable sequence $(a_n)$ of reals (in $[0,1]$) there is a sequence of nested closed intervals $\null [p_n,q_n]$ such that $a_n$ does not belong to $\null [p_n,q_n]$. In some sense, the proof of the “for every” part always seems to be the same, while the proof of the inner “there is” part can be done in many different ways. But demonstrating the “there is” part is sufficiently simple that these different approaches don’t really feel all that different. Or perhaps the real reason is that one can describe a sort of “nondeterminate algorithm” (with steps like, “If $a_n$ is less than $q_n$ then pick $p_{n+1}$ and $q_{n+1}$ such that $a_n<p_{n+1}<q_{n+1}\leq q_n$“) and the “different approaches” that we want to identify consist in making this algorithm determinate — or, if you prefer, in different proofs that the indeterminate steps can be specified precisely.
Going back to the original question of Manuel, I think it would be very interesting to show that any proof, or perhaps any “sensible” proof, of the uncountability of the reals must involve a diagonal argument of some sort. It certainly seems to be the case. Recasting it in the terms of the last paragraph makes it almost obvious that it is the case, but I haven’t quite demonstrated that you have to use the nested closed intervals property. However, one does have the basic problem, “Given a sequence $(a_n)$, construct a real that is not an element of the sequence.” Since the reals are defined as limits of sequences in one way or another, one does seem to be forced to construct another sequence in such a way that its limit is not one of the $a_n$. And then one appears to be forced to deal with each $a_n$ in turn, and the obvious way of constructing a sequence that does not have $a_n$ as a limit is to make sure that eventually it lives inside a closed interval that does not contain $a_n$. Actually, that’s equivalent of course. One could do stupid things like not ensuring that the sequence avoids $a_n$ until you reach the $2^n$th term of the sequence. This would be like taking not the obvious diagonal but something similar that did the job equally well. I’d say that this paragraph is a fairly convincing informal argument that there is essentially only one proof of the uncountability of the reals, but I’m not sure how I’d go about making it more formal.
23. gowers Says:
October 8, 2007 at 11:02 am | Reply
One more thought. Here’s a proof of uncountability of the reals that probably ends up being the same, but it might take a bit of thought to see how. You first prove that there is a countably additive measure on the reals that gives the usual length to open and closed intervals. Then the measure of a countable set is zero while the measure of $[0,1]$ is 1.
A good place to start if one wants to see why this is the same proof is the exercise (the significance of which I understand after reading Imre Leader’s forthcoming article on measures for the Princeton Companion) of showing that if $I_n$ is an interval of length $a_n$ and if the $a_n$s sum to less than 1, then the intervals $I_n$ cannot cover all of $[0,1]$. You need this to get going with measure theory and it depends fundamentally on the completeness of the reals.
24. gowers Says:
October 8, 2007 at 2:28 pm | Reply
Actually, here’s a more detailed answer to whether the “measure-theoretic proof” is just the usual proof in disguise. The Lebesgue measure of a set X is the infimum of the sum of lengths of a countable collection of intervals that covers X. It’s easy to prove that the measure of a countable set is zero, but how do you prove that the measure of the interval $[0,1]$ is not zero? You take any countable collection of open intervals with lengths adding up to less than 1/2, say, and prove that the union of this collection is not the whole of $[0,1]$. The reason it isn’t is that if $U_n$ is the union of the first $n$ open intervals, then the complement $F_n$ of $U_n$ is a finite union of closed intervals of total length at least 1/2. (This is because if you write $[0,1]$ as a finite union of intervals, whether open or closed, then their lengths must add up to at least 1, a fact that is easy to prove.) Therefore, the sets $F_n$ are a nested collection of non-empty closed sets, so they have a non-empty intersection.
This appears to be a slightly more general argument, as it involves nested intersections of closed sets that are more general than intervals. Moreover, that is necessary: if you try to find your point that is not in the union of the open intervals by constructing an intersection of closed intervals, you can’t keep the $n$th closed interval large enough to guard against one of the later open intervals containing it. And perhaps the reason we have a somewhat different proof is that we have proved something stronger: instead of merely showing that a countable union of intervals of length 0 cannot be all of $[0,1]$ we’ve shown it for a countable union of intervals of any lengths that add up to less than 1 (if we’re prepared to modify the argument a tiny bit).
But even this doesn’t make it a genuinely different proof of the uncountability of the reals , any more than it would be a genuinely different proof if we first used the diagonal argument and then said, “Oh, and by the way, here’s a proof of Fermat’s last theorem thrown in.” If we just want to prove that the set $\{a_1,a_2,\dots\}$ isn’t all of $[0,1]$ then we can choose any old open intervals around the $a_i$ and use the measure-theoretic argument to prove that their union is not all of $[0,1]$. But if we’re allowed to do that, then we can make their sizes go to zero sufficiently rapidly that it is possible to use nested intervals for the argument, rather than more general nested closed sets.
Despite all that, maybe one could say that there is in fact a very slight difference between the two approaches. If you prove the result using nested intervals, then there is a tiny effort needed to construct those intervals. If you use more general closed sets, then you can simply take the complement of the open sets you are considering, so constructing the closed sets no longer takes any effort, but now you are looking at more complicated sets.
The proof that a nested intersection of non-empty closed subsets of $[0,1]$ is non-empty is basically the same as the proof for intervals: their infs tend to a limit that belongs to the intersection. However, maybe this shouldn’t quite make us think that the proofs are essentially the same because if you generalize to higher dimensions it does seem to be genuinely harder to prove that a nested intersection of arbitrary closed sets (or even of finite unions of closed boxes) is non-empty than it is if those sets are just boxes: the inf trick no longer works.
25. Scott Carnahan Says:
October 8, 2007 at 6:47 pm | Reply
The Kazhdan-Lusztig conjecture was proved by translating the question through several areas of mathematics, until it reached a form that could be treated by Deligne’s theory of weights. At the end of Bernstein’s D-module notes, he mentions that one wouldn’t need to go through the business of varieties over finite fields if one had a satisfactory theory of mixed Hodge modules (does this exist now? I’m behind on recent developments). Let’s suppose such a theory existed, and we had a proof using it. Would such a proof be equivalent? If we had a good theory of mixed motives, I suppose one could claim that both proofs were equivalent, since we would be using two realizations of the same machine. At least, I imagine this is what Bernstein had in mind. My question is, are two proofs equivalent if the equivalence requires a conjecture?
26. Scott Carnahan Says:
October 8, 2007 at 7:08 pm | Reply
Here’s another example: three proofs of the Weil conjectures. I think Deligne’s original proof used a lot of l-adic monodromy on Lefschetz pencils, Laumon’s proof used some kind of sheafy Fourier transform, but still revolved around the l-adic cohomology theory set out in SGA, and Kedlaya’s proof also used a sheafy Fourier transform, but in the language of p-adic differential equations. Kedlaya said that he was guided by strong analogies between the l-adic and p-adic worlds, but formalizing this as an equivalence of proofs seems a bit ambitious with our current state of knowledge.
27. gowers Says:
October 9, 2007 at 12:48 am | Reply
My gut feeling (but no more than that) about whether two proofs can be equivalent if the equivalence is just a conjecture is that they can. It seems to me to be a bit like the two-dimensional equivalent of asking whether a mathematical statement can be provable if it is just a conjecture. Also, it seems clear that two proofs can be equivalent in a highly non-trivial way, so why not in a way that is so non-trivial that it actually depends on an unsolved conjecture? However, coming up with good examples of such a phenomenon is very interesting.
On a more mundane note, a fairly simple example of a result that seems to have genuinely different proofs is Fermat’s little theorem. I know of three: the deduction from Lagrange’s theorem, the deduction from the fact that multiplying by $a$ permutes the non-zero elements of $\mathbb{F}_p$ and the inductive proof based on the fact that the binomial coefficients $\begin{pmatrix} p\\ k\\ \end{pmatrix}$ are multiples of $p$ when $1\leq p<k$. Can anyone identify any two of those proofs? (It might be possible for the first two, perhaps.)
Another question for algebraic number theorists out there. Gauss’s law of quadratic reciprocity famously has a vast number of proofs. How many genuinely distinct proofs are there? Even a rough estimate would interest me.
28. Emmanuel Kowalski Says:
October 9, 2007 at 6:29 am | Reply
Coincidentally, I was just reading yesterday one of Euler’s proof of Fermat’s little theorem in a translation of his original paper. It’s essentially the first one mentioned above, tailored to the case of a cyclic group (and with unusual style and notation as one would expect) but clearly recognizable. What’s interesting is that at the end Euler specifically states that this is a completely different proof from the one with binomial coefficients, because it leads to many other results. (Those are essentially his words, though I don’t have the book handy just now to quote it).
Concerning quadratic reciprocity, this
site of F. Lemmermeyer lists 224 published ones, and classifies them roughly (and has references), which seems to indicate that the number of fairly distinct ones is maybe 40 or 50. I think one can identify some of these (e.g., many arguments are based on Gauss sums or cyclotomic fields, and these can be certainly made to correspond in some cases), but that would probably leave at least 10 genuinely different arguments.
29. James Cranch Says:
October 9, 2007 at 11:07 am | Reply
I guess a lot of things traditionally called “metatheorems” supply examples of transformation techniques for proofs.
Category theory has several. There’s the “duality principle”: that the axioms of category theory are unchanged by reversing the directions of all the morphisms and all the compositions, and so every time you prove something about the situation one way around, you can rewrite the proof to prove a dual statement.
I suppose that, logically, that’s a different thing: the rewriting process here yields a proof of a different statement. But there’s no reason not to be interested in such things.
For an honest example, there’s Freyd’s embedding theorem. The textbook statement is that any abelian category can be embedded fully and faithfully into a category of modules. But the point, philosophically, is that it allows you to prove things in abelian categories by imagining you’re working in a category of modules. In particular, you can “choose elements” of your objects, even though they’re just objects of an abstract category, and don’t really have elements. There is a rewriting technique which replaces such a proof using “elements” with one that doesn’t abuse notation in this manner.
30. Gil Kalai Says:
October 9, 2007 at 7:14 pm | Reply
This is a very interesting post! I noticed that assertively claiming that two very different proofs of the same theorem are essentially equal is something which can gain you a lot of respect. It shows not only a complete understanding of the two different proofs but understanding deep connections that nobody else see. Maybe one can gain even more respect by claiming that two different proofs for two entirely different theorems are essentially the same but here some caution is advised.
Concerning the question of when two proofs are equal or similar or different, one formal aspect regarding proofs of existence (like fixed point theorems) is the complexity classes of Christos Papadimitriou. These classes can be regarded as a rough (computational complexity) classification of mathematics’ basic tricks and this may lead to a formal notion of different proofs. (In this classification 2 proofs leading to polynomial algorithms are equivalent.)
Here are a few test cases for equality and inequality of proofs:
1) I have a friend who always tell me that the inductive proof of the crossing number inequality (recently featured in Terry’s blog) given by Ajtai Chvatal Newborn and Szemeredi is equivalent to the probabilistic proof. This always surprised me since to me they look different (and in particular I feel that I understand much better the probabilistic one.)
2) There are many proofs of Cayley formula for the number of trees on n labeled vertices. (A book of J W Moon lists quite a few and there are many more.) Can you classify them?
3) Turan’s theorem for the number of edges for a graph on n vertices required to guarantee a triangle has many proofs, and it almost looks that any strategy to prove is doomed to succeed. Which of these proofs are different?
31. Gil Kalai Says:
October 10, 2007 at 1:06 am | Reply
Regarding the related question when two proofs are entirely different, there are some cases where there is a clear difference between two types of proofs e.g. the distinction between constructive/non constructive proofs; effective/non-effective proofs; Proofs based on randomized construction/ proof based on explicit construction. In these three cases one type of proofs is usually much harder than the other.
This reminds me also of the long-time mathematical endeavor of finding an “elementary proof” of the prime number theorem. The precise meaning of “elementary proof” in this case was not entirely clear to me. (At some points I convinced myself that the distinction is manifested in that the “elementary proofs” apply to certain formulations of the PNT when you replace all primes by a dense subset of primes; but I am not sure this is correct.)
In all the cases above entirely different proofs reflect a different mathematical task to start with. It can be interesting to bring more examples of entirely different proofs for (really) the same theorem, when the distinction is not of the nature that one proof (to start with) tries to achieve some much harder goal (effective; constructive; explicit; elementary; bijective, not using CH; computerized; not computerized; etc..)
32. Gil Kalai Says:
October 10, 2007 at 5:03 am | Reply
I forgot another case of distinction between proofs which is the case of simple proofs vs difficult proofs. Unlike the distinction between effective/non-effective; constructive/non constructive, etc., there is something paradoxical about the (all important) endeavor of finding simple proofs for difficult theorems. To start with, there is a strong positive correlation between the difficulty of the proof and the difficulty of finding it. But finding simple proofs for difficult theorems is usually very difficult.
It is like the well known paradox:
Rare things tend to be more expensive;
Cheap horses are rare;
Therefore, cheap horses tend to be more expensive.
(Finding difficult proofs for easy theorems is also quite interesting from time to time.)
33. Ryan Reich Says:
October 10, 2007 at 5:49 am | Reply
One rather famous distinction between types of proof is in algebraic geometry: the proof via “complex analytic methods” versus the proof via “algebraic methods”. The former includes using just about any facet of the ordinary topology on a complex variety, so for example, one has Poincare duality proven either as Poincare might have, or the way Artin did. Of course, Artin proved it for etale cohomology, but he also proved a comparison theorem that relates that to singular cohomology, so it seems there are two proofs which cannot, even in principle, be reduced to the same thing (since the etale cohomology theorem applies to varieties in positive characteristic too!). I am no doubt revealing my ignorance by trying to talk about this, however…
Actually, it seems to me that anything that can be proven with complex analysis or with something else will have two essentially different proofs; complex analysis is good like that. There is, for example, an amusing computation of the Fresnel integral $\int_0^\infty \frac{\sin x}{x} dx$ using differentiation under the integral sign: let $I(t) = \int_0^\infty e^{-tx} \frac{\sin x}{x} dx$, observe that the derivative can be computed by elementary means, and that I(t) vanishes as t approaches infinity. That gives you a differential equation which you can solve and then plug in t = 0. You can also do it using the residue theorem, which involves integrating over a triangular contour; I have trouble seeing how these could be the same proof. Perhaps they are, though.
34. Emmanuel Kowalski Says:
October 10, 2007 at 8:15 am | Reply
In the case of the Prime Number Theorem, the stated goal (e.g., in one of Hardy’s lectures on Ramanujan’s work, if I remember right) was to find a proof that does not use complex analysis. It was hoped that this would imply significant new insight into the study of prime numbers, especially with respect to those problems that were then still unsolved, such as the Goldbach, prime \$k\$-tuples, conjectures. However, the proof of Selberg (and Erdös) didn’t really have such an effect, and elementary proofs have not yet recovered the same results provided by the standard complex-analytic proofs (with respect to remainder terms, and uniformity in arithmetic progressions).
35. im Says:
October 10, 2007 at 5:51 pm | Reply
With all this discussion of elementary vs non-elementary proofs, is there scope for a loop back to your wonderful earlier post on Cauchy’s theorem as generalization of f’=0 => f = constant?
36. Scott Carnahan Says:
October 10, 2007 at 10:58 pm | Reply
I’ve seen two proofs that $\int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$. The first one is the standard proof using Fubini and polar coordinates. The second involves constructing two functions $F(t) = \int_0^t e^{-x^2} dx$ and $G(t) = \int_0^1 \frac{e^{-t^2(1+x^2)}}{1+x^2} dx$, and noting with derivatives that $(F(t))^2 + G(t)$ is constant. It seems quite beautiful, and I learned it from W. Luxemburg.
I’m curious if there is an analytic reason for someone to come up with the function G, because it seems to come out of nowhere. It might also suggest some way that these two proofs are equivalent.
37. gowers Says:
October 11, 2007 at 11:47 am | Reply
Scott, that’s an interesting challenge. It looks to me as though there might be a perspective from which those two proofs could be similar in a way that has been discussed in earlier comments: that they have the same basic strategy, but that that strategy involves constructing something that can be constructed in many different ways. However, in this case the construction is a sufficiently non-trivial process for it to be questionable whether one would want to regard the proofs as the same. I haven’t worked this out properly but let me say why I think it looks like a possibility.
The quantity $(F(t))^2$ can be regarded as the integral of $e^{-x^2-y^2}$ over the square $\null [0,t]^2$. Thus, one could regard $(\sqrt\pi/2)-G(t)$ as a guess for what that integral ought to be (but at the moment I don’t see how the guess was made) and the differentiation trick as just a verification that the guess is correct. The first proof could be regarded as using the formula $H(t)=2\pi\int_0^t re^{-r^2}dr$ as a guess for what the integral of $e^{-x^2-y^2}$ ought to be over a circle of radius $t$. Probably that can be proved correct by differentiation as well.
Thus, it looks to me as though the basic strategy is, “Find a two-dimensional set on which you can work out the integral of $e^{-x^2-y^2}$.” (Of course, I really mean a set that grows with a parameter $t$.) Once you’ve had that idea, it turns out that there is more than one such set.
Having said that, the formula for $G(t)$ does seem rather non-obvious and seems to involve a distinctly “new idea.” Most notably, unlike in the circle case, it gives an integral that (as far as I can see) is just as hard to calculate as the original one, but which happens to behave very nicely when $t$ is 0 or infinity. So it would be very interesting to know where this comes from, and also whether there are any other proofs that fit into this general scheme.
38. gowers Says:
October 11, 2007 at 2:09 pm | Reply
On second thoughts, perhaps it doesn’t take an idea to come up with $G(t)$ so much as a leap of faith that the integral over the square will be easy to calculate, coupled with the thought (which is fairly standard, so perhaps not quite an idea in the true sense) that it might be worth differentiating it to see what one gets. If you do, then you want to calculate $2F(t)F'(t)$, which works out to be $2\int_0^t e^{-t^2-x^2}dx$. It’s quite natural to make the substitution $x=ty$, which leads to $\int_0^1 2te^{-t^2(1+y^2)}dy$. Then it’s quite natural to notice that the integrand can be antidifferentiated with respect to $t$, so this is $-\frac{d}{dt}\int_0^1\frac{e^{-t^2(1+y^2)}}{1+y^2}dy$. And now we obviously want to use the fundamental theorem of calculus, so we want to evaluate this integral for $t=0$ and $t=\infty$ … and we can.
The initial leap of faith isn’t really a piece of magic, since the square and the circle are the two simplest two-dimensional sets that one might try. So the above seems to me to be a plausible account of how the function $G$ might have been discovered.
39. David Speyer Says:
October 11, 2007 at 2:11 pm | Reply
I think I see why $G(t)=F(\infty)^2-F(t)^2$. The right hand side is the integral of $e^{-x^2-y^2}$ over the complement of the square $\null [0,t]^2$. By the symmetry of the problem over the line $x=y$, this is
$2 \int_{t \leq y \leq x} e^{-x^2-y^2} dx dy$.
Now, make the substitution $y=mx$ to get
$2 \int_{x=t}^{\infty} \int_{m=0}^1 e^{-x^2 (1+m^2)} x dm dx$. The integral on x is now elementary (substitute $u=x^2$) and I think the remaining integral on m is your formula for $G(t)$.
This can be seen as similar to, but not quite the same as, the standard proof. In the standard proof, we change to polar coordinates $\null (x,y)=(r \sin \theta, r \cos \theta)$. Then the integral on r is elementary and the integral on $\theta$ is trivial. In the new proof, we change to “(slope, intercept)-coordinates”: $(x,y)=(x,mx)$. The integral on x becomes elementary for the same reason that the integral on r did previously.
This suggests to me a general strategy of proof. Choose any parametrized convex curve $(f(u), g(u))$ cutting off a corner of the positive orthant. Make the substitution $(x,y)=(r f(u), r g(u))$. The integral on r will always be elementary; hope that you can deal with the integral on u. So the standard proof is to cut off the corner with a quarter-circle, and the modified proof is to cut it off with a square.
40. David Speyer Says:
October 11, 2007 at 2:14 pm | Reply
Gaah! I can’t get my TeX back to see what went wrong! The first formula that didn’t parse should be $\null [0,t]^2$ — I was just making sure we all knew what I meant by “the square”. The second formula that didn’t parse was just the expression for polar coordinates $(x,y)=(r \cos \theta, r \sin \theta)$.
41. gowers Says:
October 11, 2007 at 2:20 pm | Reply
David, I’ve noticed that for some strange reason WordPress doesn’t seem to recognise closed intervals if they occur at the beginning of a LaTeX expression, even if you’ve done everything completely correctly. But I’ve discovered a stupid trick that deals with the problem: instead of writing “latex [0,1]” inside the dollars I write “latex \null [0,1]“. Ridiculously, that seems to cure it. I’ve edited your two comments above so that they do parse, partly using that trick and partly removing a backslash that had attached itself to the letter r in “r \sin \theta”.
42. Scott Carnahan Says:
October 12, 2007 at 12:17 am | Reply
This seems to be a pretty satisfying answer. David’s general strategy also works well if you decompose of the lower half of the quadrant into hyperbolas and do a hyperbolic trig substitution. However, the square version seems to be special in that it produces a one-dimensional proof in a way that the other two do not, in the sense that a first-term calculus student could parse it without knowning Fubini.
I’ve encountered a similar vanishing problem in LaTeX itself, when I tried to start some aligned equations with a Lie bracket, so it might not be a problem with WordPress. By the way, you can find the original TeX in the alt text of the image.
43. gowers Says:
October 12, 2007 at 10:54 am | Reply
Here are two proofs that a dense subset of $\{1,2,\dots,N\}$ must contain a “Hilbert cube” of dimension $k$: that is, a subset of the form $\{x+\epsilon_ia_i:\epsilon\in\{0,1\}^k\}$. More precisely, for any $\delta>0$ there exists an $N$ such that any subset $A$ of $\{1,2,\dots,N\}$ of size at least $\delta N$ must contain such a set (with the $a_i$ not equal to zero). I’ll just sketch the arguments very briefly.
The first approach is to find (using an averaging argument) some $a_1$ such that the intersection of $A$ with $A-a_1$ is reasonably large. (It’s not too hard to get $\delta^2$ times an absolute constant.) Then you are left needing to find a $(k-1)$-dimensional cube in this intersection, which you do by induction.
The second approach is to prove a stronger result, namely that you can find $a_1,\dots,a_k$ such that for a significant proportion of $x$ the cube defined above is a subset of $A$. This you can prove as follows. By induction the result is true for $k-1$. This gives you $a_1,\dots,a_{k-1}$ and a big set $B$ of $x$ such that the corresponding cube is a subset of $A$. Since $B$ is dense, some difference occurs frequently in $B$, and you can make this difference your $a_k$.
This illustrates a phenomenon that occurs frequently. You want to prove a statement $\forall k P(k)$ by induction. You can either use a simple argument to deduce $P(k)$ from $P(k-1)$ or you can use $P(k-1)$ to deduce $P(k)$ from $P(1)$ (which itself is proved by a simple argument). In other words, you can think of $k$ either as $1+(k-1)$ or as $(k-1)+1$.
Often the work needed is more or less the same, as the above example illustrates. But sometimes one approach seems to be rather easier than the other (unfortunately I haven’t managed to think of any examples yet), and I’m not sure that I believe in a general principle that says that one can always convert one sort of inductive proof into the other. So perhaps we have here a grey area, where proofs can be similar but it’s not obvious whether we should regard them as the same.
44. davidspeyer Says:
October 12, 2007 at 3:43 pm | Reply
Oh! I know that LaTeX bug. It occurs when your TeX code is being fed as a parameter into a macro somewhere, and LaTeX, failing to “quote” the square bracket, interprets it as an additional optional parameter to the macro. The most common way I know to produce this bug is to have an item in a list that starts with a square bracket (arguably, this case is a feature.) For example, this LaTeX should cause trouble:
There are two types of parenthesis used in some-programming-language-or-other:
\begin{enumerate}
\item (), used to control the order of operations
\item [], used to call a function or index into an array
\end{enumerate}
The standard solution is to write {[}.
45. davidspeyer Says:
October 12, 2007 at 4:16 pm | Reply
Here is an example, in elementary number theory, of the phenomenon Prof. Gowers discusses in his previous post. We write ${[}a_1,...,a_n]$ for the continued fraction $a_1+1/(a_2+1/(a_3+...+1/a_n)...))$. Define $P_n(a_1, ..., a_n)$ by the reucrrence
$P_n(a_1, ..., a_n)=a_n P_{n-1} (a_1, ..., a_{n-1})+P_{n-2}(a_1, ... a_{n-2})$
with initial conditions $P_{-1}=1$, $P_{0}=0$. Define $Q_n$ by the same recurrence with $Q_{-1}=0$ and $Q_{0}=1$. Then a basic result about continued fractions is
${[}a_1, ..., a_n]=P_n(a_1, ..., a_n)/Q_n(a_1,...,a_n).$
Every student that I have ever seen try to prove this (I taught at PROMYS for four years) tries to use induction, starting from
${[}a_1,...,a_n]=a_1+1/[a_2,...,a_n]$.
This proof can be made to work, but it is hard. However, there is a very easy proof if you start from
${[}a_1,...,a_{n-1}, a_n]=[a_1,...,a_{n-1}+1/a_n]$.
I think there is a mental block against doing it this way, because the terms in the continued fraction will no longer be integers, but it produces a much simpler proof.
46. Adam Chmaj Says:
October 19, 2007 at 7:54 am | Reply
To me, the first proof is analytical, the other three algebraic.
Besides, good problems have unique solutions.
• observer Says:
January 22, 2011 at 12:37 pm
Number theoretic uses Bezout: Let w=sqrt2=a/b. (a,b)=1 so there are integers c,d, such that ad-bc=1.
0=(a-bw)(c+dw)=ac+adw-bcw-bdw^2=ac-2bd+w,
thus w is an integer.
47. gowers Says:
October 23, 2007 at 10:39 am | Reply
Here’s another example of a result with two proofs that look different and are both quite simple. Perhaps they can be unified — I have not tried.
The statement to be proved is that every finite subgroup of the multiplicative group of non-zero complex numbers must be the group of all $n$th roots of unity, for some $n$. This question came up on a Cambridge question sheet for a first-year course on groups, and this term I am what we call a “supervisor” for the course (that is, I see students in pairs and discuss their solutions and attempted solutions).
Here’s the proof that occurred to me. Let $G$ be such a group. Then any element of $G$ has finite order (something that had been proved in an earlier question via the usual pigeonhole-principle argument), and therefore must be a root of unity. Now choose an element of $G$ with minimal argument. If that doesn’t generate the whole of $G$ then a Euclid-algorithm style calculation shows that its argument isn’t minimal — contradiction.
This was the very first question sheet of the course, so I think that was the intended proof, but one of my students suggested using Lagrange’s theorem instead. That quickly leads to a short solution: every element of $G$ must be a $|G|$th root of unity, and there are only $|G|$ of them, so we are done.
These two proofs certainly have quite significant surface differences: the second uses a piece of algebra that can’t immediately be read out of the first, and the first uses the minimal argument, which doesn’t seem to come into the second. So any unification would have to be from a deepish perspective. Or perhaps they really are different proofs.
48. E.Bz. Says:
October 23, 2007 at 11:42 pm | Reply
I think you can get enough of Lagrange from the first argument in this situation. Your Euclidean algorithm/ minimal argument is basically exactly how one proves that subgroups of cyclic groups are cyclic – it is then pretty easy to prove that you have Lagrange for the cyclic groups so I suppose there is some similarity. Beyond that I guess they are intrinsically different.
49. Simon Morris Says:
October 24, 2007 at 10:16 am | Reply
The second proof generalises to (for example) the quaternions, but I can’t see how the first proof would. More convincingly (to me), the first proof also shows that the group is cyclic, whereas the second proof doesn’t. I don’t think I’m missing some small variation of the second proof that would show this, because the transliteration of the second proof to the quaternions certainly couldn’t.
But I’m not sure which subset, if any, of those observations really shows that the proofs are different.
50. gowers Says:
October 25, 2007 at 6:29 pm | Reply
Simon, I don’t think I understand what you are saying about quaternions. For instance, $G$ could be the group $\null \{\pm 1, \pm i, \pm j, \pm k\}$, which is not cyclic. The second proof would break down because there are more than eight 8th roots of unity, and the first hardly gets started. But perhaps you had a different generalization in mind.
I also don’t quite understand what it means to say that the second proof doesn’t show that the group is cyclic: if it shows that it’s the group of $n$th roots of unity then surely it does.
51. Simon Morris Says:
October 25, 2007 at 7:43 pm | Reply
Sorry – the “generalization” I had imprecisely in mind was just that $G$ consists of $n$ th roots of unity, not that it is all of them, which isn’t a generalization of the result you gave.
I think the first proof shows in passing, as part of its argument, that the group is cyclic, whereas with the second proof, you have to go back at the end and look at the group you’ve proved you have, and check that it’s cyclic. I’m not sure whether that’s a real distinction between the proofs. My rather poor point about the quaternions was just that $G$ needn’t be cyclic there, so there can’t be any generalization of the first proof there that uses some other definition of minimal element. Now that I’ve looked at some of your web pages, I see that I was thinking vaguely along the lines of “quaternions are another model for the complex numbers”, in a way slightly reminiscent of your page about why it isn’t obvious that the integers are bounded in the reals.
52. Maurizio Says:
November 29, 2007 at 5:07 am | Reply
I’m surprised by the amount of insight about algebraic number theory that i gained reading your proof of the irrationality of $\sqrt{2}$, i mean it really helped me to better understand many things that i was supposed to already know.
Actually the first two proofs are quite similar if you think about valuations: a valuation is a multiplicative function from a field to $\mathbb{R}^+$ that respects the triagular inequality, ie $v(ab)=v(a)v(b)$ and $v(a+b) \leq v(a)+v(b)$, and $v(a)=0$ only if $a=0$. It can be proved that all such valuations over the rational number are (up to equivalence) the usual absolute value ($v_\infty(a)=|a|$) and given by prime numbers ($v_p(p^k\frac{a}{b})=\frac{1}{p^k}$, where $a$ and $b$ are prime with $p$), so they can be thought of as a suitable generaliation of prime numers and ideals, providing additionally a “prime at infinity”. All valuations share important properties, for instance they induce a topology with respect to which you can take the completion, etc.
In the first proof of the irrationality of $\sqrt{2}$, you are taking a fraction $\frac{p}{q}$ where $p$ and $q$ minimize the valuation at the prime 2 (by dividing the numerator and denominator by the biggest common power of 2), and you get an absurd because $v_2(\frac{p^2}{q^2})$ cannot possibly be equal to $v_2(2)$. In the second proof, you are choosing $p$ and $q$ with smallest absolute value (and so smallest valuation at $\infty$), and then apply a tranformation (an element of $SL_2(\mathbb{Z})$) that leaves the value of the fraction, supposed to be equal to $\sqrt{2}$, unchanged, but decreases the valuation at $\infty$ of the denominator, absurd by minimality. So, by what i’m understanding, most of the difference in the proofs comes by how the “non-archimedean” valuation at 2 and the “archimedean” one at $\infty$ work, possibly an experienced number theorist may give us a better explaination of this fact.
53. uri Says:
December 17, 2007 at 12:26 am | Reply
Hi,
Can’t leave it unsaid that the second proof mentioned for the irrationality of \$\sqrt{2}\$ can be seen in a beautiful manner:
Take an actual triangular piece of paper of sizes p,p,q (assuming \$p^2=2q^2\$). You can now fold one of the sides on the diagonal, and get yourself a smaller triangle of sizes \$p-q,p-q,2q-p\$, revealing that \$2(p-q)^2=(2q-p)^2\$, and contradicting the minimality of \$p,q\$ if you assumed that in advance.
54. Joe Says:
January 1, 2008 at 8:41 am | Reply
Even more fundamentally, I have often reflected on the assertion that two mathematical statements are equivalent. Certainly if one proves that A is equivalent to B, where neither A or B is known to be true, the statement has practical value in that if one resolves the validity of one of the statements she also resolves the validity of the other. However one often finds that an author remarks that two statements known to be true are equivalent. The first example that comes to mind is: “The Prime Number Theorem is equivalent to the non-vanishing of the zeta function on the line Re(s)=1.”
So what does this mean? One might say that this means that if we assume the first we can prove the second and vice verse. But along these lines any two true statements are equivalent. I would protest this by saying that “The Pythagorean theorem is equivalent to the Prime Number Theorem.” Now one might try to correct this by saying that we mean that A and B are equivalent if the proof of B uses A and the proof of A uses B. But, in the Prime Number Theorem example, certainly there are proofs of the Prime Number Theorem that do not make use of the fact that Re(s)=1. Now if I had tried to point to an example (say the Selberg-Erdos “Elementary” proof, or Wiener’s using his Tauberian Theorem) someone might dive into a discussion about how the zeta function is really lurking in the background of these. But even if someone presented a case that every known proof of the prime number theorem relied on that fact that the zeta function doesn’t vanish on the line Re(s)=1, this certainly doesn’t imply that there doesn’t exist a proof the proceeded without use of the fact. Moreover while I’m not sure what we do mean when we say “A is equivalent to B” I’m pretty sure it should be a mathematical assertion and not a statement about the literature in existence on the result.
55. elicaraq Says:
March 8, 2008 at 7:18 am | Reply
I don’t know if the following two ways to explain the coverup method for partial fraction decomposition are “interestingly different” or “based on the same underlying idea”. I would like to know.
Here is an elementary explanation:
$\frac{1}{(z-1)(z-i)(z+i)} = \frac{R}{z-1} + \frac{S}{z-i} + \frac{T}{z+i}$
Multiply both sides by $(z-1)(z-i)(z+i)$,
$1 = R (z-i)(z+i) + S (z-1)(z+i) + T (z-1)(z-i)$
By setting $z = 1, i, -i$, we get
$R = \left. \frac{1}{(z-i)(z+i)} \right|_{z=1} = \frac{1}{2}$
$S = \left. \frac{1}{(z-1)(z+i)} \right|_{z=i}= -\frac{1}{2+2i}$
$T = \left. \frac{1}{(z-1)(z-i)} \right|_{z=-i} =\frac{1}{-2+2i}$
Now here is a complex explanation:
Consider
$\frac{1}{2 \pi i} \int_C \frac{dz}{(z-1)(z-i)(z+i)}$
Where $C$ is a closed contour to be determined.
We know
$\frac{1}{2 \pi i} \int_C \frac{dz}{(z-1)(z-i)(z+i)} = \frac{1}{2 \pi i} \left(\int_C \frac{R dz}{z-1} + \int_C \frac{S dz}{z-i} + \int_C \frac{T dz}{z+i} \right)$
Set $C$ be a circle of radius $1/2$ around $1, i, -i$ respectively, and use the Cauchy’s integral formula. Again, by the same process, we get:
$R = \left. \frac{1}{(z-i)(z+i)} \right|_{z=1} = \frac{1}{2}$
$S = \left. \frac{1}{(z-1)(z+i)} \right|_{z=i}= -\frac{1}{2+2i}$
$T = \left. \frac{1}{(z-1)(z-i)} \right|_{z=-i} =\frac{1}{-2+2i}$
56. Konstantin Ziegler’s Weblog Says:
May 7, 2008 at 12:17 pm | Reply
[...] Gowers asks When are two proofs essentially the same? For example, it is often possible to convert a standard inductive proof into a proof by [...]
57. Paul Says:
May 13, 2008 at 12:48 am | Reply
A good problem to think of in this context would be the classic theorem
“Whenever a rectangle is tiled by rectangles each of which has at least one integer side, then the tiled rectangle has at least one integer side.”
Wagon offers 14 different proofs in “Fourteen proofs of a result about tiling of a rectangle” and he even classifies them according to how they generalize…
http://www.jstor.org/stable/2322213?seq=1
58. Qiaochu Yuan Says:
August 7, 2008 at 4:16 am | Reply
I seem to have missed the bulk of the good discussion on this, but this is an issue that has been very much on my mind lately. Awhile ago, I considered two different proofs for the general form of a homogeneous linear recurrence: ordinary generating functions and partial fraction decomposition vs. the Jordan normal form theorem. Whether these two proofs are essentially equivalent has been bothering me for some time. They are sufficiently similar that I suspected for awhile that the first proof can in fact be adapted to a proof of the Jordan normal form theorem, but when I tried to actually write such a proof I couldn’t quite make the jump. This is either because I haven’t tried hard enough or because the second proof offers something “extra.” I think what’s going on is that partial fraction decomposition is essentially a special case of the Jordan normal form theorem (in particular, on companion matrices), but again, I’m not sure. I’d be very glad if anyone could clarify my ignorance on this matter!
More generally, I think Joe brings up a very good point: the definition of “proof homotopy” can be contingent on the current state of mathematics as opposed to anything that could be considered an independent truth. If, for example, two proofs of a result like the Fundamental Theorem of Algebra come from two distinct branches of mathematics and do not appear to relate easily to each other, then perhaps the reason they appear so different is that some deep correspondence between those two branches has not yet been uncovered, and this hypothetical correspondence would reveal those two proofs as essentially the same by placing them in context we don’t yet have. So I guess what I’m suggesting is that if a result appears to have genuinely distinct proofs, then perhaps we do not really understand it!
59. Timothy Chow Says:
September 4, 2008 at 2:41 am | Reply
Again I’m late to this discussion because I only just discovered it, but I agree with Kenny that although proof theorists have studied a variety of notions of proof equivalence, they do not come anywhere near to capturing what we intuitively mean by “these two proofs are essentially the same.” So while it’s a very important mathematical activity to think about whether two particular proofs of an important mathematical result are essentially the same, I think it’s highly premature to expect to be able to formalize the notion.
The situation is somewhat better with “these two theorems are equivalent.” As Terry said, in reverse mathematics one posits a very weak “base theory” (RCA_0 is a common choice) and then sometimes you can show that two theorems A and B are equivalent in the sense that (1) both A and B are independent of the base theory and (2) their equivalence can be proved in the base theory. There are many successful examples of this. On the other hand, it doesn’t always capture our intuitions perfectly either. Intuitively one might argue that Sperner’s lemma and Brouwer’s fixed-point theorem are equivalent, but Sperner’s lemma is provable in RCA_0 while Brouwer’s theorem isn’t.
By the way, here’s another proof of the irrationality of $\sqrt 2$. Let $a$ be the smallest positive integer such that $a \sqrt 2$ is an integer. Let $b = a(\sqrt 2 - 1)$. Then $b\sqrt 2$ is an integer yet $b$ is a strictly smaller positive integer than $a$; contradiction. Is this the same proof as one of those already given? Note that this proof immediately generalizes to show that the square root of an integer $n$ is either an integer or an irrational number (i.e., it can’t be a non-integral rational number), if you let $b = a(\sqrt n - \lfloor \sqrt n\rfloor)$. The traditional proof of the irrationality of $\sqrt 2$ does not generalize so quickly without a detour into unique factorization or something.
60. How can one equivalent statement be stronger than another? « Gowers’s Weblog Says:
December 28, 2008 at 1:22 am | Reply
[...] to discuss and that are better discussed here. I have already written about one of my favourites: when are two proofs essentially the same? Another that is closely related is the question of how it can be that two mathematical statements [...]
61. Tom Says:
December 28, 2008 at 10:53 pm | Reply
To what extent does the Curry-Howard correspondence capture “sameness” of proof?
It would say that two proofs are the same if they induce the same function between the proposition types.
Tom
• dasuxullebt Says:
January 21, 2010 at 2:40 pm
I think the curry-howard-isomorphism is exactly the way one has to look at this problem. The Terms extracted from proofs are all type-correct and thus have a normal-form which can be calculated by beta-eta-evaluation.
62. Maexe Says:
March 22, 2009 at 5:01 pm | Reply
I just started to read Gowers’s and Terry’s blogs a few days ago, and find them very interesting. Sorry to come much too late with the following remark: about the 2 proofs of uncountability of the reals in, say, $[0,1]$.
- Cantor’s original proof, as quoted by Jonah Sinick above, is taking the real interval as they are i.e. a compact and complete metric space, and then detects at least one missing (i.e. not contained in the counting) real number by sequential compactness i.e. Bolzano-Weierstrass. Viewing the real numbers in binary expansion with base 2, the two binary representations of all binary rationals coincide, e.g. $(0.0111...)=(0.1000...)=1/2$
- Whereas the diagonal argument models the reals set-theoretically as a space of sequences of digits: In the example of base 2, the sequences (0,0,1,1,1,..) and (0,1,0,0,0,…) are not the same. This is no coincidence as the topology is now very different: taking a natural base $n>1$, the space of sequences of natural numbers $(k_1,k_2,...), k_i<n$ has the desired cardinality $c=n^\mathbb(N)$ and is a compact and totally disconnected space ${0,1,...,n-1}^\mathbb(N)$ in the product (Tychonoff) topology. For base $n=2$ this space of 0-1 sequences is homeomorphic to the classical Cantor set.
Both spaces are compact, metrizable and complete in their metric topology. But they are far from being homeomorphic: the real interval $[0,1]$ is connected, and the Cantor set is totally disconnected.
Thus the a priori purely set-theoretical statement of uncountability of the reals is proven by two methods familiar from the categories of topological and metric spaces, but based on two models of the set of reals which are topologically quite different.
63. none Says:
November 1, 2009 at 9:28 am | Reply
Identity of proofs is a booming topic in proof theory, apparently. I don’t understand it at all, but here the blurb of a paper I looked at:
What Is a Logic, and What Is a Proof?
Lutz Straßburger
AbstractI will discuss the two problems of how to define identity between logics and how to define identity between proofs. For the identity of logics, I propose to simply use the notion of preorder equivalence. This might be considered to be folklore, but is exactly what is needed from the viewpoint of the problem of the identity of proofs: If the proofs are considered to be part of the logic, then preorder equivalence becomes equivalence of categories, whose arrows are the proofs. For identifying these, the concept of proof nets is discussed.
23 October 2006
Logica Universalis—Towards a General Theory of Logic, pp. 135–152, Birkhäuser, 2007
PDF of the paper is here: http://www.lix.polytechnique.fr/~lutz/papers/WhatLogicProof.pdf
Site that I got this link from, that has a ton of related stuff: http://alessio.guglielmi.name/res/cos/index.html
64. Grupo fundamental de las esferas. « Coloquio Oleis Says:
June 3, 2010 at 10:12 am | Reply
[...] definir que es que dos pruebas sean la misma o no (por una discusión sobre esto, recomiendo este post). Luego de dar la prueba, voy a comentar porque me parece (al menos un poco) [...]
65. Are these the same proof? « Gowers's Weblog Says:
September 18, 2010 at 10:36 pm | Reply
[...] I have no pressing reason for asking the question I’m about to ask, but it is related to an old post about when two proofs are essentially the same, and it suddenly occurred to me while I was bathing my two-year-old [...]
66. A Theorem About Infinite Cardinals Everybody Should Know | Combinatorics and more Says:
January 17, 2012 at 10:00 pm | Reply
[...] and infinite cardinals equally well. (For the finite case it looks that Cantor’s proof is genuinly different than the ordinary proof by [...]
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http://mathoverflow.net/questions/114484?sort=votes
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## subgroups of a $p$-solvable group and complete reducibility
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
1.
Let $G$ be a $p$-solvable group and $V$ be a finite dimensional faithful $kG$-module, where the characteristic of $k$ is $p$. But $V$ is not a semisimple $kG$-module. For every $n\geq 0$, we define $End_{k}^{0}(V)=V$ and $End_{k}^{n}(V)=End_{k}(End_{k}^{n-1}(V))$ when $n>0$.
Let $L$ be a subgroup of $G$ such that $End_{k}^{n}(V)$ is a semisimple $kL$-module for every $n\geq 0$.
Is $L$ a $p'$-subgroup?
2.
Let $V$ be a 2-dimensional vector space over a field $k$ of characteristic $p$ and $G$ be a $p$-solvable subgroup of $GL(V)$, where $p$ is a prime number larger than than 5. But $V$ is not a semisimple $kG$-module.
Let $L$ be a subgroup of $G$ such that $V$ is a semisimple $kL$-module.
Is $L$ a $p'$-subgroup?
-
## 1 Answer
The answer to the second question is yes: As $G$ is reducible, this is true for $L$ even more. As $V$ is a semisimple $L$-module, $V$ is a sum of two $1$-dimensional $L$-modules. So $L$ is conjugate to a group of diagonal matrices. But elements of finite multiplicative order in $k$ have $p'$-order.
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http://mathoverflow.net/revisions/100230/list
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## Return to Answer
2 added 193 characters in body
The answer was given by Samuel Blank in his Brandeis 1967 phd dissertation, on which Poenaru gave a Bourbaki seminar.
Then Peter Shor and C. J. Van Wyk gave a polynomial time algorithm to decide if there is an extension.
EDIT: Blank's method already gave a polynomial algorithm, but with an exponent too large to make it practical, which is needed for applications (for instance to integrated circuit design).
The answer is in term of existence of a chain of "reductions" of a certain kind for the cyclically reduced word in the free group $F_n$ on $n$ generators determined by the immersion, where $n$ is the number of bounded components of the complement of the curve (assumed to have only transverse double points).
1
The answer was given by Samuel Blank in his Brandeis 1967 phd dissertation, on which Poenaru gave a Bourbaki seminar.
Then Peter Shor and C. J. Van Wyk gave a polynomial time algorithm to decide if there is an extension.
The answer is in term of existence of a chain of "reductions" of a certain kind for the cyclically reduced word in the free group $F_n$ on $n$ generators determined by the immersion, where $n$ is the number of bounded components of the complement of the curve (assumed to have only transverse double points).
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http://unapologetic.wordpress.com/2009/12/17/iterated-integrals-ii/
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# The Unapologetic Mathematician
## Iterated Integrals II
Let’s get to proving the assertions we made last time, starting with
$\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
where $f$ is a bounded function defined on the rectangle $R=[a,b]\times[c,d]$.
We can start by defining
$\displaystyle F(x)={\int\limits^-}_c^df(x,y)\,dy$
And we easily see that $\lvert F(x)\rvert\leq M(d-c)$, where $M$ is the supremum of $\lvert f\rvert$ on the rectangle $R$, so this is a bounded function as well. Thus the upper integral
$\displaystyle\overline{I}={\int\limits^-}_a^bF(x)\,dx={\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx$
and the lower integral
$\displaystyle\underline{I}={\int\limits_-}_a^bF(x)\,dx={\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx$
are both well-defined.
Now if $P_x=\{x_0,\dots,x_m\}$ is a partition of $[a,b]$, and $P_y=\{y_0,\dots,y_n\}$ is a partition of $[c,d]$, then $P=P_x\times P_y$ is a partition of $R$ into $mn$ subrectangles $R_{ij}$. We will define
$\displaystyle\begin{aligned}\overline{I}_{ij}&={\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\\underline{I}_{ij}&={\int\limits_-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\end{aligned}$
Clearly, we have
$\displaystyle{\int\limits^-}_c^df(x,y)\,dy=\sum\limits_{j=1}^n{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy$
and so we find
$\displaystyle\begin{aligned}{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx&={\int\limits^-}_a^b\sum\limits_{j=1}^n{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\&\leq\sum\limits_{j=1}^n{\int\limits^-}_a^b{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\&=\sum\limits_{j=1}^n\sum\limits_{i=1}^m{\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\end{aligned}$
That is
$\displaystyle\overline{I}\leq\sum\limits_{j=1}^n\sum\limits_{i=1}^m\overline{I}_{ij}$
and, similarly
$\displaystyle\underline{I}\geq\sum\limits_{j=1}^n\sum\limits_{i=1}^m\underline{I}_{ij}$
We also define $m_{ij}$ and $M_{ij}$ to be the infimum and supremum of $f$ over the rectangle $R_{ij}$, which gives us the inequalities
$\displaystyle m_{ij}(y_j-y_{j-1})\leq{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\leq M_{ij}(y_j-y_{j-1})$
and from here we find
$\displaystyle m_{ij}\mathrm{vol}(R_{ij})\leq{\int\limits_-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\leq{\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\leq M_{ij}\mathrm{vol}(R_{ij})$
Summing on both $i$ and $j$, and sing the above inequalities, we get
$\displaystyle L_P(f)\leq\underline{I}\leq\overline{I}\leq U_P(f)$
and since this holds for all partitions $P$, the assertion that we’re trying to prove follows.
The second assertion from last time can be proven similarly, just replacing $F(x)$ by the lower integral over $[c,d]$. And then the third and fourth assertions are just the same, but interchanging the roles of $[a,b]$ and $[c,d]$. Finally, the last assertion is a consequence of the first four. Indeed, if the integral over $R$ exists, then the upper and lower integrals are equal, which collapses all of the inequalities into equalities.
### Like this:
Posted by John Armstrong | Analysis, Calculus
## 1 Comment »
1. [...] copy of these three for each index between and . The proofs of these are pretty much identical to the proofs in the two-dimensional case, and so I’ll just skip [...]
Pingback by | December 21, 2009 | Reply
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## 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.
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http://www.physicsforums.com/showthread.php?p=4276763
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Physics Forums
Quick question about Dirac delta functions
What does the square of a Dirac delta function look like? Is the approximate graph the same as that of the delta function?
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Quote by black_hole Is the approximate graph the same as that of the delta function?
yes
(only more so! )
Mentor
Quote by black_hole What does the square of a Dirac delta function look like?
The square of a Dirac delta function is not defined.
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Quick question about Dirac delta functions
George is right. There is no way that we can make sense of squares of the Dirac Delta function. Distributions are very strange things and their properties are subtle. Nice operations like squaring don't always make sense.
So the square doesn't exist and so the approximate graph also doesn't exist.
often this question comes up in an electrical engineering class when one is faced with convolving an impulse $\delta(t)$ with another impulse. i.e. what would happen if you had a linear, time-invariant system with impulse response $h(t) = \delta(t)$ and you input to that system $x(t) = \delta(t)$. obviously, the output should be $y(t) = \delta(t)$, but how do you get that from the convolution integral?
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Quote by rbj often this question comes up in an electrical engineering class when one is faced with convolving an impulse $\delta(t)$ with another impulse. i.e. what would happen if you had a linear, time-invariant system with impulse response $h(t) = \delta(t)$ and you input to that system $x(t) = \delta(t)$. obviously, the output should be $y(t) = \delta(t)$, but how do you get that from the convolution integral?
It's important for students to realize that the convolution of two delta functions can be well-defined, as the integral contains both a dummy variable and a free variable: ##\int d\tau~\delta(\tau)\delta(t-\tau) = \delta(t)##. One can interpret the integral as being done over one of the delta functions to give the result (there's probably a much more formal and rigorous way to do this). The problem with something like ##\int d\tau~\delta(\tau)\delta(\tau)## is that the result is naively ##\delta(0)##, which doesn't have a well-defined interpretation (though in some contexts, like quantum field theory, for example, you might see it taken to mean the volume a system which is to be taken to grow infinitely large at the end of the calculation).
Although it looks like the convolution integral will generate the squared delta function integral when t = 0, one has to remember that 'functions' like the delta function are meant to exist under integrals, so t has to remain a free variable. Fixing its value doesn't really make much sense.
Of course, this is a physicist's way of looking at the issue, so there are some gaps in the formality and rigor, and mathematicians should feel free to shore it up (or tear it down, as the case may be) with the appropriate rigor.
Quote by Mute Of course, this is a physicist's way of looking at the issue, so there are some gaps in the formality and rigor, and mathematicians should feel free to shore it up (or tear it down, as the case may be) with the appropriate rigor.
Well, first you need to know the actual definition of convolution of distributions
First, let ##f(x)## and ##g(y)## be distributions on ##\Omega \subset \mathbb{R}##. We define the two variable distribution ##f \oplus g## as
##(f \oplus g, \, \phi(x,y)) := (f(x), \, (g(y), \, \phi(x,y)))##.
We now define ##(f * g, \phi) := (f(x) \oplus g(y), \phi(x+y))##.
Now this is well-defined in any of the following cases:
1. f or g has compact support,
2. both have their support bounded from below,
3. both have their support bounded from above.
Dirac has compact support therefore the convolution exists.
Edit: Now that I think about it, the simples method would be a approximation to identity argument. You would still need to show that the identity holds irrespective of your choice of representative though.
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http://hilbertthm90.wordpress.com/2010/03/11/the-grothendieck-spectral-sequence/
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# The Grothendieck Spectral Sequence
March 11, 2010 by | 1 Comment
Well, I meant to do lots more examples building up some more motivation for how powerful spectral sequences can be in some simple cases. But I’m just running out of steam on posting about them. Since we’ve done spectral sequences associated to a double complex, we may as well do the Grothendieck Spectral Sequence, then I might move on to another topic for a bit (I admit it is sort of sad to not prove the Kunneth formula using a SS).
I haven’t scoured the blogs to see whether these topics have been done yet, but I’m thinking about either basics on abelian varieties a la Mumford, or some curve theory, possibly building slowly to and culminating in Riemann-Roch.
In any case, we have the tools to do the Grothendieck Spectral Sequence (GSS) quite easily. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories with enough injectives. Let $\mathcal{A}\stackrel{G}{\to}\mathcal{B}\stackrel{F}{\to}\mathcal{C}$ be functors (and $FG:\mathcal{A}\to\mathcal{C}$ the composition). Suppose that $F$ and $G$ are left exact and for every injective $J\in\mathcal{A}$ we have $G(J)$ is acyclic. This just means that $R^iF(J)=0$ for all positive $i$.
Then there exists a spectral sequence (the GSS) with $E_2^{pq}\simeq (R^pF)(R^qG)(X)\Rightarrow R^{p+q}(FG)(X)$ with differential $d_{r}:E_r^{pq}\to E_r^{p+r, q-r+q}$.
The proof of this is just to resolve $X$ using the injectives that we know exist. This gives us a double complex. From a double complex, the way to get the $E_2$ term is to take vertical then horizontal homology, or horizontal and then vertical. Both of these will converge to the same thing. One way completed collapses to the “0 row” due to the fact that the exact sequence remained exact after applying the functor except at the 0 spot. Thus it stabilizes at this term and writing it out, you see that it is exactly $R^{p+q}(FG)(X)$. Taking homology in the other order gives us exactly $(R^pF)(R^qG)(X)$ by definition of a derived functor. This completes the proof.
I probably should write the diagram out for clarity, but really they are quite a pain to make and import into wordpress. The entire outline of the proof is here, so if you’re curious about the details, just carefully fill in what everything is from the previous posts.
This is quite a neat spectral sequence. It is saying that just by knowing the derived functors of $F$ and $G$ you can get to the derived functors of the composition of them. There are two important spectral sequence consequences of this one. They are the Leray SS and the Lyndon-Hochschild-Serre SS. The later computes group cohomolgy.
I promised early on to do the Leray SS for all the algebraic geometers out there. The Leray SS gives a way to compute sheaf cohomology. Let $\mathcal{A}=Sh(X)$ and $\mathcal{B}=Sh(Y)$ be the category of sheaves of abelian groups on X and Y. Let $\mathcal{C}=Ab$ the category of abelian groups. Let $f:X\to Y$ be a continuous map, then we have the functor $F=f_*$ and the two global section functors $\Gamma_X$ and $\Gamma_Y$.
Applying the GSS to these functors, we get that $H^p(Y, R^qf_*\mathcal{F})\Rightarrow H^{p+q}(X, \mathcal{F})$.
There are a few things to verify to make sure that the GSS applies, and we need the fact that $\Gamma_Y\circ f_*=\Gamma_X$. It would also be nice to have an example to see that this is useful. So maybe I’ll do those two things next time.
### Like this:
Categories: algebra, algebraic geometry, algebraic topology | Tags: cohomology, double complex, grothendieck spectral sequence, leray spectral sequence, spectral sequence | Permalink.
### Author: hilbertthm90
I am a mathematics graduate student fascinated in how all my interests fit together.
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http://mathhelpforum.com/calculus/144768-constrained-optimisation-problem.html
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# Thread:
1. ## constrained optimisation problem
(a) Write down the objective function and the constraint equation for aconstrained optimisation problem which is to minimize the total surface area of a rectangular box with the constraint that its volume is fixed to be 1. (Hint: Let the box have length x, width y and height z.)
(b) Solve the problem using the method of Lagrange multipliers.
Is anyone teach me how to do this question? thanks
2. Originally Posted by lin.13579
(a) Write down the objective function and the constraint equation for aconstrained optimisation problem which is to minimize the total surface area of a rectangular box with the constraint that its volume is fixed to be 1. (Hint: Let the box have length x, width y and height z.)
(b) Solve the problem using the method of Lagrange multipliers.
Is anyone teach me how to do this question? thanks
To start you off, you are required to:
$\textrm{Minimise }\,2xy + 2xz + 2yz$
$\textrm{subject to }\,xyz = 1$.
3. To minimize F(x,y,z), subject to G(x,y,z)= constant, use $\nabla F= \lambda \nabla G$ where $\lambda$ is the "Lagrange multiplier". Setting components equal gives three equations in the four unknowns, x, y, z, and $\lambda$. The constraint G(x,y,z)= constant is a fourth equation.
Since the value of $\lambda$ is not part of the solution, I find that dividing one equation by another to eliminate $\lambda$ is often a good first step.
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http://en.wikipedia.org/wiki/Arrhenius_equation
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# Arrhenius equation
Arrhenius' equation is a simple, but remarkably accurate, formula for the temperature dependence of reaction rates. The equation was proposed by Svante Arrhenius in 1889, based on the work of Dutch chemist J. H. van't Hoff who had noted in 1884 that van't Hoff's equation for the temperature dependence of equilibrium constants suggests such a formula for the rates of both forward and reverse reactions. Arrhenius provided a physical justification and interpretation for the formula.[1] Currently, it is best seen as an empirical relationship.[2] It can be used to model the temperature variation of diffusion coefficients, population of crystal vacancies, creep rates, and many other thermally-induced processes/reactions. The Eyring equation, developed in 1935, also expresses the relationship between rate and energy.
A historically useful generalization supported by Arrhenius' equation is that, for many common chemical reactions at room temperature, the reaction rate doubles for every 10 degree Celsius increase in temperature.
## Equation
Arrhenius' equation gives the dependence of the rate constant $k$ of a chemical reaction on the absolute temperature $T$ (in kelvin), where $A$ is the pre-exponential factor (or simply the prefactor), $E_a$ is the activation energy, and $R$ is the Universal gas constant.
$k = A e^{-E_a/(R T)}$
Alternatively, the equation may be expressed as
$k = A e^{-E_a/(k_B T)}$
The only difference is the energy units of $E_a$: the former form uses energy per mole, which is common in chemistry, while the latter form uses energy per molecule directly, which is common in physics. The different units are accounted for in using either $R$ = Gas constant or the Boltzmann constant $k_B$ as the multiplier of temperature $T$.
The units of the pre-exponential factor $A$ are identical to those of the rate constant and will vary depending on the order of the reaction. If the reaction is first order it has the units s−1, and for that reason it is often called the frequency factor or attempt frequency of the reaction. Most simply, $k$ is the number of collisions that result in a reaction per second, $A$ is the total number of collisions (leading to a reaction or not) per second and $e^{{-E_a}/{(RT)}}\ \ \$ is the probability that any given collision will result in a reaction. It can be seen that either increasing the temperature or decreasing the activation energy (for example through the use of catalysts) will result in an increase in rate of reaction.
Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature. Similarly, under a wide range of practical conditions, the weak temperature dependence of the pre-exponential factor is negligible compared to the temperature dependence of the $\exp(-E_a/RT)\ \$ factor; except in the case of "barrierless" diffusion-limited reactions, in which case the pre-exponential factor is dominant and is directly observable.
## Arrhenius plot
Main article: Arrhenius plot
Taking the natural logarithm of Arrhenius' equation yields:
$\ln(k)= \frac{-E_a}{R}\frac{1}{T} + \ln(A)$
This has the same form as an equation for a straight line:
$y = m x + b$
So, when a reaction has a rate constant that obeys Arrhenius' equation, a plot of ln(k) versus T −1 gives a straight line, whose gradient and intercept can be used to determine Ea and A . This procedure has become so common in experimental chemical kinetics that practitioners have taken to using it to define the activation energy for a reaction. That is the activation energy is defined to be (-R) times the slope of a plot of ln(k) vs. (1/T )
$\ E_a \equiv -R \left[ \frac{\partial \ln k}{\partial ~(1/T)} \right]_P$
## Modified Arrhenius' equation
The modified Arrhenius' equation[3] makes explicit the temperature dependence of the pre-exponential factor. If one allows arbitrary temperature dependence of the prefactor, the Arrhenius description becomes overcomplete, and the inverse problem (i.e., determining the prefactor and activation energy from experimental data) becomes singular. The modified equation is usually of the form
$k = A (T/T_0)^n e^{{-E_a}/{(RT)}}$
where T0 is a reference temperature and allows n to be a unitless power. Clearly the original Arrhenius expression above corresponds to n = 0. Fitted rate constants typically lie in the range -1<n<1. Theoretical analyses yield various predictions for n. It has been pointed out that "it is not feasible to establish, on the basis of temperature studies of the rate constant, whether the predicted T½ dependence of the pre-exponential factor is observed experimentally."[2] However, if additional evidence is available, from theory and/or from experiment (such as density dependence), there is no obstacle to incisive tests of the Arrhenius law.
Another common modification is the stretched exponential form:[citation needed]
$k = A \exp \left[-\left(\frac{E_a}{RT}\right)^{\beta}\right]$
where β is a unitless number of order 1. This is typically regarded as a fudge factor to make the model fit the data, but can have theoretical meaning, for example showing the presence of a range of activation energies or in special cases like the Mott variable range hopping.
## Theoretical interpretation of the equation
### Arrhenius' concept of activation energy
Arrhenius argued that for reactants to transform into products, they must first acquire a minimum amount of energy, called the activation energy Ea. At an absolute temperature T, the fraction of molecules that have a kinetic energy greater than Ea can be calculated from statistical mechanics. The concept of activation energy explains the exponential nature of the relationship, and in one way or another, it is present in all kinetic theories.
The calculations for reaction rate constants involve an energy averaging over a Maxwell-Boltzmann distribution with $E_a$as lower bound and so are often of the type of incomplete gamma functions, which turn out to be proportional to $\ e^{\frac{-E_a}{RT}}$.
### Collision theory
Main article: Collision theory
One example comes from the "collision theory" of chemical reactions, developed by Max Trautz and William Lewis in the years 1916-18. In this theory, molecules are supposed to react if they collide with a relative kinetic energy along their lines-of-center that exceeds Ea. This leads to an expression very similar to the Arrhenius equation.
### Transition state theory
Another Arrhenius-like expression appears in the "transition state theory" of chemical reactions, formulated by Wigner, Eyring, Polanyi and Evans in the 1930s. This takes various forms, but one of the most common is
$\ k = \frac{k_BT}{h}e^{-\frac{\Delta G^\ddagger}{RT}}$
where $\Delta G^\ddagger$ is the Gibbs free energy of activation, $k_B$ is Boltzmann's constant, and $h$ is Planck's constant.
At first sight this looks like an exponential multiplied by a factor that is linear in temperature. However, one must remember that free energy is itself a temperature dependent quantity. The free energy of activation is the difference of an enthalpy term and an entropy term multiplied by the absolute temperature. When all of the details are worked out one ends up with an expression that again takes the form of an Arrhenius exponential multiplied by a slowly varying function of T. The precise form of the temperature dependence depends upon the reaction, and can be calculated using formulas from statistical mechanics involving the partition functions of the reactants and of the activated complex.
### Limitations of the idea of Arrhenius activation energy
Both the Arrhenius activation energy and the rate constant k are experimentally determined, and represent macroscopic reaction-specific parameters that are not simply related to threshold energies and the success of individual collisions at the molecular level. Consider a particular collision (an elementary reaction) between molecules A and B. The collision angle, the relative translational energy, the internal (particularly vibrational) energy will all determine the chance that the collision will produce a product molecule AB. Macroscopic measurements of E and k are the result of many individual collisions with differing collision parameters. To probe reaction rates at molecular level, experiments are conducted under near-collisional conditions and this subject is often called molecular reaction dynamics.[4]
## See also
• Accelerated aging
• Eyring equation
• Q10 (temperature coefficient)
• Van 't Hoff equation
• Clausius-Clapeyron relation
• Gibbs-Helmholtz equation
• Cherry blossom front - predicted using the Arrhenius equation
## References
1. Laidler, K. J. (1987) Chemical Kinetics,Third Edition, Harper & Row, p.42
2. ^ a b Kenneth Connors, Chemical Kinetics, 1990, VCH Publishers
3. Levine, R.D. (2005) Molecular Reaction Dynamics, Cambridge University Press
## Bibliography
• Laidler, K. J. (1987) Chemical Kinetics, Third Edition, Harper & Row
• Laidler, K. J. (1993) The World of Physical Chemistry, Oxford University Press
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http://math.stackexchange.com/questions/32108/how-is-riemannstieltjes-integral-a-special-case-of-lebesguestieltjes-integral
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# How is Riemann–Stieltjes integral a special case of Lebesgue–Stieltjes integral?
Thanks for reading! My questions are based on the following quotes from Wikipedia:
1. About the existence of Lebesgue–Stieltjes integral:
The Lebesgue–Stieltjes integral $\int_a^b f(x)\,dg(x)$ is defined when ƒ : [a,b] → R is Borel-measurable and bounded and g : [a,b] → R is of bounded variation in [a,b] and right-continuous, or when ƒ is non-negative and g is monotone and right-continuous.
I was wondering if this is the right condition for its existence?
2. About the existence of Riemann–Stieltjes integral:
The best simple existence theorem states that if f is continuous and g is of bounded variation on [a, b], then the integral exists. A function g is of bounded variation if and only if it is the difference between two monotone functions. If g is not of bounded variation, then there will be continuous functions which cannot be integrated with respect to g. In general, the integral is not well-defined if f and g share any points of discontinuity, but this sufficient condition is not necessary.
On the other hand, a classical result of Young (1936) states that the integral is well-defined if f is α-Hölder continuous and g is β-Hölder continuous with α + β > 1.
For the question in the part 3, I was wondering for Riemann–Stieltjes integral $\int_a^b f(x) \, dg(x)$ to exist, must g be nondecreasing? It looks like not the case quoted above.
3. Specialization from Lebesgue–Stieltjes integral to Riemann–Stieltjes integral:
Where f is a continuous real-valued function of a real variable and g is a non-decreasing real function, the Lebesgue–Stieltjes integral is equivalent to the Riemann–Stieltjes integral,
I was wondering why it only mentions the case when g is nondecreasing? Is this the necessary condition for existence of Riemann-Stieltjes integral?
4. Do Lebesgue–Stieltjes integral and Riemann–Stieltjes integral generally use the same notation $\int_a^b f(x)\,dg(x)$? How does one know which one the notation refers to?
Thanks for helping!
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## 1 Answer
I don't think my understanding is completely correct and should have posted as a comment, but it has length restrictions.
1. It seems to me that when $g$ is BV and right continuous, and $f$ is Borel measurable, $f$ does not have to be bounded. There are unbounded Lebesgue integrable functions, so the same should be true for Lebesgue-Stieltjes integral, which is just Lebesgue integral w.r.t. the signed measure $\mu_g$ on $\mathcal{B}(\mathbb{R})$ induced by $g$.
2. It should be OK for function $g$ with bounded variation. We can write $g$ as the difference of two non-decreasing functions.
3. I don't think Riemann-Stieltjes integral requires the integrator to be non-decreasing. It might be BV or possibly an even broader class of functions. I guess the author of Wikipedia entry mentions only nondecreasing functions because s/he has CDF of a random variable in mind and wants to discuss its application in probability theory. I also have the impression that these two integrals agree whenever the Riemann-Stieltjes integral exists. (Just like the relation between the Lebesgue integral and the Riemann integral.)
4. If these two notions agree, there's no danger of using the same notation. Otherwise, I've seen authors using prefix to distinguish different types of integrals, e.g., $(R)\int_a^b...$ for Riemann(-Stieltjes) integrals.
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Thanks! For 2, I was wondering if $g$ being non-decreasing implies $g$ having bounded variation? Do I need to specify both are over an interval $[a,b]$? – Tim Apr 10 '11 at 21:43
(1) $g$ nondecreasing implies it has bounded variation, because the total variation of $g$ on $[a,b]$ is always $g(b)-g(a)$ regardless of the partition. In fact, "A function $g$ is of bounded variation if and only if it is the difference between two monotone functions." (2) Yes, we are talking about functions on $[a,b]$. – GWu Apr 10 '11 at 22:24
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http://math.stackexchange.com/questions/243171/is-mathematics-the-only-language-that-is-not-subject-of-interpretation?answertab=votes
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# Is mathematics the only language that is not subject of interpretation?
Do you know any other "language" that is used by people except mathematics and is not subject of interpretation? By subject of interpratation I mean e.g. that 1 000 000 people will undertand that 1 + 1 = 2 and never 3. E.g. when you say that "This flower is red". Some people will think that it is darker, some lighter etc.
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There are, for instance, several flavors of formal logic. Do you count them as 'mathematics'? – Gregor Bruns Nov 23 '12 at 13:37
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Computer programming – Thomas Andrews Nov 23 '12 at 13:37
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Secretly, I always replace $n$ in my head by $2n$. So I interpret $1+1=2$ as $2+2=4$. My mind has no place for odd numbers.... – Michael Greinecker Nov 23 '12 at 13:48
Exactly, it is also relative. So, probably there is sadly no common language at all ;(. At least no language we have invented or find out so far. What do you think? Do you know some book about that subject? Something from math-linquistics section maybe? Any advice? – Derfder Nov 23 '12 at 13:52
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If you're looking for largely unambiguous languages, take a look at lojban. Mathematics is ambiguous, but usually clear by context. – akkkk Nov 23 '12 at 14:00
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## 5 Answers
In this answer, I will address the question: are mathematicians certain of everything they say?
In retrospect I do not think this is the question that was asked (I seem to be skilled in answering the wrong questions), but I'll leave it in place nonetheless.
This is of course not possible. The absolute truth is some platonic ideal we are all reaching for. There is nothing mathematicians are really sure of, but for many things it's so very hard to doubt it, that everyone agrees - indeed, proofs for $1+1=2$ (this doesn't mean what you think it does: $2$ is defined as $1+1$, so we have to prove something else) are so very hard to debate that we agree that it is right.
And in fact, for many proofs (mathematics is all about proving things) we can systematically expand it in simpler terms, and even have a computer check them. Sure, it is possible that everyone along the line made a mistake, so that everyone thinks a proof is right but it really isn't, but this chance is so small that we say it is proven.
There have been fake proofs in mathematics. Many mathematicians claimed to have found a proof or counterexample for $P\stackrel?=NP$. Most of these "proofs" are relatively easy to refute. And in any case, unlike in the physical sciences, all data is in the proof itself, so there is nothing else you need to check the validity of an argument. In physics and the social sciences, the huge datasets that papers are based on are usually not published, and that makes it hard to check validity.
Mathematics is as much a debatable science as physics and medicine. However, we always have a hope of answering critical questions, and in this sense mathematics is "sure" of things. But in the end, even mathematics is about convincing other mathematicians that you are right.
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Yes, I agree. I have hard time to find some other "thing" or "language" that is in principle same as "formal" math for almost all people who knows the symbols and basic operations like adding, subtracting etc. – Derfder Nov 23 '12 at 13:50
@Derfder: i don't really get your comment, what are you trying to say? – akkkk Nov 23 '12 at 13:53
I like your answer the most, because you put the more complex explanaition. Thanks for your help. – Derfder Nov 23 '12 at 14:06
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"all data is in the proof itself," -- I disagree. There is also the data needed to interpret and understand the proof. It is, after all, expressed in some language which must be read to evaluate the validity of reasoning. And even for the most basic "facts" I think you could argue that their obviousness might require some weak anthropocentric assumptions. – tomasz Nov 23 '12 at 23:38
@tomasz: ugh, no, the language of mathematics is not explained in every single paper or a mathematical subject. but that is of course not the point that I was trying to make :-/ – akkkk Nov 24 '12 at 11:10
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First, mathematics notation is subject to interpretation. Looking at your example, $1+1=0$ in modulo-$2$ arithmetic. Most mathematics requires context to be understood.
I would say that programming languages are less subject to interpretation than math; after all, machines can understand them and produce consistent results.
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+1 Thanks for pointing out what I wanted to point out: that math is subject to interpretation, and context is crucial. – amWhy Nov 23 '12 at 13:41
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I assumed he was talking about "formal" math, where all the notation, symbols, etc are locked in up front, so you wouldn't have any "overloading" of "+", for example. Real mathematics done by real people is not formal, however, and context is key in understanding all communications with other people. – Thomas Andrews Nov 23 '12 at 13:43
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@Derfder: please, don't think you understand string theory if you don't even get the uncertainty principle. – akkkk Nov 23 '12 at 13:57
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I don't think your example of computer programming language is good. Undefined behavior means that results might not be consistent, and I believe that it would require solving the halting program to remove. – soandos Nov 23 '12 at 15:23
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@soandos: the halting problem is in the same league as the undecidability in math, so in that sense math is not good either. – Martin Argerami Nov 23 '12 at 15:36
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During the intuitive creative process, mathematics is subject to interpretation. For example, during the invention of calculus, there were various intuitive interpretations of the notion of infinitesimals. These notions could not be made completely rigorous at the time (primarily due to lack of a precise formal language such as first-order logic). However, by and large, they produced correct results when employed by competent practitioners. Only a few centuries later was this intuition faithfully rigorized, when Abraham Robinson discovered a rigorous interpretation of infinitesimals in his nonstandard analysis.
For a more recent example, see the Wikipedia page on the notion of a field with one element, which begins "in mathematics, the field with one element is a suggestive name for an object that should behave similarly to a finite field with a single element, if such a field could exist".
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What is room for interpretation after all?
Statements are always subject to interpretation. To make sure there's no ambiguity in a statement you need to build a more or less complex system of axioms and corollaries that prove your statement true. The same can be applied to the flower you point at. If you state precisely which aspect of the flower you are talking about and what measure of color you apply then you can disambiguate such statement just in the same way.
While it might not always be feasible to disambiguate anything it is not always necessary or desired either. It is interesting that you came up with the example of constructivism. Indeed every person has different pictures in mind that are used to describe their sensory perceptions conceptually. This diversity is usually not a problem as there are a lot of norms and conventions "socialized" into each individual that disambiguate each others actions (including verbal utterings). Admittedly these are somewhat implicit and vary along the time and across subcultures but they provide the solid ground without which a society could not function.
As someone has mentioned computer programming to be free from interpretation, I would argue that. The programmer carefully makes explicit decisions about features and functionality, edge cases and error handling. But he usually starts with a problem statement, which he interprets. He then surrenders his code to some other application which again interprets it, and finally bring it to execution. I would argue that the average programmer knows all of the default settings that are implicitly being used by the compiler, nor is he aware of the implications of all its "optimizations". That is, the same dependency on mutually agreed conventions as in mathematics and in any other field here.
As for your example of $1+1=2$ there probably are millions of people that interpret this equation as you would expect. Such fact might be commonly accepted without any doubt, but as this question shows this is only true if the prerequisites are indeed defined accordingly. Therefore, a mathematician when being asked about the validity of $1+1=2$ should reply a fuzzy it depends, just as the attorney when being asked for legal advice would.
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Only by assuming ("pretending"?) that truth and meaning reside entirely outside the human mind can one assert that "mathematics is not subject [to] interpretation". Such a view reflects a remarkably useful working consensus, broadly embraced by professional mathematicians, that, if one puts forth a set of axioms, definitions and agreed-upon rules of inference, then some propositions will be inescapably true, others will be inescapably false, and still others will be undecidable.
This point of view will often appear inconsistant with a view based on the assumption that truth and meaning are products of living, breathing individuals. In the latter view, it becomes reasonable to ask:
-how does person A interpret this set of mathematical ideas and propositions?
-how do students come to understand the basic ideas and representations of algebra?
-how can one help individuals grasp the distinction between mathematical objects (the real line, a triangle, a function) and the tokens we employ to represent them (a number line drawn on paper, a picture of a triangle, notation for an infinite set of ordered pairs)?
The mathematics educator has to be able to shift constantly between these two perspectives. In addition, s/he has to help students become adept at the former approach while remaining aware of the latter.
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You need to use bullet points. I think you are saying in your 1st paragraph, 2nd sentence, POV_1 = {if A then (B or C or D)}. In paragraph 2, POV_1 is inconsistent with POV_2, where POV_2 = a set of three questions E, F and G, where G is the difference between E and F (I think). The reason I resorted to this was because I was trying to understand your final paragraph. What two perspectives are you referring to in your penultimate sentence, POV_1 v. POV_2, or E v. F (essentially the same as G)? In the final sentence, what are "the former" and "the latter" each referring to? – Feral Oink Jan 4 at 3:52
Also, in your final paragraph, are you distinguishing between "perspectives" and "approaches" in a definition-type way? Or perhaps you are using the words as synonyms? I am not trying to be gratuitously difficult or contrary. I am just confused. I'm sorry. – Feral Oink Jan 4 at 3:55
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@Feral Regarding your second point, in the present context, the terms "perspective" and "approach" are not terms of art with well-defined and agreed upon meanings. Nonetheless, you can think of a "perspective" as a way of looking at or understanding something, in this case, mathematics. An approach (to mathematics) is a characteristic way of dealing with situations involving mathematics. In mathematics education, both are invariably in play, and they may be hard to distinguish in practice. – David Carraher Jan 4 at 10:43
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You can think of the first point of view as one captured in the following statement:"mathematics" is a body of knowledge that would exist even if there were no people. The second point of view can be summarized, although only roughly, by the statement that mathematics is a body of knowledge that does not exist without people. These points of view often go by other names such as Platonism and (social) constructivism, but it's better to avoid such labels because they are subject to varying interpretations and would require much space to pin down. – David Carraher Jan 4 at 10:54
David, thank you so much! That helps a lot, especially your second comment, which specifically addressed my painfully belabored first comment ;o) Yes, I agree that it is better to avoid labels such as Platonism and (social or other) constructivsim, especially here, in the context of the question. We have such a difficult time (collective we, beyond even SE) with interpretation and nuance, even for supposedly defined terms. Thanks again for your promptness and detail! – Feral Oink Jan 5 at 18:33
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http://math.stackexchange.com/questions/221402/how-can-we-make-mathbbrn-into-a-multiplicative-group
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# How can we make $\mathbb{R}^n$ into a multiplicative group?
Are there any sorts of multiplicative group structures we can put on set $\mathbb{R}^n$? By multiplicative, I mean that it should be compatible with the natural scalar multiplication on $\mathbb{R}^n$; that is, for any $x,y \in \mathbb{R}^n$ and $c \in \mathbb{R}$, we should have $$c(x \star y) = cx \star y$$
For $n > 2$, can such structures be abelian? The only example I'm familiar with is $n = 4$, in which we have the quaternions; this is clearly non-abelian.
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I'm guessing you want some group structure compatible with the vector space structure. In which case, 0 having an inverse should cause problems. If you want $\mathbb{R}^n$ to be a field, $\mathbb{R}^2 \simeq \mathbb{C}$ as a vector space, and there cannot be any other. – ronno Oct 26 '12 at 7:23
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As I recall the "best" you can do in this regard gets worse and worse as you go along. The quaternions in $\mathbb{R}^4$ lose commutativity, the octonions in $\mathbb{R}^8$ lose associativity. – AsinglePANCAKE Oct 26 '12 at 7:30
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– jspecter Oct 26 '12 at 8:17
I have updated my question to be a little bit more specific about what I mean about "multiplicative" group structure. – Christopher A. Wong Oct 26 '12 at 18:12
## 3 Answers
I suppose you are searching for a finite-dimensional, associative division algebra over the reals of dimension $n>2$? According to the Frobenius theorem none of those is abelian and the only such division algebra that exists, is given by the (non-abelian) Quaternions.
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For the question, as asked, one may simply define:
$(a_1,\dots,a_n) * (b_1,\dots,b_n) = (a_1+b_1,\dots,a_n+b_n)$.
This is a multiplicative structure on $\mathbb{R}^n$, turning it into an abelian group.
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Sorry about the ambiguity; I updated my question to mention that I meant that the group structure should be compatible with scalar multiplication. – Christopher A. Wong Oct 27 '12 at 1:15
To answer the modified question, the elementwise 'multiplication' achieves $c(x\star y) = cx \star y$, in contrast to $cx \star cy$ as here for the elementwise addition. But of course 'elementwise' is somehow trivial, since it simply represents $n$ separated $1$-dimensional operations. – cubic lettuce Oct 27 '12 at 8:22
It is not entirely clear what you mean by a multiplicative group structure. Notice that the product of quaternions only turns the set $\mathbb{R}^4\setminus\{(0,0,0,0)\}$ into a group - not all of $\mathbb{R}^4$.
In an attempt to coerce a more precise question out of you, I lead off by proffering the following multiplicative structure on $\mathbb{R}^3$. Identify the vector $(a,b,c)$ with the upper triangular matrix $$\left( \begin{array}{ccc} 1&a&b\\ 0&1&c\\ 0&0&1 \end{array}\right).$$ The usual matrix product turns this copy of $\mathbb{R}^3$ into a multiplicative group, doesn't it?
Using larger matrices, we can cover other real vector spaces.
If you want the group operation to be distributive over the usual vector addition, you should say so! Sorry about being a bit cranky.
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http://mathoverflow.net/revisions/62053/list
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## Return to Answer
2 added 1481 characters in body
I think the answer to the question is yes, but it is very unlikely that it can be proved without using the classification of finite simple groups.
Note that $A_5$ of order 60 is the only simple group order for which this statement is true, because for all higher order simple groups $G$, there will be groups $L_2(p)$ with order divisible by $|G|$ that do not contain $G$ as subgroup.
Let's quickly look at all families if finite simple groups. I hope someone will correct any mistakes I make!
The Suzuki groups (Lie type $^2B_2$) are not divisible by 3, so we can forget them. All other finite simple groups have order divisible by 12, so their order is divisible by 60 if and only if it's divisible by 5.
The claim is clearly true for $A_n$, $n \ge 5$.
It is well-known that $L_2(q)$ contains $A_5$ if and only if $q \equiv \pm{1} \bmod 5$, which is equivalent to $|G|$ divisible by 5.
$U_3(q)$, $L_3(q)$, $G_2(q)$, $^3D_4(q)$ all contain $L_2(q)$ and also have order divisible by 5 if and only $q \equiv \pm{1} \bmod 5$.
$^2F_4(2^{2e+1})$ contains $^2F_4(2)$, which contains $A_5$.
$^2G_2(3^{2e+1})$ never has order divisible by 5.
$S_4(q)$ contains $L_2(q^2)$, which always contains $A_5$ for all $q$.
All remaining groups of Lie type contain $S_4(q)$ and hence contain $A_5$.
It is easily checked, for example by looking at their lists of maximal subgroups in the ATLAS or on
http://brauer.maths.qmul.ac.uk/Atlas/v3/
that the sporadic groups contain $A_5$.
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I think the answer to the q
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http://mathhelpforum.com/math-puzzles/168742-divided-square.html
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# Thread:
1. ## Divided square
Hey guys i found a problem in the MENSA book, i've tried to use basic algebra to this this problem but i think i'm on the wrong track.
Can anyone suggest any techniques to do this problem?
The answers are at the back of the book although i do want to know how to do it.
THE IMAGE IS ATTACHED.
This field measures 177m x 176m. It has been split up into 11 squares that exactly equal the total area. The squares are only roughly drawn to scale. All new squares are in whole yards. Can you calculate the size of each square ?
- much appreciated.
Attached Thumbnails
2. It's a question of setting up the right system of equations. At least, that's how I solved it. Let each variable name be the length of a side of the corresponding square. Then the equations I have are the following:
$A+B=177$
$A+D=176$
$C+F=B$
$E+G=D$
$I+K=F$
$J+H=G$
$B+F+K=176$
$K+J+G+D=177$
$H+I=J$
$D+E+F=177$
$A+C+F=177$
$D=J+H+E$
$B+C+E+G=176$
At this point, admitting laziness, I turned the solution over to Mathematica, which spit out the result:
Spoiler:
A=99, B=78, C=21, D=77, E=43, F=57, G=34, H=9, I=16, J=25, K=41.
In principle, though, you would probably employ Gaussian elimination with back substitution to solve the system. You have a decently sparse system there, so it might not be all that messy, actually.
Cheers.
3. Thank you.
4. You're very welcome. Have a good one!
5. To mention, Beiler covers this type of problem in his book on number theory recreations.
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http://unapologetic.wordpress.com/2008/06/27/the-rank-nullity-theorem/?like=1&source=post_flair&_wpnonce=fe7f791e1e
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# The Unapologetic Mathematician
## The Rank-Nullity Theorem
Today we start considering how a given linear transformation acts on a vector space. And we start with the “rank-nullity” theorem. This sounds really fancy, but it’s actually almost trivial.
We already said that $\mathbf{Vect}(\mathbb{F})$ (also $\mathbf{FinVect}(\mathbb{F})$) is a abelian category. Now in any abelian category we have the first isomorphism theorem.
So let’s take this and consider a linear transformation $T:V\rightarrow W$. The first isomorphism theorem says we can factor $T$ as a surjection $E$ followed by an injection $M$. We’ll just regard the latter as the inclusion of the image of $T$ as a subspace of $W$. As for the surjection, it must be the linear map $V\rightarrow V/\mathrm{Ker}(T)$, just as in any abelian category. Then we can set up the short exact sequence
$\mathbf{0}\rightarrow\mathrm{Ker}(T)\rightarrow V\rightarrow V/\mathrm{Ker}(T)\rightarrow\mathbf{0}$
and the isomorphism theorem allows us to replace the last term
$\mathbf{0}\rightarrow\mathrm{Ker}(T)\rightarrow V\rightarrow \mathrm{Im}(T)\rightarrow\mathbf{0}$
Now since every short exact sequence splits we have an isomorphism $V\cong\mathrm{Ker}(T)\oplus\mathrm{Im}(T)$. This is the content of the rank-nullity theorem.
So where do “rank” and “nullity” come in? Well, these are just jargon terms. The “rank” of a linear transformation is the dimension of its image — not the target vector space, mind you, but the subspace of vectors of the form $T(v)$. That is, it’s the size of the largest set of linearly independent vectors in the image of the transformation. The “nullity” is the dimension of the kernel — the largest number of linearly independent vectors that $T$ sends to the zero vector in $W$.
So what does the direct sum decomposition above mean? It tells us that there is a basis of $V$ which is in bijection with the disjoint union of a basis for $\mathrm{Ker}(T)$ and a basis for $\mathrm{Im}(T)$. In the finite-dimensional case we can take cardinalities and say that the dimension of $V$ is the sum of the dimensions of $\mathrm{Ker}(T)$ and $\mathrm{Im}(T)$. Or, to use our new words, the dimension of $V$ is the sum of the rank and the nullity of $T$. Thus: the rank-nullity theorem.
### Like this:
Posted by John Armstrong | Algebra, Linear Algebra
## 13 Comments »
1. [...] Let’s go back and consider a linear map . Remember that we defined its rank to be the dimension of its image. Let’s consider this a little more [...]
Pingback by | July 1, 2008 | Reply
2. [...] be the dimension of this subspace, which we called the nullity of the linear transformation . The rank-nullity theorem then tells us that we have a relationship between the number of independent solutions to the system [...]
Pingback by | July 14, 2008 | Reply
3. [...] system has no solutions at all. What’s the problem? Well, we’ve got a linear map . The rank-nullity theorem tells us that the dimension of the image (the rank) plus the dimension of the kernel (the nullity) [...]
Pingback by | July 18, 2008 | Reply
4. [...] the dimension of the cokernel add up to the dimension of the target space. But notice also that the rank-nullity theorem tells us that the dimension of the kernel and the dimension of the image add up to the dimension of [...]
Pingback by | July 22, 2008 | Reply
5. [...] the rank-nullity theorem says that , and similarly for all other linear maps. So we get — which expresses as the [...]
Pingback by | July 23, 2008 | Reply
6. [...] the time we get to the th power, where . Instead of repeating everything, let’s just use the rank-nullity theorem, which says for each power that . Now if then we [...]
Pingback by | February 18, 2009 | Reply
7. [...] is, the kernel of is trivial. Since is a transformation from the vector space to itself, the rank-nullity theorem tells us that the image of is all of . That is, must be an invertible [...]
Pingback by | July 17, 2009 | Reply
8. [...] this means that we can read off the rank of from the number of rows in , while the nullity is the number of zero columns in [...]
Pingback by | August 18, 2009 | Reply
9. [...] count the dimensions — if has dimension then each space has dimension — and use the rank-nullity theorem to see that they must be isomorphic. That is, every -multilinear functional is a linear combination [...]
Pingback by | October 22, 2009 | Reply
10. [...] this were a linear system, the rank-nullity theorem would tell us that our solution space is (generically) dimensional. Indeed, we could use [...]
Pingback by | November 19, 2009 | Reply
11. [...] we already know that their dimensions are equal, so the rank-nullity theorem tells us all we need is to find an injective linear map from one to the [...]
Pingback by | October 13, 2010 | Reply
12. [...] that this is onto, so the dimension of the image is . The dimension of the source is , and so the rank-nullity theorem tells us that the dimension of the kernel — the dimension of the space that sends back to [...]
Pingback by | November 9, 2010 | Reply
13. [...] to a constant function is automatically zero. Thus we conclude that . In fact, we can say more. The rank-nullity theorem tells us that the dimension of and the dimension of add up to the dimension of , which of course [...]
Pingback by | April 26, 2011 | Reply
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## 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.
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http://mathoverflow.net/questions/69488?sort=votes
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## Are surface bundles over a surface with non-zero signature necessarily complex (or algebraic)?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
By "surface bundle over a surface" I mean a compact, oriented 4-manifold $X$ which is the total space of an oriented fiber bundle $X\to B$ over an oriented 2-manifold $B$. Assume that the signature of the 4-manifold $X$ is non-trivial.
Conjecture 1: $X$ and $B$ can be given complex structures such that the map $X \to B$ is holomorphic.
Conjecture 2: $X$ and $B$ can be given the structure of complex algebraic varieties such that the map $X \to B$ is algebraic.
Question: Are the above conjectures true? If not, can you provide a counter example?
Background: The bundle $X\to B$ induces a map $f:B \to M_g$ where $M_g$ is the moduli space of curves and $g$ is the genus of the fiber. The signature of $X$ is given by
$$\sigma(X) = 4\int _B f^*(\lambda)$$
where $\lambda$ is the first Chern class of the Hodge bundle. Since the signature of $X$ is non-zero, the map $f$ is non-trivial in homology and one might hope that an argument along the following lines is true: Find a representative $\tilde{f}$ in the same homotopy class as $f$ which has minimal energy. Using the fact that $M_g$ has a hyperbolic metric ($g>2$ since $\sigma(X)\neq 0$), prove this minimal energy map is holomorphic or even algebraic. $\tilde{f}$ provides $X$ with the desired structure.
This question is a variant on this recent question: http://mathoverflow.net/questions/69344/a-four-dimensional-counterexample the counter example given there has zero signature.
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## 2 Answers
I seem to have found an answer to my own question.
The first observation is that a holomorphic fibration is automatically algebraic. This is because $M_g$ is quasi-projective and so the map $f:B\to M_g$, which trivially extends to a map to $\overline{M}_g$ is algebraic by GAGA (perhaps one should first pass to some finite cover to take care of stack issues).
However, my conjectures are false --- here is a counterexample. Let $X \to B$ be a surface bundle having non-trivial signature $\sigma$, base genus $h$, and fiber genus $g$. Let $X_N$ be the surface bundle obtained by taking the fiber connect sum of $X$ with the trivial bundle of fiber genus $g$ and base genus $N-h$. So for each $N\ge h$, we get a surface bundle $X_N \to B_N$ of signature $\sigma$ and base genus $N$. If $X_N \to B_N$ were all holomorphic as in my conjecture, then we would obtain a family of complete smooth curves $B_N$ in $M_g$ of fixed degree and unbounded genus, a contradiction. (The degree here is taken with respect to $\lambda$ which is ample on $M_g$ and fixed by the signature formula).
What happens in my minimal energy argument is that as the map $B_N \to M_g$ becomes harmonic, it bubbles off a collapsing component.
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### 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 have a question about the first posting of Jim Bryan.
what does it mean that f tilda has minimal energy?
And, how is it relevant to the nontriviality of f_{*} on the second homology?
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http://math.stackexchange.com/questions/131919/eigenfunctions-of-the-adjoint-of-an-operator/131956
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# eigenfunctions of the adjoint of an operator
If the eigenfunctions of a linear operator are known, is there a way to calculate the eigenfunctions of the corresponding adjoint operator based on the known eigenfunctions? In other words, what's the relation between the eigenfunctions of an operator and its adjoint?
Thanks!
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Also, how about the eigenfunctions of the inverse operator if it's invertible? – chaohuang Apr 16 '12 at 18:23
## 2 Answers
Jim answered the finite-dimensional case very nicely; let me address the infinite-dimensional case.
If you don't specify some conditions on your operator, the answer is no. Consider $H=\ell^2(\mathbb{N})$, and let $T$ be the "reverse shift" operator, given by $$T(a_1,a_2,\ldots)=(a_2,a_3,\ldots).$$ Then it is easy to see that every $\lambda\in\mathbb{C}$ with $|\lambda|<1$ is an eigenvalue with eigenvector $(\lambda,\lambda^2,\lambda^3,\ldots)$. More properly, one is free to choose the first coordinate, but that is irrelevant here.
Now, $T^*$ is the usual shift $$T^*(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots)$$ and it has no eigenvalues (and so no eigenvectors).
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In the finite-dimensional case, the eigenfunctions for the adjoint are the dual basis to the basis of eigenfunctions. That is, if $f_1,\ldots,f_n$ are the eigenfunctions for an operator $T$, then the eigenfunctions $g_1,\ldots,g_n$ for the adjoint are defined by the equations $$\langle g_i,f_j\rangle = \delta_{ij},$$ where $\langle-,-\rangle$ is the inner product on the space of functions.
If we write $g_i = a_{i1}f_1 + a_{i2}f_2 + \cdots + a_{in}f_n$, then we can use the above equations to solve for the coefficients $a_{ij}$. The solution is that the matrix of coefficients $a_{ij}$ is the inverse of the matrix whose entries are $\langle f_i,f_j\rangle$.
I imagine that similar statements are true for certain types of non-self-adjoint operators on a Hilbert space, but I'm not an expert on functional analysis.
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http://math.stackexchange.com/questions/294132/which-problems-do-you-recommend-it-to-me-to-solve-it
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# which problems do you recommend it to me to solve it? [closed]
i study abstract algebra from dummit and foote .
i started to solve some problems in section 3 in chapter 4
there is 36 problems
i study the subject myself , so there is no proffesor to recommend or instructor !
so , i hope that you help me to recommnd some of therse exercises because solving all of them will require lot's of time !!
so , i wait your suggestions !
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1
I think this question should be closed as too localized - it's not going to be relevant to anyone but you. And if there's no professor, why not at least try every problem? What's the hurry? The more problems you think about, the more you'll learn... – Zev Chonoles♦ Feb 4 at 1:15
I just go through the problems and do the ones that either look interesting, or focus on what I consider the trickier parts of the chapter. This also might mean going back to the problem set later if the text does more with a subject than you've practiced for. Essentially all the problems in there are good, so exactly which ones you do should be based on what you personally find interesting and/or difficult. – Robert Mastragostino Feb 4 at 1:20
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@ZevChonoles Isn't this question potentially useful to anyone studying this particular section of Dummit & Foote? A great deal of people study Dummit & Foote to learn abstract algebra, so potentially many people could benefit from a list of good problems from this section. – JSchlather Feb 4 at 1:34
@Jacob: I don't think it is useful - are any of these 36 problems so terrible / hard / uninteresting that they should be explicitly disrecommended to people? – Zev Chonoles♦ Feb 4 at 1:57
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@ZevChonoles I don't think that by recommending certain problems over other problems, you're saying these problems are without merit, simply that other problems have more merit. I've frequently found that certain problems can be much more illuminating than other exercises. Particularly in cases where Dummit & Foote put an important example or theorem as a series of exercises. For instance I recall Dummit & Foote asking you to show that the Prufer $p$-group has no maximal subgroups in an exercise. Surely this is a better exercise than counting the number of conjugacy classes of $A_4$? – JSchlather Feb 4 at 2:13
show 8 more comments
## closed as too localized by Austin Mohr, Robert Mastragostino, 5PM, rschwieb, Brett FrankelFeb 4 at 1:39
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.
## 2 Answers
Usually textbooks list problems/exercises in the order in which they progress from easier to solve to greater and greater levels of difficulty. So your best bet, if you don't want to solve them all!, is to pick a few earlier problems, to get warmed up, then some middles problems, and build up to selecting some of the problems from the last third.
A good approach would be to do all even problems, or all odd problems, or, if you aren't struggling to much with a particular section of the text, perhaps every third problem. That way you are following a progression from starting with easier exercises, and progressing to more difficult ones.
I'd suggest scattering a few of the earlier problems (every third problem, say), but putting most of your energy into the last half of the exercises: Those are the exercises that will ensure (or require) that you "really know your stuff"!
At any rate, save a few problems for giving your self "examinations" periodically, for the sake of personal accountability: In a typical semester-length (4-5 months or so), there is at least one midterm, and a final exam. So you might aim, say, for testing to return every month or two, first spending a little time reviewing the material covered during that time (studying for your exam), and then returning to some of the problems you left "unsolved" to "test yourself". If you find yourself struggling to complete those exercises, you might want to spend a bit of time reviewing the relevant material in order to recall (and/or deepen your understanding of) what you need to know to solve them, before moving on.
Then, "back to the books" and moving forward from where you left off!
Good luck!
ADDED: Here is a syllabus from an abstract algebra class taught last fall at Standford University. It includes assigned exercises from the class text (Dummit and Foote), and most of the assignments have solutions available for downloading.
You can also try googling "Dummit and Foote: .edu" to search for other links to course syllabi in which Dummit and Foote is the class text.
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I hope this helps, MrWhy! If you'd like, we can occasionally open a private chat to discuss some of Dummit and Foote, as you're studying it. I can't do that daily, but we can figure out some time each week to do so. Any way, let me know if you still have a question on the last grouptheory question you posted! the maximal sugroup question...If you still would like clarification, I can post something by tomorrow. If you're okay with it now (the question/answers), that's great. – amWhy Feb 4 at 3:35
i'm ok with it now :) , yes this disscussion will be great ! , i'll googling as you mentioned in your answer , if you found other things likes notes , videos , exercises ! this may hepl . i noticed that the course in stanford didn't cover chapter 5 or 6 in dummnit and foote ! this was strange for me !! do you have an idea about the reason of this ? on the disscusion , can we make first one after finishing chapter 4 - group actions - ?? what do you think ? if you agree , i'll contact with you here in a comment or on the way you like to determine the time to disscus . i wait your reply – Maths Lover Feb 5 at 3:03
Yes, I'll look for some more syllabi with exercises, maybe even solutions. This may have been the first semester of a year long class, so I'll see if there's a follow-up site for that class this spring. Yes, feel free to comment when you'd like to "meet" in chat. Feel free, to, to accept my answer here ;-). – amWhy Feb 5 at 3:06
last thing , i want to thank you very much for stanford course's link :) i think it'll be useful for me :) also , i want to thank you for the disscusion :) I'm Maths Lover , but i changed my name , it's similar to yours , but i liked it as one of the greatest mathematician - godel if my memory is good ! - was called Mr Why when he was a child ! if this is annoying for you , i will change it – Maths Lover Feb 5 at 3:07
Not at all annoying...I like "Why" and "MrWhy"! Just be sure to accept my answer ;-) – amWhy Feb 5 at 3:10
show 2 more comments
What I do in general is that I try to solve a couple of the problems which are more mechanic (like the ones on the euclidean algorithm and the totient function at the beginning) and then try to tackle the difficult ones at the end.) Try all of the hard ones. Especially because in that book they tell you a lot of stuff in the problems that they don't in the general text.
I save some of them for later so I can go back and consolidate what I have learned. Some of the harder problems make the text more juicy and give clues to what will be developed in future pages.
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hello , thanks for your advices , you look in high school like me and study the same subject , maybe you are intersted in knowing people interested in the same subjects and in the same age level - i'm also a high school student - . if you are interested in communicating with me to be a friends or help each other , that will be nice for me ! what do you think ?! – Maths Lover Feb 4 at 2:34
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http://crypto.stackexchange.com/questions/2887/difference-between-encrypting-something-and-hashing-something/2898
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# Difference between encrypting something and hashing something
What is the difference between encrypting something and hashing something? in what situations would I want one or the other?
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1
@moderation: please, make this question a sugested similar question to every new question that asks about encryption or hashing. We're full of "how to encrypt with MD5" and so on. – woliveirajr Jun 15 '12 at 13:53
## 4 Answers
(I am new to cryptography, so hopefully I wouldn't say anything wrong. The following I hope should help give an idea of differences).
In general you always have a plaintext or something that you want to keep secret. This might be a message or a file or a password.
Encrypting: enrypting something is a way to transform the plaintext into something that cannot be understood - something that doesn't have any meaning. Something unintelligible. You encrypt the message/plaintext by using an algorithm and a key. The output (the ciphertext) always depends on the key. The essential thing about encryption is that you can always recover the plaintext (if you know the key/password), and given a key/password you have only one plaintext. So you can think of encryption/decryption as a way to move back and forward between you plaintext and your cipher text (using a key). Uses: So encryption is good if you say have a message to send to someone. You encrypt the message with a key and the recipient decrypts with the same (or maybe even a different) key to get back the original message.
Hashing: hashing something is a way of turning something (usually a key or a password) into a (usually fixed length string of characters. So for example your hashing algorithm might always produce a string that is 8 bytes long. So as with encryption you transform something intelligible into something unintelligible. One might call that the product of the hashing the hash codes of the hash sums. One difference is that hashing two different messages might produce the same hash values. So you cannot decrypt something that you have hashed. But even though you can't decrypt a hashed value, it is (or it should) in general hard to make two messages that have the same hash value. Another difference is that hashing doesn't require a key. Uses: So one use of hashing is if you want to send someone a file. But you are afraid that someone else might intercept the file and change it. So a way that the recipient can make sure that it is the right file is if you post the hash value publicly. That way the recipient can compute the hash value of the file received and check that it matches the hash value.
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I wanted to note a couple of KEY (no pund intended).
Encryption doesnt collide because you are essentially generated a "messed" up copy of the same length of your original message. A collide is the name for what Thomas describes in his hash description. What this means is essentially:
Encrypt input to output is n bits in, n bits out, and data is reversible given the key.
Hashing (if we exemplify SHA 512 for example) takes in 1024 bit chunks at a time (but can generate a HASH for multiple blocks of 1024, but for simplicity we will only talk about a single block). Since the output is ALWAYS 512 bits (in this case), there could be as many as 2^(1024 - 512) repeated input values that result in the exact same output hash value (digest). Hashing can be keyed or not keyed.
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More formally, encryption is a random permutation by virtue of being invertible, so two distinct plaintexts cannot result in the same ciphertext under a given key (obviously) – Thomas Jun 14 '12 at 10:25
Actually, it is not a random permutation, it is a VERY DETERMINISTIC permutation (normally) based on a random input (i.e. the key). By definition, if it were truly a "random" permutation, it would not be reversible. – trumpetlicks Jun 14 '12 at 13:14
Oh yes, indeed, that was a horrible abuse of terminology! Pseudorandom, low-entropy permutation it is, of course. – Thomas Jun 14 '12 at 18:54
An encrypted document can be decrypted by those, who know the secret.
A cryptographic hash is a checksum that allows someone to proof that he knows the original input (e. g. a password) and that the input (e. g. a document) has not been modified. A cryptographic secure hash does not allow to recover the original input or any other matching input.
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An encryption scheme $E$ is a triple of algorithms $E=(\mathsf{KGen},\mathsf{Enc},\mathsf{Dec})$.
The key generation algorithm $\mathsf{KGen}$ is a probabilistic algorithm that on input a so called security parameter $\kappa$ (usually the length of the key) outputs a random key $k$.
The encryption algorithm $\mathsf{Enc}$ is a (possibly probabilistic) algorithm that on input a key $k$, and a message $m$ outputs a ciphertext $c$.
The decryption algorithm $\mathsf{Dec}$ is a deterministic algorithm that on input a key $k$ and a ciphertext $c$ outputs the message $m$.
The correctness property of an encryption scheme is that for all security parameters $\kappa$, all keys $k \gets \mathsf{KGen}(\kappa)$, all messages $m$ (that are in the schemes message space), and all ciphertexts $c \gets \mathsf{Enc}(k,m)$ it holds that $\mathsf{Dec}(k,c)=m$.
The minimal security notion we usually want the encryption scheme to fulfill, is called ciphertext indistinguishability under adaptively chosen message attacks (IND-CPA). The security notion basically means that we have an attacker, he can ask arbitrary messages and we will encrypt them and give the ciphertext to the attacker. At some point the attacker will give us two messages $m_0,m_1$, we flip a coin and encrypt only one of them (depending on the coin toss) and give that ciphertext back to the attacker. Now even though the attacker can ask us to encrypt arbitrary messages, he should not be able to decide which of his messages was encrypted.
A cryptographic hash function $H$ (actually to define security we need to speak about families of hash functions, but that might get a bit complicated) is a function that maps bitstrings of arbitrary length into some finite set. The most common case is a mapping to fixed length bit-strings. That is we have a function $H: \{0,1\}^* \rightarrow \{0,1\}^n$ for some length $n$.
The only correctness condition is that $H$ must be surjective, that is, the same input is always mapped to the same output.
The security property of a cryptographic hash function is called collision resistance. Collision resistance means, that an attacker given access to the hash function should be unable to produce two messages $m_0,m_1$ such that $H(m_0)=H(m_1)$. (This also implies a few other properties often defined separately, such as first and second preimage resistance.)
Note, however, that hash functions do have collisions. A large set is mapped onto a smaller set, so collisions are bound to happen. However, an attacker should not be able to find such collisions. (which is also why we actually need hash function families to define security)
So in summary: An encryption scheme enables you to encrypt a message, so that anybody who knows the key can decrypt it and therefore recover the message. You can use this to send or store messages while hiding their content from others.
Contrary to that, with a hash function, it is impossible recover the message, given only the hash-value. And not only that but you cannot even find a different message, that will produce the same hash value. This can be used for example to check the integrity of files (given that the hash is from a trusted source) as an attacker will not be able to produce a different file with the same hash. They are further used as a building block of most signature schemes and in password based authentication schemes.
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http://meta.math.stackexchange.com/questions/1652/what-do-we-do-with-users-who-post-numerous-unlabeled-homework-questions/1661
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# What do we do with users who post numerous unlabeled homework questions?
User student has, quite impressively, recognized that all of user6560's questions are homework questions from the ongoing course Math 620 at the University of Buffalo. I just went through, downvoted all the questions and left an explanatory comment on all of them.
It is theoretically possible that user6560 is not enrolled in the course, but simply is independently studying from the course website. In that case, I would say that user6560 hasn't done anything wrong, although they1 would have done better to explain the situation. However, in the more likely event that user6560 is enrolled, they are violating our policies on how to ask homework questions. Moreover, if they are not disclosing to their professor that they are seeking help here, then they are most likely in violation of the University of Buffalo's plagiarism policies.
My question is, should we take further action? Is it appropriate to retag the questions? Close them? E-mail professor Badzioch? UPDATE: User student has e-mailed Badzioch. Pointing this out here so that we don't flood his inbox.
On Mathoverflow, I would immediately close these down, and quite likely e-mail the professor, but I am not sure what the community norms are here. None of the meta discussion on homework seems to have a consensus on how far we will go on this issue.
Finally, let me give a word of warning from my Mathoverflow experience. We had a lot of unpleasant flame wars early on because user X would ask a question, user Y would answer it, and user Z would vote down Y's answer because Z thought that X's question looked like homework. Let's focus on X's misbehavior here, not on whether Y was acting as a sufficiently vigilant cop.
1 Yes, I sometimes use "they" for a singular person of indeterminant gender. If it is good enough for the King James Bible, it's good enough for me.
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– Yuval Filmus Feb 13 '11 at 20:06
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It seems that most of the traffic we get is HW anyway, so there's no need for an explicit tag... – Yuval Filmus Feb 13 '11 at 20:07
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@David: is there a reasonable argument against emailing the professor teaching the course? If it were me, I would be at least interested to know that someone was posting my homework problems, whether or not "they" are formally attending my course. – Pete L. Clark Feb 13 '11 at 20:24
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@David: Downvoting the questions is not an optimal solution since it is not the question itself that is the problem. This will cause good answers to undeservedly get little exposure (e.g. if they are merged when a duplicate question is asked in the future). – Gone Feb 13 '11 at 20:33
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@Yuval: I disagree with the lack of need; at least I try to behave differently when facing a homework question than when facing a curiosity-driven/self-study question. Just like I behave differently when a student comes to ask me for help with a homework problem (mine or someone else's), vs. when they come to ask me about something they are wondering about. – Arturo Magidin Feb 13 '11 at 23:19
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@David, @Peter: I have, on rare occasions, contacted professors when students were posting homework questions on `sci.math`; I've always been thanked (apparently sincerely). I don't think there is any reasonable argument against emailing the professor, we just don't want to have several dozen users e-mailing him all about the same thing at the same time. A single person e-mailing and pointing is likely a good idea. – Arturo Magidin Feb 13 '11 at 23:35
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Just for my better understanding: His behavior would have been ok, had he tagged his questions as homework and shown that he tried to do it on his own first, right? – Stefan Walter Feb 14 '11 at 0:03
@Stefan Walter: as far as I'm concerned, tagged it, made it clear that the problems came from homework assignments. Two questions from each assignment does not seem excessive to me, though if we'd had, say, four or five in quick succession, that might have been a problem for me even if clearly labeled. – Arturo Magidin Feb 14 '11 at 0:09
@Arturo: acknowledged. Don't worry -- I didn't email the professor myself. – Pete L. Clark Feb 14 '11 at 1:25
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@Arturo: I believe Yuval was just joking. But there is some truth: we seem to have been overrun by homework problems... – Aryabhata Feb 15 '11 at 2:28
I was joking, but at the same time I do believe many of the questions derive from homework, and most of the rest are "coursework related"; that would explain the apparent drop in question rate in the off-months. – Yuval Filmus Feb 15 '11 at 4:05
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@Yuval Filmus, I am sorry that you didn't like my answer to 21630, as you indicated above; I thought actually it led to some interesting mathematical issues in the comments, and I think the OP learned from my answer. So why do you object? Do you really think it would improve the site to have fewer such answers? (Please see my answer to this question below.) – JDH Feb 15 '11 at 4:50
Another indication of course-related questions is when waves of group or ring theory questions arrive from group-first or ring-first algebra courses. – Gone Feb 15 '11 at 4:59
@JDH My feeling is that something like double induction can only be "really" understood if one works it out by oneself. A hint like "the two variables should be the two digits" is probably enough (if not too much), leaving the student with some missing details. A complete solution, on the other hand, is something that can be copied without full understanding. – Yuval Filmus Feb 15 '11 at 6:53
In this particular instance, the students were supposed to put it in some formal form, so perhaps even a complete solution wouldn't do without the requisite understanding. Moreover, that particular course is "enhanced", i.e. intended for the serious students, and so we shouldn't worry so much about students doing as much as they can to avoid learning anything. This could be hinted by the uncommon remarks accompanying the actual question. – Yuval Filmus Feb 15 '11 at 6:54
show 1 more comment
## 5 Answers
This is definitely not what MSE is for. I've emailed the user asking them to stop. If the behavior continues, they will be suspended and their questions closed or locked.
To David and student: thank you for your diligence.
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be sure to use the moderator contact facility provided from the mod menu on the user, unless you are comfortable sharing your personal direct email address with this user. – Jeff Atwood♦ Feb 14 '11 at 6:59
@Jeff: Dear Jeff, that's what I did (though I've made my personal email address public). – Akhil Mathew Feb 14 '11 at 12:52
My opinion is that there is nothing wrong at all with posting homework questions here, particularly interesting ones, and I find much of the negative reaction to homework-question posters to be somewhat strange, alien to my way of learning mathematics in a give-and-take exchange of mathematical ideas. Surely posting questions here and studying the answers is not much different than studying hard in the library, talking mathematics with one's colleagues at math tea or talking to one's professor, which are all excellent ways to learn mathematics. In particular, I expect that students who post questions here might learn just as much if not more from the resulting answers as from their professors---we have a number of talented mathematicians, who are very good at explaining things---and that math.SE provides a valuable service to students having unapproachable professors, having professors who do not explain well, or who have few colleagues able to help them. Furthermore, the math.SE community strongly benefits from the questions and the insightful answers that might be posted.
So my opinion is that there is no homework issue to speak of.
In particular, I hereby give all of my own students complete permission to post any and all their homework problems here, and indeed I encourage them to post their questions here and to study the answers well and thereby to learn some mathematics. I will be testing them on their understanding at the exam.
I would also encourage all mathematics professors to adopt a policy of encouraging collaboration on homework among their students, as talking about mathematics with one's colleagues is assuredly one of the best ways to learn mathematics. Indeed, I recommend that all professors should actively encourage their students to form study groups in order to work on their homework problems together. Learning as a group, they will go very far.
Finally, let me say that the policy of encouraging weasily half-answers to questions that have been deemed to be homework, consisting of obscure hints only, amounts to an annoying policy of encouraging bad answers here at math.SE, and I am completely opposed to it. For this reason, I think we should abandon or ignore the homework tag. If we are to answer mathematics questions, then let us answer them well, with solutions exhibiting such clarity and elegance as we can muster.
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You're ignoring the fact that many math classes are taken by unwilling students of other disciplines. Those students are after a solution to the exercise, not understanding. If not forced to tackle the questions by themselves (or with fellow students), they will never learn anything. This is especially true in regard to questions "phrased in the imperative". – Yuval Filmus Feb 15 '11 at 4:03
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I'm not ignoring that fact. Such students will benefit from understanding the answers posted here. How is it different from learning from a book or from talking to their professor? Would you object to such students going to the library for the same reason? – JDH Feb 15 '11 at 4:09
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@Joel I disagree that incomplete answers or hints are automatically bad answers. I have often given hints that were subsequently accepted and I claim that the students got more out of the exercises. Here are some examples: tinyurl.com/5szyuqc, tinyurl.com/62h5z8x, tinyurl.com/6huofk8. Also, your comparison with a conversation with a professor gives further support to the policy of "no complete answers to HW", since most professors I know would much prefer giving a hint a letting the students struggle with the problem, instead of handing them a solution on a silver plate. – Alex B. Feb 15 '11 at 6:18
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Alex, I stand by my remarks. As a mathematician, I could never prefer an obscure half-answer to a mathematics question here, when an insightful discussion of general strategies or extensions and an enlightening solution might be available. – JDH Feb 15 '11 at 14:12
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@JDH: I think whether or not a hint is a proper answer heavily depends on what is being asked. If the question is asking for hints or suggestions as to how to start a problem or how to progress from a particular point in a solution, a hint isn't an obscure half-answer. That is, the question being asked and the problem being discussed are not necessarily the same. – Isaac Feb 15 '11 at 14:24
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@JDH: Hi, Joel. I think there are some good points you make; but I will point out some differences between learning from a book or getting a problem worked out here. The student who goes to the library will have to be somewhat active, locate applicable results, realize that they are applicable, adapt them, and then use them, except perhaps on the rare occasion in which he finds the exact assigned problem worked out in a book. If the specific problem is worked out here, this requires much less initiative and action on the part of the student. Likewise with group work or asking a prof. – Arturo Magidin Feb 15 '11 at 14:34
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@JDH: Dear Joel, I see nothing wrong with collaboration (which has encouraged in most, if not all, of the math and science courses that I've taken, and many others whose syllabi I've read). However, in many of these courses, it is emphasized that students' write-ups should be their own, and for that reason I think writing a fully fleshed out (and easily copiable) answer to a question that is homework is sub-optimal: the student both gains an unfair advantage over her classmates (especially in classes where the grading is based almost entirely on homework) and does not get the same practice... – Akhil Mathew Feb 15 '11 at 14:51
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...of thinking through a solution in sufficient detail oneself to produce a coherent submission. As @Arturo observes, it requires less effort for a student to copy (with perhaps some touch-ups) a solution posted here than to flesh out a hint provided by anyone else. I understand that the current policy of encouraging people to post hints to questions that are definitely homework is not perfect, and there may be both false positives and false negatives to people's guesses whether a question is homework, but I do not see how giving full solutions to homework questions would be a better one. – Akhil Mathew Feb 15 '11 at 14:55
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(Incidentally, the student may even find herself in an awkward position --- at least according to course policies --- if she asks for a "hint" to a question and gets a full answer. Granted, the OP in the present situation made no such request.) – Akhil Mathew Feb 15 '11 at 14:56
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The answers can always be edited to provide the fully clear and elaborate answers after the student has done the work. I have seen people willing to put in the effort to guide the student and then later modify the answer with a fuller better version of the answer. From what I have seen, people who provide hints are usually almost always willing to clarify any comments from the students (in fact, I have seen other people jump in too!). It would not be too much of an additional burden on them if we encouraged them to add the full answer later, IMO. – Aryabhata Feb 15 '11 at 17:40
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@Akhil: I dislike the argument that students who are able to easily copy homework answers are given an “unfair advantage”. Presumably homework is assigned to give those who work through and understand it an advantage over those who don’t do it or just copy it, right? – ghshtalt Feb 15 '11 at 17:53
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@ghsh: What about a take home exam? To give an example of the unfairness, and not just to other students: I wouldn't want to be operated upon by a doctor who got through college without actually understanding the subjects. Of course, this might be a failure of the education system, but that is not our concern right now. – Aryabhata Feb 15 '11 at 20:35
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YES. +1 for a great answer, +1 for being reasonable as a professor, +1 for "test them at the exam", +1 for knowledge doesn't come out of thin air and discussion is a great way to learn. All too often, a professor will adopt a DADT policy on how his students get the assignments done, (i.e. "turn a blind eye" to forums altogether). Your perspective is extremely healthy and I'm sure your students will be a lot more comfortable with this straightforward policy. – bobobobo Feb 16 '11 at 16:36
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As a teacher I generally encouraged collaboration on homework, albeit with the proviso that students write up their solutions in their own words. However, your final paragraph is throwing the baby out with the bath. Well-crafted hints can be excellent answers. It would make more sense to encourage good hints, and perhaps to encourage answerers to go back after a few days to flesh out their hints, if they've not already done so. – Brian M. Scott May 8 '12 at 22:21
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+1: To commenters including @YuvalFilmus, who suggests we should concern ourselves with students who don't have willingness to learn. I don't agree. I prefer to assume that a student does have a willingness to learn and deal with this case properly - for these students, a complete answer is far more helpful than a hint that may or may not hit the source of their misunderstanding. Educationally, it's much preferable to provide a complete answer. – Ronald May 10 '12 at 11:44
show 5 more comments
Mainly directed at user6560's answer.
The answer on when using someone's hints or ideas becomes frowned upon is up to the professor on your class. The problem here is not whether or not you are using the hints in your write-ups, but the fact that you failed to disclose to us that you were seeking answers (or hints, or help, or whatever) to homework. My main problem is that it makes any action I take on your question a potential accessory to academic dishonesty, while keeping me in the dark about it (and, to some degree, makes me feel used). Had you disclosed ahead of time that these are homework problems you are working on for a course, I doubt it would have caused any problems.
Let me also say that I think you are misunderstanding the point of giving credit; it's not because we need or want the credit. You aren't insulting us by not giving credit. The problem is that you are potentially appropriating the work of others as your own. Even setting aside the issues of academic honesty, let me point out a few of the problems with this that you may not even have thought about:
Say I'm teaching a course, assign homework, including some problems that I know are challenging. My students turn in wonderful solutions, well-written, clear, often insightful, sometimes even clever, without having to ask me for help, essentially indicating that they are coming up with these solutions on their own. Not only will I get the impression that I have really good students (good for them), I will also get the impression that I am doing a truly wonderful job teaching the material (which may not be the case at all!), and that the material is easy for the students, so I can go faster, cover more advanced material, and spend some time on the more obscure but interesting bits since the students are getting the basics so well. If it turns out that the solutions are being obtained by asking for help from elsewhere, that the students are not getting the material as well as I think they are, then they are going to be very ill-served by a course in which I am going faster, with less detail, and not going over the basics as well as they need.
In a sense, the homework is not just for the students, it's also for me to gauge how things are going. If my students turn in work obtained from others without acknowledgement, then they are giving me a false impression of how things are going, which can be very bad for them in the long run.
In that same vein, I may construct exams that are too hard for the students, because I think the problem was so well understood given the results I was given. During the exam, they will not have the benefit of coming to the website to ask for questions, resulting in bad grades for the student. Not a good outcome at all.
So it's not about me expecting credit for the help, it's about you giving your professor an accurate picture of how things are going. In fact, if I were to give you an answer I would not require, request, or expect to be thanked by name in your write-up, though in order to abide by the plagiarism policy of your school you would need to mention that you obtained help/key ideas from this website (you could do so without mentioning me by name). I'm not sensitive because I'm annoyed at "missing a citation" (I don't even report citations of my work in journals when I file my annual work report). It's about being unhappy at being part of actions that will make someone else's job that much more difficult (a job with which I can sympathize, since I also have it).
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I see, I had not looked at it from that point of view. I did, actually, get something out of this though. If one reads my responses to the questions, I was not just trying to "get the answer." I guess I'm learning more than just math on here. – user6560 Feb 14 '11 at 12:17
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This is well said. I wonder what percentage of modern grad students do their own work. – Carl Brannen Mar 23 '11 at 1:28
At what point does using someone's hints or ideas become frowned upon? I have been made WELL AWARE that my recent actions are not accepted here. I am trying to understand why. Do not get angry with me please. I am not trying to steal someone elses original ideas here... These exercises are somewhat elementary to those who have far surpassed this material, just as the calculus questions tutors who help with homework in a math help room run into.
The solutions to these questions must have been around for some time, and there is little possibility any of the "hints" given to such questions are original. It was even stated so in one of the questions. I'm am NOT saying they are not original, and I appreciate all the help I have gotten, and I WAS helped. The responses have aided in my understanding. And I thank you for that.
In addition, the material is elementary to many of you here, and I do not expect credit for helping someone with a proof in elementary set theory. I did not even THINK that something would be so sensitive here, but I am wrong, and again, am trying to understand why.
I am sorry for offending this community.
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– Willie Wong♦ Feb 13 '11 at 22:59
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+1: Thanks, user6560, for posting here for to ask for clarification of the problem. – Jonas Meyer Feb 14 '11 at 0:03
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@user, here we have the luxury of italics... Writing words in CAPS is usually a bad way to emphasize, seen by many as a for of shouting. – Mariano Suárez-Alvarez♦ Feb 14 '11 at 3:19
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Arturo has given a great explanation. Here is another aspect: many teachers spend a considerable amount of time coming up with nice problems that are supposed to make their students think and enjoy finding the solution after a serious expenditure of time and effort. If you simply ask for a solution on the web, you ruin your professor's work. While I encourage you to not do that at all (it's much better to spend several days stuck, than to ask for a solution), the problem would be much smaller if you described the situation, since we could take care not to spoil your professor's work. – Alex B. Feb 14 '11 at 7:23
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@Jonas: for future reference, moderators do not have the power to migrate comments to another thread. In this particular case certain (not generally available) work around was used. User6560 likely could not comment on Arturo's post because he/she has fewer than 50 rep points at the moment. – Willie Wong♦ Feb 14 '11 at 12:20
@Willie: Thanks. I asked because it looked like the answer had been turned into a comment (which I've seen done before), but on the wrong thread. I realized the rep$\lt50$ thing, but didn't know what powers you have for moving comments. – Jonas Meyer Feb 14 '11 at 12:54
Just a comment more than an answer, professors need to change the way they teach and not award marks for take-home assignments, unless the assignment can be marked by some kind of stamp of uniqueness (like ISU type assignments).
This will change in coming years, as professors brought up in the Web age start teaching.
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any suggestions as to -how- to do this change? no grade at all for take-home assignments? – Mitch Apr 19 '11 at 21:04
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In my opinion non-unique take-home assignments shouldn't be graded at the university level. Frequent in-class quizzing can be used instead (for the purpose of checked participation). – bobobobo Apr 23 '11 at 19:33
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http://math.stackexchange.com/questions/256864/solve-the-sde-by-applying-ito-formula
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# Solve the SDE by applying Ito formula
Solve the SDE $d\delta_{t}=\mu\delta_{t}dt+\sigma\delta_{t}dW_{t}$ by applying the Ito formula and the final solution should be in the form of $\delta_{t}=\delta_{0}\exp\left((\mu-\frac{1}{2}\sigma^{2})t+\sigma W_{t}\right)$
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## 1 Answer
The useful form of Ito's lemma is that:
$$f(W_b,b)-f(W_a,a)=\int_a^b\partial_wf(W_t,t)dW_t+\int_a^b\left(\frac{1}{2}\partial^2_wf(W_t,t)+\partial_tf(W_t,t)\right)dt$$
which is equivalent to:
$$df=\partial_tfdt+\partial_wfdW_t+\frac{1}{2}\sigma\partial^2_wfdt$$
for a brownian motion $W_t$. You have then that $\sigma\delta_t=\partial_w\delta(W_t,t)$ and that $\mu\delta_t=\partial_t\delta(W_t,t)+\frac{1}{2}\partial^2_w\delta_t(W_t,t)$. The first equation gives $\delta(W_t,t)=C(t)\exp(\sigma W_t)$, so plugging this into the second equation and dividing by $\exp(W_t)$ gives: $\mu C(t)=C'(t)+\frac{\sigma^2}{2}C(t)$. This is again easy to solve and gives $C(t)=D\exp\left[\left(\mu-\frac{\sigma^2}{2}\right)t\right]$. Combining, we have $\delta(W_t,t)=D\exp\left((\mu-\frac{\sigma^2}{2})t+\sigma W_t\right)$. At $t=0$, $\delta(W_0,0)=\delta_0=D$, so we finally end up with:
$$\delta(W_t,t)=\delta_0\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t\right)$$
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Is there anyway that I can set up a substitution to solve it and transfer it into the final answer. – Tony Dec 12 '12 at 5:41
– Alex Dec 12 '12 at 5:43
Can you give out the substitution directly? – Tony Dec 12 '12 at 5:48
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http://mathhelpforum.com/math-challenge-problems/94701-crazy-integral-print.html
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# Crazy integral
Printable View
• July 9th 2009, 12:10 AM
simplependulum
Crazy integral
Determine which integral is more crazy ...
$\int \frac{dx}{1 + \sin^3{x}}$
and
$\int \frac{dx}{1+\sin^6{x}}$
3 or 6 ? The answer is 3 ! It is very interesting that although 6 is greater than 3 , the integral $\int \frac{dx}{1+\sin^6{x}}$ is actually easier than the other .
I suggest to try $\int \frac{dx}{1+\sin^6{x}}$ first , finally finish the rest (Evilgrin)
• July 9th 2009, 06:26 PM
halbard
If I were to attempt to find $\int\frac1{1+\sin^3 x}\mathrm dx$ then I'd be crazy. I have attempted to find it, and I am crazy ... now.
Thanks a lot, simplependulum! You have determined which member is more crazy.
• July 9th 2009, 07:48 PM
AlephZero
Just use the vastly underrated method of Weierstrass Substitution... then both integrals are easy as pie...
• July 9th 2009, 08:46 PM
simplependulum
Quote:
Originally Posted by halbard
If I were to attempt to find $\int\frac1{1+\sin^3 x}\mathrm dx$ then I'd be crazy. I have attempted to find it, and I am crazy ... now.
Thanks a lot, simplependulum! You have determined which member is more crazy.
Yes , people can prove they are crazy or not in this question .
I was wondering which method you have made use , I first deduced a general solution for $\int \frac{ dx}{ \sin^2{x} + b \sin{x} + c } ~~$ and if the module of the root $u^2 + bu + c = 0$ is 1 , it will become easier to solve !
• July 10th 2009, 02:51 AM
halbard
Are you by chance referring to the identity
$\frac3{1+\sin^3 x}=\dfrac1{1+\sin x}+\frac1{1-\omega_1\sin x}+\frac1{1-\omega_2\sin x}$
where $\omega_1$, $\omega_2$ are the complex cube roots of $-1$, together with the substitution $t=\tan{\textstyle\frac12}x$ for $\int\dfrac1{1-a\sin x}\mathrm dx$ which can be found when $a^2\neq 1$ using $\int\frac1{1-2at+t^2}\mathrm dt=\frac{\arctan\left(\displaystyle\frac{t-a}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}}$? No, I haven't tried it this way (;->)
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http://mathhelpforum.com/advanced-statistics/115471-chi-square-test.html
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# Thread:
1. ## [SOLVED] Chi-Square Test
Two different teaching procedures were used in two different groups of students. Each group had 100 students of similiar ability. These are the results:
Group A B C D F
I 15 25 32 17 11
II 9 18 29 28 16
Data is independent from two respective multinomial distribution with k = 5. Test at the 5% level the hypothesis that the two teaching methods are equally effective.
So let X be the students in group 1, and Y students in group 2.
So the test statistic would be: $Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}$.
Since it's a multinomial distribution, I know that the expectation is np_i, but how would I find p_i?
2. Originally Posted by statmajor
Two different teaching procedures were used in two different groups of students. Each group had 100 students of similiar ability. These are the results:
Group A B C D F
I 15 25 32 17 11
II 9 18 29 28 16
Data is independent from two respective multinomial distribution with k = 5. Test at the 5% level the hypothesis that the two teaching methods are equally effective.
So let X be the students in group 1, and Y students in group 2.
So the test statistic would be: $Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}$.
Since it's a multinomial distribution, I know that the expectation is np_i, but how would I find p_i?
Under the null hypothesis the maximum likelihood estimate of the expected value of cell frequency $n_{ij}$ for a contigency table is $\frac{r_i \cdot c_j}{n}$ where $r_i$ is the row total (for row i) and $c_j$ is the column total (for column j) and n is the total frequency.
3. Got it; thanks.
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http://www.physicsforums.com/showthread.php?p=3924927
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Physics Forums
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## Classic Papers in Astronomy and Cosmology
Hellfire requested that I post a list of classic papers that one could use for reference when trying to understand the current state of astronomy and cosmology. I'll do so here, but if I could politely ask the following:
- Please do not debate the veracity of the papers in this thread. In fact, I think it would be best just to use this thread as a reference, so make a new thread if you want to discuss the content of any of the papers.
- If you would like to contribute a paper, please ensure that it's mainstream and that it has lot of citations (>~ 300). Citation counts can be found here.
- When contributing a paper, please include a brief description in more pedagogical terms than given in the abstract.
That should do it. I'll add new papers as time permits and as I think of them. The sample that I provide will certainly be biased and incomplete, as my exposure is limited and a lot of the classic papers come from the institution at which I'm studying, so please do not take exception to any of my choices. I'll list them in categories (like cosmology, stellar astronomy, etc.) for ease of reference.
Recognitions: Gold Member Science Advisor Staff Emeritus Cosmology Press & Schechter 1974 - This paper describes, with a simple analytical model, the collapse of matter "halos" (a term that includes any overdensity, including galaxies and clusters of galaxies), which arise from the perturbations seen in the microwave background. The methodology has been embraced by cold dark matter theorists because it describes a simple bottom-up scenario, where the smaller halos form before the larger ones, exactly as one would expect in a CDM universe. PS74 is indispensible for the pencil and paper cosmologist because it gives a very simple means of describing the distribution of matter in the universe. Gunn & Gott 1972 - Unlike PS74, which treats the whole spectrum of "halos" and their distribution, this paper follows the collapse of an individual matter halo, demonstrating along the way the importance of accretion in performing such calculations. As a bonus, the end of the paper presents an early model of "ram-pressure stripping" for removing the gas from galaxies embedded in clusters, one of the requirements for reproducing the observed morphology-density relationship. Spergel et al. 2003 - Not a particularly old paper, but certainly one of the most important in cosmology today, the WMAP first-year release demonstrates how the CMB is used as a powerful constraint on the fundamental parameters of the universe. Not only did it independently confirm the values of many cosmological parameters that had been measured by other means, but it measured them more precisely than had ever been done before, ushering in what some would call an age of "precision cosmology". Guth 1982 - The paper that introduced the idea of inflation to the world and gave us the most likely explanation for why we lived in what seemed to be a perfectly flat universe. Many variations on the theme have since been proposed, but the original idea remains intact and it still seems to be a pretty good fit to the data. Reiss et al. 1998 - It had long been assumed that we lived in a decelerating universe because, after all, its theorized constituents (matter and radiation) could only slow the expansion. The above paper, analyzing data from Type Ia supernovae, gave us the first hard observational evidence that this assertion was incorrect and that there was some extra component (or "dark energy") driving the universe to accelerate. Later confirmed by WMAP, this discovery is still an enigma to theorists and suggests that there is still much we don't know about the universe in which we live. Alcock et al. 2000 - This paper doesn't quite make the 300 citation limit I listed above, but that's because it is the combination of many years of observing, with many, many papers along the way. The result that came out of the MACHO survey was extremely important because it ruled out an entire class of dark matter candidates and convinced most folks that dark matter would, most likely, turn out to be in particle form. There are still debates raging about whether or not massive objects can make up a non-negligible fraction of the dark matter in our galaxy, but it's relatively clear now that it can't be the majority. Zeldovich 1970 - It has long been known that galaxies tend to distribute themselves in filament and pancake-like structures, but it is hard to reproduce analytically because the mathematics of non-linear gravitational collapse are tricky. The model presented in this paper is a simple one in which gravitating bodies are shown to collapse along each of their principal axes in succession, producing pancake-like and filament-like structures along the way. Although originally created for different purposes, the model now provides an excellent means of deriving the initial conditions to cosmological N-Body simulations.
Recognitions: Gold Member Science Advisor Thank you SpaceTiger for those references, this site is as good as having a very versed library service! It is a pity the Guth paper cannot be downloaded free, does anyone know where it may be obtained? Garth
Recognitions:
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## Classic Papers in Astronomy and Cosmology
Quote by Garth It is a pity the Guth paper cannot be downloaded free, does anyone know where it may be obtained?
Unfortunately, I don't, and I suspect some of the other papers I post will be the same way. The links I provide will only be to the ADS abstract, since I'm assuming that service won't be going dead any time soon, but I can't promise that the paper will be freely available to those outside of academic institutions. If the paper is within the past 10 years, it may be available on astro-ph, but otherwise make note of the reference and look it up in the library.
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Thank you SpaceTiger. I would add one paper:
A Universal Density Profile from Hierarchical Clustering
Julio F. Navarro, Carlos S. Frenk, Simon D.M. White
http://arxiv.org/abs/astro-ph/9611107
It discusses mainly results of simulations for the density profile of halos. I have seen this paper cited many times (the universal "NFW profile") and I assume that it's results are still considered as relevant. To skip the description of the simulations just go to section 5 where the main results are discussed. The main result is summarized in the paper as follows:
...our primary goal, namely the presentation of a simple and apparently general theoretical result: hierarchical clustering leads to a universal halo density profile just as it leads to universal distributions of halo axial ratios and halo spins; none of these properties depends strongly on power spectrum, on $$\Omega$$, or on $$\Lambda$$
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Quote by hellfire Thank you SpaceTiger. I would add one paper
Thanks hellfire, given enough time to think of it, I probably would have added that one as well.
By the way, I'm adding papers to sections by editing the posts, so what you see above is not the complete and final list for cosmology.
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Quote by Garth Thank you SpaceTiger for those references, this site is as good as having a very versed library service! It is a pity the Guth paper cannot be downloaded free, does anyone know where it may be obtained? Garth
Garth, I cannot promise that the SLAC link will work for you. I have a SLOW dial-up connection and I couldn't get it to download, but hopefully you can. Good luck!
http://www.slac.stanford.edu/pubs/sl...-pub-2576.html
Recognitions: Gold Member Science Advisor An excellent selection, ST. I have a fair collection of my own. Will sift, sort and nominate with your approval.
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Quote by turbo-1 Garth, I cannot promise that the SLAC link will work for you. I have a SLOW dial-up connection and I couldn't get it to download, but hopefully you can. Good luck! http://www.slac.stanford.edu/pubs/sl...-pub-2576.html
Thank you turbo-1, it didn't work last night but when I tried this morning it downloaded straight away.
Garth
Recognitions: Gold Member Science Advisor Staff Emeritus This could be the "A&C reference library v.2". :) stickifying it...
Recognitions: Gold Member Science Advisor From my private collection of favorites over the past 10 years: http://arxiv.org/abs/astro-ph/0402512 Type Ia Supernova Discoveries at z>1 From the Hubble Space Telescope: Evidence for Past Deceleration and Constraints on Dark Energy Evolution 397 citations Easily the most important paper of 2004 to date. It further solidifies the case for dark energy. Heavily cited already for being relatively recent. http://arxiv.org/abs/astro-ph/0302431 Primordial Nucleosynthesis in Light of WMAP 81 citations Not hugely cited to date, but one of the most important papers of 2003. It uses WMAP data to independently corroborate BBN elemental abundance predictions. http://arxiv.org/abs/astro-ph/0305008 Cosmological Results from High-z Supernovae 332 citations This high z supernova data was crucial in ruling out possible systematic effects in low redshift data, bolstering the case for dark energy. http://arxiv.org/abs/astro-ph/0205387 Cosmological Parameters from Cosmic Background Imager Observations and Comparisons with BOOMERANG, DASI, and MAXIMA 163 citations The CBI study, while rather insensitive compared to WMAP, was the first sensitive enough to detect the structures that gave rise to galactic clusters. It also provided important confirmation of prior studies. http://arxiv.org/abs/astro-ph/0104455 The Farthest Known Supernova: Support for an Accelerating Universe and a Glimpse of the Epoch of Deceleration 289 citations One of the early, high z supernova observations that provided important evidence in favor of the accelerating universe hypothesis. http://arxiv.org/abs/astro-ph/0004404 A Flat Universe from High-Resolution Maps of the Cosmic Microwave Background Radiation 897 citations These are the COBE results, the first definitive study supporting the LCDM model. WMAP later confirmed these findings to high precision. http://arxiv.org/abs/astro-ph/0006053 A Fundamental Relation Between Supermassive Black Holes and Their Host Galaxies 515 citations Among the most startling finds in the past ten years, IMO. It immediately attracted a huge amount of interest. http://arxiv.org/abs/astro-ph/9911476 Reverberation Measurements for 17 Quasars and the Size-Mass-Luminosity Relations in Active Galactic Nuclei 284 citations This paper relates an important discovery regarding the luminosity function of AGN's. http://arxiv.org/abs/astro-ph/9812133 Measurements of Omega and Lambda from 42 High-Redshift Supernovae 2144 citations This is the original Perlmutter paper. One of the most astonishing finds of the 20th century. No doubt already mentioned, but worth mentioning again. http://arxiv.org/abs/astro-ph/9712020 The Cosmic Baryon Budget 437 citations This important paper was among the early results that began severly constraining the permissible fraction of baryonic matter in the universe. WMAP further confirmed this finding. http://arxiv.org/abs/astro-ph/9607060 The Four-Year COBE Normalization and Large-Scale Structure 364 citations The preliminary COBE results. It was quite exciting at the time. http://arxiv.org/abs/astro-ph/9508025 The Structure of Cold Dark Matter Halos 924 citations A companion paper to the one referenced by Hellfire, also a good read. http://arxiv.org/abs/astro-ph/9510034 Galaxy Harassment and the Evolution of Clusters of Galaxies 279 citations Not quite as well known as the NFW papers, but another important paper on the evolution of galactic clusters. It took awhile to compile this, sorry for any redundancies.
Recognitions: Gold Member Science Advisor Staff Emeritus Thanks Phobos, guess I'll have to make this count, huh? Eggen, Lynden-Bell, & Sandage 1962 - One of the pioneering papers in the theory of galaxy formation, ELS62 tries to explain the observed properties of stars in terms of a rapidly collapsing galaxy model. Although not identical to currently favored theories in the details, the basic idea sticks with us and has inspired a great deal of theoretical interest on the topic. A combination of this rapid collapse and the hierarchical build-up of CDM seems to provide a good fit to most of the data. Schwarzschild 1979 - In addition to the formation mechanism, we're also interested in what constitutes a stable configuration for a post-collapse galaxy. Making use of some very primitive (by modern standards) computational resources, Martin Schwarzschild constructed a model of a triaxial galaxy, solving for the orbits of stars as well as the overall distribution function in position-momentum space. Many of the fundamentals of this original model are used to this day, despite having much more powerful computers to work with. Oort 1932 - One of the most debated quantities in modern astrophysics is the "Oort limit", which respresents the mean density of mass in the solar neighborhood. The reason this quantity is so important is that allows us to determine whether or not dark matter is significant contributor to the dynamics of the galactic disk. The answer is still unknown, but the methods of determining it are still very similar to that adopted by Jan Oort in this 1932 paper. Sandage 1961 - Don't bother trying to download this online (at least from ADS) because it's not available; in fact, it's not even a paper. The Hubble Atlas is old-fashioned astronomy at its best, demonstrating in both pictures in text how astronomer's classify galaxies. The classification scheme was originally designed by Edwin Hubble himself and has since been extended to include types that had not been in his original sample. An easy read, I would recommend this book even for the layman. Toomre & Toomre 1972 - A good follow-up to the Hubble Atlas, this 1972 paper demonstrates that the simulated world can be almost as pretty as the real one. The cause of the curious "bridges" and "tails" that seemed to extend from many galaxies (often classified as peculiar galaxies) had long been a subject of debate, but with the release of this paper, the issue was practically settled overnight. Using a simple N-body code, TT72 demonstrated that two galaxies on a parabolic collision course will interact in such a way as to produce extended streams of stars gas very similar to those observed in peculiar galaxies. Perhaps even more importantly, the paper suggests for the first time (look carefully on page 662) that elliptical galaxies are actually the remnants of mergers. To this day, the origins of morphology have not been completely settled, but that off-hand suggestion has morphed into the leading theory of elliptical galaxy formation.
Blog Entries: 9 Recognitions: Homework Help Science Advisor -- Synthesis of the Elements in Stars -- E. Margaret Burbidge, G. R. Burbidge, William A. Fowler, and F. Hoyle Rev. Mod. Phys. 29, 547 - 650 (1957) http://prola.aps.org/pdf/RMP/v29/i4/p547_1
On the Gravitational Field of a Mass Point According to Einstein’s Theory K. Schwarzschild, Sitzungsber.Preuss.Akad.Wiss.Berlin (Math.Phys.) 1916 (1916) 189-196 Full Text On a Stationary System with Spherical Symmetry Consisting of Many Gravitating Masses. A. Einstein, Annals of Mathematics Vol 40. No. 4 Oct 1939 First Page
A Relation Between Distance and Radial Velocity Among Extra-Galactic Nebulae Edwin Hubble, P.N.A.S., March 15, 1929 vol. 15 no. 3 168-173 Acrobat pdf document While the Big Bang may have started 13.5 billion years ago, recognition of the Big Bang arguably started here.
In general, I would suggest to people new in a field to go through a review paper, like the ones published in ARAA (Annual Reviews in Astronomy and Astrophysics), which repeat the same topics every few years, whenever appropriate. Authors of such papers put their contributions on astro-ph most of the time!
Quote by Ophiolite A Relation Between Distance and Radial Velocity Among Extra-Galactic Nebulae Edwin Hubble, P.N.A.S., March 15, 1929 vol. 15 no. 3 168-173 Acrobat pdf document While the Big Bang may have started 13.5 billion years ago, recognition of the Big Bang arguably started here.
does anybody understands this paper? how does he derive that velocity is prpotional linearly to the distanse of the nebulae?
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http://mathforum.org/mathimages/index.php?title=Prime_spiral_(Ulam_spiral)&diff=32702&oldid=14449
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# Prime spiral (Ulam spiral)
### From Math Images
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| - | {{Image Description | + | {{Image Description Ready |
| - | |ImageName=Ulam spiral | + | |ImageName=Ulam Spiral |
| | |Image=Ulam_spiral.png | | |Image=Ulam_spiral.png |
| | |ImageIntro=The Ulam spiral, or prime spiral, is a plot in which <balloon title="load:defprime">prime numbers</balloon><span id="defprime" style="display:none">A prime number is a natural number greater that is divisible only by 1 and itself. The first few prime numbers are :<math>2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, \dots</math></span> are marked among positive integers that are arranged in a counterclockwise spiral. The prime numbers show a pattern of diagonal lines. | | |ImageIntro=The Ulam spiral, or prime spiral, is a plot in which <balloon title="load:defprime">prime numbers</balloon><span id="defprime" style="display:none">A prime number is a natural number greater that is divisible only by 1 and itself. The first few prime numbers are :<math>2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, \dots</math></span> are marked among positive integers that are arranged in a counterclockwise spiral. The prime numbers show a pattern of diagonal lines. |
| - | |ImageDescElem=The prime spiral was discovered by Stanislaw Ulam (1909-1984) in 1963 while he was doodling on a piece of paper during a science meeting. Starting with 1 in the middle, he wrote positive numbers in a grid as he spiraled out from the center, as shown in <i>Image 1</i>. He then circled the prime numbers, and the prime numbers showed patterns of diagonal lines as shown by the grid in <i>Image 2</i>. The grid in <i>Image 2</i> is a close-up view of the center of the main image such that the green line segments and red boxes in the center of the main image line up with those in <i>Image 2</i>. | + | |ImageDescElem={{Anchor|Reference=1|Link=[[Image:primegrid2.jpg|Image 1|thumb|250px|left]]}} |
| - | [[image:primegrid2.jpg|left|thumb|Image 1|200px]] | + | {{Anchor|Reference=2|Link=[[Image:ulamgrid.jpg|Image 2|thumb|400px]]}} |
| - | [[image:ulamgrid.jpg|right|thumb|Image 2|400px]] | + | The prime spiral was discovered by Stanislaw Ulam (1909-1984) in 1963 while he was doodling on a piece of paper during a science meeting. Starting with 1 in the middle, he wrote positive numbers in a grid as he spiraled out from the center, as shown in [[#1|Image 1]]. He then circled the prime numbers, and the prime numbers showed patterns of diagonal lines as shown by the grid in [[#2|Image 2]]. The grid in [[#2|Image 2]] is a close-up view of the center of the main image such that the green line segments and red boxes in the center of the main image line up with those in the grid. |
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| | A larger Ulam spiral with 160,000 integers and 14,683 primes is shown in the main image. Black dots indicate prime numbers. In addition to diagonal line segments formed by the black dots, we can see white vertical and horizontal line segments that cross the center of the spiral and do not contain any black dots, or prime numbers. There are also white diagonal line segments that do not contain any prime numbers. Ulam spiral implies that there is some order in the distribution of prime numbers. | | A larger Ulam spiral with 160,000 integers and 14,683 primes is shown in the main image. Black dots indicate prime numbers. In addition to diagonal line segments formed by the black dots, we can see white vertical and horizontal line segments that cross the center of the spiral and do not contain any black dots, or prime numbers. There are also white diagonal line segments that do not contain any prime numbers. Ulam spiral implies that there is some order in the distribution of prime numbers. |
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| | |ImageDesc=From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is no known simple formula that yields all the primes. The diagonal patterns in the Ulam spiral gives some hint for formulas of primes numbers. | | |ImageDesc=From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is no known simple formula that yields all the primes. The diagonal patterns in the Ulam spiral gives some hint for formulas of primes numbers. |
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| | Many half-lines in the Ulam spiral can be described using <balloon title="load:quadratic"> quadratic polynomials</balloon><span id="quadratic" style="display:none"> A quadratic polynomial is a polynomial in which the exponent of the variables do not exceed 2. A quadratic polynomial typically has the form <math>ax^2+bx+c</math> where <math>x</math> is a variable and <math>a, b, c</math> are coefficients.</span>. A ''half-line'' is a line which starts at a point and continues infinitely in one direction. In this page, we only consider half-lines that are horizontal or vertical, or have a slope of -1 or +1. The half-lines that can be described using quadratic polynomials are the ones in which each entry of the diagonal is positioned on a different ''ring'' of the spiral. | | Many half-lines in the Ulam spiral can be described using <balloon title="load:quadratic"> quadratic polynomials</balloon><span id="quadratic" style="display:none"> A quadratic polynomial is a polynomial in which the exponent of the variables do not exceed 2. A quadratic polynomial typically has the form <math>ax^2+bx+c</math> where <math>x</math> is a variable and <math>a, b, c</math> are coefficients.</span>. A ''half-line'' is a line which starts at a point and continues infinitely in one direction. In this page, we only consider half-lines that are horizontal or vertical, or have a slope of -1 or +1. The half-lines that can be described using quadratic polynomials are the ones in which each entry of the diagonal is positioned on a different ''ring'' of the spiral. |
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| - | The ''ring'' of a spiral can be considered as the outermost layer of a concentric square or a rectangle centered around the center of the spiral, 1, as defined by the blue line in the grid. Although there is no exact place where we can determine the beginning and end of a ring, we will assume that entries that are positioned on squares or rectangles of the same sizes to be on the same ring. For instance, 24, 25, 26, 27, 28, which are shown in red boxes in <i>Image 3</i>, are on the same ring, whereas 47, 48, 49, 50, 51, 52, as shown in blue boxes, are on a different ring because they are positioned on a bigger square. | + | The ''ring'' of a spiral can be considered as the outermost layer of a concentric square or a rectangle centered around the center of the spiral, 1, as defined by the blue line in the grid. Although there is no exact place where we can determine the beginning and end of a ring, we will assume that entries that are positioned on squares or rectangles of the same sizes to be on the same ring. For instance, 24, 25, 26, 27, 28, which are shown in red boxes in [[#3|Image 3]], are on the same ring, whereas 48, 49, 50, 51, 52, as shown in blue boxes, are on a different ring because they are positioned on a bigger square. |
| | | + | {{Anchor|Reference=3|Link=[[Image:Halfline10.jpg|Image 3:Diagonal half-lines|thumb|250px]]}} |
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| - | [[image:Halfline10.jpg|thumb|Image 3: Diagonal half-lines|250px]] | + | The green lines in the [[#3|Image 3]] qualify as diagonal half-lines whereas the red lines do not because the red lines cross the corner of the grid in a way that two entries of the red diagonal line are positioned on the same ring. |
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| - | For instance, the green lines in the <i>Image 3</i> qualify as diagonal half-lines whereas the red lines do not because the red lines cross the corner of the grid in a way that two entries of the red diagonal line are positioned on the same ring. | + | |
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| | Thus, diagonal line segments that are composed of prime numbers can also be expressed as outputs of quadratic polynomials. To learn more about the relation between the diagonal lines and quadratic polynomials, click below. | | Thus, diagonal line segments that are composed of prime numbers can also be expressed as outputs of quadratic polynomials. To learn more about the relation between the diagonal lines and quadratic polynomials, click below. |
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| | | + | '''Half-lines and quadratic polynomials''' |
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| | | + | First, we need to know the relation between quadratic polynomials and the difference table of sequences. Let's choose the diagonal |
| | | + | |FullText= |
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| - | '''Half-lines and quadratic polynomias''' | + | '''Half-lines and quadratic polynomials''' |
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| - | First, we need to know the relation between quadratic polynomials and the {{EasyBalloon|Link=difference table|Balloon=The difference table lists the terms of a sequence and its differences. It can include other higher order differences such as the second-differences and the third-differences.}} of sequences. Let's choose the diagonal <math> 5, 19, 41, 71, 109, 155</math>, the diagonal indicated with green dotted line in <i>Image 3</i>, as our original sequence and create a difference table. | + | First, we need to know the relation between quadratic polynomials and the {{EasyBalloon|Link=difference table|Balloon=The difference table lists the terms of a sequence in one row, and the differences between consecutive terms in the next row. It can include other higher order differences such as the second-differences and the third-differences.}} of sequences. Let's choose the diagonal <math> 5, 19, 41, 71, 109, 155</math>, the diagonal indicated with green dotted line in [[#3|Image 3]], as our original sequence and create a difference table. |
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| | [[image:diff.jpg]] | | [[image:diff.jpg]] |
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| | As we can see from the table, the 2nd differences are constant, and this implies that the original sequence can be described by a second degree polynomial. For more information about difference tables and about finding specific polynomials for sequences, go to [[Difference_Tables|difference tables]]. | | As we can see from the table, the 2nd differences are constant, and this implies that the original sequence can be described by a second degree polynomial. For more information about difference tables and about finding specific polynomials for sequences, go to [[Difference_Tables|difference tables]]. |
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| - | [[image:primeani.gif|Image 4: Constant 2nd differences|left|thumb|250px]] | + | {{Anchor|Reference=4|Link=[[Image:primeani.gif|Image 4:Constant 2nd differences|thumb|250px|left]]}} |
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| | Now, we will see that the second differences are always constant for any segment of a diagonal half-line. Let's choose the same diagonal with entries <math>5, 19, 41, 71, 109, 155</math>. (For a better understanding of the following paragraphs, the reader should wait until the animation shows three different blue rings and then watch the rest of the animation.) | | Now, we will see that the second differences are always constant for any segment of a diagonal half-line. Let's choose the same diagonal with entries <math>5, 19, 41, 71, 109, 155</math>. (For a better understanding of the following paragraphs, the reader should wait until the animation shows three different blue rings and then watch the rest of the animation.) |
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| - | In <i>Image 4</i>, the number of the innermost light blue boxes, which is 14, indicate the distance from <math>5</math> to <math>19 </math>, or the difference between these two numbers. The number of darker blue boxes in the middle indicate the difference between the numbers <math> 19</math> and <math>41</math>. The darkest blue boxes on the outside indicate the difference between the entries <math> 41</math> and <math>71</math>. The number of all these blue boxes correspond to the first differences in the difference table. | + | In [[#4|Image 4]], the number of the innermost light blue boxes, which is 14, indicate the distance from <math>5</math> to <math>19 </math>, or the difference between these two numbers. The number of darker blue boxes in the middle indicate the difference between the numbers <math> 19</math> and <math>41</math>. The darkest blue boxes on the outside indicate the difference between the entries <math> 41</math> and <math>71</math>. The number of all these blue boxes correspond to the first differences in the difference table. |
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| - | In <i>Image 4</i>, the light blue boxes are divided into pieces, and each piece is moved to on top of the darker blue boxes. We can see that there are <math>8</math> more darker blue boxes than the innermost light blue boxes. Similarly, the blue boxes in the middle are divided into pieces and are moved to on top of the darkest blue boxes. | + | In [[#4|Image 4]], the light blue boxes are divided into pieces, and each piece is moved to on top of the darker blue boxes. We can see that there are <math>8</math> more darker blue boxes than the innermost light blue boxes. Similarly, the blue boxes in the middle are divided into pieces and are moved to on top of the darkest blue boxes. |
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| - | Indeed, as <i>Image 4</i> illustrates, regardless of the exact number of the blue boxes, there are <math>8</math> more boxes in any given ring than there are in the ring that is one layer inwards from it. This means that the the second differences, or the differences between the first differences is always going to be constant at <math>8</math>. Thus, the sequence for any diagonal half-line can be described using a quadratic polynomial. | + | Indeed, as [[#4|Image 4]] illustrates, regardless of the exact number of the blue boxes, there are <math>8</math> more boxes in any given ring than there are in the ring that is one layer inwards from it. This means that the the second differences, or the differences between the first differences is always going to be constant at <math>8</math>. Thus, the sequence for any diagonal half-line can be described using a quadratic polynomial. |
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| | }} | | }} |
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| | '''Examples of quadratic polynomials for half-lines''' | | '''Examples of quadratic polynomials for half-lines''' |
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| - | For instance, prime numbers <math> 5, 19, 41, 71, 109</math>, which are aligned in the same diagonal, can be described through the output of the polynomial <math>4x^2+10x+5</math> for <math>x=0, 1, 2, 3, 4</math>. Similarly, numbers <math> 1, 3, 14, 31, 57, 91 \dots </math>, which are also aligned in a green diagonal in <i> Image 5</i> starting from the center and continuing to the upper right corner, can be expressed by: | + | For instance, prime numbers <math> 5, 19, 41, 71, 109</math>, which are aligned in the same diagonal, can be described through the output of the polynomial <math>4x^2+10x+5</math> for <math>x=0, 1, 2, 3, 4</math>. Similarly, numbers <math> 1, 3, 14, 31, 57, 91 \dots </math>, which are also aligned in a green diagonal in [[#5|Image 5]] starting from the center and continuing to the upper right corner, can be expressed by: |
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| | {{EquationRef2|Eq. (1)}}<math>4x^2-2x+1</math> | | {{EquationRef2|Eq. (1)}}<math>4x^2-2x+1</math> |
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| | for <math>x=0, 1, 2, 3, \dots</math>. (We will refer back to this polynomial in a later section) In fact, the part of the green diagonal that starts from the center and continues to the bottom left corner can also be described through {{EquationNote|Eq. (1)}} for <math>x=0, -1, -2, -3, \dots</math>. | | for <math>x=0, 1, 2, 3, \dots</math>. (We will refer back to this polynomial in a later section) In fact, the part of the green diagonal that starts from the center and continues to the bottom left corner can also be described through {{EquationNote|Eq. (1)}} for <math>x=0, -1, -2, -3, \dots</math>. |
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| - | [[image:hvdline.jpg|thumb|Image 5|250px]] | + | {{Anchor|Reference=5|Link=[[Image:hvdline.jpg|Image 5|thumb|250px]]}} |
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| - | In fact, even horizontal and vertical line segment in the grid can be described by quadratic polynomials, as long as the lines satisfy the condition that no two entries are positioned on the same ring. For instance, the blue horizontal line segment in <i>Image 5</i>, <math> 10, 27, 52, 85</math> can be described by a quadratic polynomial. However, the sequence <math> 9, 10, 27, 85</math> on the same horizontal line cannot be described by a polynomial because the entries 9 and 10 are on the same ring. | + | In fact, even horizontal and vertical line segment in the grid can be described by quadratic polynomials, as long as the lines satisfy the condition that no two entries are positioned on the same ring. For instance, the blue horizontal line segment in [[#5|Image 5]], <math> 10, 27, 52, 85</math> can be described by a quadratic polynomial. However, the sequence <math> 9, 10, 27, 85</math> on the same horizontal line cannot be described by a polynomial because the entries 9 and 10 are on the same ring. |
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| - | Moreover, it is not hard to show that the green diagonal line from <i>Image 5</i> that goes through the center and has a slope of +1 is the only line on which the Ulam numbers in both directions can be described by the same polynomial. Even the red diagonal line that goes through the center and has a slope of -1 cannot be described by one polynomial. | + | Moreover, it is not hard to show that the green diagonal line from [[#5|Image 5]] that goes through the center and has a slope of +1 is the only line on which the Ulam numbers in both directions can be described by the same polynomial. Even the red diagonal line that goes through the center and has a slope of -1 cannot be described by one polynomial. |
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| | Half of the red diagonal, the segment going up from the center, <math> 5, 17, 37, 67, \dots</math> can be described by <math>x^2+1</math> for inputs that are even numbers, while the diagonal going down from the center, <math>1, 9, 25, 49, \dots </math> is a sequence of perfect squares of odd numbers. Thus we cannot find one polynomial that generates both all entries of the red diagonal. | | Half of the red diagonal, the segment going up from the center, <math> 5, 17, 37, 67, \dots</math> can be described by <math>x^2+1</math> for inputs that are even numbers, while the diagonal going down from the center, <math>1, 9, 25, 49, \dots </math> is a sequence of perfect squares of odd numbers. Thus we cannot find one polynomial that generates both all entries of the red diagonal. |
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| | Euler's polynomial generates distinct prime numbers for each integer <math>x</math> from <math>x=1</math> to <math> x=40</math>. | | Euler's polynomial generates distinct prime numbers for each integer <math>x</math> from <math>x=1</math> to <math> x=40</math>. |
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| - | [[image:Eulerpgrid1.jpg|thumb|left|Image 6|400px]] | + | {{Anchor|Reference=6|Link=[[Image:Eulerpgrid1.jpg|Image 6|thumb|400px|left]]}} |
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| - | As we can see in <i>Image 6</i>, we can start an Ulam's spiral with 41 at the center of the grid and get a long, continuous diagonal with 40 prime numbers. One interesting fact is that the 40 numbers in the diagonal line segment are the first 40 prime numbers that are generated through Euler's polynomial. Moreover, the prime numbers are not aligned in order of increasing values. In fact, with 41 in the center, other prime numbers alternate in position between the upper right and lower left part of the diagonal. | + | As we can see in [[#6|Image 6]], we can start an Ulam's spiral with 41 at the center of the grid and get a long, continuous diagonal with 40 prime numbers. One interesting fact is that the 40 numbers in the diagonal line segment are the first 40 prime numbers that are generated through Euler's polynomial. Moreover, the prime numbers are not aligned in order of increasing values. In fact, with 41 in the center, other prime numbers alternate in position between the upper right and lower left part of the diagonal. |
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| - | We will show why an Ulam's spiral that starts with 41 generates entries aligned in a diagonal that can also be descried as outputs of Euler's polynomial. {{HideShowThis|ShowMessage=Click here to show more|HideMessage=Click here to hide|HiddenText= | + | We will show why an Ulam's spiral that starts with 41 generates entries aligned in a diagonal that can also be descried as outputs of Euler's polynomial. |
| | | + | First, we found out in the previous section that Eq. (1) is the polynomial for the diagonal that goes through the center 1 and has a slope of +1 in the Ulam's spiral. That is, |
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| - | First, we found out in the previous section that {{EquationNote|Eq. (1)}} is the polynomial for the diagonal that goes through the center 1 and has a slope of °+1 in the Ulam's spiral. That is, | + | |FullText= |
| | | + | First, we found out in the previous section that {{EquationNote|Eq. (1)}} is the polynomial for the diagonal that goes through the center 1 and has a slope of +1 in the Ulam's spiral. That is, |
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| | :<math> 4x^2-2x+1</math> | | :<math> 4x^2-2x+1</math> |
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| - | describes the upper half of the diagonal as we plug in <math>x=0, 1, 2, 3, \dots</math> and the lower half of the diagonal as we plug in <math>x=0, -1, -2, -3, \dots</math>. Thus, <math>x</math>'s are positioned in the diagonal as if the diagonal was a number line that starts with 0 at the center, with positive integers on the upper right and negative integers on the lower left direction, as shown in <i>Image 8</i>. | + | describes the upper half of the diagonal as we plug in <math>x=0, 1, 2, 3, \dots</math> and the lower half of the diagonal as we plug in <math>x=0, -1, -2, -3, \dots</math>. Thus, <math>x</math>'s are positioned in the diagonal as if the diagonal was a number line that starts with 0 at the center, with positive integers on the upper right and negative integers on the lower left direction, as shown in [[#7|Image 7]]. |
| - | [[image:picposition.jpg|250px|thumb|Image 8|none]] | + | {{Anchor|Reference=7|Link=[[Image:picposition.jpg|Image 7|thumb|250px|none]]}} |
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| | Then, if we start the Ulam's spiral at 41 and thus make 41 the center, each entry on the grid, including the entries on the diagonal segment, will increase by 40. Then, the diagonal can be described by the polynomial : | | Then, if we start the Ulam's spiral at 41 and thus make 41 the center, each entry on the grid, including the entries on the diagonal segment, will increase by 40. Then, the diagonal can be described by the polynomial : |
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| | {{EquationRef2|Eq. (2)}}<math>4x^2-2x+41</math>. | | {{EquationRef2|Eq. (2)}}<math>4x^2-2x+41</math>. |
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| - | In fact, we will see that this polynomial and Euler's polynomial are describing the same entries along the diagonal once we make some changes in numbering the position of <math>x</math>'s. Euler's polynomial can be found by reassigning the position of the entries of the diagonals so that <math>x'=1</math> is at the center, <math>x'=2, 4, 6, 8, \dots </math> are the positions up the diagonal line, <math>x'=1, 3, 5, 7, 9, \dots</math> are positions of entries down the diagonal line from the center, as shown in <i>Image 9</i>. Thus, starting with <math>x'=1</math> at the center, we alternate between the right and left of the center as we number the positions <math>x'=1, 2, 3, 4, 5, \dots</math>. | + | In fact, we will see that this polynomial and Euler's polynomial are describing the same entries along the diagonal once we make some changes in numbering the position of <math>x</math>'s. Euler's polynomial can be found by reassigning the position of the entries of the diagonals so that <math>x'=1</math> is at the center, <math>x'=2, 4, 6, 8, \dots </math> are the positions up the diagonal line, <math>x'=1, 3, 5, 7, 9, \dots</math> are positions of entries down the diagonal line from the center, as shown in [[#8|Image 8]]. Thus, starting with <math>x'=1</math> at the center, we alternate between the right and left of the center as we number the positions <math>x'=1, 2, 3, 4, 5, \dots</math>. |
| - | [[image:eulerposition1.jpg|250px|thumb|Image 9|none]] | + | {{Anchor|Reference=8|Link=[[Image:eulerposition1.jpg|Image 8|thumb|250px|none|center]]}} |
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| - | Then, for the upper half of the diagonal in <i>Image 9</i>, we can use <math>x'=2, 4, 6, 8, \dots</math> instead of <math>x=1, 2, 3, 4, \dots</math> from <i>Image 8</i>. Then, | + | Then, for the upper half of the diagonal in [[#8|Image 8]], we can use <math>x'=2, 4, 6, 8, \dots</math> instead of <math>x=1, 2, 3, 4, \dots</math> from [[#7|Image 7]]. Then, |
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| | :<math>x=\frac{1}{2} x'</math> | | :<math>x=\frac{1}{2} x'</math> |
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| | which has the same form as Euler's polynomial. | | which has the same form as Euler's polynomial. |
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| - | The same method works for the bottom half as well. We use <math>x'=1, 3, 5, 7, \dots</math> in <i>Image 9</i> while we used <math>x=0, -1, -2, -3, \dots</math> in <i>Image 8</i>. Then, | + | The same method works for the bottom half as well. We use <math>x'=1, 3, 5, 7, \dots</math> in [[#8|Image 8]] while we used <math>x=0, -1, -2, -3, \dots</math> in [[#7|Image 7]]. Then, |
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| | :<math>x=\frac{-x'+1}{2}</math>. | | :<math>x=\frac{-x'+1}{2}</math>. |
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| | ==Sacks Spiral== | | ==Sacks Spiral== |
| - | [[image:sackspiral.jpg|right]] | + | {{Anchor|Reference=9|Link=[[Image:sackspiral.jpg|Image 9|thumb|right|600px]]}} |
| | | + | {{Anchor|Reference=10|Link=[[Image:sack.gif|Image 10|thumb|250px|left]]}} |
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| - | Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places <math>0</math> in the center and places nonnegative numbers on an {{EasyBalloon|Link=Archimedean spiral|Balloon= An Archimedean spiral is a spiral traced by a point that is moving away from the center of the spiral with a constant speed along a line which is rotating with constant speed, or angular velocity. An Archimedean spiral typically has the form <math> r= a+b\theta</math>, where <math> a, b</math> are real numbers. The image below is an example of an Archimedean spiral. [[image:Archimedean.png]]}}, whereas the Ulam spiral places <math>1</math> in the center and places other numbers on a square grid. Moreover, Sacks spiral makes one full counterclockwise rotation for each square number <math> 4, 9, 16, 25, 36, 46, \dots </math>, as shown in the image below. The darker dots indicate the prime numbers. | + | Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places <math>0</math> in the center and places nonnegative numbers on an {{EasyBalloon|Link=Archimedean spiral|Balloon= An Archimedean spiral is a spiral traced by a point that is moving away from the center of the spiral with a constant speed along a line which is rotating with constant speed, or angular velocity. An Archimedean spiral typically has the form <math> r= a+b\theta</math>, where <math> a, b</math> are real numbers. The image below is an example of an Archimedean spiral. [[image:Archimedean.png]]}}, whereas the Ulam spiral places <math>1</math> in the center and places other numbers on a square grid. Moreover, Sacks spiral makes one full counterclockwise rotation for each square number <math> 4, 9, 16, 25, 36, 46, \dots </math>, as shown in [[#10|Image 10]]. The darker dots indicate the prime numbers. |
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| - | [[image:sack.gif|250px]] | + | We can also see in [[#10|Image 10]] that numbers that have blue check marks and are aligned in the left side of the spiral are <balloon title="load:pronicnum"> pronic numbers</balloon><span id="pronicnum" style="display:none">A pronic number is the product of two consecutive integers. Pronic numbers have the form <math>x(x+1)</math></span>. For example, <math> 2=1(1+1), 6=2(2+1), 12=3(3+1),\dots</math>. |
| - | | + | |
| - | We can also see that numbers that have blue check marks and are aligned in the left side of the spiral are <balloon title="load:pronicnum"> pronic numbers</balloon><span id="pronicnum" style="display:none">A pronic number is the product of two consecutive integers. Pronic numbers have the form <math>x(x+1)</math></span>. For example, <math> 2=1(1+1), 6=2(2+1), 12=3(3+1),\dots</math>. | + | |
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| | | + | {{Anchor|Reference=11|Link=[[Image:41prime.jpg|Image 11|thumb|250px|left|none]]}} |
| | Moreover, these numbers are aligned in positions that are a little less than half of one full rotation from one perfect square to the next perfect square, for instance, from 4 to 9, or 9 to 16. Let the <math>n</math>th perfect square be <math>n^2</math>. Then, going from the <math>n</math>th perfect square to the <math>(n+1)</math>st perfect square, the difference between the two numbers will be <math>2n+1</math>. We can show this by calculating the difference between two consecutive perfect squares, | | Moreover, these numbers are aligned in positions that are a little less than half of one full rotation from one perfect square to the next perfect square, for instance, from 4 to 9, or 9 to 16. Let the <math>n</math>th perfect square be <math>n^2</math>. Then, going from the <math>n</math>th perfect square to the <math>(n+1)</math>st perfect square, the difference between the two numbers will be <math>2n+1</math>. We can show this by calculating the difference between two consecutive perfect squares, |
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| - | :<math>(n+1)^2-n^2=n^2+2n+1-n^2=2n+1</math> | + | :<math>(n+1)^2-n^2=n^2+2n+1-n^2=2n+1</math>. |
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| | Because the pronic numbers are positioned a little less than half of one full rotation from the perfect squares, their position is a little less than <math>n+1/2</math> from the <math>n</math>th perfect square as we go around the spiral. Indeed, any pronic number <math>n(n+1)=n^2+n</math>, and the pronic number with the form <math>n(n+1)</math> is aligned at the <math>(n^2+n)</math>th position from the origin. From this, we can see that pronic number that has the form <math>n(n+1)</math> appears as the <math>n</math>th number along the spiral from the <math>n</math>th perfect square. | | Because the pronic numbers are positioned a little less than half of one full rotation from the perfect squares, their position is a little less than <math>n+1/2</math> from the <math>n</math>th perfect square as we go around the spiral. Indeed, any pronic number <math>n(n+1)=n^2+n</math>, and the pronic number with the form <math>n(n+1)</math> is aligned at the <math>(n^2+n)</math>th position from the origin. From this, we can see that pronic number that has the form <math>n(n+1)</math> appears as the <math>n</math>th number along the spiral from the <math>n</math>th perfect square. |
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| - | An interesting pattern can be discovered when we start the spiral at 41. As the image below shows, the red dots are the first 40 prime numbers generated by Euler's polynomial, and they are aligned in the center and positions where pronic numbers used to be in the original Sacks spiral. | + | An interesting pattern can be discovered when we start the spiral at 41. As [[#11|Image 11]] shows, the red dots are the first 40 prime numbers generated by Euler's polynomial, and they are aligned in the center and positions where pronic numbers used to be in the original Sacks spiral. |
| - | [[image:41prime.jpg|250px]]}} | + | |
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| | ==Other Numbers and Patterns== | | ==Other Numbers and Patterns== |
| | ===Triangular Number=== | | ===Triangular Number=== |
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| | A number <math>n</math> is a triangular number if <math>n</math> number of dots can be arranged into an equilateral triangle evenly filled with the dots. As shown in theimage below, the sequence of triangular numbers continue as <math>1, 3, 6, 10, 15, 21, \dots</math>. | | A number <math>n</math> is a triangular number if <math>n</math> number of dots can be arranged into an equilateral triangle evenly filled with the dots. As shown in theimage below, the sequence of triangular numbers continue as <math>1, 3, 6, 10, 15, 21, \dots</math>. |
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| - | [[image:triangular.jpg]] | + | {{Anchor|Reference=12|Link=[[Image:triangular.jpg|Image 12|thumb|none|300px]]}} |
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| | The <math>n^{\rm th}</math> triangular number, <math>T_n</math>, is given by the formula : | | The <math>n^{\rm th}</math> triangular number, <math>T_n</math>, is given by the formula : |
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| | When we mark the triangular numbers in a Ulam spiral, a set of spirals are formed as shown in the image below. | | When we mark the triangular numbers in a Ulam spiral, a set of spirals are formed as shown in the image below. |
| - | [[image:triangularnumbers.gif|300px]]}} | + | {{Anchor|Reference=13|Link=[[Image:triangularnumbers.gif|Image 13|thumb|300px|none]]}} |
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| | | + | }} |
| | ===Prime numbers in lines=== | | ===Prime numbers in lines=== |
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| - | [[image:Irisprime.jpg|700px|thumb|Image 10|none]] | + | {{Anchor|Reference=14|Link=[[Image:Irisprime.jpg|Image 14|thumb|700px|none]]}} |
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| - | The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. First, we create a table that has 30 columns and write all the natural numbers starting from 1 as we go from left to right. Thus, each row will start with a multiple of 30 added by 1, such as 1, 31, 61, 91, 121, ... . When we mark the prime numbers in this table, we get <i>Image 10</i>. | + | The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. First, we create a table that has 30 columns and write all the natural numbers starting from 1 as we go from left to right. Thus, each row will start with a multiple of 30 added by 1, such as 1, 31, 61, 91, 121, ... . When we mark the prime numbers in this table, we get [[#14|Image 14]]. |
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| | We can see that prime numbers appear only on certain columns that had 1, 7, 11, 13, 17, 19, 23, 29 on their first row. This image shows that all prime numbers have the form: | | We can see that prime numbers appear only on certain columns that had 1, 7, 11, 13, 17, 19, 23, 29 on their first row. This image shows that all prime numbers have the form: |
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| - | | + | |FullText= |
| - | <i>Image 11</i> shows the process of eliminating multiples of prime numbers from the table. First, we eliminate multiples of 2, and then we eliminate multiples of 3. Since 4 is a multiple of 2, any multiple of 4 was already eliminated. Thus, we next eliminate multiples of 5. We continue this process until we eliminate multiples of 19. (We do not eliminate numbers bigger than 19 because we only have numbers up to 390 in this table.) Note that when we eliminate multiples of 2, 3 or 5, the entire column that comes below any multiple of 2, 3, 5 on the first row get eliminated as well. | + | [[#15|Image 15]] shows the process of eliminating multiples of prime numbers from the table. First, we eliminate multiples of 2, and then we eliminate multiples of 3. Since 4 is a multiple of 2, any multiple of 4 was already eliminated. Thus, we next eliminate multiples of 5. We continue this process until we eliminate multiples of 19. (We do not eliminate numbers bigger than 19 because we only have numbers up to 390 in this table.) Note that when we eliminate multiples of 2, 3 or 5, the entire column that comes below any multiple of 2, 3, 5 on the first row get eliminated as well. |
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| - | [[image:irisgrid3.gif|700px|thumb|none|Image 11]] | + | {{Anchor|Reference=15|Link=[[Image:irisgrid3.gif|Image 15|thumb|700px|none]]}} |
| | | | |
| - | We can see from <i>Image 11</i> that eliminating multiples of 2, 3, 5 leaves us with columns where prime numbers appear. All of these multiples of 2, 3, and 5, or all the numbers that are not in the prime-concentrated column, have the form | + | We can see from [[#15|Image 15]] that eliminating multiples of 2, 3, 5 leaves us with columns where prime numbers appear. All of these multiples of 2, 3, and 5, or all the numbers that are not in the prime-concentrated column, have the form |
| | | | |
| | :<math>30n + 2^{a} 3^{b} 5^{c}</math> | | :<math>30n + 2^{a} 3^{b} 5^{c}</math> |
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| | }} | | }} |
| | }} | | }} |
| - | | | |
| | |AuthorName=en.wikipedia | | |AuthorName=en.wikipedia |
| | |Field=Number Theory | | |Field=Number Theory |
| - | |WhyInteresting= | + | |WhyInteresting={{Anchor|Reference=16|Link=[[Image:prime1.gif|Image 16|thumb|300px]]}} |
| - | [[image:prime1.gif|300px|thumb|Image 12]] | + | |
| | Not much is discovered about the Ulam spiral. For instance, the reason for the diagonal alignment of prime numbers or the vertical and horizontal arrangement of non-prime numbers is not clear yet. Indeed, Ulam spiral is not heavily studied by mathematicians. However, Ulam spiral's importance lies on the fact that it shows a clear pattern among prime numbers. | | Not much is discovered about the Ulam spiral. For instance, the reason for the diagonal alignment of prime numbers or the vertical and horizontal arrangement of non-prime numbers is not clear yet. Indeed, Ulam spiral is not heavily studied by mathematicians. However, Ulam spiral's importance lies on the fact that it shows a clear pattern among prime numbers. |
| | | | |
| - | Some might suspect that we are seeing diagonal lines in the Ulam spiral because the human eye seeks patterns and groups even among random cluster of dots. However, we can compare <i>Image 12</i> and <i>Image 13</i> and see that the prime numbers actually have a distinct pattern of diagonal lines that random numbers do not have. <i>Image 12</i> is a Ulam spiral where the black dots denote for the prime numbers, and <i>Image 13</i> is a Ulam spiral of random numbers. | + | Some might suspect that we are seeing diagonal lines in the Ulam spiral because the human eye seeks patterns and groups even among random cluster of dots. However, we can compare [[#16|Image 16]] and [[#17|Image 17]] and see that the prime numbers actually have a distinct pattern of diagonal lines that random numbers do not have. [[#16|Image 16]] is a Ulam spiral where the black dots denote for the prime numbers, and [[#17|Image 17]] is a Ulam spiral of random numbers. |
| | | | |
| - | [[image:randomintegers.jpg|300px|thumb|Image 13]] | + | {{Anchor|Reference=17|Link=[[Image:randomintegers.jpg|Image 17|thumb|300px]]}} |
| | People are interested in the pattern among prime numbers because the pattern might give enough information for us to discover a new polynomial that will generate more prime numbers than previously-discovered polynomials. The discovery of formula for prime numbers can lead us to have better understanding of other mysterious conjectures and theories involving prime numbers, such as {{EasyBalloon|Link=twin prime conjecture|Balloon=Twin prime conjecture states that there are infinitely many primes <math>p</math> such that <math>p+2</math> is also a prime.}} or <balloon title="The Goldbach's conjecture states that every even integer greater than 2 can be written as a sum of two primes.">Goldbach's conjecture</balloon>. For more information about the twin prime conjecture or Goldbach's conjecture, go to [http://mathworld.wolfram.com/TwinPrimeConjecture.html Wolfram Math World :Twin Prime Conjecture] or [http://mathworld.wolfram.com/GoldbachConjecture.html Wolfram Math World :Goldbach Conjecture]. | | People are interested in the pattern among prime numbers because the pattern might give enough information for us to discover a new polynomial that will generate more prime numbers than previously-discovered polynomials. The discovery of formula for prime numbers can lead us to have better understanding of other mysterious conjectures and theories involving prime numbers, such as {{EasyBalloon|Link=twin prime conjecture|Balloon=Twin prime conjecture states that there are infinitely many primes <math>p</math> such that <math>p+2</math> is also a prime.}} or <balloon title="The Goldbach's conjecture states that every even integer greater than 2 can be written as a sum of two primes.">Goldbach's conjecture</balloon>. For more information about the twin prime conjecture or Goldbach's conjecture, go to [http://mathworld.wolfram.com/TwinPrimeConjecture.html Wolfram Math World :Twin Prime Conjecture] or [http://mathworld.wolfram.com/GoldbachConjecture.html Wolfram Math World :Goldbach Conjecture]. |
| | | + | |References=Pickover, Clifford A. (2009). The Math Book : From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics. London : Sterling Publishing |
| | | + | |
| | | + | Wikipedia (Ulam Spiral). (n.d.). Ulam Spiral. Retrieved from http://en.wikipedia.org/wiki/Ulam_spiral. |
| | | + | |
| | | + | Wikipedia (Sacks Spiral). (n.d.). Sacks Spiral. Retrieved from http://en.wikipedia.org/wiki/Sacks_spiral. |
| | | + | |
| | | + | Weisstein, Eric W. "Prime Spiral." In MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/PrimeSpiral.html. |
| | | | |
| | | + | Sacks, Robert. (2007) NumberSpiral.com. Retrieved from http://www.numberspiral.com/index.html. |
| | | | |
| - | |InProgress=Yes | + | Weisstein, Eric W. "Prime-Generating Polynomial." In MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html |
| | | + | |ToDo=*An explanation for the patterns appearing among triangular numbers in [[Prime_spiral_%28Ulam_spiral%29#Triangular_Number|Triangular number]] section |
| | | + | *An helper page for Archimedean Spiral in [[Prime_spiral_%28Ulam_spiral%29#Sacks_Spiral| Sacks Spiral]] section |
| | | + | *An explanation of why the vertical and horizontal lines through the center do not contain any prime numbers |
| | | + | *A dynamic development of the spiral with e.g. red = all products of 2; green = all products of 3; nextcolour = all products of nextprime (except for those that are already coloured). If we start with all numbers = black, then at the end, black=prime |
| | | + | |InProgress=No |
| | }} | | }} |
## Current revision
Ulam Spiral
Field: Number Theory
Image Created By: en.wikipedia
Ulam Spiral
The Ulam spiral, or prime spiral, is a plot in which prime numbersA prime number is a natural number greater that is divisible only by 1 and itself. The first few prime numbers are :$2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, \dots$ are marked among positive integers that are arranged in a counterclockwise spiral. The prime numbers show a pattern of diagonal lines.
# Basic Description
Image 1
Image 2
The prime spiral was discovered by Stanislaw Ulam (1909-1984) in 1963 while he was doodling on a piece of paper during a science meeting. Starting with 1 in the middle, he wrote positive numbers in a grid as he spiraled out from the center, as shown in Image 1. He then circled the prime numbers, and the prime numbers showed patterns of diagonal lines as shown by the grid in Image 2. The grid in Image 2 is a close-up view of the center of the main image such that the green line segments and red boxes in the center of the main image line up with those in the grid.
A larger Ulam spiral with 160,000 integers and 14,683 primes is shown in the main image. Black dots indicate prime numbers. In addition to diagonal line segments formed by the black dots, we can see white vertical and horizontal line segments that cross the center of the spiral and do not contain any black dots, or prime numbers. There are also white diagonal line segments that do not contain any prime numbers. Ulam spiral implies that there is some order in the distribution of prime numbers.
# A More Mathematical Explanation
[Click to view A More Mathematical Explanation]
From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is [...]
[Click to hide A More Mathematical Explanation]
From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is no known simple formula that yields all the primes. The diagonal patterns in the Ulam spiral gives some hint for formulas of primes numbers.
Definition of Half-lines
Many half-lines in the Ulam spiral can be described using quadratic polynomials A quadratic polynomial is a polynomial in which the exponent of the variables do not exceed 2. A quadratic polynomial typically has the form $ax^2+bx+c$ where $x$ is a variable and $a, b, c$ are coefficients.. A half-line is a line which starts at a point and continues infinitely in one direction. In this page, we only consider half-lines that are horizontal or vertical, or have a slope of -1 or +1. The half-lines that can be described using quadratic polynomials are the ones in which each entry of the diagonal is positioned on a different ring of the spiral.
The ring of a spiral can be considered as the outermost layer of a concentric square or a rectangle centered around the center of the spiral, 1, as defined by the blue line in the grid. Although there is no exact place where we can determine the beginning and end of a ring, we will assume that entries that are positioned on squares or rectangles of the same sizes to be on the same ring. For instance, 24, 25, 26, 27, 28, which are shown in red boxes in Image 3, are on the same ring, whereas 48, 49, 50, 51, 52, as shown in blue boxes, are on a different ring because they are positioned on a bigger square.
Image 3:Diagonal half-lines
The green lines in the Image 3 qualify as diagonal half-lines whereas the red lines do not because the red lines cross the corner of the grid in a way that two entries of the red diagonal line are positioned on the same ring.
Thus, diagonal line segments that are composed of prime numbers can also be expressed as outputs of quadratic polynomials. To learn more about the relation between the diagonal lines and quadratic polynomials, click below.
Half-lines and quadratic polynomials
First, we need to know the relation between quadratic polynomials and the difference table of sequences. Le [...]
Half-lines and quadratic polynomials
First, we need to know the relation between quadratic polynomials and the difference tableThe difference table lists the terms of a sequence in one row, and the differences between consecutive terms in the next row. It can include other higher order differences such as the second-differences and the third-differences. of sequences. Let's choose the diagonal $5, 19, 41, 71, 109, 155$, the diagonal indicated with green dotted line in Image 3, as our original sequence and create a difference table.
As we can see from the table, the 2nd differences are constant, and this implies that the original sequence can be described by a second degree polynomial. For more information about difference tables and about finding specific polynomials for sequences, go to difference tables.
Image 4:Constant 2nd differences
Now, we will see that the second differences are always constant for any segment of a diagonal half-line. Let's choose the same diagonal with entries $5, 19, 41, 71, 109, 155$. (For a better understanding of the following paragraphs, the reader should wait until the animation shows three different blue rings and then watch the rest of the animation.)
In Image 4, the number of the innermost light blue boxes, which is 14, indicate the distance from $5$ to $19$, or the difference between these two numbers. The number of darker blue boxes in the middle indicate the difference between the numbers $19$ and $41$. The darkest blue boxes on the outside indicate the difference between the entries $41$ and $71$. The number of all these blue boxes correspond to the first differences in the difference table.
In Image 4, the light blue boxes are divided into pieces, and each piece is moved to on top of the darker blue boxes. We can see that there are $8$ more darker blue boxes than the innermost light blue boxes. Similarly, the blue boxes in the middle are divided into pieces and are moved to on top of the darkest blue boxes.
Indeed, as Image 4 illustrates, regardless of the exact number of the blue boxes, there are $8$ more boxes in any given ring than there are in the ring that is one layer inwards from it. This means that the the second differences, or the differences between the first differences is always going to be constant at $8$. Thus, the sequence for any diagonal half-line can be described using a quadratic polynomial.
Examples of quadratic polynomials for half-lines
For instance, prime numbers $5, 19, 41, 71, 109$, which are aligned in the same diagonal, can be described through the output of the polynomial $4x^2+10x+5$ for $x=0, 1, 2, 3, 4$. Similarly, numbers $1, 3, 14, 31, 57, 91 \dots$, which are also aligned in a green diagonal in Image 5 starting from the center and continuing to the upper right corner, can be expressed by:
$4x^2-2x+1$
for $x=0, 1, 2, 3, \dots$. (We will refer back to this polynomial in a later section) In fact, the part of the green diagonal that starts from the center and continues to the bottom left corner can also be described through Eq. (1) for $x=0, -1, -2, -3, \dots$.
Image 5
In fact, even horizontal and vertical line segment in the grid can be described by quadratic polynomials, as long as the lines satisfy the condition that no two entries are positioned on the same ring. For instance, the blue horizontal line segment in Image 5, $10, 27, 52, 85$ can be described by a quadratic polynomial. However, the sequence $9, 10, 27, 85$ on the same horizontal line cannot be described by a polynomial because the entries 9 and 10 are on the same ring.
Moreover, it is not hard to show that the green diagonal line from Image 5 that goes through the center and has a slope of +1 is the only line on which the Ulam numbers in both directions can be described by the same polynomial. Even the red diagonal line that goes through the center and has a slope of -1 cannot be described by one polynomial.
Half of the red diagonal, the segment going up from the center, $5, 17, 37, 67, \dots$ can be described by $x^2+1$ for inputs that are even numbers, while the diagonal going down from the center, $1, 9, 25, 49, \dots$ is a sequence of perfect squares of odd numbers. Thus we cannot find one polynomial that generates both all entries of the red diagonal.
## Euler's Prime Generator
Many have come up with polynomials in one variable that generate prime numbers, although none of these polynomials [...]
Many have come up with polynomials in one variable that generate prime numbers, although none of these polynomials can generate all the prime numbers. One of the most famous polynomials is the one discovered by Leonhard Euler(1707-1783), which is :
$x^2-x+41$
Euler's polynomial generates distinct prime numbers for each integer $x$ from $x=1$ to $x=40$.
Image 6
As we can see in Image 6, we can start an Ulam's spiral with 41 at the center of the grid and get a long, continuous diagonal with 40 prime numbers. One interesting fact is that the 40 numbers in the diagonal line segment are the first 40 prime numbers that are generated through Euler's polynomial. Moreover, the prime numbers are not aligned in order of increasing values. In fact, with 41 in the center, other prime numbers alternate in position between the upper right and lower left part of the diagonal.
We will show why an Ulam's spiral that starts with 41 generates entries aligned in a diagonal that can also be descried as outputs of Euler's polynomial.
First, we found out in the previous section that Eq. (1) is the polynomial for the diagonal that goes through the center 1 and has a slope of +1 in the [...]
First, we found out in the previous section that Eq. (1) is the polynomial for the diagonal that goes through the center 1 and has a slope of +1 in the Ulam's spiral. That is,
$4x^2-2x+1$
describes the upper half of the diagonal as we plug in $x=0, 1, 2, 3, \dots$ and the lower half of the diagonal as we plug in $x=0, -1, -2, -3, \dots$. Thus, $x$'s are positioned in the diagonal as if the diagonal was a number line that starts with 0 at the center, with positive integers on the upper right and negative integers on the lower left direction, as shown in Image 7.
Image 7
Then, if we start the Ulam's spiral at 41 and thus make 41 the center, each entry on the grid, including the entries on the diagonal segment, will increase by 40. Then, the diagonal can be described by the polynomial :
$4x^2-2x+41$.
In fact, we will see that this polynomial and Euler's polynomial are describing the same entries along the diagonal once we make some changes in numbering the position of $x$'s. Euler's polynomial can be found by reassigning the position of the entries of the diagonals so that $x'=1$ is at the center, $x'=2, 4, 6, 8, \dots$ are the positions up the diagonal line, $x'=1, 3, 5, 7, 9, \dots$ are positions of entries down the diagonal line from the center, as shown in Image 8. Thus, starting with $x'=1$ at the center, we alternate between the right and left of the center as we number the positions $x'=1, 2, 3, 4, 5, \dots$.
Image 8
Then, for the upper half of the diagonal in Image 8, we can use $x'=2, 4, 6, 8, \dots$ instead of $x=1, 2, 3, 4, \dots$ from Image 7. Then,
$x=\frac{1}{2} x'$
If we plug this value in to $4x^2-2x+41$ of Eq. (2), we get :
$4(\frac{1}{2}x')^2-2(\frac{1}{2}x')+41$
$={x'}^2-x'+41$,
which has the same form as Euler's polynomial.
The same method works for the bottom half as well. We use $x'=1, 3, 5, 7, \dots$ in Image 8 while we used $x=0, -1, -2, -3, \dots$ in Image 7. Then,
$x=\frac{-x'+1}{2}$.
By substituting the $x$'s in the polynomial $4x^2-2x+41$ the same way we did above, we get:
${x'}^2-x'+41$
Thus, we can see that Eq. (2), in fact, have the same output as Euler's polynomial after we reassign the position of $x$'s. We have shown that Eq. (1), which describes the diagonal that goes through the center and has a slope of +1, can be transformed to Euler's polynomial that describes the same diagonal that has 41 as its center.
## Sacks Spiral
Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places 0 in th [...]
Image 9
Image 10
Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places $0$ in the center and places nonnegative numbers on an Archimedean spiralAn Archimedean spiral is a spiral traced by a point that is moving away from the center of the spiral with a constant speed along a line which is rotating with constant speed, or angular velocity. An Archimedean spiral typically has the form $r= a+b\theta$, where $a, b$ are real numbers. The image below is an example of an Archimedean spiral. , whereas the Ulam spiral places $1$ in the center and places other numbers on a square grid. Moreover, Sacks spiral makes one full counterclockwise rotation for each square number $4, 9, 16, 25, 36, 46, \dots$, as shown in Image 10. The darker dots indicate the prime numbers.
We can also see in Image 10 that numbers that have blue check marks and are aligned in the left side of the spiral are pronic numbersA pronic number is the product of two consecutive integers. Pronic numbers have the form $x(x+1)$. For example, $2=1(1+1), 6=2(2+1), 12=3(3+1),\dots$.
Image 11
Moreover, these numbers are aligned in positions that are a little less than half of one full rotation from one perfect square to the next perfect square, for instance, from 4 to 9, or 9 to 16. Let the $n$th perfect square be $n^2$. Then, going from the $n$th perfect square to the $(n+1)$st perfect square, the difference between the two numbers will be $2n+1$. We can show this by calculating the difference between two consecutive perfect squares,
$(n+1)^2-n^2=n^2+2n+1-n^2=2n+1$.
Because the pronic numbers are positioned a little less than half of one full rotation from the perfect squares, their position is a little less than $n+1/2$ from the $n$th perfect square as we go around the spiral. Indeed, any pronic number $n(n+1)=n^2+n$, and the pronic number with the form $n(n+1)$ is aligned at the $(n^2+n)$th position from the origin. From this, we can see that pronic number that has the form $n(n+1)$ appears as the $n$th number along the spiral from the $n$th perfect square.
An interesting pattern can be discovered when we start the spiral at 41. As Image 11 shows, the red dots are the first 40 prime numbers generated by Euler's polynomial, and they are aligned in the center and positions where pronic numbers used to be in the original Sacks spiral.
## Other Numbers and Patterns
### Triangular Number
A number n is a triangular number if n number of dots can be arranged into an equilateral triangle ev [...]
A number $n$ is a triangular number if $n$ number of dots can be arranged into an equilateral triangle evenly filled with the dots. As shown in theimage below, the sequence of triangular numbers continue as $1, 3, 6, 10, 15, 21, \dots$.
Image 12
The $n^{\rm th}$ triangular number, $T_n$, is given by the formula : $T_n=\frac{n(n+1)}{2}$
When we mark the triangular numbers in a Ulam spiral, a set of spirals are formed as shown in the image below.
Image 13
### Prime numbers in lines
The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. Firs [...]
Image 14
The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. First, we create a table that has 30 columns and write all the natural numbers starting from 1 as we go from left to right. Thus, each row will start with a multiple of 30 added by 1, such as 1, 31, 61, 91, 121, ... . When we mark the prime numbers in this table, we get Image 14.
We can see that prime numbers appear only on certain columns that had 1, 7, 11, 13, 17, 19, 23, 29 on their first row. This image shows that all prime numbers have the form:
$30n+(1, 7, 11, 13, 17, 19, 23, 29)$
Note that not all numbers generated by this form are prime numbers. Another point to notice in the picture is that the nonprime numbers that appear on prime-concentrated columns are all multiples of prime numbers larger than or equal to 7. For instance, $49=7\times 7, \quad 77=7\times 11, \quad 91=7\times 13, \quad 119=7\times 17, \quad 121=11\times 11, \quad 133=7\times 19 \dots$. In fact any combination of two prime numbers larger than or equal to 7 all appear on these prime-concentrated columns.
Image 15 shows the process of eliminating multiples of prime numbers from the table. First, we eliminate multiples of 2, and then we eliminate
Image 15 shows the process of eliminating multiples of prime numbers from the table. First, we eliminate multiples of 2, and then we eliminate multiples of 3. Since 4 is a multiple of 2, any multiple of 4 was already eliminated. Thus, we next eliminate multiples of 5. We continue this process until we eliminate multiples of 19. (We do not eliminate numbers bigger than 19 because we only have numbers up to 390 in this table.) Note that when we eliminate multiples of 2, 3 or 5, the entire column that comes below any multiple of 2, 3, 5 on the first row get eliminated as well.
Image 15
We can see from Image 15 that eliminating multiples of 2, 3, 5 leaves us with columns where prime numbers appear. All of these multiples of 2, 3, and 5, or all the numbers that are not in the prime-concentrated column, have the form
$30n + 2^{a} 3^{b} 5^{c}$
or
$(2\times 3 \times 5)n+2^{a} 3^{b} 5^{c}$,
where $a \geq 0, b \geq 0, c \geq 0$.
Thus, for a given number with the form of Eq. (3) we are able to factor out 2 or 3 or 5, or any combination of the three, which means that this number is divisible by 2, 3, or 5.
However, we know that prime numbers are not multiples of 2 or 3 or 5. Then, these prime numbers will not have the same form as Eq. (3), or else the prime numbers will be divisible by 2 or 3 or 5. Indeed, we can see that the prime numbers are positioned at :
$(2 \times 3 \times 5)+ (1, 7, 11, 13, 17, 19, 23, 29)$
Thus, prime numbers have to be positioned on the blue columns that do not have the same form as Eq. (3).
Next, we look at nonprime numbers that are products of multiplication of two prime numbers or a perfect square of a prime number. These products will be divisible by the prime numbers they are composed of, but they will not be divisible by 2 or 3 or 5.
For instance, 77, which is a nonprime number that appears on the column of prime numbers 117, 47, 107, 137, ..., is a product of two prime numbers, 7 and 11. Because each of these prime numbers do not have any divisors other than 1 and themselves, 77 is not divisible by 2 or 3 or 5. Thus, 77 and all other products of prime numbers also cannot have the same form as Eq. (3), and they have to appear on the blue columns that are concentrated with prime numbers.
# Why It's Interesting
Image 16
Not much is discovered about the Ulam spiral. For instance, the reason for the diagonal alignment of prime numbers or the vertical and horizontal arrangement of non-prime numbers is not clear yet. Indeed, Ulam spiral is not heavily studied by mathematicians. However, Ulam spiral's importance lies on the fact that it shows a clear pattern among prime numbers.
Some might suspect that we are seeing diagonal lines in the Ulam spiral because the human eye seeks patterns and groups even among random cluster of dots. However, we can compare Image 16 and Image 17 and see that the prime numbers actually have a distinct pattern of diagonal lines that random numbers do not have. Image 16 is a Ulam spiral where the black dots denote for the prime numbers, and Image 17 is a Ulam spiral of random numbers.
Image 17
People are interested in the pattern among prime numbers because the pattern might give enough information for us to discover a new polynomial that will generate more prime numbers than previously-discovered polynomials. The discovery of formula for prime numbers can lead us to have better understanding of other mysterious conjectures and theories involving prime numbers, such as twin prime conjectureTwin prime conjecture states that there are infinitely many primes $p$ such that $p+2$ is also a prime. or Goldbach's conjecture. For more information about the twin prime conjecture or Goldbach's conjecture, go to Wolfram Math World :Twin Prime Conjecture or Wolfram Math World :Goldbach Conjecture.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# References
Pickover, Clifford A. (2009). The Math Book : From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics. London : Sterling Publishing
Weisstein, Eric W. "Prime Spiral." In MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/PrimeSpiral.html.
Weisstein, Eric W. "Prime-Generating Polynomial." In MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html
# Future Directions for this Page
• An explanation for the patterns appearing among triangular numbers in Triangular number section
• An helper page for Archimedean Spiral in Sacks Spiral section
• An explanation of why the vertical and horizontal lines through the center do not contain any prime numbers
• A dynamic development of the spiral with e.g. red = all products of 2; green = all products of 3; nextcolour = all products of nextprime (except for those that are already coloured). If we start with all numbers = black, then at the end, black=prime
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_1&diff=29929&oldid=29928
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# User:Michiexile/MATH198/Lecture 1
### From HaskellWiki
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| | ===What do we require?=== | | ===What do we require?=== |
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| | | + | Our examples will be drawn from discrete mathematics, logic, Haskell programming and linear algebra. I expect the following concepts to be at least vaguely familiar to anyone taking this course: |
| | | + | * Sets |
| | | + | * Functions |
| | | + | * Permutations |
| | | + | * Groups |
| | | + | * Partially ordered sets |
| | | + | * Vector spaces |
| | | + | * Linear maps |
| | | + | * Matrices |
| | | + | * Homomorphisms |
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| | ==Category== | | ==Category== |
## 2 Introduction
### 2.1 Why this course?
An introduction to Haskell will usually come with pointers toward Category Theory as a useful tool, though not with much more than the mention of the subject. This course is intended to fill that gap, and provide an introduction to Category Theory that ties into Haskell and functional programming as a source of examples and applications.
### 2.2 What will we cover?
The definition of categories, special objects and morphisms, functors, natural transformation, (co-)limits and special cases of these, adjunctions, freeness and presentations as categorical constructs, monads and Kleisli arrows, recursion with categorical constructs.
Maybe, just maybe, if we have enough time, we'll finish with looking at the definition of a topos, and how this encodes logic internal to a category. Applications to fuzzy sets.
### 2.3 What do we require?
Our examples will be drawn from discrete mathematics, logic, Haskell programming and linear algebra. I expect the following concepts to be at least vaguely familiar to anyone taking this course:
• Sets
• Functions
• Permutations
• Groups
• Partially ordered sets
• Vector spaces
• Linear maps
• Matrices
• Homomorphisms
## 3 Category
### 3.1 Graphs and paths
A graph is a collection G0 of vertices and a collection G1 of arrows. The structure of the graph is captured in the existence of two functions, that we shall call source and target, both going from G1 to G1. In other words, each arrow has a source and a target.
We denote by [v,w] the collection of arrows with source v and target w.
We extend the notation, and denote by Gi the collection of all paths of length i. Such a path is a sequence $f_1,\dots,f_i$ of arrows, such that for each j, target(fj − 1) = source(fj).
### 3.2 Definition of a category
A category is a graph with some special structure:
• Each [v,w] is a set and equipped with a composition operation $[u,v] \times [v,w] \to [u,w]$. In other words, any two arrows, such that the target of one is the source of the other, can be composed to give a new arrow with target and source from the ones left out.
We write $f:u\to v$ if $f\in[u,v]$.
$u \to v \to w$ => $u \to w$
• The composition of arrows is associative.
• Each vertex v has a dedicated arrow 1v with source and target v, called the identity arrow.
• Each identity arrow is a left- and right-identity for the composition operation.
The composition of $f:u\to v$ with $g:v\to w$ is denoted by $gf:u\to v\to w$. A mnemonic here is that you write things so associativity looks right. Hence, (gf)(x) = g(f(x)). This will make more sense once we get around to generalized elements later on.
### 3.3 Examples
• The empty category with no vertices and no arrows.
• The category 1 with a single vertex and only its identity arrow.
• The category 2 with two objects, their identity arrows and the arrow $a\to b$.
• For vertices take vector spaces. For arrows, take linear maps. This is a category, the identity arrow is just the identity map f(x) = x and composition is just function composition.
• For vertices take finite sets. For arrows, take functions.
• For vertices take logical propositions. For arrows take proofs in propositional logic. The identity arrow is the empty proof: P proves P without an actual proof. And if you can prove P using Q and then R using P, then this composes to a proof of R using Q.
• For vertices, take data types. For arrows take (computable) functions. This forms a category, in which we can discuss an abstraction that mirrors most of Haskell. There are issues making Haskell not quite a category on its own, but we get close enough to draw helpful conclusions and analogies.
• Suppose P is a set equipped with a partial ordering relation <. Then we can form a category out of this set with elements for vertices and with a single element in [v,w] if and only if v<w. Then the transitivity and reflexivity of partial orderings show that this forms a category.
Some language we want settled:
A category is concrete if it is like the vector spaces and the sets among the examples - the collection of all sets-with-specific-additional-structure equipped with all functions-respecting-that-structure. We require already that [v,w] is always a set.
A category is small if the collection of all vertices, too, is a set.
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http://physics.stackexchange.com/questions/53338/confusion-regarding-photons/53339
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# Confusion regarding photons?
I'm still in high school, and while I can't complain about the quality of my teachers (all of them have done at least a bachelor, some a masters) I usually am cautious to believe what they say straight away. Since I'm interested quite a bit in physics, I know more about it than other subjects and I spot things I disagree with more often, and this is the most recent thing:
While discussing photons, my teacher made a couple of statements which might be true but sound foreign to me:
• He said that under certain conditions, photons have mass. I didn't think this was true at all. I think he said this to avoid confusion regarding $E=mc^2$, however, in my opinion it only adds to the confusion since objects with mass can't travel with the speed of light, and light does have a tendency to travel with the speed of light.. I myself understand how photons can have a momentum while having no mass because I lurk this site, but my classmates don't.
• He said photons don't actually exist, but are handy to envision. This dazzled my mind. Even more so since he followed this statement by explaining the photo-electric effect, which to me seems like a proof of the existence of photons as the quantum of light. He might have done this to avoid confusion regarding the wave-particle duality.
This all seems very odd to me and I hope some of you can clarify.
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– Nivalth Feb 7 at 19:28
Photons do have a mass inside a superconductor. Which is why, inside a superconductor, the electromagnetic force becomes short-range. Perhaps that's what your teacher meant. – Dmitry Brant Feb 7 at 20:07
@DmitryBrant I know this is to some extent just a matter of semantics, but I personally feel it's somewhat misleading to call the effective mass of a photon inside of a superconductor its mass. – joshphysics Feb 7 at 21:10
– joshphysics Feb 7 at 21:12
I would definitely ignore anything that your teacher or classmates have to say about physics that you can not find in a good text book. Until you reach topics of quantum gravity and quantum information theory, these things are well understood within the physics community (although not by all its members). I would also ignore most media releases until you can sift the good from the bad. A good list of freely available books can be found athttp://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books – Hal Swyers Feb 8 at 11:45
show 1 more comment
## 5 Answers
1. Photons are massless. This should not cause confusion with $E=mc^2$ because the expression for the relativistic energy of the photon is $E = h \nu$, there $\nu$ is the photon's frequency and $h$ is Planck's constant. You can also understand the relativistic energy of the photon by noting that $E = pc$ where $p=\hbar k$ is the magnitude of its momentum, and photons possess momentum, as you point out. $k$ is the photon's wavevector and $\hbar = h/2\pi$ is the reduced Plank's constant. To tie this all together, we have the formula $E=\sqrt{m^{2}c^{4}+p^{2}c^{2}}$.
2. Photons exist! As you point out, the existence of photons has physical consequences that can be measured. Perhaps, as you also mention, your teacher is trying to insist that you resist thinking of photons purely as particles (which is probably not such a bad idea), but the statement "photons don't exist," in my opinion, should be considered just as fallacious as "the keyboard on which I'm typing doesn't exist.
You're being prudent by taking everything your teacher says with a large grain of salt. In my experience in physics (heck in life as a whole), it's good to take everything that anyone ever says with a grain of salt (including my response for that matter).
Cheers!
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It depends on the definition of mass, obviously. Photons have zero rest mass. But physicists working in relativity also use "mass" as a synonim of "energy". – Bzazz Feb 7 at 22:48
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@Bzazz Nuclear and particle physicists use the work "mass" to mean only the rest mass. Every time. While there is no difficulty in defining the "relativistic mass" the concept obscures and confuses rather than clarifying. – dmckee♦ Feb 7 at 22:57
Yes, that's why I specified physicists working on relativity. In all the courses of GR I had, you put c=1 and talk of mass and energy equivalently. Of course, one seldom talks about the mass of a photon, because, we are more familiar with speaking of photon energy. But if we want to look for "the mass of the photon" as in the question, this is what comes to my mind first. Remember that relativistic mass is effectively mass, in the sense that also experiences gravity. – Bzazz Feb 7 at 23:05
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– Hal Swyers Feb 8 at 10:57
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I definitely agree with @HalSwyers on this one. – joshphysics Feb 8 at 16:22
He is doing a good job trying to communicate the weirdness of Photons, but a poor one at being consistant. It's difficult to communicate how strange they are, without using the relationships you described.
First of all photons have no mass. At rest. The concept of describing a photon at rest is a bit weird even as there is no such thing.
Once your talking about photons in motion it is tempting to say they have mass, but most physicists these days don't speak of it this way. Rather its best to say it has ENERGY. While people like to say things can have relativistic mass, this is incorrect, mass is a constant: the mass a thing has never changes no matter how fast you travel: rather the energy to accelerate it increases (inertia).
So a photon has no mass: but it does have momentum. This is the fascinating thing about light. And leads to the fact that momentum is not always related to mass as you mentioned.
Examine $E=mc^2$ in more detail and the correct equation actually is $E^2 = m^2 c^4 + p^2 c^2$. As photons have momentum related to its energy/frequency this is fine.
The best way to think of light is as photons. Light is a photon, that is the fact, BUT light waves are only a model. The appearance of wave phenomenon is because quantum mechanically light interacts probabilistically, and this nature allows it to display wave like properties.
The science of Quantum Electro-Dynamics examines this strange behavior.
For now think of light as a photon and a wave, it allows every day behavior to be modeled well and is perfectly fine.
But in reality light is a photon particle (this is why the photo-electric effect works) that when described using Schrodinger's Equations (a description of probabilites) can be transformed into a description of a wave of Electro-Magnetism.
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There are two reasons why a photon can't be described by the Schrödinger equation: A single particle theory for a photon is non-sense and Schrödinger's equation is non-relativistic. – Nivalth Feb 7 at 20:08
For modeling interactions between photons to show that they display wave characteristics, Shrodinger's equation works as a good approximation to demonstrate his equations and Maxwells are equivalent. Nontheless like I mentioned QED manages it better: see Feynman's books for good explanations without equations. – Eric_ Feb 7 at 20:41
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Eric, Schrödinger's equation is explicitly built on the Newtonian relationship between kinetic energy, momentum and mass (the rest mass for those that insist). It really, really doesn't do to use it with photons. If you must do non-field-theoretical QM with light, use the Klein-Gordon equation. However, you've got the right relationship in your post: $E^2 = m^2c^4 + p^2c^2$ is the correct answer. Don't detract from it by using the wrong wave equation. – dmckee♦ Feb 7 at 22:45
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In any case, welcome to Physics.SE. WE have the MathJax rendering engine active on the site which lets you write latex-alike math inside pairs of `$`'s (for inline) or `$$` (for block typesetting). I've done this post for you. – dmckee♦ Feb 7 at 22:47
I have sat in a college class and watched step by step as a Schrodinger equation is transformed into a wave equation with a form identical to maxwells. The point is not that it explains light, but that probabilistic quantum mechanics explains how a wave nature can appear for ANY and ALL particle phenomenon. There is wave particle duality in everything, not just light. – Eric_ Feb 8 at 0:00
show 1 more comment
He said that under certain conditions, photons have mass.
Massless particles move at the speed of light $c$ in vacuum. By his statement your teacher may have been alluding to the fact that photons travel at speeds slower than $c$ when they travel through media, like glass for example. However, I would rather phrase this as something like "in the transmission of a photon through a medium, the photon's transit time through that medium is such that it travels as if it had a mass." The travel of photons through media is a rather complex affair, which I don't fully understand, involving interactions with charged particles in the medium and quasiparticle states, and I'm not sure to what extent the incident photon even retains its identity whilst in the medium (maybe that's another question).
He said photons don't actually exist, but are handy to envision.
There is no question that photons exist. Particle physicists deal with "hard" (i.e high energy) photons, which behave like particles - scatter off other charged particles etc. Good evidence that "soft" (low energy) photons exist comes from antibunching experiments.
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1) I would refer your high school teacher to some of the good answers found in this physics stack exchange question. The answer with the highest votes is really good and is referring to the electroweak theory which governs the electroweak section of the Standard Model.
General Relativity respects special relativity, and therefore respects that the invariant mass of the photon is zero, since the photon is described by a null-like vector in its rest frame.
In particle physics, as discussed in the links above, the current understanding is that mass is a measure of the relative interaction strength of a particle with the Higgs field as mediated by the Higgs boson. The photon does not directly interact with the Higgs boson, and therefore has no mass.
2) As far as visualizing the photon, I would venture the easiest way is to think in terms of classical EM theory (which is a gauge theory btw) where we consider the orthogonal oscillating electric and magnetic fields as representing the photon.
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$E = mc^2$ is a popular formula but only valid in a special case when "the total additive momentum is zero for the system under consideration" $^1$
The more general "energy-momentum relation" is: $E^2 = (mc^2)^2 + (pc)^2$
(Also, here's a neat little video for your classmates ;) -> http://www.youtube.com/watch?v=NnMIhxWRGNw )
$^1$ http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence
Even today in the time of quantum optics, there are still quite a lot of papers, that try to grasp what a "photon" is, e.g.:
1. What is a photon - David Finkelstein
2. Light reconsidered - Arthur Zajonc
3. The concept of the photon - revisited - Ashok Muthukrishan, Marlan O. Scully, M.Suhail Zubairy
and many more all together in this nice review: http://www.sheffield.ac.uk/polopoly_fs/1.14183!/file/photon.pdf
Or in the words of Roy Glauber: "A photon is what a photodetector detects"
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## protected by Qmechanic♦Feb 8 at 17:42
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://math.stackexchange.com/questions/255590/spherical-coordinates-to-cartesian-coordinates
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# Spherical coordinates to cartesian coordinates.
I want to find out the distance between the centers of $2$ circles. Say, circle $1$ $(\theta,\phi)$ circle $2$ $(\theta,\phi)$
The radius of this circle is found using $d\tan(\theta)$ where $d$ is the range (different from radius)
$$cd=\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$$ is the formula I'm going to use to find out the distance between the $2$ points.
But can anyone help me in defining $x$, $y$ and $z$?
I know to convert it to Cartesian coordinates if say the points are $(r,\theta)$ where $r$ is the radius [$2$ dimensional]
cd is a euclidean distance between the centre of circle#1 (formed due to angles az(t,n),el(t,n) in space from transmitter t, at trial n and distance d away from origin) and circle#2 (formed due to angles az(r,n),el(r,n) in space from receiver r, at trial n and distance d away from origin)
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1
It is difficult to understand what you are asking. – copper.hat Dec 10 '12 at 16:44
the centers of the circles are defines by angles say (theta, phi) I need to find out the distance between the centers of the 2 circles using cd=sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2) but I do not know what x, y and z are – Nidhi Dec 10 '12 at 16:45
but r1 and r2 are the radius. I have 2 angles as my center. will the formula still work? – Nidhi Dec 10 '12 at 16:49
my problem is 3 dimensional. :( – Nidhi Dec 10 '12 at 16:49
I am not sure I understand exactly what you are asking. What do you mean by saying that you have a circle given by $(\theta, \phi)$? – Thomas Dec 10 '12 at 16:50
show 3 more comments
## 1 Answer
In spherical coordinates, and generally in $\mathbb R^3$, it takes three coordinates to specify a point. If you only have $\phi, \theta$ you have a ray from the origin. You need the distance from the origin to get a point. Then, given $(r,\theta,\phi)$ for each point you can convert to Cartesian coordinates with $x=r \sin \theta \cos \phi, y=\sin \theta \sin \phi, z=r \cos \theta$
Added, based on your comment: The easiest is to erect a coordinate system at each transmitter. From transmitter 1, you know the point is at $(r_1,\theta_1,\phi_1)$, which you can convert to $(x_1,y_1,z_1)$. Then from transmitter 2, you know the vector is $(r'_2,\theta'_2,\phi'_2)$ where the primes indicate that the coordinate system is centered on transmitter 2. Hopefully your axes are aligned, that makes it easier. Convert this to $(x'_2,y'_2,z'_2)$, still referenced to transmitter 2. Then if transmitter 2 is at $(t2_x,t2_y,t2_z)$ in the transmitter 1 frame, the second point is at $(x'2+t2_x,y'_2+t2_y,z'_2+t2_z)$. Now you can use the usual Cartesian distance function.
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No I have tried this but the problem is will r be the radius or the range. I have 2 different parameters. I'll try to attach a figure – Nidhi Dec 10 '12 at 16:59
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http://mathoverflow.net/questions/114963/tangent-space-of-the-stack-overline-mathcalm-g-nx-beta
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## Tangent space of the stack $\overline{\mathcal{M}}_{g,n}(X,\beta)$.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\overline{\mathcal{M}}_{g,n}(X,\beta)$ be the moduli stack of stable maps $f$ from genus $g$, $n$-marked curve $C$ to a variety $X$ i nth curve class $\beta \in H^2(X,\mathbb{Z})$. I would like to know why the hyperext group $$Ext^1_C(f^*\Omega_X\rightarrow \Omega_C(\sum_{i=1}^n),\mathcal{O}_C)$$ represents tangent space of the moduli and the obstruction lies in
$$Ext^2_C(f^*\Omega_X\rightarrow \Omega_C(\sum_{i=1}^n),\mathcal{O}_C)?$$ I am aware that the deformation-obstrcution of a map $f:C\rightarrow X$ with a fixed curve $C$ is governed by $H^i(C,f^*T_X)$ for $i=1,2$ and the automorphism-deformation of a $n$-marked points $(C;p_1,\dots,p_n)$ is governed by $Ext^i_C(\Omega_C(\sum_{i=1}^n),\mathcal{O}_C)$ for $i=0,1$. However, I don't know how to combine them into one package in the hyperext groups above.
I would appreciate it if someone could kindly explain how to obtain and understand the hyperext groups above.
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Are you looking for a reference? – Jason Starr Nov 30 at 11:44
I didn't mean reference request as I thought the claim would follow from my observation about two deformation theory (but I don't know how to combine). Of course I would appreciate any good reference. – Zheng Nov 30 at 17:56
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In general it is quite difficult to "combine" deformation theories. There is an article by Jack Hall about the problem of combining deformation theories in general. On the other hand, the late sections of Behrend-Fantechi together with Behrend's paper on Gromov-Witten invariants do this for your moduli problem. – Jason Starr Nov 30 at 21:19
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@jason: do you think you could expand your comments into a full answer? I personally would very much appreciate a detailed explanation of how to compute the cotangent complex of such a space. (actually, now that I think of it, there should be a formula for the cotangent complex of a mapping stack, is there not?) – Jacob Bell Nov 30 at 21:35
Note that the variety must be smooth above. Otherwise there are additional obstructions to deforming the map. – Jonathan Wise Dec 2 at 23:31
## 2 Answers
The classical reference is Illusie's PhD thesis "complexe cotangent et deformations".There the result without the marked points is proven in a very, very general context using the cotangent complex; then one needs to know that for a smoooth variety, or for a nodal curve (or, again, in many more cases) the cotangent complex is isomorphic in the derived category to the cotangent sheaf.
Fixing the case of the marked points can easily be done by hand (they're smooth points after all): I can expand on this step if it helps. Or, one can go modern and use logarithmic geometry.
If you want to get a feeling for why such a result is true, a very short reference is a paper by Ziv Ran on deformations of morphisms: Algebraic Curves and Projective Geometry Lecture Notes in Mathematics Volume 1389, 1989, pp 246-253 Deformations of maps
Another approach would be to sit down and prove it yourself using as only references a few classical facts about deformation theory (say, what you learn from the first ten pages of Artin's Tata lectures). Whether this is a useful exercise or a waste of time depends on your personality. Again, details upon request.
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### 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 $L$ be the complex $[ f^\ast \Omega_X \rightarrow \Omega_C(D) ]$, concentrated in degrees $[-1,0]$ on $C$.
One way to understand the obstructions is to understand the deformations for affine curves first. Observe first that there is a canonical equivalence of categories between the category of deformations of a pointed map $f : C \rightarrow X$, with $C$ not-necessarily proper, and the category of extensions
$G(C) = \mathrm{Ext}(L, \mathcal{O}_C)$.
The meaning of the category on the right is the category of extensions $E$ of $\Omega_C(D)$ by $\mathcal{O}_C$ together with a trivialization of the induced extension of $f^\ast \Omega_X$ by $\mathcal{O}_C$. One checks explicitly that the category of extensions of $\Omega_C(D)$ by $\mathcal{O}_C$ is canonically equivalent to the category of deformations of the pointed curve $C$ and that splitting the induced extension of $f^\ast \Omega_X$ is the same as extending the map $f : C \rightarrow X$ to a map from the deformed curve to $X$.
Note that $G$ is a commutative group stack (what is sometimes called a "Picard stack"). It is easy to check that $G$ is non-empty when $C$ is an affine curve (use the fact that deformations of nodal curves are unobstructed and the formal criterion of smoothness for $X$), which means that any obstructions to deforming a proper curve come from the failure of local deformations to glue. There is a standard way to compute these local obstructions---e.g., by Cech cohomology, noting that mutatis mutandis, Cech cohomology still works for group stacks---and this yield a class in $H^1(C, G)$. Cohomology of a commutative group stack is the same as cohomology of the associated complex, so we get an obstruction in
$H^1(C, L^\vee[1]) = H^2(C, L^\vee) = \mathrm{Ext}^2(L, \mathcal{O}_C)$.
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http://www.appropedia.org/Pulser_pump
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Appropriate technology • Construction • Energy • Food & Agriculture • Green living • Health & Safety • Medical Devices • Projects • Service learning • Solar • Transport • Water
# Pulser pump
From Appropedia
Status
This OSAT has been designed but not yet tested - use at own risk.
This OSAT has been modeled.
This OSAT has been prototyped.
This OSAT has been deployed by: Neil White at least: one pump used seasonally times in: Rathvilly, Co. Carlow, Republic of Ireland since 1989.
You can help Appropedia by contributing to the next step in this OSAT's Category:Status
I give permission to reuse and adapt any and all pictures and animated gifs that I have produced in the past for this project. Brian White, 3rd May 2010
Diagram of a Pulser Pump
The pulser pump is a simple, water powered mechanical device, also known as a bubble pump. Components of this pump have been used for various purposes, including the extraction of oil or in refrigeration cycles. Heat driven bubble pumps are most common, but this particular design of a pulser pump using the turbulent flow in a stream to trap air has yet to become common. The two main benefits of this pump are that it has no mechanical or moving parts, and that it doesn't use any chemicals, only the water from a stream. Once installed near a stream, the pump can lift water using only the energy from the stream.
## Background Information
### Overview
The pulser pump is a combination of a trompe and an air lift pump. Installed near a stream, the pulser pump can pump water to a height above the level of the stream. This allows streams in difficult-to-reach locations to be easily accessed, or water from a stream to be piped to a different location, for irrigation or drinking water purposes.
The pulser pump simply uses the trompe part to power the airlift part. A video explaining the pump can be seen here.
### Benefits
As mentioned in the Introduction section, the main benefits of the pump are twofold. First, this design of pulser pump has no chemical components, as are common in heat-driven bubble pumps, which operate under similar principles (see Heat-Driven Bubble Pumps below). This allows the pump to be used for a large variety of tasks that require uncontaminated water to be pumped, such as irrigation and to pump drinking water. In addition, not requiring any chemicals, and using the stream water available as the pumping fluid greatly reduces the cost of the pump.
Secondly, the pulser pump has no moving components. Once installed, this pump uses the turbulent flow of the stream to trap air, and gravity to compress it (see Principle of Operation section below) to pump a portion of the water to a height above that of the stream. No mechanical components, which are typically more costly and more difficult to install, are required.
In addition to these benefits, it has also been claimed that pulser pumps have a positive effect on the quality of water by increasing the oxygen content. [1] The basic idea is that by mixing air and water together in the intake pump, the increased surface area between them allows more oxygen to be transmitted to the water than is typical in a stream. Further research should be done to confirm this idea.
### History
Trompes were used before hydroelectric turbines to pump air into mines, to provide air for the pneumatic machines that made some of the first alpine tunnels and to provide air to drive motors that lit the wealthy parts of Paris in the late 19th century. Air lift pumps are still widely used by water utilities to pump water from very deep wells. They use compressors to push air down into the wells and the air erupts through a second wider pipe carrying water with it.
## Principle of Operation
Pulser Pump in Action
Split process Pulser Pump has several advantages
Airlift also works in an inclined pipe. Wind creates waves! even in a pipe! I have used this to lift water at least 10 meters from the pulser pump. (I could not make a vertical tower 10 meters high beside the stream) but with waves in a pipe, there is no need to
A pulser pump (aka bubble pump) uses a hydraulic head of water to compress air, which displaces water, pushing water “pulses” to a higher height than before. This works on the same principles as a trompe and an airlift pump.
## Construction of a Model
Based on the success of other similar pilot projects,[2] a proof of concept model was constructed. The materials listed here are for the construction of a model, which can be used for small scale applications, such as those described in the Existing Pumps section below, or for further testing. The plastic tubing described here are flexible plastic tubing, which is useful for testing, as it can be bent to the correct height and reused for different tests, however solid PVC piping could also be used, and would be more practical to implement in a non-test situation. (See the Testing section.)
### Materials
These materials are intrinsic components of this model pulser pump. Aside from scissors to cut the tubing, no additional tools were used.
Fig 1: 2 pieces of 3/4”Inner Diameter Clear Plastic Tubing
Fig 1a: 1 piece of 3/8”Inner Diameter Clear Plastic Tubing
Fig 1b: 2 @ 3/4”plastic tube connectors that fit inside the plastic tubing on one end and have threads on the other end
Fig 1c: 1 @ 3/8”compression fitting that fits around the tube on one end, compressing it into place, and has threads on the other end
Fig 1d: 1 @ 1 1/2” three way connector with two friction fit openings across from one another and one above
Fig 1e: 2 @ 1 1/2”to 3/4”connectors, threaded inside one end and friction fit on one end to connect to the three way connector and the plastic tube connector
Fig 1f: 1 @ 1 1/2”to 3/8”connector, threaded inside one end and friction fit on one end to connect to the three way connector and the plastic tube connector
In addition to the above items, these additional materials were required to setup this model and test it:
Fig 1: 1 hose to provide water in controlled manner
Fig 1a: supports and connection mechanisms to hold up tubing
Fig 1b: buckets and sink for draining excess water
### Costs
The cost of this prototype were somewhat costly, however if compared to the costs of most other means of pumping water, it is very low. Additionally, if this design is constructed on a larger scale, it is likely that it could be built for a much lower cost. The costs of this prototype are approximated below:
Item Price
Plastic Tubing \$40
Tubing Connectors (all) \$10
Three Way Connector \$10
Total \$60
### Building and Setting Up the Model
The process is very straight-forward to build this model, as it has very few components. The tricky part comes with setting up the tubes so they stay as vertical as possible.
A video of the model I made, when working, is shown below. Note the bubbles in the tubes that clearly demonstrate that the inlet tube is in the Bubble Flow Regime, and the pumping tube is in Slug Flow.
The process used to build this model is described in detail below.
Image Step
Cut the Tubes First, the tubes must be cut to the desired length. In this model, the inlet 3/4” tube was cut to be approximately 2.1m long. The outlet tube was also cut to be 2.1m long, so it could be varied to perform different tests The pumping tube was kept long in order to be able to vary the height in the tests described below.
Connect the Separation Container In this model, the three way connector is used as the separation container for the pulser pump. The 1 1/2”to 3/4”connectors were fit snugly into the directly opposing sides of the three way connector. Epoxy can be applied to ensure a tightly sealed fit. The 1 1/2”to 3/8”connector was fit into the three way connector hole that is 90 degrees from either of the other two holes.
Connect the Tubing Connectors First, screw in the compression fitting for the 3/8” connection to the 1 1/2”to 3/8”connector Then, screw in the plastic tube fittings for the 3/4” connection to the 1 1/2”to 3/4”connectors By connecting these before connecting the tubing, the tubing won’t become tangled.
Connect the Tubing The 3/4” tubing should fit tightly over the 3/4” connector. It can be further held in place by tightening a metal clamp around it. The 3/8” compression has a component that goes around the outside of the tubing, a small piece fits inside the tubing to hold it open, then the piece that is around the outside of the tubing can be screwed into the compression fitting already connected to the 1 1/2”to 3/8”connector, holding the tube in place.
Assembly Now all the main components of the pump are connected. The next stage is to set up the pump. First, determine a method of holding the tubes in place. For this, a wooden backboard can be used and the tubes can be nailed to it. In order to have an easily adjustable device, in this experiment the tubes were attached with tape to the support system and wall. Make sure the tubes are vertical, and that the heights of the tubes are as desired. A hose was connected to the top of the inlet tube by inserting both the 3/4” diameter tube and the hose nozzle into a short 1” diameter scrap tube. The hose was used to simulate the flow of the stream. The outlet tube was set to drain into the sink.
### Testing
After setting up the model, a few preliminary tests were run to show that the pump can indeed produce useful work. Much more detailed testing is required before this design of a pulser pump will gain more widespread acceptance and use.
In this test, two variables were changed, the hydraulic head on the pump, and the height of the pumping tube.
It is expected that as the hydraulic head increases, the flow rate will also increase. This effect was demonstrated by the pump, and the results are shown below.
As the height of the pumping tube increases, more energy is required for the pulses to reach the top of the tube. The corresponding decrease in flow rate expected can be seen below.
When the effects of both these variables are combined, the result is a plot showing how the flow rate depends on both hydraulic head and the height of the pumping tube. This plot can be used to demonstrate the relationship between both these quantities and flow rate. Even at large hydraulic heads, the flow rate could be small if the height of the pumping tube is large. Additionally, even when the height of the pumping tube is small, a small head will reduce the amount of water pumped through the pumping tube. This is shown below.
At peak performance (large head and short pumping tube), this model pumped at a rate reaching nearly 100mL/s, or 1L every 10 seconds! Despite multiple tests at each value, the reproducibility of these results is still questionable. Regardless of the exact rate of flow, this experiment shows that there is a huge potential for the pulser pump to be widely used to pump water. Further testing should be done to better evaluate the exact relationship between flow, head and the height of the pumping tube.
## Scientific Model
Despite the fact that pulser pumps, or at least various similar designs of this sort of pump, have existed for a fairly long time, there is not a good explanation or model that describes them. Similar problems, such as for Heat-Driven Bubble Pumps, often are closed systems, which don't require an outlet pipe. In this section, some of the scientific principles behind how this pump functions are presented, and two different models are developed. The first model is the simple Manometer Model, and the second is the more complicated Pressure Model.
### Two Phase Flow
An important concept for this design is two phase flow, which is when there is a liquid and a gas separated by a meniscus. There are at least seven different regimes of two phase flow, [3] several of which are exhibited during the operation of the pulser pump.
#### Slug Flow
The lifting action of the pump up the pumping tube occurs mostly in the slug flow regime. In slug flow, the liquid and gas separate into different layers occupying nearly the entire cross section of the tube, as shown below.
For flow in the slug regime, the diameter of the tube allowed depends on the speed of the flow and its viscosity. The velocity in the pump tube is very difficult to describe, even with several simplifying assumptions. For example, if the tube is not vertical, the bubbles are no longer symmetric resulting in changes in the speed of the bubbles. Several dimensionless terms including the Froude number, the Eotvos number and the Reynolds number are required to fully describe the flow.[4] Although there is still not clear agreement on the most appropriate model, several have been proposed.[5]
The full determination of the fluid properties during slug flow and derivation of a model including the full effects of this flow regime are beyond the scope of this model. Instead, a much simpler model is considered, assuming completely vertical tubes and steady flow, etc. as discussed further below.
### Manometer Model
Equal Height of Straws
With multiple tubes in the same liquid, the maximum height of the liquid in each tube is given by conservation of mass. Using the same principle as a manometer, the external pressure, density, and diameter of the tube determine the height of the liquid. This means that the density, ρ, times the cross sectional area of the tube, A, times the height of the fluid, h, is the same for each tube when they exit to the same pressure, as shown in the diagram to the right. This means that if a closed container is full of water with two of the same straws at the top that are open to the air, the liquid in the straws will rise to the same height, i.e. there will not be more liquid in one straw than the other.
That is to say: $\ (\rho Ah)_1 =(\rho Ah)_2$
This makes sense if all the variables are constant. In a pulser pump, the problem is more complicated. There is an inlet tube and an outlet tube, with the same cross section, but different heights, and then there is the pumping tube with a smaller area and higher height. The inlet and outlet tubes are almost completely filled with water, and so the density can be approximated as that of water, however, at a given time, the pumping tube has a large portion of the tube full of air, not water. The above equation then becomes:
$\ \rho_{water} (Ah)_{inlet} = \rho_{water} (Ah)_{outlet} + [\rho_{water} %_{water} + \rho_{air} %_{air}](Ah)_{pump}$
Or, if the inlet and outlet pipes have the same cross sectional area:
$\ \rho_{water} A (h_{inlet}-h_{outlet}) = [\rho_{water} %_{water} + \rho_{air} %_{air}](Ah)_{pump}$
Where hinlet − houtlet is the hydraulic head. The height that the water can be pumped to can then be determined by solving for hpump
$h_{pump} =\frac{ \rho_{water} A (h_{inlet}-h_{outlet})}{[\rho_{water} %_{water} + \rho_{air} %_{air}]A_{pump}}$
The main problem with this model is that it ignores the velocity of the fluid as it travels through the pump. This is a non-negligible quantity, as if the motion of the fluid was negligible, the air in the outlet tube would separate from the water, and the pump would lose its ability to move water. This manometer model does illustrate the basic principle of the pump, however, and disproves the most common criticism for the pulser pump, which is that it is physically impossible for the smaller pumping tube to pump water above the initial height of the stream. The above argument shows that this is only true if the fluid in the pumping tube is stationary, or if the amount of air present is negligible.
### Pressure Model
#### Theory
Important Points Used in Pressure Model
To account for the velocity, the flow needs to be evaluated in greater detail, using conservation of energy and Bernoulli’s Equation. This method models the change in pressure of the flow at the different points.
At location 4, the pressure from location 1 is given by:
$\ P_4 = P_1 + \rho_4 g h_1 - \frac{\rho_4 (v_4^2 - v_1^2)}{2}$
Similarly, at 5 from 2:
$\ P_5 = P_2 + \rho_5 g h_2 - \frac{\rho_5 (v_5^2 - v_2^2)}{2}$
And at 6 from 3:
$\ P_6 = P_3 + \rho_6 g h_3 - \frac{\rho_6 (v_6^2 - v_3^2)}{2}$
The velocity at 3 is assumed to be zero, in order to determine the maximum height that the water can be pumped to. This will provide an upper limit on the height the water can reach. Also, the pressures at 1, 2, and 3 can be assumed to be approximately one atmosphere, as ‘3’ exits to the atmosphere and ‘1’ and ‘2’ only have a very small hydrostatic pressure, associated with the depth of the stream. Performing a control volume analysis on the bottom container between 4, 5, and 6, conservation of mass dictates:
$\ \rho_4 A_4 v_4 = \rho_5 A_5 v_5 + \rho_6 A_6 v_6$
The areas of all the tubes are known from the geometry of the experiment. The outlet tube has minimal air content, so the density can be assumed to equal that of water.
Building on the work of one source, [6] which analysed a heat-driven bubble pump in a closed system, the change in pressure from 4 and 5 to 6 can be described as follows:
$\ P_6 = P_4 - \rho_6 v_4 (v_6-v_4)-P_5 + \rho_6 v_5 (v_6-v_5)$
Up to this point in the derivation, all the assumptions have been for a fairly general case. The following assumptions in the model provide a more specific, simplified model. The first assumption is that the velocity between 4 and 5 is approximately constant. As only a small portion of the fluid is diverted to exit out the pumping tube, it is assumed the majority of the fluid maintains its momentum while continuing though the outlet tube.
Secondly, the gas content in the slug regime within the pumping tube is assumed to be 70%, which is an average value for slug flow. Also, the gas in the inlet tube is assumed to be in the bubble flow regime, where the gas content is on average 30%.[7] These mean that:
$\ \rho_6 = 0.7 \rho_{air} + 0.3 \rho_{water}$
$\ \rho_4 = 0.3 \rho_{air} + 0.7 \rho_{water}$
#### Testing the Theory
Diagram of a Pulser Pump
With these seven equations, and accompanying assumptions, the Pressure Model still has one more unknown than equation, which means that a reasonable height must be determined through iteration. The inputs are the geometric parameters of the system, and the pressure and velocity at ‘4’ where the highest pressure is expected to occur, should be output. If these values are reasonable, the system can be set up to function; if not, then another iteration should occur.
A first pass at modeling this system was done using EES software. The above equations were defined, and some initial conditions were set. As can be seen in the image, the height and diameters of the tubes and the velocity of the stream at 1 and 2 must be entered, in addition to the assumptions listed above. The program then outputs the velocity throughout the tube, and the pressures. In order to better assess the validity of this model, this EES program or a similar one can be used.
## Existing Pulser Pumps
Pulser pumps lack credibility because there has been no peer review, although they are starting to be researched further (see the External Links below). Several models have been built however, and videos are available online to show how they work, and them working. This design of pulser pump is not patented, and the designs are in the public domain. [8]
#### Working Pulser Pump
An example of a 20 year old working pulser pump is available here, if it does not load below. This pump is powered by a little stream with 300 litres of water falling 0.5 meters producing the power. Figures from that pump show that an apparent water speed down the tromp part of between 0.32 meters per second and 0.68 meters per second is fast enough to send the air bubbles down the pipe. The apparent air speed up the airlift section seems to work best between 0.7 ms-1 and and 1.5 meters per second. This was when using 12 mm and 19 mm pipes and pumping straight up. A lower apparent air speed worked best when pumping up an incline. (Apparent speed is the speed of the water or air through the pipes assuming that only one fluid was in the pipe) It is a good guide if you make your own. Brian
```Pulser pumps can work with much larger flows and heads than that according to Gaiatechnician.
```
Because of their extreme simplicity, they can be of great value to waterside communities. The tiny pulser pump in the video can pump about 5 tonnes of water per day to a storage container. [9]
A second example (shown here) uses the pulser pump to provide water for animals. It has a supply flow rate of approximately 30 litres/min through a 40mm waste pipe. It can lift 30 mL/min to 3m or 1 L/min to 1m. [10]
Mike Donevan of practical farm ideas has allowed me to use pictures and text from when the pump was in his magazine in exchange for a link. http://www.farmideas.co.uk/ I think they will be valuable to anyone who wants ball park figures to guess at good pipe sizes for their projects
Brian white
### Heat-Driven Bubble Pumps
Diagram of a Heat Driven Bubble Pump
Heat-Driven Bubble pumps are the most common type of pulser pump found. They use a similar principle of operation as this pulser pump design, but in a closed system. In general, a refrigerant with a boiling point below that of water is mixed with the fluid. After the mixture is compressed, it is heated causing bubbles to form from the refrigerant in the working fluid. The bubbles of refrigerant then push the water up the pump tube, as in the pulser pump. The mixture then enters a separation chamber, where the liquid is sent to an absorber and the refrigerant to a condenser. [11]
## Conclusions
This particular design for a pulser pump is extremely simple to build, and has the potential to make a large impact in how water is pumped. It does not use chemicals to pump the water, so the water let back into the stream is not contaminated. On the contrary, claims have been made that indicate the water returned to the stream contains more oxygen, providing a better environment for underwater creatures. The water that is not returned to the stream can be used to irrigate land or provide drinking water. The added elevation that the pump provides allows water to be transported further than the stream alone would be able to move it.
The pump itself is made of very few materials, only simple tubes and connections, and after optimum design is found, it could be made inexpensively. Almost no maintenance is required once the pump is set up, so aside from the initial installation and equipment costs, the pump can provide water cheaply and easily to those nearby.
In this analysis, both an experimental model and a theoretical model were developed. The experimental model was based off a previous design, and acted as a proof of concept. The model clearly indicated the trends expected, showing highest flow rates for a large hydraulic head and a short pumping tube. Two theoretical models were proposed, the first based on the principle of conservation of mass, like a manometer. This model provided a rough estimate for the height of the second tube, but the validity is very limited because it assumes a negligible velocity, however it does serve to disprove the idea that this design for a pulser pump is physically impossible. The second theoretical model used both conservation of mass and conservation of energy to evaluate the velocities and pressures at each stage of the pump. This second model requires the user to iterate on the values produced in order to determine a reasonable geometry under specific conditions. Using Engineering Equation Solver, a template was formed that would allow the user to iterate on the assumptions made in order to determine the optimum geometry.
## Recommendations
Both the experimental and the theoretical model would benefit greatly from a peer review and further research. The experimental model should be tested using a more rigid testing scheme, with many more tests at each height and head. Other variables such as the size of the container, the diameter of the tubes and the flow velocity should also be tested to determine their effects on the pulser pump, and a wider range of values should be used. The main setback encountered during this experiment was leakage though the original container and hose. By using the methods described above, these problems can easily be overcome and more time can be spent performing more detailed tests.
The theoretical model proposed could be further expanded and tested using software such as the EES program described above. In addition, there are several factors that were not accounted for that must be included in the model for it to accurately predict the flow rate through the pump. These include:
• Analysis of two phase slug flow in pumping tube
• Determination of air content and flow regime in inlet and outlet tubes
• Friction/Viscous losses
• Turbulent Flow
• Pressure gradient across separation container
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# Tagged Questions
Questions on fractals, irregular-looking mathematical objects that display the property of self-similarity.
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### fractal structure of the sum of squares function
The sum of squares function came up at a job interview, corrected for signs and symmetry. $d_2(n)=\#\{(x,y): x^2 + y^2 = n\}$ However, want $(x,y)\sim (\pm x, \pm y) \sim (y,x)$. The first ...
1answer
106 views
### Buddhabrot Sewing machine [closed]
The Buddhabrot fractal traces the orbits of the points outside the Mandelbrot set. What design considerations need to be taken into account to create a computerised sewing machine that traces out ...
2answers
117 views
### Why should Gaussian noise have fractal dimension of 1.5?
In a paper I'm trying to understand, the following time series is generated as "simulated data": $$Y(i)=\sum_{j=1}^{1000+i}Z(j) \:\:\: ; \:\:\: (i=1,2,...,N)$$ where $Z(j)$ is a Gaussian noise with ...
2answers
256 views
### Is a 3D Mandelbrot-esque fractal analogue possible?
I understand that (unlike complex numbers) there's no consistent 3 dimensional number system (even 4D loses some nice properties). Regardless, I'm wondering if there might be a 'trick' to create a 3D ...
1answer
176 views
### Is an Inverse Menger Sponge a fractal?
Is the Inverse of a Menger Sponge a fractal? I know a Menger sponge is fractal in nature, and it seems to me that the inverted form of it would be fractal as well, but I don't know.
1answer
112 views
### Integration on fractals [closed]
Who can explain the proof of the formula (2.12) given here: J. Phys. A: Math. Gen. 20 (1987) 3861-3875. Printed in the UK ...
1answer
87 views
### Classification of points in the Mandelbrot set
I am trying to understand the classification of points in the Mandelbrot set. There are an infinite number of baby Mandelbrots, each associated with a defined set of landing rays. There are the pre ...
1answer
2k views
### Odd and even numbers in Pascal's triangle-Sierpinski's triangle
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed. I recently learned that when the Pascal's triangle is reduced ...
1answer
74 views
### Unexplainable noise graph function.
I'm sorry for the ambiguity here but I've recently discovered a function which plots, what seems to be either a fractal or simply noise in a selected area. Can anyone explain this function: ...
1answer
44 views
### Mandelbrot precision target the center of a pixel?
I read this question and I don't understand the answer: http://stackoverflow.com/questions/8381675/how-to-perform-simple-zoom-into-mandelbrot-set?rq=1. Especially how can I aim for the center of the ...
0answers
21 views
### Density of a multifractal distribution
I am trying to grasp the concept of density of a multifractal. So I start from the simple case of a line. Let's assume I have a uniform distribution of points on a line and I center a cubic box in the ...
1answer
151 views
### Every basin of attraction contains a critical point?
Years and years ago, back when I first became interested in fractals [but didn't know much about anything], I vaguely remember coming across an interesting theorem. The gist of it was that "every ...
2answers
182 views
### Quadratic Julia sets and periodic cycles
Consider the function $f_c(z) = z^2 + c$. Applying this function repeatedly, we get the familiar quadratic Julia sets that fractal enthusiasts burn compute cycles plotting. Infinity is always one ...
1answer
72 views
### Mandelbrot bulb's countable?
Are the Mandelbrot set's bulb's countably infinite? My daughter asked me this question, after I pointed out that some Julia sets are a Cantor dust. For a point not in the Mandelbrot set, the ...
0answers
56 views
### fractal string problem
Let the zeta function of a fractal string be $Z(s)= \sum_{n}g(n)(l_{n})^{s}$ Here $g(n)$4 is the degeneracy of the strings and $l_{n}$ are the lengths of the string. In order to evaluate the ...
2answers
306 views
### Undiscovered fractal sets?
I'm interested in the topic of fractals, such as those created by the borders of the Mandelbrot and Julia sets. My question is if there are other, not yet discovered fractal sets, which one could ...
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http://nrich.maths.org/2666/note
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### Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Great Squares
Investigate how this pattern of squares continues. You could measure lengths, areas and angles.
### Square Areas
Can you work out the area of the inner square and give an explanation of how you did it?
# Isosceles Triangles
### Why do this problem?
Paul Andrews, a respected mathematics educator based at Cambridge University, explains why he likes this problem :
"I first saw it used as a homework question more than ten years ago with a year seven class (11 year old students) in Budapest and have used it with every group I have taught since - initial teacher training, mathematics education research students, in-service; everyone gets it. I love it because it forces an acknowledgement of so many different topics in mathematics and is sufficiently challenging to keep almost any group meaningfully occupied within a framework of Key Stage Three mathematics content. This, for me, is the key to a good problem; Key Stage Three content alongside non-standard and unexpected outcomes. For example, notwithstanding the obvious problem solving skills necessary for managing such a non-standard problem, I think it requires an understanding of coordinates, isosceles triangles and the area of a triangle. It requires an awareness of the different factors of 18 and which are likely to yield productive solutions. It requires, also, an understanding not only of basic transformations like reflection and rotation but also an awareness of their symmetries. Moreover, the solution, which is numerically quite small, is attainable without being trivial. In short, I love this problem because of the wealth of basic ideas it encapsulates and the sheer joy it brings to problem solvers, of whatever age, when they see why the answer has to be as it is. It is truly the best problem ever and can provoke some interesting extensions."
### Possible approach
The richness of this task might best be exposed by working on the problem in small groups. The groups should be encouraged to keep asking the questions: are our solutions valid? have we found all of the solutions?
Once groups feel that they have finished, they could try to explain clearly to the class why they have found all of the solutions. Those who missed out solutions could be encouraged to think about why these solutions were overlooked.
### Key questions
• How do you know that your triangles have the correct area?
• Are there any more like that?
• Can you explain why you are certain that there are no more solutions?
### Possible extension
Repeat for isosceles triangles of areas $50$ and $15$. Which starting areas lead to more/fewer solutions? Is the number of solutions predictable?
and/or
The problem doesn't insist that one edge is horizontal/vertical - what new triangles can be found, with all sides sloping? Can you prove that you have got them all?
### Possible support
The problem can be used as a context for practising: drawing isosceles triangles, using co-ordinates accurately, calculating areas, communicating results, working with others, etc.
When the students have worked out the basic possibilities for the isosceles triangles, they could cut them out to help to search for congruent solutions.
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.
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http://math.stackexchange.com/questions/36609/what-is-the-nature-of-this-sequence
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# what is the nature of this sequence?
$3, 4, 10, 33, 136$
what will be next most appropriate value? I tried finding any relation in the sequence but i couldn't.
$a.276$
$b.539$
$c.612$
$d.685$
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2
Is this just a puzzle or do you have some context you can provide? – Aryabhata May 3 '11 at 6:15
I have got options with me. $a.276$ $b.539$ $c.612$ $d.685$ – amul28 May 3 '11 at 6:18
3
One option is $685$. $3, 3*1+1 = 4, 4*2 + 2 = 10, 10*3+3 = 33, 33*4+4 = 136, 136*5 + 5 = 685$... But such questions are nonsense as Qwirk's answer shows. I am voting to close a NARQ. – Aryabhata May 3 '11 at 6:21
2
@amul: I really don't know. There is no systematic method. Really, such questions are nonsense and solving this one likely won't help you solve other nonsense questions like this. It is unfortunate, but you have to guess what the idiot who wrote the question on the entrance test was thinking of... – Aryabhata May 3 '11 at 6:41
1
I agree that questions like this can be frustrating, especially when multiple answers might fit, but I don't agree that they're completely worthless. I think the ability to look at arbitrary data, see patterns, find possible relationships, and then identify the most likely relationship is a critical skill for a mathematician to have. The question isn't "what was the exam writer thinking when he wrote this?", it's "what's the simplest relationship you can find between these numbers?" – Kevin May 3 '11 at 15:15
show 10 more comments
## 2 Answers
I agree with all the complaints about this sort of problem, but still.... There are some techniques which work from time to time.
Try taking differences: $4-3=1$, $10-4=6$, $33-10=23$, $136-33=103$, so now we have to explain the sequence $1,6,23,103,\dots$. Hmm, that doesn't seem very helpful.
OK, subtraction didn't work, try division: $4\div3=1r1$, $10\div4=2r2$, $33\div10=3r3$, $136\div33=4r4$ - hey, that looks much better! (When I write $arb$, I mean quotient $a$, remainder $b$.)
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seems very clear for me. – amul28 May 3 '11 at 7:09
Wow! I never knew division could be applied. I always used subtraction. +1 for arb notation. Its really helpful. – Shiplu Dec 21 '11 at 7:44
I must say, I have always disliked 'find the next term in the series question'. For any sequence, it is easy to produce any number next (e.g. for a sequence of $n$ terms, pick the $n+1$ number and then fit a polynomial to those $n+1$ terms).
OEIS does not give anything useful for your sequence - how has it arisen?
Edit: For this question, as Moron has shown, the likely answer is 685, based on the sequence $3,3\times 1 + 1 = 4, 4 \times 2 + 2 = 10, 10 \times 3 + 3 = 33, 33 \times 4 + 4 = 136,$$136 \times 5 + 5 = 685$ . But in general knowing how to find the pattern in this sequence, will not help (much) in finding patterns in similar sequences.
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http://physics.stackexchange.com/questions/52081/how-do-i-correctly-express-the-work-required-to-put-an-object-with-positive-buoy
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# How do I correctly express the work required to put an object with positive buoyancy down into a certain depth (of water)?
It seems to me that I can express the work required to put an object under water in the same way that I express lifting an object up against gravity. I. e., in both cases I increase the potential energy of the object by moving it vertically against the direction of the force. Conversely, when the object is allowed to surface, it would be comparable to an object falling down (just with significantly more drag in the water, I guess).
Is that comparison viable?
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## 3 Answers
The comparison is viable, here's why:
Let's choose the positive $x$-direction to point upward, perpendicular to the water's surface. By Archimedes' principle, the magnitude of the buoyant force on an object of volume $V$ equals the weight of the displaced water;
$F_B = \rho_w V g$
where $\rho_w$ here denotes the density of water. The buoyant force points in the positive $x$-direction.
In addition to the buoyant force, an object submerged in water will experience a gravitational force
$F_g = \rho V g$
pointing in the negative $x$-direction, where $\rho$ is the density of the object. It follows that the net vertical force on the water is
$F_x = F_B - F_g = (\rho_w-\rho)V g$
Let's assume that the object is rigid, so that its volume changes negligibly with depth, and let's make the approximation that the density of water varies negligibly as a function of depth as well, then the work done in moving such an object to a depth $d$ below the water's surface along the $x$-axis will simply be
$W = F\Delta x = (\rho_w-\rho)V g(0-d) = \boxed{(\rho-\rho_w)V g d}$
Notice how similar this looks to $mgd$, the change in potential energy of moving an object upwards under the influence of gravity by an amount $d$. In fact, we could think of the water as effectively decreasing the mass of the object so that it has an effective mass $m_\mathrm{eff} = (\rho- \rho_w)V$, and then the analogy becomes clear
An important caveat to all of this is that water is viscous, so they will be an additional drag force you have to contend with that will change the answer in the event that you submerge the object quickly. However, by moving the object sufficiently slowly, you can make this contribution as small as you'd like. You can also relax the constant water density and constant object volume assumptions if you know how these things change with depth. If you'd like more details on this, then let me know!
Cheers!
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Yes, that comparison is perfectly reasonable. If you know the buoyancy of the object in water, then you know the force you have to apply to push it down into the water. The instantaneous work is always the force vector dotted into the displacement vector, and the total work is the instantaneous work integrated over the displacement, so you can easily calculate the work you have to do.
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@joshphysics, thank for pointing that out; I have made the correction. – Colin McFaul Jan 24 at 16:59
No problem dude! – joshphysics Jan 24 at 18:25
There's not much fundamental difference between an object being in water or in air. If it's heavier, it will want to move down, if it's lighter, it will want to move up.
So there's no need to compare the movement under water to "falling upward". Gravity works in the same direction, regardless.
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– Manishearth♦ Jan 24 at 16:04
Oh, so it's energy. OK. – Mr Lister Jan 24 at 17:48
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http://unapologetic.wordpress.com/2011/07/13/the-lie-derivative-on-forms/?like=1&source=post_flair&_wpnonce=18b5244a97
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The Unapologetic Mathematician
The Lie Derivative on Forms
We’ve defined the Lie derivative $L_XY$ of one vector field $Y$ by another, $X$. This worked by using the flow of $X$ to compare nearby points, and used the derivative of the flow to translate vectors.
Well now we know how to translate $k$-forms by pulling back, and thus we can define another Lie derivative:
$\displaystyle L_X\omega=\lim\limits_{t\to0}\frac{1}{t}\left((\Phi_t)^*(\omega)-\omega\right)$
What happens if $\omega$ is a $0$-form — a function $f$? We check
$\displaystyle\begin{aligned}\left[L_Xf\right](p)&=\lim\limits_{t\to0}\frac{1}{t}\left(\left[(\Phi_t)^*(f)\right](p)-f(p)\right)\\&=\lim\limits_{t\to0}\frac{1}{t}\left(f\left(\Phi_t(p)\right)-f(p)\right)\\&=X_p(f)=Xf(p)\end{aligned}$
That is, the Lie derivative by $X$ acts on $\Omega^0(M)$ exactly the same as the vector field $X$ does itself.
I also say that the Lie derivative by $X$ is a degree-zero derivation of the algebra $\Omega(M)$. That is, it’s a real-linear transformation, and it satisfies the Leibniz rule:
$\displaystyle L_X(\alpha\wedge\beta)=\left(L_X\alpha\right)\wedge\beta+\alpha\wedge\left(L_X\beta\right)$
for any $k$-form $\alpha$ and $l$-form $\beta$. Linearity is straightforward, and given linearity the Leibniz rule follows if we can show
$\displaystyle L_X(\alpha_1\wedge\dots\wedge\alpha_k)=\sum\limits_{i=1}^k\alpha_1\wedge\dots\wedge\left(L_X\alpha_i\right)\wedge\dots\wedge\alpha_k$
for $1$-forms $\alpha_i$. Indeed, we can write $\alpha$ and $\beta$ as linear combinations of such $k$- and $l$-fold wedges, and then the Leibniz rule is obvious.
So, let us calculate:
$\displaystyle\begin{aligned}L_X\left(\alpha_1\wedge\dots\wedge\alpha_k\right)=&\lim\limits_{t\to0}\frac{1}{t}\left((\Phi_t)^*\left(\alpha_1\wedge\dots\wedge\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\left((\Phi_t)^*\alpha_k\right)\right)\\&+\lim\limits_{t\to0}\frac{1}{t}\left(\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\left(\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_{k-1}\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\wedge\left((\Phi_t)^*\alpha_k\right)\right)\\&+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_k\right)-\alpha_k\right)\\=&\left(\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_{k-1}\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\right)\wedge\alpha_k\\&+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge L_X\alpha_k\\&=L_X\left(\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\wedge\alpha_k+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge L_X\alpha_k\end{aligned}$
So we see how we can peel off one of the $1$-forms. A simple induction gives us the general case.
Like this:
Posted by John Armstrong | Differential Topology, Topology
5 Comments »
1. [...] Lie derivative looks sort of familiar as a derivative, but we have another sort of derivative on the algebra of [...]
Pingback by | July 15, 2011 | Reply
2. [...] . But since it takes -forms and sends them to -forms, it has degree one instead of zero like the Lie derivative. As a consequence, the Leibniz rule looks a little different. If is a -form and is an -form, I [...]
Pingback by | July 16, 2011 | Reply
3. [...] It turns out that there is a fantastic relationship between the interior product, the exterior derivative, and the Lie derivative. [...]
Pingback by | July 26, 2011 | Reply
4. [...] Cartan’s formula in hand we can show that the Lie derivative is a chain map . That is, it commutes with the exterior derivative. And indeed, it’s easy to [...]
Pingback by | July 28, 2011 | Reply
5. [...] Armstrong: The algebra of differential forms, Pulling back forms, The Lie derivative on forms, The exterior derivative is a derivative, The exterior derivative is nilpotent, De Rham Cohomology, [...]
Pingback by | July 30, 2011 | Reply
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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.
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http://mathoverflow.net/questions/23706/reductive-lie-algebra-of-a-lie-group
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## Reductive Lie algebra of a Lie group
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In the answer of my question:
http://mathoverflow.net/questions/23085/on-the-full-reducibility-of-representations-of-reductive-lie-algebras
James E. Humphreys replied to me saying that:"the notion of "reductive" for a Lie algebra in characteristic 0 has no intrinsic interest, unless you study the Lie algebra of a Lie (or algebraic) group and relate their representations carefully."
Can please someone explain that to me or give to me any reference? thank you!
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I should let the experts answer, but my understanding is that all this means is that there is very little you can say (except structurally) about a reductive Lie algebra in characteristic zero. For example, you cannot infer much about its representation theory simply from the fact that a Lie algebra is reductive. On the other hand, if you know that it the Lie algebra of a Lie group, then at least for those representations which integrate to representations of the group you can say more. – José Figueroa-O'Farrill May 6 2010 at 13:32
## 2 Answers
What Jim means is that one naive definition of reductive Lie algebra
• $\mathfrak{g}$ is reductive if all its finite-dimensional representations are semi-simple.
already has a name: semi-simple.
Another one
• $\mathfrak{g}$ is reductive if all its representations are semi-simple.
is actually trivial; there are no Lie algebras that satisfy it.
Of course, there actually is a pretty good definition that matches better with reductive for groups:
• $\mathfrak{g}$ is reductive if its adjoint representation is semi-simple.
but it's important to keep in mind that the properties above don't follow from that.
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1
To reinforce what Ben says, the standard definition of "reductive" Lie algebra in characteristic 0 just tells you that it is the direct sum of a semisimple and an abelian Lie algebra. The semisimple ones are characterized by having a nondegenerate Killing form (and resulting rich structure), while the abelian ones are just boring vector spaces. Even for algebraic groups, "reductive" has strong representation-theoretic implications only in characteristic 0; but the structure is interesting in any characteristic and over various ground fields. – Jim Humphreys May 6 2010 at 14:33
Thank you. The fact is that I'm dealing with a Lie algebra of a connected algebraic group and I'm trying to understand their connection, also because I'd like to prove that a certainly representation is semisimple. – Michele Torielli May 6 2010 at 14:55
### 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 wouldn't exclude the possibility that some nicely-looking ("intrinsic"?) characterization of reductive Lie algebras exist, say, in homological terms. How relevant such characterization might be to the questions discussed here (connection with Lie groups and representation theory) - that's another matter (probably it will not).
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http://math.stackexchange.com/questions/263691/simple-equation-misunderstanding
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# Simple equation misunderstanding
Im trying to use an equation on this page
http://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics
The angle of lean, $\theta$, can easily be calculated using the laws of circular motion
$$\theta = \arctan\left(\frac{v^2}{gr}\right)$$
where $v$ is the forward speed, $r$ is the radius of the turn and $g$ is the acceleration of gravity. For example, a bike in a 10 m radius steady-state turn at 10 m/s must be at an angle of 45.6°.
If enter the above example values into the equation a get a value of 89.4 not 45.6. What am I doing wrong?
$$89.4 = \arctan\left(\frac{10^2}{9.81 \cdot 10}\right)$$
Cheers
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1
I don't know what you're doing wrong, but I'm getting a different value (in fact, the one you're looking for) for the arctan. How are you calculating it? – akkkk Dec 22 '12 at 12:30
Im entering arctan(10^2/9.81*10)*57.2957795 into google. I get 89.4379464 – user346443 Dec 22 '12 at 12:36
4
Presumably, you computed $\arctan (10^2/9.81\cdot10)$. But this was interpreted as $\arctan((10^2/9.81)\cdot 10)=\arctan(1000/9.81) \approx 89.4^\circ$. Enclose the denominator in parentheses to obtain the correct result: $\arctan(10^2/(9.81\cdot10))\approx 45.6^\circ$. – David Mitra Dec 22 '12 at 12:36
Oh ok, thanks a million and Merry Christmas. – user346443 Dec 22 '12 at 12:38
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http://physics.stackexchange.com/questions/19055/is-there-really-no-meaning-in-potential-energy-and-potential/19059
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# Is there really no meaning in potential energy and potential?
I have been told all my physics life that potential energy between two mass/charge has no meaning and only their difference has meaning. The same goes for electric potential, only the difference matter.
Perhaps I am not understanding it correctly, but before I talk about masses, let's talk about potential energy/potential associated with two charges.
I am not sure where had I seen it, but a long time ago I was presented with a problem like this. Let's say I have a +Q and a -Q. What is the potential energy between them?
The change in potential energy is the negative work done by the conservative force namely (vector sign and dot product got rid of, since the cosine 1)
$\Delta U = -\int_{a}^{b} k(Q)(-Q) \frac{1}{r^2} = -k(Q)(-Q) \left. \frac{-1}{r} \right |_{a}^{b} = -k(Q)(-Q) \left (\frac{1}{b} - \frac{1}{a} \right)$
Now I assume that I brought it from infinitely far, so that 1/a = 0
In that case I am left with $\delta U = \frac{kQQ}{b}$
Here are my questions
1) A long time ago, I saw a formula looking EXACTLY like what I just did there, but it doesn't concern with the change, it's just gives me the potential energy. Now the formula I remember was $U = -\frac{kQq}{r}$ where there is a negative. I thought this formula already takes care of the signs? Or am I wrong?
2) Kinda the same concept. If I tell you some charge (not telling you the sign) has a electric field and you have another charge (which is the test particle, not telling you the sign again even though it is conventional to use + charge ) somewhere in that field. I tell you that the electric potential at that point (not the difference) is K (where K is positive number). What can you conclude, if anything? What if it were negative?
Suppose I tell you suddenly that the charges are the same signs, and I give you a location in which the electric potential is positive (I think it has to be). What does it mean?
EDIT: Let me also just clarify a bit that I was taught that electric potential (not difference, I stress again) is the work that someone does to bring a charge from infinity to some point whereever. In my book however, it's defined as $V = \int_{R}^{\infty} \vec{E} \cdot \vec{ds}$
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Where are you learning electromagnetism from? Your book must explain this... – Chris Gerig Jan 3 '12 at 8:33
If the potential energy never changes, it is useless for dynamics since it does not produce any force that makes a work. And a work is the potential energy difference. – Vladimir Kalitvianski Jan 3 '12 at 9:28
A word that you may (will if you stick with the subject) encounter in this regard is "gauge", which expresses the idea of picking a basis for comparison. – dmckee♦ Jan 5 '12 at 4:07
## 2 Answers
It is indeed correct that only the difference between two potential energies is physically meaningful. An in-depth explanation follows. For the rest of this answer, forget everything you know about potential energy.
I suppose you know that when you have a conservative force $\vec{F}$ acting on an object to move it from an initial point $\vec{x}_i$ to a final point $\vec{x}_f$, the integral $\int_{\vec{x}_i}^{\vec{x}_f}\vec{F}\cdot\mathrm{d}\vec{s}$ depends only on the endpoints $\vec{x}_i$ and $\vec{x}_f$, not on the path. So imagine doing this procedure:
1. Pick some particular starting point $\vec{x}_0$
2. Define a function $U(\vec{x})$ for any point $\vec{x}$ by the equation
$$U(\vec{x}) \equiv -\int_{\vec{x}_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$
This function $U(\vec{x})$ is the definition of the potential energy - relative to $\vec{x}_0$. It's very important to remember that the potential energy function $U$ depends on that starting point $\vec{x}_0$.
Note that the potential energy function necessarily satisfies $U(\vec{x}_0) = 0$. So you can write
$$U(\vec{x}) - U(\vec{x}_0) = -\int_{\vec{x}_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$
Now why would you do that? Well, suppose you choose a different starting point, say $\vec{x}'_0$, and define a different potential energy function
$$U'(\vec{x}) \equiv -\int_{\vec{x}'_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$
(Here I'm using the prime to indicate the different choice of reference point.) Just like the original potential energy function, this one is equal to zero at the starting point, $U'(\vec{x}'_0) = 0$. So you can also write this one as a difference,
$$U'(\vec{x}) - U'(\vec{x}'_0) = -\int_{\vec{x}'_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$
The neat thing about this definition is that even though the potential energy itself depends on the starting point,
$$U(\vec{x}) \neq U'(\vec{x})$$
the difference does not:
$$U(\vec{x}_1) - U(\vec{x}_2) = U'(\vec{x}_1) - U'(\vec{x}_2)$$
Check this yourself by plugging in the integrals. You'll notice that anything depending on the starting point cancels out; it's completely irrelevant.
This is good because the choice of the starting point is not physically meaningful. There's no particular reason to choose one point over another as the starting point, just as if you're on a hilly landscape, there's no particular reason to choose any one level to be zero height. And that's why potential energy itself is not physically meaningful; only the difference is.
Now, there is a convention in (very) common use in physics which says that when possible, unless specified otherwise, the starting point is at infinity. This allows you to get away without saying "difference of potential energy" and explicitly defining a starting point every time. So when you see some formula for potential energy, like
$$U(\vec{r}) = -\frac{k q_1 q_2}{r}$$
unless specified otherwise it is actually a difference in potential energy relative to infinity. That is, you should read it like this:
$$U(\vec{r}) - U(\infty) = -\frac{k q_1 q_2}{r}$$
Note that the function $\frac{k q_1 q_2}{r}$ goes to zero as $r\to\infty$. That's not a coincidence. It was chosen that way to ensure that $U(\infty) = 0$, so that you could insert it the same way I inserted $U(\vec{x}_0)$ in the calculations above. (This is just another way of saying it was chosen to make the $\frac{1}{a}$ term in the integral you did go away, so you don't have to write it.)
There are some situations in which you can't choose the reference point to be at infinity. For example, a point charge with an infinite charged wire has an electrical potential energy of
$$U = -2kq\lambda\ln\frac{r}{r_0}$$
where $r$ is the distance between the point charge and the wire. This potential energy function decreases without bound as you go to infinite distance ($r\to\infty$), it doesn't converge to zero, so you can't use infinity as your starting point. Instead you have to pick some point at a finite distance from the wire to be your starting point. The distance of that point from the wire goes into that formula in place of $r_0$.
By the way, electrical potential (not potential energy) is something a little different: it's just the potential energy per unit charge of the test particle. For a given test particle, it's proportional to electrical potential energy. So everything I've said about applies equally well to electrical potential.
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The potential energy between two charges is $$E_{\rm pot} = \frac{Q_1Q_2}{4\pi\epsilon_0 r}$$ If the charges have the same sign, the energy is positive. If they have the opposite charge, it's negativing (the charges may be bound).
In general, additive shifts $$E \to E+ \Delta E$$ may often be unobservable, and only energy differences are physically meaningful, except that in many situations, there exists a damn good definition what the level $E=0$ means.
In relativistic theories, $E=0$ is the unique choice that preserves the Lorentz symmetry together with the zero momentum; the absolute additive constant is determined from Lorentz symmetry. In the same way, $\rho=0$ gives flat space in general relativity; other energy densities curve the space.
In electromagnetism, there is a natural additive shift coming from the condition $$E_{r=\infty}=0$$ which means that the potential energy vanishes at infinite separation of the charges. This convention was used above. Concerning your last question, $$V = \int_R^\infty \vec E\cdot \vec{ds}$$ is the work obtained by moving a unit charge from $R$ to infinity, up to the sign, so the two things you are describing are the same and there's no contradiction. In particular, $\vec E$ is the electric field $$\vec E = - \nabla \phi$$ where $\phi$ is the electrostatic potential and $-\nabla \phi\cdot \vec{ds}$ is nothing else than $-\Delta \phi$ by the definition of the derivatives (and the gradient) for an infinitesimal $\vec{ds}$. At any rate, there's no contradiction in anything you wrote.
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http://math.stackexchange.com/questions/305011/regular-compact-domains-of-a-riemannian-manifolds
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# Regular compact domains of a Riemannian manifolds
In a Riemannian manifold $M$ a regular compact domain $D$ is a compact subset of $M$ with non empty interior and such that for every $p \in \partial D$ there exists $\left(U,\varphi\right)$ coordinate neighbourhood of $p$ such that $\varphi(U\cap \partial D)\subseteq \partial \mathbb{R}^{n}_{+}$ and $\varphi (U\cap D)\subseteq \mathbb{R}_{+}^{n}$ where $\mathbb{R}^{n}_{+}$ is the set such that $x_n\geq 0$. Now let $R<R'$. My question: is there a regular compact domain $D$ such that $B_R \subset D$ and $\partial D \subset B_{R'}$ (where $B_R$ is the metric ball of $M$ )? I'm thinking that if we can construct a smooth function $\varphi$ such that $\varphi = 1$ on $B_R$, $supp(\varphi)\subset B_{R'}$ and $|\nabla \varphi |\neq 0$ on $interior( supp(\varphi)) \diagdown \overline{B_R}$ then the subset $\Omega=\{p \in M:\varphi(p)\geq \frac{1}{2}\}$ is the regular compact domain looked for. We have to assume completeness of $M$ in my argument. But if we replace the metric balls with two compact subsets $K \subset interior K'$ it should work without completeness.
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## 1 Answer
With metric balls on non-complete manifolds you have an obvious obstruction: if $\overline{B_R}$ is not compact, $B_R$ cannot be contained in a compact domain. The second version of your question is more robust: if $K$ is compact, $\Omega$ is open with compact closure, and $K\subset \Omega$, then we can find a regular compact domain pinched between $K$ and $\Omega$. Indeed, let $(\varphi_\alpha,U_\alpha)$ be a smooth partition of unity on $\Omega$, where $\overline{U_\alpha}\subset \Omega$ for all $\alpha$. Let $\psi=\sum_{U_\alpha\cap K\ne\varnothing }\varphi_\alpha$. Then $\psi$ is smooth, compactly supported in $\Omega$, and is identically $1$ on $K$. By Sard's theorem there exists $\lambda\in (0,1)$ such that the set $D=\{x:\psi(x)\ge \lambda\}$ is a regular compact domain.
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Thanks! I didn't know the Sard's theorem. It's very useful. – user55449 Feb 16 at 8:52
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– user53153 Feb 16 at 19:05
I didn't know this thing (i'm a new user for this site). How can i give a checkmark to an answer? – user55449 Feb 16 at 19:23
I've understood. Thanks – user55449 Feb 16 at 19:31
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http://physics.stackexchange.com/questions/43423/determine-when-a-light-is-going-into-the-object/43427
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# Determine when a light is going into the object [closed]
Please look at this image: (http://mypages.iit.edu/~smart/acadyear/refract.gif)
How can I determine when light is going into the object?
Actually, if there is, what is the simplest way to determine whether light is going into the object or not?
NOTE: I could not find any meaningful title. Sorry.
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@lubosMotl should I use i dot N as a guide ? – gveaf Nov 4 '12 at 12:38
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I can't understand quite correctly of what do you mean by "When light goes?" (it goes if it's switched on) You'd see the object illuminated when light propagates through it. That's a determining way. Do you require Ray tracing or something like that..? Please be a bit clear – Ϛѓăʑɏ βµԂԃϔ Nov 4 '12 at 13:08
Are you asking how to tell which is the entry and which is the exit? If so, there is no way to tell because the Physics involved is time reverisble i.e. the behaviour is the same if the light travels from the exit to the entrance. – John Rennie Nov 4 '12 at 13:22
## closed as too localized by Qmechanic♦, Manishearth♦, Emilio PisantyDec 10 '12 at 13:14
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.
## 1 Answer
Light partly penetrates transparent objects and is partly reflected. Depending on the indexes of refraction of the two media. In your case light enters from above ( first medium) it goes through the glass, (second medium) and is also partially reflected back, and goes out to the air ( third medium).
There exists a critical angle with respect to the normal when there is total internal reflection, i.e. light does not go through the glass (your third medium). It follows Snell's law, where n1 and n2 are the indexes of refraction:
$n_1 sin \theta_i = n_2sin\theta_t$
to find the critical angle we set $\theta_t$= 90 degrees.
$\theta_c = \theta_i = arcsin \frac{n_2}{n_1},$
Have a look at the wikipedia articles on refraction and total reflection.
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So `( i dotProduct N ) < 0 ray go into the glass`, is it true ? – gveaf Nov 4 '12 at 14:11
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I cannot understand your symbols, but as long as the angle to the normal is not the critical angle part of the light beam will go through when crossing a boundary between two indexes of refraction. – anna v Nov 4 '12 at 15:05
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http://mathhelpforum.com/algebra/179699-cube-root-5-a-print.html
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# Cube root of 5
Printable View
• May 6th 2011, 02:14 AM
mathguy80
Cube root of 5
Hi Guys,
Need a little hint on this one.
Expand $(1 + x)^{\frac{1}{3}}$ in ascending powers of $x$ as far as the term in $x^3$, simplifying the terms as much as possible. By substituting 0.08 for $x$ in your result obtain an approximate value of the cube root of 5, upto four decimal places.
I expanded the problem as a binomial series,
$1 + \dfrac{x}{3} - \dfrac{x^2}{9} + \dfrac{5x^3}{81}$
The rest of the question is confusing me. How do I break up 5 to include 0.08?
The only approach I could think of was,
$<br /> \sqrt[3]{5} = \sqrt[3]{8\left(1 - \frac{3}{8}\right)} = 2\sqrt[3]{1 - \frac{3}{8}}<br />$
Then using x = -3/8, to put in the above equation.
This gives me 1.7122 which is incorrect, and also does not use the suggested 0.08. Any ideas where I am going wrong?
Thanks for your help.
• May 6th 2011, 02:21 AM
alexmahone
$(1+0.08)^\frac{1}{3}=(\frac{27}{25})^\frac{1}{3}=3/\sqrt[3]{25}$
Can you proceed?
• May 6th 2011, 02:22 AM
Prove It
Surely if you're evaluating the cube root of 5, you need to let x = 4...
• May 6th 2011, 02:39 AM
mathguy80
Quote:
Originally Posted by alexmahone
$(1+0.08)^\frac{1}{3}=(\frac{27}{25})^\frac{1}{3}=3/\sqrt[3]{5}$
Can you proceed?
Thanks for the quick reply!
I am missing something here, Wouldn't that be $3/\sqrt[3]{25}$? How does that lead to $\sqrt[3]{5}$?
• May 6th 2011, 02:51 AM
alexmahone
Quote:
Originally Posted by mathguy80
Thanks for the quick reply!
I am missing something here, Wouldn't that be $3/\sqrt[3]{25}$? How does that lead to $\sqrt[3]{5}$?
Of course. I'm sorry, that was a typo.
Anyway, $3/\sqrt[3]{25}=3/(\sqrt[3]{5})^2$
• May 6th 2011, 03:09 AM
mathguy80
Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!
One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?
• May 6th 2011, 03:12 AM
mathguy80
@Prove It, I am not sure I follow your suggestion.
• May 6th 2011, 03:50 AM
Archie Meade
Quote:
Originally Posted by mathguy80
Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!
One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?
$\sqrt[3]{1.08}=\sqrt[3]{\frac{108}{100}}=\sqrt[3]{\frac{(4)27}{(4)25}}$
$\Rightarrow\sqrt[3]{5}=\sqrt[3]{\frac{500}{100}}=\sqrt[3]{\left(\frac{108}{108}\right)\frac{500}{100}}$
$=\sqrt[3]{\frac{500}{108}\left(\frac{108}{100}\right)}}$
$=\sqrt[3]{\frac{500}{108}}\sqrt[3]{1+0.08}=\sqrt[3]{\frac{(4)125}{(4)27}}\sqrt[3]{1+0.08}$
$=\sqrt[3]{\frac{125}{27}}\sqrt[3]{1+0.08}$
• May 6th 2011, 05:29 AM
Prove It
Quote:
Originally Posted by mathguy80
@Prove It, I am not sure I follow your suggestion.
Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...
• May 6th 2011, 05:32 AM
alexmahone
Quote:
Originally Posted by Prove It
Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...
But the binomial series (when the index is a fraction) converges only if |x|<1.
• May 6th 2011, 05:53 AM
Prove It
Quote:
Originally Posted by alexmahone
But the binomial series (when the index is a fraction) converges only if |x|<1.
Point taken. I'm sure, however, that a series can be found which is centred somewhere near x = 4 so that you can evaluate it at x = 4.
• May 6th 2011, 07:28 AM
mathguy80
Nice! This is why I love this site! Thanks @Archie Meade.
• May 6th 2011, 07:32 AM
mathguy80
Interesting point @Prove it. The |x| < 1 for the convergent binomial series has tripped me up as well. Don't know enough Math to know if there is such a series.. Thanks for all the help today, guys, much appreciated.
All times are GMT -8. The time now is 05:43 AM.
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http://www.abstractmath.org/Word%20Press/?tag=formula
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Gyre&Gimbleposts about math, language and other things that may appear in the wabe
Conceptual blending
2012/06/18 — SixWingedSeraph
This post uses MathJax. If you see formulas in unrendered TeX, try refreshing the screen.
A conceptual blend is a structure in your brain that connects two concepts by associating part of one with part of another. Conceptual blending is a major tool used by our brain to understand the world.
The concept of conceptual blend includes special cases, such as representations, images and conceptual metaphors, that math educators have used for years to understand how mathematics is communicated and how it is learned. The Wikipedia article is a good starting place for understanding conceptual blending.
In this post I will illustrate some of the ways conceptual blending is used to understand a function of the sort you meet with in freshman calculus. I omit the connections with programs, which I will discuss in a separate post.
A particular function
Consider the function $h(t)=4-(t-2)^2$. You may think of this function in many ways.
FORMULA:
$h(t)$ is defined by the formula $4-(t-2)^2$.
• The formula encapsulates a particular computation of the value of $h$ at a given value $t$.
• The formula defines the function, which is a stronger statement than saying it represents the function.
• The formula is in standard algebraic notation. (See Note 1)
• To use the formula requires one of these:
• Understand and use the rules of algebra
• Use a calculator
• Use an algebraic programming language.
• Other formulas could be used, for example $4t-t^2$.
• That formula encapsulates a different computation of the value of $h$.
TREE:
$h(t)$ is also defined by this tree (right).
• The tree makes explicit the computation needed to evaluate the function.
• The form of the tree is based on a convention, almost universal in computing science, that the last operation performed (the root) is placed at the top and that evaluation is done from bottom to top.
• Both formula and tree require knowledge of conventions.
• The blending of formula and tree matches some of the symbols in the formula with nodes in the tree, but the parentheses do not appear in the tree because they are not necessary by the bottom-up convention.
• Other formulas correspond to other trees. In other words, conceptually, each tree captures not only everything about the function, but everything about a particular computation of the function.
• More about trees in these posts:
GRAPH:
$h(t)$ is represented by its graph (right). (See note 2.)
• This is the graph as visual image, not the graph as a set of ordered pairs.
• The blending of graph and formula associates each point on the (blue) graph with the value of the formula at the number on the x-axis directly underneath the point.
• In contrast to the formula, the graph does not define the function because it is a physical picture that is only approximate.
• But the formula does represent the function. (This is "represents" in the sense of cognitive psychology, but not in the mathematical sense.)
• The blending requires familiarity with the conventions concerning graphs of functions.
• It sets into operation the vision machinery of your brain, which is remarkably elaborate and powerful.
• Your visual machinery allows you to see instantly that the maximum of the curve occurs at about $t=2$.
• The blending leaves out many things.
• For one, the graph does not show the whole function. (That's another reason why the graph does not define the function.)
• Nor does it make it obvious that the rest of the graph goes off to negative infinity in both directions, whereas that formula does make that obvious (if you understand algebraic notation).
GEOMETRIC
The graph of $h(t)$ is the parabola with vertex $(2,4)$, directrix $x=2$, and focus $(2,\frac{3}{4})$.
• The blending with the graph makes the parabola identical with the graph.
• This tells you immediately (if you know enough about parabolas!) that the maximum is at $(2,4)$ (because the directrix is vertical).
• Knowing where the focus and directrix are enables you to mechanically construct a drawing of the parabola using a pins, string, T-square and pencil. (In the age of computers, do you care?)
HEIGHT:
$h(t)$ gives the height of a certain projectile going straight up and down over time.
• The blending of height and graph lets you see instantly (using your visual machinery) how high the projectile goes.
• The blending of formula and height allows you to determing the projectile's velocity at any point by taking the derivative of the function.
• A student may easily be confused into thinking that the path of the projectile is a parabola like the graph shown. Such a student has misunderstood the blending.
KINETIC:
You may understand $h(t)$ kinetically in various ways.
• You can visualize moving along the graph from left to right, going, reaching the maximum, then starting down.
• This calls on your experience of going over a hill.
• You are feeling this with the help of mirror neurons.
• As you imagine traversing the graph, you feel it getting less and less steep until it is briefly level at the maximum, then it gets steeper and steeper going down.
• This gives you a physical understanding of how the derivative represents the slope.
• You may have seen teachers swooping with their hand up one side and down the other to illustrate this.
• You can kinetically blend the movement of the projectile (see height above) with the graph of the function.
• As it goes up (with $t$ increasing) the projectile starts fast but begins to slow down.
• Then it is briefly stationery at $t=2$ and then starts to go down.
• You can associate these feelings with riding in an elevator.
• Yes, the elevator is not a projectile, so this blending is inaccurate in detail.
• This gives you a kinetic understanding of how the derivative gives the velocity and the second derivative gives the acceleration.
OBJECT:
The function $h(t)$ is a mathematical object.
• Usually the mental picture of function-as-object consists of thinking of the function as a set of ordered pairs $\Gamma(h):=\{(t,4-(t-2)^2)|t\in\mathbb{R}\}$.
• Sometimes you have to specify domain and codomain, but not usually in calculus problems, where conventions tell you they are both the set of real numbers.
• The blend object and graph identifies each point on the graph with an element of $\Gamma(h)$.
• When you give a formal proof, you usually revert to a dry-bones mode and think of math objects as inert and timeless, so that the proof does not mention change or causation.
• The mathematical object $h(t)$ is a particular set of ordered pairs.
• It just sits there.
• When reasoning about something like this, implication statements work like they are supposed to in math: no causation, just picking apart a bunch of dead things. (See Note 3).
• I did not say that math objects are inert and timeless, I said you think of them that way. This post is not about Platonism or formalism. What math objects "really are" is irrelevant to understanding understanding math [sic].
DEFINITION
A definition of the concept of function provides a way of thinking about the function.
• One definition is simply to specify a mathematical object corresponding to a function: A set of ordered pairs satisfying the property that no two distinct ordered pairs have the same second coordinate, along with a specification of the codomain if that is necessary.
• A concept can have many different definitions.
• A group is usually defined as a set with a binary operation, an inverse operation, and an identity with specific properties. But it can be defined as a set with a ternary operation, as well.
• A partition of a set is a set of subsets of a set with certain properties. An equivalence relation is a relation on a set with certain properties. But a partition is an equivalence relation and an equivalence relation is a partition. You have just picked different primitives to spell out the definition.
• If you are a beginner at doing proofs, you may focus on the particular primitive objects in the definition to the exclusion of other objects and properties that may be more important for your current purposes.
• For example, the definition of $h(t)$ does not mention continuity, differentiability, parabola, and other such things.
• The definition of group doesn't mention that it has linear representations.
SPECIFICATION
A function can be given as a specification, such as this:
If $t$ is a real number, then $h(t)$ is a real number, whose value is obtained by subtracting $2$ from $t$, squaring the result, and then subtracting that result from $4$.
• This tells you everything you need to know to use the function $h$.
• It does not tell you what it is as a mathematical object: It is only a description of how to use the notation $h(t)$.
Notes
1. Formulas can be give in other notations, in particular Polish and Reverse Polish notation. Some forms of these notations don't need parentheses.
2. There are various ways to give a pictorial image of the function. The usual way to do this is presenting the graph as shown above. But you can also show its cograph and its endograph, which are other ways of representing a function pictorially. They are particularly useful for finite and discrete functions. You can find lots of detail in these posts and Mathematica notebooks:
3. See How to understand conditionals in the abstractmath article on conditionals.
References
1. Conceptual blending (Wikipedia)
2. Conceptual metaphors (Wikipedia)
3. Definitions (abstractmath)
4. Embodied cognition (Wikipedia)
5. Handbook of mathematical discourse (see articles on conceptual blend, mental representation, representation, and metaphor)
6. Images and Metaphors (article in abstractmath)
7. Links to G&G posts on representations
8. Mirror neurons (Wikipedia)
9. Representations and models (article in abstractmath)
10. Representations II: dry bones (article in abstractmath)
11. The transition to formal thinking in mathematics, David Tall, 2010
12. What is the object of the encapsulation of a process? Tall et al., 2000.
The Mathematical Definition of Function
2011/05/20 — Charles Wells
Introduction
This post is a completely rewritten version of the abstractmath article on the definition of function. Like every part of abstractmath, the chapter on functions is designed to get you started thinking about functions. It is no way complete. Wikipedia has much more complete coverage of mathematical functions, but be aware that the coverage is scattered over many articles.
The concept of function in mathematics is as important as any mathematical idea. The mathematician’s concept of function includes the kinds of functions you studied in calculus but is much more abstract and general. If you are new to abstract math you need to know:
• The precise meaning of the word “function” and other concepts associated with functions. That’s what this section is about.
• Notation and terminology for functions. (That will be a separate section of abstractmath.org which I will post soon.)
• The many different kinds of functions there are. (See Examples of Functions in abmath).
• The many ways mathematicians think about functions. The abmath article Images and Metaphors for Functions is a stub for this.
I will use two running examples throughout this discussion:
• $latex {F}&fg=000000$ is the function defined on the set $latex {\left\{1,\,2,3,6 \right\}}&fg=000000$ as follows: $latex {F(1)=3,\,\,\,F(2)=3,\,\,\,F(3)=2,\,\,\,F(6)=1}&fg=000000$. This is a function defined on a finite set by explicitly naming each value.
• $latex {G}&fg=000000$ is the real-valued function defined by the formula $latex {G(x)={{x}^{2}}+2x+5}&fg=000000$.
Specification of function
We start by giving a specification of “function”. (See the abstractmath article on specification.) After that, we get into the technicalities of the definitions of the general concept of function.
Specification: A function $latex {f}&fg=000000$ is a mathematical object which determines and is completely determined bythe following data:
• $latex {f}&fg=000000$ has a domain, which is a set. The domain may be denoted by $latex {\text{dom }f}&fg=000000$.
• $latex {f}&fg=000000$ has a codomain, which is also a set and may be denoted by $latex {\text{cod }f}&fg=000000$.
• For each element $latex {a}&fg=000000$ of the domain of $latex {f}&fg=000000$, $latex {f}&fg=000000$ has a value at $latex {a}&fg=000000$, denoted by $latex {f(a)}&fg=000000$.
• The value of $latex {f}&fg=000000$ at $latex {a}&fg=000000$ is completely determined by $latex {a}&fg=000000$ and $latex {f}&fg=000000$ .
• The value of $latex {f}&fg=000000$ at $latex {a}&fg=000000$ must be an element of the codomain of $latex {f}&fg=000000$.
The operation of finding $latex {f(a)}&fg=000000$ given $latex {f}&fg=000000$ and $latex {a}&fg=000000$ is called evaluation.
Examples
• The definition above of the finite function $latex {F}&fg=000000$ specifies that the domain is the set $latex {\left\{1,\,2,\,3,\,6 \right\}}&fg=000000$. The value of $latex {F}&fg=000000$ at each element of the domain is given explicitly. The value at 3, for example, is 2, because the definition says that $latex {F(2) = 3}&fg=000000$. The codomain of $latex {F}&fg=000000$ is not specified, but must include the set $latex {\{1,2,3\}}&fg=000000$.
• The definition of $latex {G}&fg=000000$ above gives the value at each element of the domain by a formula. The value at 3, for example, is $latex {G(3)=3^2+2\cdot3+5=20}&fg=000000$. The definition does not specify the domain or the codomain. The convention in the case of functions defined on the real numbers by a formula is to take the domain to be all real numbers at which the formula is defined. In this case, that is every real number, so the domain is $latex {{\mathbb R}}&fg=000000$. The codomain must include all real numbers greater than or equal to 4. (Why?)
Comment: The formula above that defines the function $latex G$ in fact defines a function of complex numbers (even quaternions).
Definition of function
In the nineteenth century, mathematicians realized that it was necessary for some purposes (particularly harmonic analysis) to give a mathematical definition of the concept of function. A stricter version of this definition turned out to be necessary in algebraic topology and other fields, and that is the one I give here.
To state this definition we need a preliminary idea.
The functional property
A set R of ordered pairs has the functional property if two pairs in R with the same first coordinate have to have the same second coordinate (which means they are the same pair).
Examples
• The set $latex {\{(1,2), (2,4), (3,2), (5,8)\}}&fg=000000$ has the functional property, since no two different pairs have the same first coordinate. It is true that two of them have the same second coordinate, but that is irrelevant.
• The set $latex {\{(1,2), (2,4), (3,2), (2,8)\}}&fg=000000$ does not have the functional property. There are two different pairs with first coordinate 2.
• The graphs of functions in beginning calculus have the functional property.
• The empty set $latex {\emptyset}&fg=000000$ has the functional property .
Example: Graph of a function defined by a formula
The graph of the function $latex {G}&fg=000000$ given above has the functional property. The graph is the set
$latex \displaystyle \left\{ (x,{{x}^{2}}+2x+5)\,\mathsf{|}\,x\in {\mathbb R} \right\}.&fg=000000$
If you repeatedly plug in one real number over and over, you get out the same real number every time. Example:
• if $latex {x = 0}&fg=000000$, then $latex {{{x}^{2}}+2x+5=5}&fg=000000$. You get 5 every time you plug in 0.
• if $latex {x = 1}&fg=000000$, then $latex {{{x}^{2}}+2x+5=8}&fg=000000$.
• if $latex {x =-2}&fg=000000$, then $latex {{{x}^{2}}+2x+5=5}&fg=000000$.
This set has the functional property because if $latex {x}&fg=000000$ is any real number, the formula $latex {{{x}^{2}}+2x+5}&fg=000000$ defines a specific real number. (This description of the graph implicitly assumes that $latex {\text{dom } G={\mathbb R}}&fg=000000$.) No other pair whose first coordinate is $latex {-2}&fg=000000$ is in the graph of $latex {G}&fg=000000$, only $latex {(-2, 5)}&fg=000000$. That is because when you plug $latex {-2}&fg=000000$ into the formula $latex {{{x}^{2}}+2x+5}&fg=000000$, you get $latex {5}&fg=000000$ every time. Of course, $latex {(0, 5)}&fg=000000$ is in the graph, but that does not contradict the functional property. $latex {(0, 5)}&fg=000000$ and $latex {(-2, 5)}&fg=000000$ have the same second coordinate, but that is OK.
How to think about the functional property
The point of the functional property is that for any pair in the set of ordered pairs, the first coordinate determines what the second one is. That’s why you can write “$latex {G(x)}&fg=000000$” for any $latex {x }&fg=000000$ in the domain of $latex {G}&fg=000000$ and not be ambiguous.
Mathematical definition of function
A function$latex {f}&fg=000000$ is a mathematical structure consisting of the following objects:
• A set called the domain of $latex {f}&fg=000000$, denoted by $latex {\text{dom } f}&fg=000000$.
• A set called the codomain of $latex {f}&fg=000000$, denoted by $latex {\text{cod } f}&fg=000000$.
• A set of ordered pairs called the graph of $latex { f}&fg=000000$, with the following properties:
• $latex {\text{dom } f}&fg=000000$ is the set of all first coordinates of pairs in the graph of $latex {f}&fg=000000$.
• Every second coordinate of a pair in the graph of $latex {f}&fg=000000$ is in $latex {\text{cod } f}&fg=000000$ (but $latex {\text{cod } f}&fg=000000$ may contain other elements).
• The graph of $latex {f}&fg=000000$ has the functional property. Using arrow notation, this implies that $latex {f:A\rightarrow B}&fg=000000$.
Examples
• Let $latex {F}&fg=000000$ have graph $latex {\{(1,2), (2,4), (3,2), (5,8)\}}&fg=000000$ and define $latex {A = \{1, 2, 3, 5\}}&fg=000000$ and $latex {B = \{2, 4, 8\}}&fg=000000$. Then $latex {F:A\rightarrow B}&fg=000000$ is a function.
• Let $latex {G}&fg=000000$ have graph $latex {\{(1,2), (2,4), (3,2), (5,8)\}}&fg=000000$ (same as above), and define $latex {A = \{1, 2, 3, 5\}}&fg=000000$ and $latex {C = \{2, 4, 8, 9, 11, \pi, 3/2\}}&fg=000000$. Then $latex {G:A\rightarrow C}&fg=000000$ is a (admittedly ridiculous) function. Note that all the second coordinates of the graph are in $latex {C}&fg=000000$, along with a bunch of miscellaneous suspicious characters that are not second coordinates of pairs in the graph.
• Let $latex {H}&fg=000000$ have graph $latex {\{(1,2), (2,4), (3,2), (5,8)\}}&fg=000000$. Then $latex {H:A\rightarrow {\mathbb R}}&fg=000000$ is a function.
According to the definition of function, $latex {F}&fg=000000$, $latex {G}&fg=000000$ and $latex {H}&fg=000000$ are three different functions.
Identity and inclusion
Suppose we have two sets A and B with $latex {A\subseteq B}&fg=000000$.
• The identity function on A is the function $latex {{{\text{id}}_{A}}:A\rightarrow A}&fg=000000$ defined by $latex {{{\text{id}}_{A}}(x)=x}&fg=000000$ for all$latex {x\in A}&fg=000000$. (Many authors call it $latex {{{1}_{A}}}&fg=000000$).
• The inclusion function from A to B is the function $latex {i:A\rightarrow B}&fg=000000$ defined by $latex {i(x)=x}&fg=000000$ for all $latex {x\in A}&fg=000000$. Note that there is a different function for each pair of sets A and B for which $latex {A\subseteq B}&fg=000000$. Some authors call it $latex {{{i}_{A,\,B}}}&fg=000000$ or $latex {\text{in}{{\text{c}}_{A,\,B}}}&fg=000000$.
Remark The identity function and an inclusion function for the same set A have exactly the same graph, namely $latex {\left\{ (a,a)|a\in A \right\}}&fg=000000$.
Graphs and functions
• If $latex {f}&fg=000000$ is a function, the domain of $latex {f}&fg=000000$ is the set of first coordinates of all the pairs in $latex {f}&fg=000000$.
• If $latex {x\in \text{dom } f}&fg=000000$, then $latex {f(x)}&fg=000000$ is the second coordinate of the only ordered pair in $latex {f}&fg=000000$ whose first coordinate is $latex {x}&fg=000000$.
Examples
The set $latex {\{(1,2), (2,4), (3,2), (5,8)\}}&fg=000000$ has the functional property, so it is the graph of a function. Call the function $latex {H}&fg=000000$. Then its domain is $latex {\{1,2,3,5\}}&fg=000000$ and $latex {H(1) = 2}&fg=000000$ and $latex {H(2) = 4}&fg=000000$. $latex {H(4)}&fg=000000$ is not defined because there is no ordered pair in H beginning with $latex {4}&fg=000000$ (hence $latex {4}&fg=000000$ is not in $latex {\text{dom } H}&fg=000000$.)
I showed above that the graph of the function $latex {G}&fg=000000$, ordinarily described as “the function $latex {G(x)={{x}^{2}}+2x+5}&fg=000000$”, has the functional property. The specification of function requires that we say what the domain is and what the value is at each point. These two facts are determined by the graph.
Other definitions of function
Because of the examples above, many authors define a function as a graph with the functional property. Now, the graph of a function $latex {G}&fg=000000$ may be denoted by $latex {\Gamma(G)}&fg=000000$. This is an older, less strict definition of function that doesn’t work correctly with the concepts of algebraic topology, category theory, and some other branches of mathematics.
For this less strict definition of function, $latex {G=\Gamma(G)}&fg=000000$, which causes a clash of our mental images of “graph” and “function”. In every important way except the less-strict definition, they ARE different!
A definition is a device for making the meaning of math technical terms precise. When a mathematician think of “function” they think of many aspects of functions, such as a map of one shape into another, a graph in the real plane, a computational process, a renaming, and so on. One of the ways of thinking of a function is to think about its graph. That happens to be the best way to define the concept of function. (It is the less strict definition and it is a necessary concept in the modern definition given here.)
The occurrence of the graph in either definition doesn’t make thinking of a function in terms of its graph the most important way of visualizing it. I don’t think it is even in the top three.
Templates in mathematical practice
2010/03/11 — Charles Wells
This post is a first pass at what will eventually be a section of abstractmath.org. It’s time to get back to abstractmath; I have been neglecting it for a couple of years.
What I say here is based mainly on my many years of teaching discrete mathematics at Case Western Reserve University in Cleveland and more recently at Metro State University in Saint Paul.
Beginning abstract math
College students typically get into abstract math at the beginning in such courses as linear algebra, discrete math and abstract algebra. Certain problems that come up in those early courses can be grouped together under the notion of (what I call) applying templates [note 0]. These are not the problems people usually think about concerning beginners in abstract math, of which the following is an incomplete list:
• Reasoning from axioms
• Encapsulating processes (Tall et al, 2000).
• Semantic contamination (abstractmath)
• Translating between mathematical English and logic (abstractmath)
The students’ problems discussed here concern understanding what a template is and how to apply it.
Templates can be formulas, rules of inference, or mini-programs. I’ll talk about three examples here.
The template for quadratic equations
The solution of a real quadratic equation of the form $latex {ax^2+bx+c=0}&fg=000000$ is given by the formula
$latex \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}&fg=000000$
This is a template for finding the roots of the equations. It has subtleties.
For example, the numerator is symmetric in $latex {a}&fg=000000$ and $latex {c}&fg=000000$ but the denominator isn’t. So sometimes I try to trick my students (warning them ahead of time that that’s what I’m trying to do) by asking for a formula for the solution of the equation $latex {a+bx+cx^2=0}&fg=000000$. The answer is
$latex \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2c}&fg=000000$
I start writing it on the board, asking them to tell me what comes next. When we get to the denominator, often someone says “$latex {2a}&fg=000000$”.
The template is telling you that the denominator is 2 times the coefficient of the square term. It is not telling you it is “$latex {a}&fg=000000$”. Using a template (in the sense I mean here) requires pattern matching, but in this particular example, the quadratic template has a shallow incorrect matching and a deeper correct matching. In detail, the shallow matching says “match the letters” and the deep matching says “match the position of the letters”.
Most of the time the quadratic being matched has particular numbers instead of the same letters that the template has, so the trap I just described seldom occurs. But this makes me want to try a variation of the trick: Find the solution of $latex {3+5x+2x^2=0}&fg=000000$. Would some students match the textual position (getting $latex {a=3}&fg=000000$) instead of the functional position (getting $latex {a=5}&fg=000000$)? [Note [0]). If they did they would get the solutions $latex {(-1,-\frac{2}{3})}&fg=000000$ instead of $latex {(-1,-\frac{3}{2})}&fg=000000$.
Substituting in algebraic expressions have other traps, too. What sorts of mistakes would students have solving $latex {3x^2+b^2x-5=0}&fg=000000$?
Most students on the verge of abstract math don’t make mistakes with the quadratic formula that I have described. The thing about abstract math is that it uses more sophisticated templates
• subject to conditions
• with variations
• with extra levels of abstraction
The template for proof by induction
This template gives a method of proof of a statement of the form $latex {\forall{n}\mathcal{P}(n)}&fg=000000$, where $latex {\mathcal{P}}&fg=000000$ is a predicate (presumably containing $latex {n}&fg=000000$ as a variable) and $latex {n}&fg=000000$ varies over positive integers. The template says:
Goal: Prove $latex {\forall{n}\mathcal{P}(n)}&fg=000000$.
Method:
• Prove $latex {\mathcal{P}(1)}&fg=000000$
• For an arbitrary integer $latex {n>1}&fg=000000$, assume $latex {\mathcal{P}(n)}&fg=000000$ and deduce $latex {\mathcal{P}(n+1)}&fg=000000$.
For example, to prove $latex {\forall n (2^n+1\geq n^2)}&fg=000000$ using the template, you have to prove that $latex {2^2+1\geq 1^1}&fg=000000$, and that for any $latex {n>1}&fg=000000$, if $latex {2^n+1\geq n^2}&fg=000000$, then $latex {2^{n+1}+1\geq (n+1)^2}&fg=000000$. You come up with the need to prove these statements by substituting into the template. This template has several problems that the quadratic formula does not have.
Variables of different types
The variable $latex {n}&fg=000000$ is of type integer and the variable $latex {\mathcal{P}}&fg=000000$ is of type predicate [note 0]. Having to deal with several types of variables comes up already in multivariable calculus (vectors vs. numbers, cross product vs. numerical product, etc) and they multiply like rabbits in beginning abstract math classes. Students sometimes write things like “Let $latex {\mathcal{P}=n+1}&fg=000000$”. Multiple types is a big problem that math ed people don’t seem to discuss much (correct me if I am wrong).
The variable $latex {n}&fg=000000$ occurs as a bound variable in the Goal and a free variable in the Method. This happens in this case because the induction step in the Method originates as the requirement to prove $latex {\forall n(\mathcal{P}(n)\rightarrow\mathcal{P}(n+1))}&fg=000000$, but as I have presented it (which seems to be customary) I have translated this into a requirement based on modus ponens. This causes students problems, if they notice it. (“You are assuming what you want to prove!”) Many of them apparently go ahead and produce competent proofs without noticing the dual role of $latex {n}&fg=000000$. I say more power to them. I think.
The template has variations
• You can start the induction at other places.
• You may have to have two starting points and a double induction hypothesis (for $latex {n-1}&fg=000000$ and $latex {n}&fg=000000$). In fact, you will have to have two starting points, because it seems to be a Fundamental Law of Discrete Math Teaching that you have to talk about the Fibonacci function ad nauseam.
• Then there is strong induction.
It’s like you can go to the store and buy one template for quadratic equations, but you have to by a package of templates for induction, like highway engineers used to buy packages of plastic French curves to draw highway curves without discontinuous curvature.
The template for row reduction
I am running out of time and won’t go into as much detail on this one. Row reduction is an algorithm. If you write it up as a proper computer program there have to be all sorts of if-thens depending on what you are doing it for. For example if want solutions to the simultaneous equations
| | | |
|---------|----|----|
| 2x+4y+z | = | 1 |
| x+2y | = | 0 |
| x+2y+4z | = | 5 |
you must row reduce the matrix
| | | | |
|----|----|----|----|
| 2 | 4 | 1 | 1 |
| 1 | 2 | 0 | 0 |
| 1 | 2 | 4 | 5 |
(I haven’t yet figured out how to wrap this in parentheses) which gives you
| | | | |
|----|----|----|----|
| 1 | 2 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 1 |
This introduces another problem with templates: They come with conditions. In this case the condition is “a row of three 0s followed by a nonzero number means the equations have no solutions”. (There is another condition when there is a row of all 0′s.)
It is very easy for the new student to get the calculation right but to never sit back and see what they have — which conditions apply or whatever.
When you do math you have to repeatedly lean in and focus on the details and then lean back and see the Big Picture. This is something that has to be learned.
What to do, what to do
I have recently experimented with being explicit about templates, in particular going through examples of the use of a template after explicitly stating the template. It is too early to say how successful this is. But I want to point out that even though it might not help to be explicit with students about templates, the analysis in this post of a phenomenon that occurs in beginning abstract math courses
• may still be accurate (or not), and
• may help teachers teach such things if they are aware of the phenomenon, even if the students are not.
Notes
1. Many years ago, I heard someone use the word “template” in the way I am using it now, but I don’t recollect who it was. Applied mathematicians sometimes use it with a meaning similar to mine to refer to soft algorithms–recipes for computation that are not formal algorithms but close enough to be easily translated into a sufficiently high level computer language.
2. In the formula $latex {ax^2+bx+c}&fg=000000$, the “$latex {a}&fg=000000$” has the first textual position but the functional position as the coefficient of the quadratic term. This name “functional position” has nothing to do with functions. Can someone suggest a different name that won’t confuse people?
3. I am using “variable” the way logicians do. Mathematicians would not normally refer to “$latex {\mathcal{P}}&fg=000000$” as a variable.
4. I didn’t say anything about how templates can involve extra layers of abstract. That will have to wait.
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http://math.stackexchange.com/questions/119074/system-of-4-equations-3-of-them-linear-what-can-be-said-about-the-solution-s?answertab=oldest
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# System of 4 equations (3 of them linear) - What can be said about the solution set
I have a homework at linear algebra and we have this system of linear equations:
$x+y+z+w=0$
$x+2*y+9*z+13*w=0$
$4*x+41*y+6*z+656*w = 0$
And we add this equation:
$x^3 + y^4 + 8* z^5 + 8* w^6 = 0$
What can be said about the solution set of this system?
I tried wolfram alpha and got this:
Real solutions:
• w~~0, x~~0, y~~0, z~~0
• w~~-0.0417376, x~~-0.665814, y~~0.73708, z~~-0.0295286
Doesn't this mean that the solution set is a non empty set?
And also that it is a finite set (with 2 elements? the (0,0,0,0) and (-0.0417376, -0.665814, 0.73708, -0.0295286) ?
Also, how can I find out if the solution set of the above system is a vector space?
Thank you!
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1
Over $\mathbb Q$, the $3$ linear equations give you a matrix of rank $3$. So the nullspace is of dimension $1$, with basis vector $$b = (\frac{335}{21}, \frac{-2596}{147}, \frac{-2596}{147}, 1),$$ i.e., the solution space is all scalar multiples of $b$. All solutions to the system however, should be zeros of the polynomial. I'm not sure though how to use the polynomial to pick a particular solution. – user2468 Mar 12 '12 at 0:45
## 3 Answers
Your three linear equations can be satisfied by $w = 147 t, x = 2345 t, y = -2596 t, z = 104 t$ for any $t$.
Substitute that into your fourth equation and you have a sextic equation for $t$, $$80722386956232 t^6+97332232192 t^5+45417032294656 t^4+12895213625 t^3=0$$ but you can take out a factor of $t^3$ so you have in effect three roots at $t=0$ (and so $w=y=z=0$) and a cubic: $$80722386956232 t^3+97332232192 t^2+45417032294656 t+12895213625=0.$$ This particular cubic has one real root and two complex roots, which you can translate back to values for $w,x,y,z$.
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You say "This particular cubic has one real root and two complex roots, which you can translate back to values for w,x,y,z.", the one real root is the one that wolfram gave as a result? So there are two real roots? (We are looking for real roots) – Chris Mar 12 '12 at 1:21
@Chris: $t=-0.000283929\ldots$ is the same real root as you have in your question when you multiply it by $147, 2345$ etc. – Henry Mar 12 '12 at 1:36
So all in all, is the system's solution set a finite set? We are sure that 0 vector belongs to the solution set. – Chris Mar 12 '12 at 18:10
@Chris: yes - depending on how you count, there are two distinct real solutions (one of which is zero and you might count it three times) and two complex solutions – Henry Mar 12 '12 at 21:35
So could we possibly say that the solution set is finite? – Chris Mar 13 '12 at 21:38
show 1 more comment
You can solve for any three of x, y, z, and w in terms of the other by using the first three equations. For example, from the first two, you get $y+z+w = 2∗y+9∗z+13∗w$, or $y = -8z -12 w$.
From the first and third equations, you get (multiplying the first by 4), $4y+4z+4w = 41y+6z+656w$, or $37y = -2z-652w$.
This lets you get z as a multiple of w, then y as another multiple of w, and, finally, x as another multiple of w.
Substituting these into the last equation, you will get a sextic (sixth degree) equation in w. This will have 6 roots, and since the coefficients will be real, they will be real or in complex conjugate pairs.
Looking at the equation, there will be three zero roots. Since Alpha has found a non-zero real root, the other two roots are real or complex conjugate.
Since the resulting equation, after removing the zero roots is a cubic, you can solve it of divide out the root found by Alpha to get a quadratic.
I see that a solution has been entered while I was typing.
Let's look.
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As for whether or not it's a vector space: one of the axioms of vector spaces says that if $\mathbb v$ is a vector in the space, then so is $k \mathbb v$ for any $k \in \mathbb R$. This means that $k (-0.0417376, -0.665814, 0.73708, -0.0295286)$ would have to be a solution for any $k$.
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1
But it is not: the $x^3 + y^4 + 8 z^5 + 8 w^6 = 0$ prevents it, except for $k=0,1$ or the two complex values. – Henry Mar 12 '12 at 12:43
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http://math.stackexchange.com/questions/297351/implicit-derivative-and-related-rates
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# Implicit derivative and related rates
I'm having a bit of trouble with this assignment. I had to miss a couple class, and I'm afraid the teacher talked about this. I have the notes, but they are not of much help and don't talk about related rates at all. I tried looking around, I found lots of PDF from different universities, but I still don't get how to solve this problem. If I could get some pointers, that'd be great :D
Thanks
Here's the problem (translated from French):
A particle oscillates on the $x$-axis. It's position $x(t)$ (in meters) at the time $t$ (in seconds) is related to its speed $s(t)$ by the equation:
$$x^2 + x \cdot s + s^2 = 13$$
Give the acceleration of this particle when its position is $x = 3$ meters , knowing that at that moment, the speed is positive.
Hint: This is a related rates problem in which you need to use the implicit derivative. Use the fact that $a(t) = {dv \over dt}$ and $s(t) = {dx \over dt}$
I did this, but I'm not sure if it's the right thing to do, or what to do with my result :/
$$x^2+x \cdot s + s^2 = 13$$ $$(x^2)' + (x \cdot s)' + (s^2)' = (13)'$$ $$2x + (1 \cdot s + s' \cdot x) + 2 \cdot s \cdot s' = 0$$ $$s' \cdot x + 2 \cdot s \cdot s' = -2x - s$$ $$s' ( x + 2 \cdot s) = -2x - s$$ $$s' = \frac{-2x-s}{x+2s}$$
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It looks correct. Do you wanna evaluate $a(t)$? – Babak S. Feb 7 at 17:21
@BabakSorouh ... but it isn't. Note that $\cdot'$ stands for $\frac d{dt}$, not $\frac d{dx}$ here. – Hagen von Eitzen Feb 7 at 17:25
## 1 Answer
Taking the derivative (with respect to time $t$, not $x$!) of $$x^2+xs+s^2=13,$$ you should obtain $$2xx'+(x's+xs') +2ss'=0$$ or after inserting $s,a$ for $x',s'$ $$2xs+(s^2+xa) +2sa=0$$ and hence $$a=-\frac{(2x+s)s}{x+2s}.$$ Remark: You fell for interpreting the prime $'$ as taking the derivative with respect to $x$. It is very usual in physics to denote the derivative with respect to time by a dot instead (among others, to avoid such confusion). So we aould then rather write $2x\dot x+\dot xs+x\dot s+2s\dot s=0$ and insert $\dot x=s, \dot s=a$.
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Honestly, when I saw your approach, I found out that. Yes Hagen. I am losing my information about Classical Mechanics...... Thanks +1 – Babak S. Feb 7 at 17:31
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http://mathhelpforum.com/advanced-statistics/31821-continous-random-variable.html
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# Thread:
1. ## continous random variable
Y has a distribution function:
$F(y) = \left\{ \begin{array}{rcl} 0 & \mbox{for} & y \leq 0 \\ <br /> \\<br /> \frac{y}{8} & \mbox{for} & 0 < y < 2 \\<br /> \\<br /> \frac{y^2}{16} & \mbox{for} & 2 \leq y < 4 \\<br /> \\<br /> 1 & \mbox{for} & y \geq y 4<br /> \end{array}\right.$
find $\mu$ and $\sigma^2$
since $\mu = \int yf(y) dy \therefore \int^{2}_{0} y \left( \frac{y}{8} \right) dy + \int^{4}_{2} y \left( \frac{y^2}{16} \right) dy + \int^{\infty}_{4} y(1) dy= \frac{y^3}{24} \bigg{|}^{2}_{0} + \ \ \frac{y^4}{64} \bigg{|}^{4}_{2} + \ \ \frac{y^2}{2} \bigg{|}_{4}$ $= \frac{1}{3} +\frac{256-16}{16} + 8 = \frac{145}{12}$
I know that my final solution is incorect since it's suppossed to be $\frac{31}{12}$, where did I go wrong?
2. differentiate $F(y)$ to get $f(y)$ (not integrate). Then take $\int yf(y) \ dy$ to get $\mu$.
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http://physics.stackexchange.com/questions/16386/a-wonky-gravitational-potential-and-its-critical-points
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# A wonky gravitational potential and its critical points
I have tough problem I am not sure how to solve:
For this question, we are confined to a plane. Consider a gravitational field that is proportional to $\frac{1}{r^3}$ instead of $\frac{1}{r^2}$, and consider its potential which is then proportional to $\frac{1}{r^2}$. Suppose that I put $n$ identical point masses in this plane, all at different locations and let $U(x,y)$ be the potential function.
What can we say about the critical points of $U(x,y)$? Specifically, can we show that the number of critical points is always $\leq n$?
Remark: By critical point, I mean the usual definition in vector calculus, that is a place where both partial derivatives are zero.
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So you are confined to a plane that has a gravitation field pointing to $(0,0)$ in that plane with a magnitude of $1/r^3$? I think people will be confused by the mention of a "plane", thinking that this is a planar source. Could you be more clear about the geometry? – AlanSE Oct 31 '11 at 0:50
@Zassounotsukushi: I don't understand your question at all. Imagine you put $n$ point masses at distinct locations on a 2 dimensional plane. Think of the gravitational potential function. Now imagine that for a single mass, it decayed like $\frac{1}{r^2}$ instead of $\frac{1}{r}$. – EPN Oct 31 '11 at 15:28
and $r$ is defined how? I could say $r$ is referenced to a point outside that plane, and that's not quite going to achieve your intention, but I don't think this is clear in your question. My comment is mostly establishing that if you're referencing some $U(x,y)$ then the $r$ you refer to is probably $\sqrt{x^2+y^2}$. Someone reading your question might see that this is probably the case. – AlanSE Oct 31 '11 at 15:46
@Zassounotsukushi: Yes, I mean that if there was a single point mass at the point $(x_0,y_0)$ then $$U(x,y)=\frac{C}{(x-x_0)^2+(y-y_0)^2}.$$ ($C$ is some constant that doesn't matter. Might as well make it one) – EPN Oct 31 '11 at 18:17
## 2 Answers
Take 4 points arranged in a square. The middle of the square is a critical point by symmetry, and the midpoint of the four sides of the square would be a critical point just for the two vertices it joins together, ignoring the other two. But the other two are little more than twice as far away, so eight times weaker force, so if you bring the point closer to the center of the square by an amount about 1/8 of the way to the other side, you will cancel out the force from the far pair from the force on the near pair. So there are 5 critical points for four points on a square, and this holds for all sufficiently fast-falling-off forces.
Using squares-of-squares, I believe it is easy to establish that the number of critical points is generically $n^2$. I believe it is a difficult and interesting mathematical problem to establish any sort of nontrivial bound on the number of critical points. The trivial bound is from the order of the polynomial equation you get, and it's absurdly large---- it grows like $2^n$.
### EDIT: The correct growth rate
The answer for a general configuration is almost certainly N+C critical points (this should be an upper bound and a lower bound for two different C's --- I didn't prove it, but I have a, perhaps crappy, heuristic). For the case of polygons, there are N+1 critical points. For squares where the corners are expanded to squares, and so on fractally, the number of critical points is N+C where C is a small explicit constant. The same for poygons of polygons.
I found a nifty way of analyzing the problem, and getting some good estimates, but I want to know how good the mathematicians are at this before telling the answer. Perhaps you can ask this question on MathOverflow?
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I don't think that this is true. I like the reasoning, but can you provide a proof of what you claim? I somehow suspect that for a square there is only exactly one critical point, and that it is at the center of the square. Please be careful! – EPN Oct 31 '11 at 15:25
@Eric: there are five critical points--- you can easily prove it by the fact that there are three critical points along the vertical bisector of the square. Critical points are topological, and it takes a big perturbation to get rid of them. For the critical point half-way between the two vertices of the square, the other two vertices are too far to get rid of the critical point, they just move it about 1/6 of the way to the center. – Ron Maimon Oct 31 '11 at 19:24
I believe you now since I checked via computer algebra. It gave $(\sqrt{2(\sqrt{2}-1)},0)\approx (0.91,0)$ as another location. (And of course the 3 other points given by symmetry) I see why it follows if there are 3 critical points on the vertical bisector, but why must this be true in the first place? – EPN Oct 31 '11 at 20:25
2
Seeing a plot of the potential at an axis equidistant from the sources did the trick for me. – mmc Oct 31 '11 at 22:11
@Eric: the 3 points are easy because the force is projected to 1d by symmetry, and you can count the zeros by the number of sign changes. The force is clearly down on the edge of the square, and clearly transitions to up, then to zero, then to down, then to up again. The general problem of giving a polynomial bound for the number of critical points (which I am sure exists) seems really interesting. There are topological methods, which go by the winding number of the vector field, and there are variety methods because the problem is algebraic, but mathoverflow might be best here. – Ron Maimon Nov 1 '11 at 1:42
In this answer we analyze a generalization of OP's question. In the last section 6 we will argue heuristically that one should expect an upper bound of critical points of the form
$$c(M)~\leq~~5n-11 \qquad {\rm for} \qquad n\geq 3.$$
1) Let us identify the plane $\mathbb{R}^2\cong \mathbb{C}$ with the complex plane $z=x+\mathrm{i} y$. Let
$$Z:=\{z_1, \ldots, z_n\}\subseteq \mathbb{C}$$
be a set of $n$ different punctures in the complex plane, where $n\in\mathbb{N}$. Consider the punctured complex plane $M:=\mathbb{C} \backslash Z$ and the Riemann sphere $S^2:=\mathbb{C} \cup \{\infty\}$ with Betti numbers
$$b_0(M)~=~1, \qquad b_1(M)~=~n, \qquad b_2(M)~=~0,$$ $$b_0(S^2)~=~1, \qquad b_1(S^2)~=~0, \qquad b_2(S^2)~=~1,$$
and Euler characteristics
$$\chi(M)~=~b_0(M)-b_1(M)+b_2(M)~=~ 1-n,$$ $$\chi(S^2)~=~b_0(S^2)-b_1(S^2)+b_2(S^2)~=~ 2,$$
respectively.
2) Let $p>0$ and $k_1, \ldots, k_n >0$ be $n+1$ positive constants. Let the potential $V:M \to]0,\infty[$ and its extension $\tilde{V}: S^2 \to [0,\infty]$ be
$$V(z)~:=~\sum_{i=1}^n \frac{k_i}{p|z-z_i|^p},$$
$$\tilde{V}(z)~:=~\left\{ \begin{array}{rcl} V(z) &{\rm for}& z \in M, \cr +\infty &{\rm for}& z \in Z, \cr 0 &{\rm for}& z \in\{\infty\}. \end{array} \right.$$
Let $$c_0~:=~ \#{\rm minimum~pts}, \qquad c_1~:=~ \#{\rm saddle~pts}, \qquad c_2~:=~ \#{\rm maximum~pts},$$
So
$$c_0(S^2)~=~c_0(M)+1, \qquad c_1(S^2)~=~c_1(M), \qquad c_2(S^2)~=~c_2(M)+n,$$
because $z=\infty$ is a minimum point and $Z$ are maximum points for $\tilde{V}$.
3) Define two positive functions $E, F:M \to\mathbb{R}_{+}$ as
$$E(z)~:=~\sum_{i=1}^n\frac{ k_i(x-x_i)^2}{|z-z_i|^{p+4}}~>~0,\qquad F(z)~:=~\sum_{i=1}^n\frac{ k_i(y-y_i)^2}{|z-z_i|^{p+4}}~>~0, \qquad z \in M.$$
The $2\times 2$ Hessian matrix $H$ for the potential $V$ is
$$H_{xx}~=~\sum_{i=1}^n k_i\frac{(p+1)(x-x_i)^2-(y-y_i)^2}{|z-z_i|^{p+4}} ~=~(p+1)E-F,\qquad z \in M,$$ $$H_{yy}~=~\sum_{i=1}^n k_i\frac{(p+1)(y-y_i)^2-(x-x_i)^2}{|z-z_i|^{p+4}} ~=~(p+1)F-E,\qquad z \in M,$$ $$H_{xy}~=~(p+2)\sum_{i=1}^n \frac{k_i(x-x_i)(y-y_i)}{|z-z_i|^{p+4}}, \qquad z \in M,$$
with positive trace
$${\rm tr}H ~=~H_{xx}+H_{yy}~=~p(E+F) ~=~p\sum_{i=1}^n \frac{k_i}{|z-z_i|^{p+2}}~>~0, \qquad z \in M.$$
and determinant
$${\rm det}H ~=~H_{xx}H_{yy}-H^2_{xy} ~=~\underbrace{(p^2+2p+2)EF}_{>0} - \underbrace{\left((p+1)(E^2+F^2)+H^2_{xy}\right)}_{>0}.$$
A sufficient condition for negative determinant is:
$$F> (p+1)E \qquad \vee \qquad E> (p+1)F\qquad\Rightarrow \qquad {\rm det}H<0.$$
The off-diagonal element $H_{xy}$ is bounded by
$$\frac{|H_{xy}|}{p+2} ~\leq~\sum_{i=1}^n \frac{k_i|x-x_i||y-y_i|}{|z-z_i|^{p+4}} ~\leq~\frac{E+F}{2}.$$
4) The positive trace ${\rm tr}H>0$ implies that there cannot be any maximum points in $M$,
$$c_2(M)~=~0 \qquad \Leftrightarrow \qquad c_2(S^2)~=~n .$$
A minimum point $z$ must have positive determinant ${\rm det}H(z)>0$, while a saddle point $z$ must have non-positive determinant ${\rm det}H(z)\leq 0$. Generically, one may assume that all critical points $z$ in $M$ are non-degenerate ${\rm det}H(z)\neq 0$, so that $V$ is a Morse function. This implies that all critical points in $M$ are isolated points. Morse theory yields that$^{1)}$
$$c_2(M)-c_1(M)+c_0(M)~=~\chi(M)\qquad \Leftrightarrow \qquad c_2(S^2)-c_1(S^2)+c_0(S^2)~=~\chi(S^2),$$
or equivalently,
$$c_1(M)-c_0(M)~=~n-1\qquad \Leftrightarrow \qquad c_1(S^2)-c_0(S^2)~=~n-2.$$
5) Lemma: The determinant ${\rm det}H(z)<0$ is negative sufficiently close to the set $Z$.
The open set
$$U := \{z\in M \mid {\rm det}H(z)>0\} ~=~\cup_a U_a$$
consists of a number of connected components $U_a$. Since the restriction $V|_{U}$ is concave, each connected component $U_a$ can contain at most one minimum point.
6) Up until now our analysis is rigorous. In the rest of the answer we speculate. Heuristically, there should exists a triangularization of the Riemann sphere $S^2$ with the set $Z$ as vertices, such that each face contains at most one minimum point. Assume that this is true. Let the number of faces, edges, and vertices be $f$, $e$, and $n$, respectively. Assume $n\geq 3$. Then
$$f-e+n ~=~\chi(S^2)~=~ 2, \qquad 2e~=~\sum_{j\geq 3} j f_j~\geq~3f, \qquad f~=~\sum_{j\geq 3} f_j,$$
implies that $\frac{f}{2}~=~\frac{3f}{2}-f~ \leq~ e-f~=~n-2$, and hence,
$$c_0(M)+1~=~c_0(S^2)~\leq~f~\leq~ 2(n-2)\qquad \Rightarrow \qquad c_0(M)~\leq~ 2n-5.$$
Then an upper bound for the number of critical points in $M$ becomes
$$c(M)~=~ c_{0}(M) + c_{1}(M) + c_{2}(M) ~=~ 2c_{0}(M)+n-1~\leq~ 5n - 11.$$
--
$^{1)}$ Naively, the Morse inequality implies that $c_{0}(M) \geq b_{0}(M)=1$, which is obviously wrong for $n=1$. This can be cured, if one takes boundary contributions in $M$ properly into account. The formulation in terms of the Riemann sphere $S^2$ doesn't have this problem, because it has no boundary.
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Very cool +1. I will try to thoroughly read this tomorrow. – EPN Nov 9 '11 at 20:51
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http://conservapedia.com/Golden_ratio
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# Golden Ratio
### From Conservapedia
(Redirected from Golden ratio)
The golden ratio is a number, commonly known as phi ($\varphi$), defined to be the larger of the two solutions to $\varphi^2 = \varphi + 1$. The value of phi is
$\frac{1 + \sqrt{5}}{2}$
Phi is an irrational number. Its decimal expansion begins
1.61803...
The golden ratio was first defined by Euclid of Alexandria around 300BC. His definition reads:
A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the lesser.
## The golden ratio in nature and art
The golden ratio appears in innumerable places in nature and art. The periods of the planets are all related by the golden ratio. The proportions of the bodies of many animals, including humans, dolphins, and ants are all based on the golden ratio. The Parthenon, Egyptian Pyramids, The Last Supper, Ark of the Covenant, Noah's ark, and even the common credit card are all based on the golden ratio.
A dodecahedron, a solid in which each side is a pentagon, is related to the Golden Ratio in that both the surface area and the volume of a dodecahedron of unit edge length are functions of the Golden Ratio. Plato described the dodecahedron as the solid which the gods used for embroidering the constellation on the whole heaven.
## Fibonacci Numbers and the Golden Ratio
Leonardo of Pisa (c1170 – 1250), commonly known as Fibonacci, was an Italian mathematician, who wrote an arithmetic treatise called Liber Abaci ( the book of the abacus) in 1202. In this book, Fibonacci describes a problem concerning rabbit popuations, the solution to which is the sequence:
```1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
```
The sequence can be continued by adding together the last two terms generated.
The numbers in the sequence are normally called Fibonacci numbers, and they have some remarkable properties, such as if the terms in the series are labelled with the natural numbers, then any Fibonacci number which occurs with a label which is prime, is itself prime. (for example, the 11th number in the sequence is 89. Both 11 and 89 are prime).
The relevance of this sequence here is that the ratio of the last two numbers generated at each stage in the Fibonacci sequence converges to the Golden Ratio. This can be checked:
• 1/1 = 1
• 2/2 = 2
• 3/2 = 1.5
• 5/3 = 1.6666..
• 8/5 = 1.6
• 13/8 = 1.625 etc.
It can be seen from this that successive ratios underestimate and overestimate the ratio, and the convergence is quite rapid.
For more on the sequence, see: http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibnat.html
## The golden ratio in art
It has been observed that the golden ratio is evident in the arrangement of branches along the stems of plants, and of veins in leaves. The golden ratio also has a place in art when used to determine what arrangements of a composition are aesthetically pleasing to the eye. Some would suggest that the "perfect" human face adheres to the golden ratio [1].
Leonardo da Vinci was a close friend of Luca Pacioli, who published a treatise (Divina Proportione) on the Golden Ratio in 1509. It has been claimed that he used the Golden Ratio in his composition of the Mona Lisa, but there is no documentation to prove this.
Salvador Dali deliberately used the golden ratio in his painting Sacrament of the Last Supper, incorporating also a huge dodecahedron around the supper table.
Le Corbusier designed an entire design system called the Modulor, based on the Golden Ratio, which aimed to give harmonious proportions to everything about a building.
## See also
Golden ratio in the solar system
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http://physics.stackexchange.com/questions/tagged/quantum-field-theory+general-relativity
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# Tagged Questions
1answer
95 views
### proper variation of action term
I have a term I want to vary by a field, $\phi$. $$`S' = \frac{-1}{2}\,\sqrt{-g}\,g^{\mu\,\nu}\,\delta\left[h(\phi)\,\partial_{\mu}\phi\,\partial_{\nu}\phi \right].$$ Is it correct to get this? ...
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### Massless Dirac equation is Weyl covariant
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### Dirac Equation in General Relativity
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### Showing that the Ricci scalar equals a product of commutators
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### Question on inflation
I have two particular questions regarding the inflationary scenario. They are: 1.) What is the physical origin of the inflaton field? 2.) Why has the potential of the inflation field its particular ...
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### Curiosity episode with Stephen Hawking. The Big-Bang
In an episode of Discovery's Curiosity with host Stephen Hawking, he claims the Big Bang event can be explained from physics alone, and does not require the intervention of a creator. 1) His ...
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### Wavefunction collapse and gravity
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### Extending General Relativity with Kahler Manifolds?
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### gravitational convergence of light
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### Is eternal inflation Lorentz invariant?
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### Laws of gravity for a universe that only consists of two objects?
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### Theory that gets rid of dark matter/energy
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### Are gravitomagnetic monopoles hypothesized?
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### General Relativity - Einstein field equation and quantum field theory
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### Interpretation of the Einstein-Hilbert action
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1answer
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### Hawking radiation: direct matter -> energy conversion?
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6answers
155 views
### General Relativity research and QFT in curved spacetime
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### Why you need a graviton when you have the higgs boson?
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### Why is Dirac Lagrangian in Curved Spacetime Weyl Invariant?
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### Gravitation and the QFT vacuum
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359 views
### Why is there a flux of radiation in the Hawking effect but not in the Unruh effect? (and other questions)
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3answers
605 views
### Does the Unruh effect violate Mach's principle?
Mach's principle says that it is impossible to tell if something is accelerating unless there is something else in the universe to compare that motion to, which seems reasonable. However, if you had ...
1answer
334 views
### Why is GR ghost-free?
I wonder how one can show that general relativity is ghost-free? By ghost I mean the negative norm state that breaks the unitarity. I think it is a well-known "fact" but I just couldn't find any ...
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701 views
### No hair theorem for black holes and the baryon number
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256 views
### Quantization of Gravitational Field: Quantization conditions
I'm begining to study Quantization of field with the second quantization formalism. I've studied phononic field, electromagnetic field in the vacuum and a generic relativistical scalar field. I ...
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628 views
### Do traversable wormholes exist as solutions to string theory?
There has been some heated debate as to whether the laws of physics allow for traversable wormholes. Some physicists claim we require exotic matter to construct wormholes, but then others counter the ...
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http://mathhelpforum.com/advanced-algebra/157066-prove-phi-1-w-v-isomorphism.html
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# Thread:
1. ## Prove that Phi^-1:W->V is an isomorphism
Suppose $V$ and $W$ are vector spaces over a field $F$.
Now suppose $\varphi :V\rightarrow W$ is an isomorphism. The inverse of this is defined as $\varphi^{-1}:W\rightarrow V$. Prove that $\varphi^{-1}:W\rightarrow V$ is an isomorphism, by verifying that it is linear and bijective.
So far, this is what I have available.
$\forall w\in W, \exists v\in V$ such that $\varphi (v)=w$. This is for the surjectivity of $\varphi$.
Additionally, the aforementioned $v$ is unique, which is for the injectivity of $\varphi$.
As such, I can define $\varphi^{-1}(w)=v$.
This implies that $\varphi(\varphi^{-1}(w))=\varphi(v)=w$ and $\varphi^{-1}(\varphi(v))=\varphi^{-1}(w)=v$
My guess is that I should do the same things for $\varphi^{-1}$, but just in reverse of $\varphi$.
What I mean is:
$\forall v\in V, \exists w\in W$ such that $\varphi^{-1}(w)=v$ (surjectivity of $\varphi^{-1}$).
Since the aforementioned $w$ will be unique, this implies that $\varphi^{-1}$ is injective.
Both of these together would imply that $\varphi^{-1}$ is bijective.
I haven't gotten to the linearity part yet, as I'd like to check to make sure I'm on the right track, or if I'm assuming too much.
2. Originally Posted by Runty
Suppose $V$ and $W$ are vector spaces over a field $F$.
Now suppose $\varphi :V\rightarrow W$ is an isomorphism. The inverse of this is defined as $\varphi^{-1}:W\rightarrow V$. Prove that $\varphi^{-1}:W\rightarrow V$ is an isomorphism, by verifying that it is linear and bijective.
So far, this is what I have available.
$\forall w\in W, \exists v\in V$ such that $\varphi (v)=w$. This is for the surjectivity of $\varphi$.
Additionally, the aforementioned $v$ is unique, which is for the injectivity of $\varphi$.
As such, I can define $\varphi^{-1}(w)=v$.
This implies that $\varphi(\varphi^{-1}(w))=\varphi(v)=w$ and $\varphi^{-1}(\varphi(v))=\varphi^{-1}(w)=v$
My guess is that I should do the same things for $\varphi^{-1}$, but just in reverse of $\varphi$.
What I mean is:
$\forall v\in V, \exists w\in W$ such that $\varphi^{-1}(w)=v$ (surjectivity of $\varphi^{-1}$).
Since the aforementioned $w$ will be unique, this implies that $\varphi^{-1}$ is injective.
Both of these together would imply that $\varphi^{-1}$ is bijective.
I haven't gotten to the linearity part yet, as I'd like to check to make sure I'm on the right track, or if I'm assuming too much.
A function is a bijection if and only if it has an inverse.
This map $\varphi$ is just a function. We know it has an inverse (as a function), which is just $\varphi$. We also know that its inverse has an inverse,
$(\varphi^{-1})^{-1} = \varphi$.
Thus, $\varphi^{-1}$ is a bijection because it has an inverse.
Do you understand that?
That said, you working did seem fine. It was just the long way round!
If you have any problems showing linearity, ask. But do try it first...
3. Originally Posted by Swlabr
A function is a bijection if and only if it has an inverse.
This map $\varphi$ is just a function. We know it has an inverse (as a function), which is just $\varphi$. We also know that its inverse has an inverse,
$(\varphi^{-1})^{-1} = \varphi$.
Thus, $\varphi^{-1}$ is a bijection because it has an inverse.
Do you understand that?
I'd forgotten that bit about bijections. Thanks for reminding me. Now just to get to the linearity part.
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http://mathoverflow.net/questions/71587/what-are-the-different-theories-that-the-motivic-fundamental-group-attempts-to-un
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## What are the different theories that the motivic fundamental group attempts to unify?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I must preface by confessing complete ignorance in the subject. I've read introductory texts about the theory of motives, but I am certainly no expert.
In http://www.math.ias.edu/files/deligne/GaloisGroups.pdf Deligne talks about (introduces?) the motivic fundamental group. But what is the purpose of this object?
Motives come up in cohomology in order to unify the different Weil cohomologies. But I only know of one way to define the algebraic fundamental group! After all, in cohomology you have a choice of coefficients (in a sheaf), but in the definition of the fundamental group (to my knowledge) there is no equivalent to this. Is the point that just as the algebraic fundamental group classifies etale covers, we can do this for other Grothendieck topologies as well? In what sense would the motivic fundamental group unify these?
The above was really one question: i. What theories does the motivic fundamental group unify, and in what sense does it do so?
I will add two more: ii. Is the existence of the motivic fundamental group conjectural?
and
iii. What purpose does the motivic fundamental group serve other than unifying? Deligne makes some references to conjectures that arise as prediction related to the motivic fundamental group. What insight does it provide?
I'm well aware that Deligne's text probably has all the answers to these questions, but I find it to be a hard read, so the more I know coming in the more I will take out of it.
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Perhaps some way of making sense out of some sort of monodromy action. Just a shot in the dark. – Spice the Bird Jul 29 2011 at 17:18
## 2 Answers
As in Birdman's comment, the motivic fundamental group is unifying the notion of monodromy action on the fibers of local systems of "geometric origin."
To explain this, let us start with the case of a field $K$. We have a semisimple $\mathbb{Q}$-linear Tannakian category $\operatorname{Mot}_K$ of (pure) motives over $K$ for which fiber functors are cohomology theories, i.e., it makes sense to have an $L$-valued fiber functor for a field $L$, and this is the same as a Weil cohomology theory for smooth proper $K$-varieties with values in $L$-vector spaces. A motivic Galois group, to my understanding, is attached to a cohomology theory/fiber functor $F$ of $\operatorname{Mot}_K$.
Then the motivic Galois group is the associated group scheme/$L$ whose representations are given by the category $\operatorname{Mot}_{K}\underset{\mathbb{Q}}{\otimes}L$, i.e., it is the group scheme of automorphisms of the fiber functor $F$. So it is "the group which acts on $F$-cohomology of (smooth projective) varieties." Since this category is semi-simple, the motivic Galois group is pro-reductive. E.g., the absolute Galois group (considered as a discrete group scheme) of $K$ acts on $\ell$-adic cohomology, so there is a homomorphism from $\operatorname{Gal}(K)$ to the motivic Galois group of $K$ corresponding to the fiber functor defined by $\ell$-adic cohomology.
For, say, a smooth variety $X$ over $K$, there should be a category of "motivic sheaves" on $X$, or rather, a semi-simple category of pure motivic sheaves contained in an Artinian category of mixed motivic sheaves. You should have e.g. an $\ell$-adic" fiber functor from the mixed category to $\ell$-adic perverse sheaves on $X$ which sends pure guys to (cohomologically shifted) lisse sheaves (alias local systems). E.g., if $K=\mathbb{F}_q$, then this is the category of pure (resp. mixed) perverse sheaves on $X$. If $K=\mathbb{C}$, this should be a full subcategory of pure (resp. mixed) polarizable Hodge modules on $X$. For any smooth proper (resp. just any) map $f:Y\to X$, there should an object in the category of pure (resp. mixed) motivic sheaves on $X$ corresponding to push-forward of the structure sheaf on $Y$.
The motivic fundamental group act on the fibers" of pure motivic sheaves on $X$. I.e., for a $K$-point of $X$, you should get a functor to the category of $K$-motives. This is a motivic incarnation of taking the fiber of a local system. Then given our cohomology theory $F$, we obtain a functor from pure motivic sheaves on $X$ to $L$-vector spaces, and the automorphisms of this functor will be the $F$-realization of the motivic Galois group of $X$.
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### 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.
A short answer: The different Weil cohomology theories do not only provide cohomology groups or rings but also come with extra structure; e.g. l-adic cohomology comes with an action from a Galois group and de Rham cohomology comes with a Hodge structure (and Hodge structures can also be expressed as being an action from some group). This extra structure varies with the cohomology theory.
The motivic fundamental group should unify these extra structures -- they all should be shadows of an action of the motivic fundamental group.
For a start see e.g. the "motivic Galois group" section on the wikipedia page here and these notes by Sujatha Ramdorai. The book by Yves Andre referenced there is maybe a good next step.
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The notes are great! – James D. Taylor Jul 29 2011 at 22:05
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http://physics.stackexchange.com/questions/16777/when-does-energy-turn-to-matter/16779
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# When does energy turn to matter?
I always hear about matter converting to energy - fusion, fission...
When does it go the other way around? What conditions lead to it? Are there reproducible experiments on this topic?
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I was going to write an answer about how endothermic nuclear reactions (in which energy turns into matter) are common in both fusion and fission reactors. But other people beat me to the punch. Suffice it to say they are, although they can't be the majority in a net-power reactor, obviously. – AlanSE Nov 10 '11 at 5:01
## 3 Answers
It is annoying to talk about mass-energy conversion, because it is too often misinterpreted to mean that energy doesn't weigh on a scale before it "turns into matter". So I will preface the answer by saying that if you heat up a gas, the heat weighs on a scale, if you burn some paper and let the heat escape, the escaping heat makes the combustion products weigh less than if they stayed hot, and if you seal up a nuclear bomb, let it explode, and keep all the products and heat inside the box, the box has the same total mass before and after the explosion.
The conversion of energy, kinetic or electromagnetic, to particles with rest mass is experimentally observed only when the energy comes in big enough clumps to create a massive particle, when the energy is low, this is going to be the lightest charged particle, the electron. Electrons can only be created along with a positively charged particle, to conserve charge, and this is almost always a positron (in rare weak interactions you can make an electron, a proton, an antineutron and an electron-antineutrino). The production of electron positron pairs only happens for very hard X-rays, or particles moving with a comparable kinetic energy, and you don't have this much energy in a single particle even in an atomic explosion. You need to accelerate particles specially.
Because of this gap between the energy of particles and the energy of the lightest charged particle, you don't usually see conversion of energy to mass in day-to-day life.
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Sigh... yet another ignorant downvote. What can you do? – Ron Maimon Nov 10 '11 at 7:50
+1, maybe the down vote is not due to ignorance but due to jealousy? I can even imagine who that person might be! ;-) – user1355 Nov 10 '11 at 15:40
@sb1: No, that wasn't it--- I had a genuine error--- I said you make a proton a neutron, an electron, and an antineutrino, which is stupid. It's an electron, proton, antineutron, and antineutrino, I didn't cross the neutron! But the downvoter could have said "it's an antineutron, silly, not a neutron!". – Ron Maimon Nov 10 '11 at 15:47
Exactly! It is just a silly mistake which one could have easily mention or even fix unless s/he is on a mission to lower your score. It's the most exact answer anyway spirit wise. – user1355 Nov 10 '11 at 15:55
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All the particle accelerator in the world are continuously (when they are operating) converting kinetic energy into matter. For example, when two protons collide at the LHC the kinetic energy of the protons is converted in to tens to hundreds of particles (matter) which the experimenters then try to characterize with their detectors. Most of these particles are very short-lived and either decay into multiple less massive particles or they may interact with the detector material and create even more particles. However there will always be a significant number of electrons, protons, positrons and antiprotons (that are all stable) as the eventual result of the collision. The positrons and antiprotons will eventually annihilate with electrons and protons but in principle if they could be kept separate from ordinary matter they would all last forever. So the original two protons can create multiple protons and electrons (and their antiparticles) and this additional matter is made from the kinetic energy of the original protons.
High energy cosmic rays that strike the earth from outer space do the same thing - the kinetic energy of the cosmic ray will be converted into additional electrons, protons and their antiparticles.
On a much smaller scale, any chemical reaction that requires energy to run will convert the energy required by the reaction into a very slight additional mass. For example, photosynthesis which takes $$6\space CO_2 + 6\space H_2O +sunlight \rightarrow C_6H_{12}O_6 + 6\space O_2$$ will convert the energy of the sunlight into a very very slight additional total mass of the product chemicals when compared to the total mass of reactant chemicals. There was a question on this topic recently but I cannot find it right now.
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I wouldn't call pair production as an example of matter/mass creation from energy, because its only temporal. The pairs will disintegrate soon after, no additional particles remain. There is some asymmetry in disintegration, which is blamed for the creation (shortly after BB) of the mass we see in universe, but was that asymmetry ever seen in experiments? – Georg Nov 10 '11 at 11:33
A photon is emitted when an electron jumps from a higher energy state to a lower energy state in an atom.
http://en.wikipedia.org/wiki/Atomic_spectral_line
In a process called pair production, a photon can turn into an electron and a positron.
http://en.wikipedia.org/wiki/Pair_production
These processes have been understood for about 50-100 years and are perfectly reproducible and measurable in a physics laboratory.
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http://math.stackexchange.com/questions/295173/fibonnaci-identity
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# Fibonnaci identity
Here's a problem that is leading me in circles.
Consider the Fibonacci number $F_n$ defined by $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n \geq 2$.
Prove that $F_{2n-1} = F_{n}^2 + F_{n-1}^2$.
Tried an induction proof that lead me nowhere fast. So far we have only been given limited identities, and they are all basically summation formulas for the odd terms, or even terms, or all terms up to n. I also though that maybe if it was a difference, the difference of squares could factor to something, but I'm just not getting it.
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This is impossible. $$F_{2n-1}^2 > F_{n+1}^2 = F_n^2 + 2F_nF_{n-1} + F_{n-1}^2 > F_n^2 + F_{n-1}^2$$ – EuYu Feb 5 at 5:52
@EuYu Thanks, I was about to ponder. – ggg Feb 5 at 5:53
It appears that the correct identity has no square for the $F_{2n-1}$ term, i.e. $$F_{2n-1} = F_n^2 + F_{n-1}^2$$ Is that what you meant? – EuYu Feb 5 at 5:57
## 2 Answers
Using Binet's Fibonacci Number Formula,
$F_r=\frac{a^r-b^r}{a-b}$ where $a,b$ are the roots of $x^2-x-1=0\implies ab=-1,a+b=1$
So, $$F_n^2+F_{n-1}^2=\frac{(a^n-b^n)^2+(a^{n-1}-b^{n-1})^2}{(a-b)^2}$$
$$=\frac{a^{2n}+b^{2n}+a^{2n-2}+b^{2n-2}-2(ab)^{n-1}(ab+1)}{(a-b)^2}$$
$$=\frac{a^{2n-1}\left(a+\frac1a\right)+b^{2n-1}\left(b+\frac1a\right)}{(a-b)^2}\text{ as } ab=-1$$
$$=\frac{a^{2n-1}(a-b)+b^{2n-1}(b-a)}{(a-b)^2} \text{ as } \frac1a=-b,\frac1b=-a$$
$$=\frac{a^{2n-1}-b^{2n-1}}{(a-b)}=F_{2n-1}$$
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This was exactly what I needed. I noticed, buried in my notes from class that the professor had briefly gone over the formula, but failed to show its use in applications. Now I feel dumb for having been told of its existence and not figuring it out.... – Christopher Ernst Feb 9 at 5:22
The identity may be derived from the interesting fact that
$$\left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^k = \left ( \begin{array} \\ F_{k+1} & F_k\\F_k & F_{k-1} \\ \end{array} \right )$$.
From this, we may observe that
$$\begin{align} \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^m \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^n &= \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^{m+n} \\ &= \left ( \begin{array} \\ F_{m+1} & F_m\\F_m & F_{m-1} \\ \end{array} \right ) \left ( \begin{array} \\ F_{n+1} & F_n\\F_n & F_{n-1} \\ \end{array} \right )\\ &= \left ( \begin{array} \\ F_{m+n+1} & F_{m+n}\\F_{m+n} & F_{m+n-1} \\ \end{array} \right ) \end{align}$$
Therefore we can say that, for example,
$$F_{m+n-1} = F_m F_n + F_{m-1} F_{n-1}$$
Plug in $m=n$ and the desired identity follows.
EDIT
The first identity quoted above follows from putting the Fibonacci recurrence into matrix form:
$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) \left ( \begin{array} \\F_{k+1} \\ F_k \\ \end{array} \right )$$
which may be immediately verified. We may repeat this matrix multiplication $k$ times to get
$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\F_{2} \\ F_1 \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\1 \\ 1 \\ \end{array} \right )$$
Noting that $F_{k+2}= F_{k+1}+ F_k$ and $F_{k+1} = F_k + F_{k-1}$, the stated identity follows.
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How did you come up with the 2x2 matrix identity above? What basis do you have to use that in your derivation? – Christopher Ernst Feb 9 at 5:21
@ChristopherErnst: this is a well-known identity, the proof of which I outlined above. I do not understand your second question. – Ron Gordon Feb 9 at 6:35
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http://mathoverflow.net/questions/8999/discrete-harmonic-function-on-a-planar-graph/9050
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## Discrete harmonic function on a planar graph
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Given a graph $G$ we will call a function $f:V(G)\to \mathbb{R}$ discrete harmonic if for all $v\in V(G)$ , the value of $f(v)$ is equal to the average of the values of $f$ at all the neighbors of $v$. This is equivalent to saying the discrete Laplacian vanishes.
Discrete harmonic functions are sometimes used to approximate harmonic functions and most of the time they have similar properties. For the plane we have Liouville's theorem which says that a bounded harmonic function has to be constant. If we take a discrete harmonic function on $\mathbb{Z}^2$ it satisfies the same property (either constant or unbounded).
Now my question is: If we take a planar graph $G$ so that every point in the plane is contained in an edge of $G$ or is inside a face of $G$ that has less than $n\in \mathbb{N}$ edges, does a discrete harmonic function necessarily have to be either constant or unbounded?
I know the answer is positive if $G$ is $\mathbb{Z}^2$, the hexagonal lattice and triangular lattice, I suspect the answer to my question is positive, but I have no idea how to prove it.
Edited the condition of the graph to "contain enough cycles". (So trees are ruled out for example)
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Maybe I am misunderstanding the question, but you can write down (many) nonconstant bounded harmonic functions on a trivalent regular tree. – moonface Dec 15 2009 at 16:37
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I think the poster wants to require that there is a constant n such that any face of R^2 \setminus G has at most n edges. The complement of the regular trivalent tree is a single face with infinitely many edges. – David Speyer Dec 15 2009 at 16:44
@moonface: You're right, that's true for most trees. I meant a different restriction on the graph, so I edited the above. @David: Yep :), I should have been more precise. – Gjergji Zaimi Dec 15 2009 at 16:53
## 5 Answers
The answer is no.
I first describe the graph $G$. Let $N_i$ be a sequence of positive integers; we will choose $N_i$ later. Let $T$ be an infinite tree which has one root vertex, the root has $N_1$ children; the children of that root have $N_2$ children, those children have $N_3$ children and so forth. Let $V_0$ be the set containing the root, $V_1$ be the set of children of the root, $V_2$ the children of the elements of $V_1$, and so forth. To form our graph, take $T$ and add a sequence of cycles, one going through the vertices of $V_1$, one through $V_2$ and so forth. (In the way which is compatible with the obvious planar embedding of $T$.)
Every face of $G$ is either a triangle or a quadrilateral.
We will build a harmonic function $f$ on $G$ as follows: On the root, $f$ will be $0$. On $V_1$, we choose $f$ to be nonzero, but average to $0$. On $V_i$, for $i \geq 2$, we compute $f$ inductively by the condition that, for every $u \in V_{i-1}$, the function $f$ is constant on the children of $u$. Of course, we may or may not get a bounded function depending on how we choose the $N_i$. I will now show that we can choose the $N_i$ so that $f$ is bounded. Or, rather, I will claim it and leave the details as an exercise for you.
Let $a_i$ be a decreasing sequence of positive reals, approaching zero. Take $N_i = 6/(a_{i+1} - a_i)$. Exercise: If $f$ on $V_1$ is taken between $-1+a_1$ and $1-a_1$, then $f$ on $V_i$ will lie between $-1+a_i$ and $1-a_i$. In particular, $f$ will be bounded between $-1$ and $1$ everywhere.
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You can probably also take any periodic tiling of the hyperbolic plane. And probably the right condition on $G$ is that it be amenable. – Greg Kuperberg Dec 15 2009 at 17:17
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Is there a definition of amenable for general graphs? I only knew it for Cayley graphs. – David Speyer Dec 15 2009 at 17:39
Yes: No infinite Ponzi scheme. It makes sense for general metric spaces too, although that's not really different. Also, in this case amenability could be stronger than strictly necessary. The discrete Laplacian is a model of random walks, and you could possibly have a non-amenable structure that is only noticed by non-random walks. – Greg Kuperberg Dec 15 2009 at 18:19
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It quickly leaps out that all of these counterexamples are infinite Ponzi schemes, so amenability is a natural condition. It is considered for instance here arxiv.org/abs/0706.2844 – Greg Kuperberg Dec 15 2009 at 20:29
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I also stumbled upon jstor.org/stable/119840?seq=1 where they also state that nonamenability is "sort of" necessary in the analogous problem of existence of non-constant harmonic functions. – Gjergji Zaimi Dec 15 2009 at 21:56
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Benjamini and Schramm proved that an infinite, bounded degree, planar graph is non-Liouville if and only if it is transient.
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For general complete Riemannian manifolds other than the plane, one need some curvature conditions to guarantee the Liouville's theorem. Similarly, for planar graphs, one need some curvature constraints too. See Geometric analysis aspects of infinite semiplanar graphs with nonnegative curvature Hua, Jost, Liu, http://arxiv.org/abs/1107.2826, where the Liouville's theorem and recurrency of random walks are proved on semiplanar graphs (graphs that could be embedded in a 2-manifold, including planar graphs) with nonnegative Higuchi's curvature, a analogue of the sectional curvature (or Ricci curvature) of a 2-manifold.
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For instance, any regular $H_{p,q}$ tessellation of the hyperbolic space $\mathbb{H}_2$ with $\frac{1}{q}+\frac{1}{q}<4$ does the job.
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another way to ensure that bounded harmonic functions on a graph $G$ are constant is to consider a random walks $X_n$: since $M_n = V(X_n)$ is a bounded martingale, it converges almost surely. Hence, if one can prove that random walks on $G$ are recurrent, this shows that $V$ has to be constant. Of course, this can be difficult to show that random walks on $G$ are recurrent.
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http://stats.stackexchange.com/questions/26082/bootstrap-variance-of-squared-sample-mean
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# Bootstrap variance of squared sample mean
The following is question 8 of chapter 8 in Wasserman's All of Statistics:
Let $T_n = \overline{X}_n^2$, $\mu = \mathbb{E}(X_1)$, $\alpha_k = \int|x - \mu|^kdF(x)$, and $\hat{\alpha}_k = n^{-1}\sum_{i=1}^n|X_i - \overline{X}_n|^k$.
Show that $$v_{\mathrm{boot}} = \frac{4\overline{X}_n^2\hat{\alpha}_2}{n} + \frac{4\overline{X}_n\hat{\alpha}_3}{n^2} + \frac{\hat{\alpha}_4}{n^3} \>.$$
He previously defines $v_{\mathrm{boot}} = \frac{1}{B}\sum_{b=1}^B(T_{n,b}^* - \frac{1}{B}\sum_{r=1}^BT_{n,r}^*)^2$, where $T_{n,i}^*$ is the desired statistic from the $i$th bootstrap replication of the sample $X_1,...,X_n$.
It seems that the question as stated does not make sense: how can there be a formula for the bootstrap variance if the quantity requires simulation? Perhaps he meant to ask for the variance of the sampling distribution, but I get $\frac{\sigma^2}{n}$ for that. Any hints on how to intepret or solve this?
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Hi Alex, welcome to the site. Here is what I think the question is asking: Find $v_{\mathrm{boot}}$ which is the variance under the (empirical) measure $\hat F_n$. The "previously defined" version is simply the Monte Carlo estimate of $v_{\mathrm{boot}}$ rather than the quantity itself. Even so, I think what you'll find is that there are at least two other typos lurking: (1) The definition of $\hat \alpha_k$ should probably not include the modulus and (2) I believe the last term on the right-hand size is not quite correct. :) – cardinal Apr 8 '12 at 19:20
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(When I looked on that page of the book, I saw what I think was at least one other typo in the question before that as well. Also, this same problem, with the same [conjectured] errors appears to be reproduced in Wasserman's All of Nonparametric Statistics as well, on page 39. Once you've completed the exercise, you might consider sending a note to the author so that he can add it to the errata.) – cardinal Apr 8 '12 at 19:23
Thank you, @cardinal. But for $v_{boot}$ I then obtain $V_{\hat{F}_n}(\overline{X}_n^2) = \mathbb{E}_{\hat{F}_n}(\overline{X}_n^4) - \mathbb{E}_{\hat{F}_n}(\overline{X}_n^2)^2 = 0$! (I reason that $\mathbb{E}_{\hat{F}_n}(\overline{X}_n^2) = n^{-2}\sum\sum\mathbb{E}_{\hat{F}_n}(X_i^*)\mathbb{E}_{\hat{F}_n}(X_j^*) = \overline{X}_n^2$ for samples from $\hat{F}_n$. Same thinking for $\mathbb{E}_{\hat{F}_n}(\overline{X}_n^4)$. I cannot figure out my error. For the previous question, do you mean that the author doesn't distinguish between the random variable $\hat{\theta}$ and its observed value? – AlexK Apr 8 '12 at 22:24
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@whuber and cardinal, thank you each for the explanations, I understand much better now. I believe I've worked out the question (the last term is indeed different) and will notify the author once I type it up. Is it accepted to post the solution here as well? Finally, how can I credit you for the help? – AlexK Apr 10 '12 at 18:39
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Hi AlexK. Yes, please do post your solution as an answer! I was hoping this would be the outcome. I do not want to speak for @whuber too much (though I doubt he'll mind in this instance), but do not be concerned with "crediting" us. I, for one, am happy to see you've arrived at a positive result and have benefitted from the site. I hope you'll continue to frequent it and participate. Cheers. – cardinal Apr 15 '12 at 2:27
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http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries/37152
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## Classification of finite groups of isometries
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Consider the problem of classifying the finite groups of isometries of R^n. --For n=2 it is cyclic and dihedral groups. --For n=3 they are well known, probably from Kepler and are related to ade-classification. --For n=4 we can get them by taking the universal cover of SO(4) which is isomorphic to SU2 x SU2, though I do not know where the classification is available.
But my main question is for dimension n=5 and above. Does anybody knows the state of the art? A reference would be most helpful. Note that the finite subgroups of GLn(Z) are classified for n<=10.
Mathieu
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You mean linear isometries, right? - Otherwise, considering all the affine isometries, in dimension 2 you have the 17 crystallographic groups. – Qfwfq Aug 30 2010 at 11:19
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The finite subgroups of SO(4) are listed in the book by Conway and Smith on quaternions and octonions. I have recently checked that it is correct and recovered it from the classification of finite subgroups of Spin(4), which we needed for a separate project: see the paper arxiv.org/abs/1007.4761 . – José Figueroa-O'Farrill Aug 30 2010 at 11:29
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You may also look at the references in this answer to a previous MO question: mathoverflow.net/questions/17072/… – José Figueroa-O'Farrill Aug 30 2010 at 12:46
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@unknown: If $G \subset GL(n,\mathbb{R})$ is finite, then we may change coordinates so that the origin is at the centre of mass of some orbit $Gx$ (for an arbitrary $x\in \mathbb{R}^n$). Then the origin is fixed by every element of $G$, so linearity follows from finiteness. On the other hand, if you want to consider discrete subgroups, then you do need to allow for the possibility that some of your isometries may be affine transformations. – Vaughn Climenhaga Aug 30 2010 at 19:35
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It's also worth pointing out that if G is any finite subgroup of GL(n,R), then one can introduce an inner product with respect to which G acts by isometries; in particular, there is an inner automorphism of GL(n,R) that maps G to a finite subgroup of Isom(n,R). Thus classifying finite groups of isometries is equivalent to classifying finite groups of linear transformations. – Vaughn Climenhaga Aug 30 2010 at 19:38
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## 4 Answers
This is one of the problems that just gets hopelessly messy beyond a few small dimensions. The reason is that asking for all finite subgroups of isometries of Euclidean space is essentially the same as asking for all orthogonal representations of all finite groups, and since irreducible representations have dimension at most the square root of the order of the group, you have to use all groups of order up to at least n2 to find groups of isometries of Rn. A major problem in doing this is that there are huge numbers of nilpotent groups of order pn once n is larger than about 5; for example there are several hundred groups of order 64, all of whose irreducible representations have dimension at most 8. So my guess would be that classifying all groups of isometries in dimensions greater than about 10 will require a lot of obstinacy and a big computer.
(Added later) On checking the literature, I find that people classifying such subgroups usually make some simplifying assumptions, by only looking for ones that are irreducible, maximal, and that act on an integral lattice. With these extra simplifications one can get a bit further: the state of the art seems to be around 30 dimensions.
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Another simplifying assumption is that of primitivity, which means that the representation is irreducible and not induced from some proper subgroup. That excludes in one fell swoop all nilpotent groups. There are still a lot of them left however. – Torsten Ekedahl Aug 30 2010 at 17:54
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I recommend using nilpotent groups as an excuse to give up. Otherwise you shortly afterwards run into solvable groups, which are much worse. – Richard Borcherds Aug 31 2010 at 4:31
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There is a vast literature on the classification of finite linear groups over various fields. Over the complex or real fields, all finite linear groups are conjugate to subgroups of the respective unitary or orthogonal group, so as remarked in one of the comments above, studying finite groups of isometries in this context is the same as studying all the finite subgroups of ${\rm GL}(n,\mathbb{C})$ or ${\rm GL}(n,\mathbb{R}).$ As Richard Borcherds remarked, this soon becomes a complicated problem. But strategies have evolved since the birth of representation theory to tackle the problem (for general fields) difficult as it is, in a systematic way. I'll discuss the real and complex cases. Generally speaking, we want to concentrate attention on linear groups which can't be described in some "obvious" way in terms of linear groups in smaller dimensions. The first reduction, then, is to concentrate on irreducible groups, those which leave no proper non-zero subspace invariant. Maschke's Theorem tells us that no information is lost in the reduction. Another question, for real representations, is what changes if we extend scalars to the complex field, where life is generally easier. An irreducible real linear group may become reducible when the scalars are extended to the complex numbers (this only happens when its character has squared-norm $2$ or $4$). In each case, the real finite linear group is isomorphic to a finite complex linear group in half the original dimension. So now I only speak of finite complex linear groups. As remarked in someone's earlier comment, the next natural reduction is to the case of primitive linear groups, those which (up to equivalence) be induced from linear groups of smaller dimension. There are strong restrictions on normal subgroups of finite primitive linear groups. In particular, the structure of primitive solvable finite linear groups is very tight, and is well-understood. Having reduced to the primitive case (back to the general finite group), the next question is whether the underlying module is a tensor product of two non-trivial modules of smaller dimension. At this point, it may be necessary to take (still finite) central extensions of the group you started with. If there is a non-trivial tensor factorization, then we are reduced to questions in smaller dimension. If there is no such factorization (even allowing for central extensions), then the structure of the residual groups is very restricted indeed. The given representation may be "tensor induced" from a representation (of smaller dimension) of a proper subgroup. Tensor induction was introduced by Serre. If it can't be tensor induced from a lower dimensional representation (again, even allowing for central extensions), then the only possibility that remains is subgroup of a central extension of the automorphism group of a finite simple group (containing all inner automorphisms). Many mathematicians, for example, Guralnick, Tiep, Zalesski, have calculated (relatively) low dimensional complex representations of (central extensions of) finite simple groups in recent years. My answer is therefore: yes, it is a difficult question, but one which can be addressed systematically in any given case, and for which much hard-won theory is available in the mathematical literature. Addendum: Just as it becomes impractical to list all groups of a given finite order relatively soon, and we have to content ourselves with understanding the "building blocks", that is, the finite simple groups, so it is with finite linear groups. There are three types of building blocks for finite complex linear groups: a) 1-dimensional cyclic linear groups. b) Finite complex linear groups $G$ of dimension $p^{n}$, for some prime $p$ and integer $n > 0$, which have an irreducible normal $p$-subgroup $E$ (extraspecial of order $p^{2n+1}$ and exponent $p$ when $p$ is odd; either extraspecial or the central product of an extraspecial group of order $p^{2n+1}$ with a cyclic group of order $4$ when $p = 2.$). In this case, $G/EZ(G)$ is isomorphic to an irreducible subgroup of the finite symplectic group ${\rm Sp}(2n,p)$. c) Finite complex linear groups $G$ of degree $m$ which have an irreducible quasisimple subgroup $S$ ( this means that $S = S^{\prime}$ and $S/Z(S)$ is a non-Abelian simple group). Then $G/SZ(G)$ is a subgroup of the outer automorphism group of $S/Z(S)$. The third type of building block naturally does not occur for solvable linear groups.
In both cases b) and c), the respective subgroups $E$ and $S$ are minimal subject to being normal, but not central.
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***@Richard and Geoff***: Why should we consider "irreducible representations" of groups? In language of representation theory, the question will be simply "To find faithfull representations of finite groups over $\mathbb{R}$"; not necessarily irreducible. – joseph Sep 10 2011 at 7:44
At least over the complex numbers, and for finite groups, all finite dimensional representations are equivealent to unitary representations, and are hence completely reducible. Hence trhey are direct sums of irreducible representations. So if we can understand irreducible representations, we can undersatnd al representations. – Geoff Robinson Sep 10 2011 at 14:37
There are a few papers by Gabriele Nebe and Wilhelm Plesken on this topic, eg:
Nebe, Gabriele Finite subgroups of ${\rm GL}_{24}(\mathbb Q)$. Experiment. Math. 5 (1996), no. 3, 163--195.
Nebe, Gabriele Finite subgroups of ${\rm GL}_n(\mathbb Q)$ for $25\leq n\leq 31$. Comm. Algebra 24 (1996), no. 7, 2341--2397.
Nebe, G.; Plesken, W. Finite rational matrix groups. Mem. Amer. Math. Soc. 116 (1995), no. 556, viii+144 pp.
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1. Surprisingly, I found explicit lists of discrete subgroups of the orthogonal group O(n) for up to n=8 dimensions on the wikipedia page for point groups, with rather unspecific references, however. Point groups is another name for discrete subgroups of O(n). [UPDATE+CORRECTION: For dimensions n=4 and larger, only the point groups which are generated by reflections (Coxeter groups) are listed. In particularly, subgroups of SO(n) (which include no matrix of determinant $-$1) are missing.]
2. There is an old sequence of two long papers by Threlfall and Seifert, part I Mathematische Annalen 1931, Volume 104, Issue 1, pp. 1-70, part II 1933, Volume 107, Issue 1, pp. 543-586, where they apparently do the classification of discrete subgroups of SO(4) by associating to each element of SO(4) a pair of rotations from SO(3). (Although my native language is German, I had a hard time reading (through) this, because I am not used to the terminology that was used at that time.) [Addition: These results are mentioned in the book by Conway and Smith on quaternions and octonions; Conway and Smith say that the list is complete, but contains duplicates.]
3. I have a rather wild conjecture (true up to three dimensions).
Every discrete point group in n dimensions is the symmetry group of an n-dimensional polytope which is the Cartesian product of regular polytopes, or a subgroup thereof.
[UPDATE: Norman Johnson pointed out counterexamples: The symmetries of the root lattices E6, E7, E8 in 6, 7, and 8 dimensions. (I could not yet fully convinced myself that they are indeed counterexamples.) So dimensions 4 and 5 remain open. If I extend my conjecture to include the polytopes which have those E6, E7, or E8 symmetries, in addition to the regular polytopes, in which dimension would the next counterexamples be?]
For example, the symmetries of an $m$-gonal anti-prism in 3-space are contained in the symmetries of the $2m$-sided prism, which is the 1-simplex $\times$ the regular $2m$-gon.
Since the regular polytopes are known in all dimensions, this would give an easy way to obtain all finite point groups. (at least in principle).
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http://math.stackexchange.com/questions/178940/proofs-that-every-mathematician-should-know/219528
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# Proofs that every mathematician should know? [closed]
There are mathematical proofs that have that "wow" factor in being elegant, simplifying one's view of mathematics, lifting one's perception into the light of knowledge etc.
So I'd like to know what mathematical proofs you've come across that you think other mathematicans should know, and why.
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I think this should be made CW. – Peter Tamaroff Aug 5 '12 at 1:10
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@PeterTamaroff so do I but my reputation isn't high enough – user10389 Aug 5 '12 at 1:45
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There is a lot to be said for not clever arguments. – André Nicolas Aug 5 '12 at 2:50
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– t.b. Aug 5 '12 at 12:56
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@All Links to proof references, as far as possible, would be very cool! Thanks the ones given so far... – draks ... Aug 5 '12 at 17:34
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## 23 Answers
Here is my favourite "wow" proof .
Theorem
There exist two positive irrational numbers $s,t$ such that $s^t$ is rational.
Proof
If $\sqrt2^\sqrt 2$ is rational, we may take $s=t=\sqrt 2$ .
If $\sqrt 2^\sqrt 2$ is irrational , we may take $s=\sqrt 2^\sqrt 2$ and $t=\sqrt 2$ since $(\sqrt 2^\sqrt 2)^\sqrt 2=(\sqrt 2)^ 2=2$.
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14
Actually $\sqrt 2^\sqrt 2$ is irrational, but this is a hard result, not needed in the elementary proof above. – Georges Elencwajg Aug 5 '12 at 8:10
2
This proof (together with some other related facts) is mentioned here and here. – Martin Sleziak Aug 5 '12 at 10:04
1
Is this proof exist the other way around? like s and t irrational and s pow(1/t) is rational – Nicolas Manzini Aug 6 '12 at 7:59
@NicolasManzini consider t = 1 / sqrt 2 – Jimmy Aug 7 '12 at 0:31
Haha that is a pretty awesome proof – ZafarS Jan 5 at 0:41
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I think every mathematician should know the following (in no particular order):
1. Summing $\sum_{k = 1}^{n} k$ using Gauss' triangle trick.
2. Irrationality of $\sqrt{2}$ by proof without words.
3. Niven's proof of the irrationality of $\pi$.
4. Uncountability of the Reals by Cantor's Diagonal Argument.
5. Denumerability of the Algebraics by Heights and Counting Roots.
6. Infinitude of primes by both Euclid's proof and Euler's proof.
7. Constructibility of the Regular 17-gon by Gauss' explicit construction.
8. Binomial Theorem by Induction.
9. FLT $n = 4$ by Fermat's Infinite Descent.
10. Every PID is a UFD.
11. The $\lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e$ by L'Hôpital's Rule.
12. Pick's Theorem by reduction to triangles and squares.
13. Fibonacci numbers in terms of the Golden Ratio by recurrence relations.
14. $\mathbb{R}^{n}$ is a metric space in more than one way.
15. Euler's Formula $e^{i \theta} = \cos \theta + i \sin \theta$ by differentiation.
16. Summing $\sum_{k \geq 1} \frac{1}{k^{2}}$ by Fourier series.
17. Quadratic reciprocity by Eisenstein's proof (counting lattice points).
18. $(\mathbb{Z}/n \mathbb{Z})^{\times}$ is a group (of units) for $n \in \mathbb{N}$, and $\mathbb{Z} / p \mathbb{Z}$ is a field for prime $p$.
19. Euler's formula $v - e + f = 2$ for planar graphs.
20. Fundamental Theorem of Algebra by Liouville's Theorem.
This is, of course, my opinion....
NB: When I write "by ..." I mean that particular proof methodology (as opposed to another perhaps easier method), because of the pedagogical benefit of that route of proof.
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16
Euclid's proof of the infinitude of primes is not a proof by contradiction. (It does not even talk of infinity explicitly; the statement is that given any list of primes, we can extend the list by adding a prime not in the list.) And the version without contradiction is in fact easier for students to understand, and IMHO more elegant. – ShreevatsaR Aug 5 '12 at 3:44
4
Absolutely, Euclid's proof is not by contradiction. Dirichlet said Euclid's proof was by contradiction, and so have zillions of otherwise respectable mathematicians. They're wrong. See my joint paper with Catherine Woodgold in the Fall 2009 issue of the Mathematical Intelligencer debunking this falsehood. Euclid's actual proof was simpler and better than the proof by contradiction conventionally attributed to him. – Michael Hardy Aug 5 '12 at 4:33
63
If this list is a general consensus, then I am a terrible mathematician... Anyone else? – Arthur Fischer Aug 5 '12 at 6:40
14
I don't think 11 is appropriately proved by L'Hospital. $\lim(1+1/n)^n$ is just the definition for $e$, and we only need to prove that it converges! – Frank Science Aug 5 '12 at 13:50
3
Hmm. I think that items 11 and 15 depend on hiding the imprecisions in the definitions of $e^x$ as well as the trigonometric functions, so while they look sleek, they wouldn't be on my list. I quite like using an infinitesimal generator for the rotation group for proving 15 anyway. Thanks for the link to Niven's proof. A gem, indeed. 14 and 18 are too trivial unless I misunderstood? – Jyrki Lahtonen Aug 5 '12 at 13:51
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Cantor's Theorem: There is no surjection from $A$ onto $\mathcal P(A)$.
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17
+1: For my entire adult life I have felt that this is the number one theorem and proof that every mathematician should know. The value of $\frac{\operatorname{bang}}{\operatorname{buck}}$ is simply off the charts. – Pete L. Clark Aug 5 '12 at 13:46
My Russian analysis professor even gave a "Soviet Poli Sci" proof of this result in the first lecture. – M Turgeon Aug 5 '12 at 16:08
@Pete Totally agree and that's why I refuse to give up until I can reproduce this result several ways on demand. I think I'm missing an enzyme or something in my neural pathways preventing it...... – Mathemagician1234 Mar 16 at 5:43
To remember the proof, the idea behind is diagonalisation process. Surprise! – enoughsaid05 Mar 22 at 15:08
2
When one of my professors proved this, he said: "You need to be able to write this prove in the snow with pee even at night when drunk" – Laugerizor May 4 at 0:42
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.... and of course the neat proof that $$\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}.$$
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7
Agreed. I suppose you know the story that some mathematician was talking to Grothendieck, alluded to this formula, and it turned out that Grothendieck was of the opinion that he had never seen it before? – Pete L. Clark Aug 5 '12 at 8:08
1
@PeteL.Clark : no, actually I didn't! :) It goes well together with the Grothendieck's prime (57?) anecdote. I believe that was Lord Kelvin to claim that how to compute the above integral had to be known by "everyone", but a quick Google search didn't produce any reference. – Andrea Mori Aug 5 '12 at 8:21
8
– Gregor Bruns Aug 5 '12 at 15:09
2
Andrea: The quote that you're looking for is probably "A mathematician is one to whom that is as obvious as that twice two makes four is to you." – Eric Stucky Aug 5 '12 at 20:21
11
@EricStucky My least favorite mathematical quote by some margin. Also, I never learned a reason why every mathematician should know this proof. – user31373 Aug 5 '12 at 21:42
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Proofs from THE BOOK is a brilliant compilation of such beautiful succinct proofs.
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24
If I said that my reaction to much of "Proofs from THE BOOK" was meh, then that would definitely be a bit harsh. But I am tempted to say it anyway, because by averaging out meh and brilliant I think one comes closer to the truth. – Pete L. Clark Aug 5 '12 at 13:33
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I had better add at least some substance to my previous comment. Someone once said that there is no beautiful proof of a theorem that is not itself beautiful. In line with this I cannot help but judge the book based on the theorems it includes rather than just the proofs. For instance, as a working number theorist I find their selection of theorems in the number theory chapter more than a little off-center. For instance they do not mention quadratic reciprocity but they discuss which binomial coefficients are perfect powers? Weird. – Pete L. Clark Aug 5 '12 at 13:41
3
On the other hand, I wrote an entire paper generalizing the first proof they give of Fermat's Two Squares Theorem, so there's definitely some good stuff in there. (I was not struck by the proof of F2ST via Thue's Lemma at the time I read it there, but only later when a colleague brought it to my attention in a different context. But that sounds like my mistake...) – Pete L. Clark Aug 5 '12 at 13:43
Here is one strategy for proving Fermat's Last Theorem: suppose you could show that $x^n + y^n = z^n$ has no nontrivial solutions ${}\bmod p$ for infinitely many primes $p$. Then any nontrivial solution over $\mathbb{Z}$ necessarily reduces to a nontrivial solution mod a sufficiently large prime, so you've proven FLT.
Unfortunately, this is false: for fixed $n$, the Fermat equation has nontrivial solutions ${}\bmod p$ for all sufficiently large primes $p$! This was first proven by Schur (I am told), and the proof uses Ramsey theory and very little actual number theory. I think this proof teaches the following valuable lessons:
• What seems like a problem in one field might be best thought of as a problem in another.
• Sometimes the way to solve a problem is to ignore a lot of its structure.
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1
– Andres Caicedo Nov 18 '12 at 7:49
Gödel's theorem was definitely a "wow" for me.
Also interesting, the proof around the non-Enumerability of $\mathbb{R}$.
In the same area, the Fact that $\mathbb{Q}$ is a dense subset of $\mathbb{R}$, despite the fact that $\mathbb{Q}$ is numerable while $\mathbb{R}$ is not. It kind of suggests that $n(\mathbb{Q}) = n(\mathbb{R}-\mathbb{Q})$ while at the same time $\mathbb{R}-\mathbb{Q}$ is infinitely bigger than $\mathbb{Q}$...
Church numerals if you're into computer science.
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4
This has received several upvotes, so perhaps we should mention that some of these proofs have names. Gödel's theorem stands alone, but it's generally Cantor's Diagonalization argument that proves the uncountability of the reals and Dirichlet's Box Principle that proves that the rationals are dense in the reals. – mixedmath♦ Aug 6 '12 at 1:36
I would say the proof of the Brouwer Fixed Point Theorem for $D^n$ using the fact that $H_n(S^n) \cong \Bbb{Z}$ and $H_n(D^n) = 0$ is nice. The idea of the proof by contradiction that if for no $x$ is $f(x) = x$, we can draw a straight line through these points, that for me was very elegant.
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4
The essence of this proof can also be found in Milnor's (p.13ff) where the crucial lemma is attributed to Hirsch, , Proc. Amer. Math. Soc. 14 (1963), 364-365. Note: The simplicial version of Hirsch's original argument was recently subjected to criticism. – t.b. Aug 5 '12 at 12:09
I'm particularly fond of Ramsey's Theorem.
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Are you fond of the theorem or the proof? – user10389 Aug 7 '12 at 22:18
Both. But the proof moreso because the R(3,3)≤6 proof is so cool and then it's basically "Well what if we just tried to do *more* of that?" – Eric Stucky Aug 7 '12 at 23:19
The proof of the Fundamental Theorem of Algebra via Liouville's theorem is short and sweet.
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1
– Jesse Madnick Nov 18 '12 at 6:28
Ok,you HAVE to tell me how the heck to do THAT,Jesse.......... – Mathemagician1234 Mar 23 at 22:47
I would have to include (at least) one of the proofs available for quadratic reciprocity. My personal preference would be for the proof due to Eisenstein presented in Ireland and Rosen, but there are so many others to choose from.
A second one I would include would be Minkowski's lattice point theorem, as proved in Hasse's "Number Theory".
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The ultrafilter proof of Tychonoff's theorem.
The proof is simple, show the power of working with filters and incorporats a good deal of what "everyone should know about compactness".
The strategy-stealing argument for why the first player can force a win in hex.
The argument is simple, elegant, clever and there is eesentially no effort in learning it.
The proof of Zorn's lemma by way of ordinals.
Too many people believe that Zorns lemma is an inherently incomprehensible black box. It is not.
Heine-Borel by "induction."
The argument is very neat and shows exactly where the completeness of $\mathbb{R}$ matters.
The visual argument for finding the area of a circle, given radius and circumference.
It's simply beautiful.
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– Qiaochu Yuan Aug 10 '12 at 0:50
+1 for some subtly sophisticated and original proofs of some very standard results that are not common knowledge among graduate students and up. They are also very good examples of why a comprehensive working knowledge of such taken-for-granted basics as point set topology, modern logic and axiomatic set theory can give some very powerful tools to the mathematican for new takes on old results. – Mathemagician1234 Mar 16 at 5:39
$$2+2=4$$
Of course I am talking in the context of Peano Axioms (or some other reasonable theory of arithmetics).
Indeed most mathematicians could come up with the proof in a matter of minutes, after seeing the axioms, the trick of course is to understand what is there to prove here anyway?
We defined $+$ by induction. We denote by $2=S(1)$ and $4=S(S(S(1)))$. Now we need to prove that the terms $S(1)+S(1)$ and $S(S(S(1)))$ are equal, because there is no axiom tell us that directly.
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The Weyl Character formula is an excellent example of a deep result with a clever proof. The result states (in one form) that the irreducible characters of a compact, connected Lie group are parametrized uniquely by their heighest weight vectors! The intuition is that characters on a compact, connected Lie group $G$ are class functions on $G$ and their restrictions to a maximal torus of $G$ are $W$-invariant functions on $T$ where $W$ is the Weyl group of $T$. The $W$-invariance of a character on $T$ allows you to parametrize it by a heighest weight vector using the theory of roots and weights. However, the clever point of the proof is that one studies $W$-anti-invariant functions on $T$ rather than $W$-invariant functions on $T$! The quotient of two $W$-anti-invariant functions on $T$ is a $W$-invariant function on $T$. I think that this is a deep and extremely important result in mathematics with a clever proof.
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I personally believe some of the proofs of Pythagoras' theorem can be both beautiful and elegant, though it is unfortunate that it is not taught in school (at least as far as I am aware).
Take any square with sides of length $x+y$. Then $x$ units from each corner, connect to the next corner, again $x$ units away. Call this distance $z$. Therefore you have a square with side length $x+y$ with four triangles with base and height $x$ and $y$ and a smaller square in the middle with a side length of $z$.
$$(x+y)^2=4\frac{1}{2}xy+z^2$$ $$x^2+y^2+2xy-2xy=z^2$$ $$x^2+y^2=z^2$$
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1
I'm not sure why this got downvoted so hard, since it hasn't been edited, and it gives an honest answer to the question. I can see people not agreeing with it, but… – Eric Stucky Aug 7 '12 at 23:24
@EricStucky I can understand why it was downvoted as it was possibly a bit simplistic for the site, but believe that it is elegant, and unfortunate that (as far as I know from my friends) is not massively widely known. – jClark94 Aug 8 '12 at 10:20
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@Eric: I didn't downvote this answer, but...I believe the culture of downvoting answers to community wiki questions is a bit different than that of other questions. For an "ordinary" question one should downvote an answer only when there is something objectively incorrect there. For an answer to a cw question the general sentiment seems to be that you may downvote to indicate a difference of opinion or taste. In any case, one should take such downvotes less personally... – Pete L. Clark Aug 9 '12 at 4:03
I actually don't really understand the difference between the "main site" and the community wiki; would you mind explaining (or an appropriate link)? – Eric Stucky Aug 11 '12 at 3:19
@EricStucky As I understand it, CW questions have too many different answers (like this one), or several linked yet separate answers (such as a question on tikz libraries and their uses), so no definite answer can be given, or no "right" answer, such as a code golf question. I'm probably wrong, but if anyone can explain where I'm wrong in this theory, I'd be grateful – jClark94 Aug 11 '12 at 18:10
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As a beginner (and far from being a mathematician) there are two proofs that I have come across that I would say, for me, were "symphonic" capers of some areas of study - showing what can be done with material you've studied.
An additional point is that the actual presentation of the proofs themselves were instructive in their elegance. Sort of like a virtuoso performer.
The Stone-Weierstrass Theorem - Vaughan Jones https://sites.google.com/site/math104sp2011/lecture-notes
The Sylow Theorems - Benedict Gross http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra
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The full classical proof of the classification theorem of compact surfaces has always been-and remains-one of my favorite proofs. Despite it's tediousness, it demonstrates to the beginner how important it is to be able to prove results constructively,using very little beyond the definitions of a surface and the fundamental group.
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Agreed; also, Massey's text does a great job with this proof, imo! – james Aug 11 '12 at 20:30
2
@james So does John Lee in his terrific INTRODUCTION TO TOPOLOGICAL MANIFOLDS. In fact,the version of the proof in the second edition is even nicer! – Mathemagician1234 Aug 31 '12 at 22:40
Perhaps geometric and algebraic proofs of the fundamental theorem of calculus.
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1
Where would one find those? – Peter Tamaroff Nov 18 '12 at 5:59
@PeterTamaroff en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/… ? – Argon Nov 18 '12 at 15:09
Euclid's proof. Very simple, very elegant
Theorem
There are more primes than found in any finite list of primes.
Proof
Call the primes in our finite list $p_1, p_2, ..., p_r$. Let P be any common multiple of these primes plus one (for example, $P = p_1p_2...p_r+1$). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime, call it p. Notice p can not be any of $p_1, p_2, ..., p_r$, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete.
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This appeared on the list already. – Asaf Karagila Mar 16 at 0:33
@AsafKaragila Sorry, I didn't notice that. – Anil Baseski Mar 16 at 0:55
When I did my first analysis course I found the proof the Lebesgue differentiation theorem using maximal functions and covering lemma arguments to be very beautiful.
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My all time favorite proof?
Furstenberg's proof of the infinity of primes by point-set topology. I know, a lot people think it's not a big deal. I think:
a) It's an immensely clever way to use point set topology to prove a result in a seemingly unrelated field, namely number theory.
b) I used it as the beginnings of my first research in additive number theory; looking to generalize this result to create similar proofs of results for sumsets and arithmetic progressions. Sadly, my health failed again,but I hope to return to this research soon.
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It is propably not something which everybody should know, nevertheless it is simply beautiful!
The stable Hurewicz theorem using that the sphere spectrum is a compact generator of the stable homotopy category. In particular Serre's theorem that the rational stable homotpy groups of spheres are trivial for degrees bigger than 1.
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I really like the simple and nice proof of the 5-color theorem (i.e. that for every planar graph there exists a vertex coloring with not more than 5 colors) and how surprisingly difficult it is to proof the sharper 4-color theorem.
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## protected by Asaf KaragilaAug 6 '12 at 0:27
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http://math.stackexchange.com/questions/tagged/differential-operators+differential-geometry
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# Tagged Questions
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### Complex differential geometric form of the Grothendieck–Hirzebruch–Riemann–Roch theorem
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### differential operator on manifold
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### Are there n-th roots of differential operators?
In analogy to a Dirac operator, it seems to me that formally, the equation $$\frac{\partial^n}{\partial x^n}f(x,y)=D_yf(x,y)$$ is solved by $$f(x,y)=\exp{(x \sqrt[n]{D_y})}\ g(y).$$ Is there a ...
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On a complex manifold, if we are dealing with the $d$ operator, there's a pretty easy way of showing some form is not $d$-exact, simply by integrating in a closed loop. If you can find a loop that is ...
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### What does ad$f$ mean, for $f$ a smooth function?
I am currently reading Nicole Berline "Heat Kernels and Dirac Operators". On page 64 Differential Operators are introduced that are generalized from operators acting on scalar functions to vector ...
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http://physics.stackexchange.com/questions/32178/liquid-crystal-polarizes-light-reflection-question
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# Liquid crystal polarizes light reflection question
I was hoping someone could help me with understanding why a row of polarizes reflects a light wave when the whole row is the same length as the wavelength of the light.
I pretty much get the physics behind the rest, just don't understand this little part. Why only the light of that wavelength and does this also apply for light coming in at a different phase than the crystals?
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## 2 Answers
The picture is misleading because it makes it look as if all the reflection occurs at the surface. In fact, reflection is occurring throughout the volume, and only for a particular wavelength will all the reflections add "in phase", for other wavelengths they will partially cancel each other.
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But how exactly? The light with the correct wavelength passes trough unafected, while the rest gets polarized and whatnot. After that most of these crystals have a black layer to absorb everything that comes trough, but why doesn't the light of the correct length get absorbed? – Coolcrab Jul 16 '12 at 18:58
Each layer partially reflects light, partially transmits. If the wavelength and polarization are right, the reflections enhance each other. It's kind of like Bragg reflection. – user1631 Jul 16 '12 at 20:24
Thanks, I understand now! – Coolcrab Jul 17 '12 at 8:37
actually, reflection only happens at surfaces, namely at borders of two phases with different refractive indices. – gigacyan Jul 17 '12 at 17:13
The question is somewhat mixed so I will try to restate it as I understood it and you should correct me if I got it wrong.
As I see, the question here is: why does a thin film of liquid crystals changes its color as it gets colder or warmer?.
The answer to this question is interference. The light is reflected from both the top and the bottom surface of the film and the two reflections will interfere with each other. If for some wavelength accumulated phase difference between reflections is close to a multiple of $2\pi$ then the interference is constructive and this color will be enhanced in reflecting light.
The phase difference difference depends on the path length, and when the film gets thicker, the wavelength of reflected light will shift towards red.
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What you say would be correct for reflection from a film of homogeneous material, but that is the case under consideration here. The resonant condition in question is dependent on the pitch of the cholesteric liquid crystals, not the total thickness of the sample. – user1631 Jul 17 '12 at 21:25
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http://physics.stackexchange.com/questions/tagged/geometry+string-theory
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# Tagged Questions
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### Why is the world sheet of an open string a cylinder?
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### Generalized Complex Geometry and Theoretical Physics
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### Question about associative 3-cycles on G2 manifolds
Let $X$ be a manifold with $G_2$ holonomy and $\Phi$ be the fundamental associative 3-form on $X$. Let $*\Phi$ be the dual co-associative 4-form on $X$. Now consider a particular associative 3-cycle ...
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http://mathhelpforum.com/calculus/157763-derivative-problem.html
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# Thread:
1. ## Derivative Problem
Would someone tell me how would you solve the following:
1) Find $\frac{\partial z}{\partial y}$
$z=sin(x^3 e^{y}-4y^5)$
this is what i have done, but i don't think its correct.
$\frac{\partial z}{\partial y}=x^3 e^y-20y^4cos(x^3 e^{y}-4y^5)$
P.S
2. $z = \sin{(x^3e^y - 4y^5)}$.
Let $u = x^3e^y - 4y^5$ so that $z = \sin{u}$.
$\frac{\partial u}{\partial y} = x^3e^y - 20y^4$.
$\frac{dz}{du} = \cos{u} = \cos{(x^3e^y - 4y^5)}$.
Therefore $\frac{\partial z}{\partial y} = (x^3e^y - 20y^4)\cos{(x^3e^y - 4y^5)}$.
I agree with your answer, as long as you put the brackets where they're needed.
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http://scicomp.stackexchange.com/questions/1982/problems-that-can-be-reduced-to-the-traveling-salesman-problem
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# Problems that can be reduced to the Traveling Salesman Problem
Which search/optimization problems can be reduced to the famous "Traveling Salesman Problem"? For instance, I have a collection of N particles, in 3D, and there is a function (Van der Waals energy) which depends on their coordinates. I want to find which configuration of the system minimizes the value of the function. Can this problem or type of problems by reduced to TSP? Could you point me to some bibliography?
PS: How could this problem be reduced to TSP?
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If you are looking for a technique that can be used to solve the TSP problem then one popular method is simulated annealing. It will give you the global minimum but is very expensive unless you can come up with a cheaper version of your problem (i.e., solve an approximate problem instead of the actual one). If your problem has a single minima then any gradient based method will work and will be much cheaper. Having said that, all global optimization problems involve searching for a global minima or maxima. In that sense all such problems are like the TSP. – stali Apr 20 '12 at 12:27
Is the function (and its derivatives) continuous? Are there any restrictions on how you select the N points? For example, can you select all N points at the global minimum? Do the N particles have to be located at distinct (x,y,z) points? – Paul♦ Apr 20 '12 at 13:23
It doesn't seem apparent to me that your problem is a variant of TSP... As @stali noted, your problem seems to be one of global optimization. Do you have any need to find the shortest path that connects to every point and returns to its point of origin? – Paul♦ Apr 20 '12 at 13:27
3
The general procedure for determining if a problem is in the class of so-called NP problems is to prove bidirectional equivalency through polynomial transformations. It's a little hard to even guess whether this is possible without more information about the function you're considering. – Aron Ahmadia Apr 20 '12 at 15:46
## 2 Answers
Depending on which version of the Traveling Salesman Problem (TSP) you're looking at, it's either NP-hard (finding the shortest possible route), or NP-complete (called the decision version of TSP; given a length $L$, determine if any tour of a given graph has length less than or equal to $L$).
The difference between the two is subtle, but important: NP-hard problems are at least as hard as any problem in NP. The way you show a problem is at least as hard as another problem is to use what's called a reduction. A reduction is a way of transforming your given problem into the problem you'd like to compare it with. In this case, if you want to compare TSP to, say, a known NP problem like 3-SAT (satisfiability of a Boolean statement consisting of clauses containing 3 literals), you would try to reduce TSP to 3-SAT. The idea is to convert an instance of a problem $a$ that we don't yet know how to solve into a problem $b$ that we do know how to solve, such that the solution to $b$ can be used to determine a solution to $a$. Then, by converting $a$ to $b$ and solving $b$, we've constructed an algorithm to solve $a$. If we know how long the conversion process -- that is, the reduction -- takes, and we know how long it takes to solve problem $b$, then we can figure out an upper bound on how long it takes to solve the problem $a$.
The most common and useful kind of reduction I saw in my introductory graduate-level theory of computation course was called a polynomial-time reduction, which is a reduction process whose algorithm takes polynomial time, as a function of some metric of problem size, to produce its output. This concept is then used to define what NP-complete means. A problem $p$ is NP-complete if it is in NP, and for every problem in NP, there exists a polynomial-time reduction from that problem to problem $p$. NP-hard can be defined a similar way: a problem $p$ is NP-hard if there exists a polynomial time reduction from an NP-complete problem to $p$.
Now that all of that terminology is out of the way, to answer your question:
• If you're talking about the NP-complete decision version of TSP, then any problem in NP can be reduced in polynomial time to TSP. Global optimization problems tend to be NP-hard (though I don't know for sure that all of them are, I do know that nonconvex optimization problem are NP-hard). Since NP-hard problems by definition are polynomial time reductions of NP-complete problems, TSP can be polynomial time reduced to NP-hard global optimization problems.
• If you're talking about the NP-hard version of TSP, then no statements can be made about reductions of NP-hard global optimization problems.
• There is nothing to suggest that NP-hard optimization problems can be reduced to either version of TSP. However, it is known that 0-1 integer programming is an NP-complete problem, so it can be reduced to TSP.
A good, but dense, reference is Michael Sipser's Introduction to the Theory of Computation, 2nd edition. (Disclaimer: Professor Sipser taught my theory of computation class.)
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What you're trying to compute, if I understood your question correctly, is usually known as Lennard-Jones cluster minimization. A quick search in Google Scholar will give you a list of the most relevant publications on the topic.
Having worked on this problem myself years ago, I don't think that this can be modelled as a TSP. If you consider a path over all particles using the Lennard-Jones/Van der Waals potential as the distance measure, then, for each particle, only the energy to two other particles is taken into account, although it interacts with all neighbours. Furthermore, your interaction potential may permit negative values, leading to negative distances, which may cause problems for most TSP algorithms.
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http://mathoverflow.net/questions/92119/duality-of-eta-product-identities-a-new-idea
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## Duality of eta product identities: a new idea?
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Looking at the collection of Eta Function Product Identities by Michael Somos, it seems like generally those identities come in pairs: let's call two eta product identities $\sum\limits_{i=1}^r a_iP_i=0$ and $\sum\limits_{i=1}^r b_iQ_i=0$ dual if the eta products $Q_i$ can be re-labeled and the factors of each product re-arranged such that for all $i$, we can transform $P_i$ into $Q_i$ by replacing each argument $q^k$ or $k\tau$ of an eta function ($k\in\mathbb N$) by $q^{n/k}$ or $(n/k)\tau$. Note that the scalar coefficients $a_i$ and $b_i$ need not be the same, but they should be non-zero of course.
Example: The first two identities on the n=14 page, both of degree $12$, are
````q14_12_44 = +1*u1^7*u2*u7^3*u14 +7*q*u1^3*u2*u7^7*u14 -1*u2^8*u7^4 -7*q^3*u1^4*u14^8 ;
q14_12_64 = +1*u1^8*u14^4 +7*u2^4*u7^8 -56*q^2*u1*u2^3*u7*u14^7 -8*u1*u2^7*u7*u14^3 ;
````
re-arranged and written in a more humanly-readable form, putting $s_k=\eta(k\tau)$, we have
q14_12_44$\iff\qquad s_1^7s_2^{\ }s_7^3s_{14}^{\ } +7s_1^3s_2^{\ }s_7^7s_{14}^{\ } -s_2^8s_7^4 -7s_1^4s_{14}^8=0$
q14_12_64$\iff -56s_{14}^7s_7^{\ }s_2^3s_1^{\ }-8s_{14}^3s_7s_2^7s_1^{\ } +7s_7^8s_2^4+s_{14}^4s_1^8=0$
so both identities are duals of each other.
It should be easy to see that an eta product identity (edit: which is not dual to itself) cannot have two (linearly independant) duals. Is that really easy to see?
An eta product identity can also be self-dual, e.g. the known linear identities mentioned in my recent MO thread or the one equivalent to the well-known theta identity $\theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)$, which becomes, when written in terms of $\eta$'s, $$s_2^{24}=s_1^{16}s_4^8 +16s_4^{16}s_1^8 .$$
This concept of duality seems so logical that I wonder: Has there already been any research on that? Has each eta product identity a dual counterpart, unless it is self-dual ?
In the collection mentioned above, for $n=22$ there is only one identity listed, and that one is not self dual, it is even highly asymmetrical. Its dual should have the label x22_19_124. Maybe that one is missing just because the computer search had to stop at some point.
Given that self-dual identities have a higher degree of symmetry: can they always be expressed in terms of Ramanujan's psi, phi, and/or chi functions?
I would think the answer is yes, as there are even some not self-dual ones that can be expressed in that way.
Edit concerning the first question: Even this may be trickier than expected. For example, there are two self-dual identities, labeled q6_14_36a and q6_14_36b. They are:
$s_1^9s_3^{\ }s_6^4 +\ \ 6s_1^4s_2^{\ }s_6^9 +2s_2^9s_3^4s_6^{\ } -3s_1^{\ }s_2^4s_3^9=0$
$s_1^9s_3^{\ }s_6^4 +12s_1^4s_2^{\ }s_6^9 -4s_2^9s_3^4s_6^{\ } +3s_1^{\ }s_2^4s_3^9=0$.
They contain exactly the same $\eta$-products and so can also be considered as dual to each other (and to linear combinations of them). Moreover, by linear combinations we may eliminate any one of the four terms, resulting in four 3-term identities with coefficients
```` 0 1 -1 1
-1 0 -8 9
1 8 0 -1
-1 -9 1 0.
````
Note that each of those has exactly one dual counterpart again, as they come in two dual pairs.
So the above conjecture of a unique dual doesn't necessarily hold if the given identity is already self-dual.
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## 2 Answers
This looks like the Fricke involution to me. Given any positive integer $M$, define $W_M$ to be the operator given by slashing with the matrix $$\left(\begin{array}{cc} 0 & -1 \newline M & 0 \end{array}\right),$$ or, equivalently, if $f(z)$ is weight $k$ modular, by $f(z)\mid_k W_M = (-iz\sqrt{M})^{-k}f\left(\frac{-1}{Mz}\right)$. This is the weight $k$, level $M$ Fricke involution. It preserves $M_k(\Gamma_0(M))$, albeit changing the character.
Now, using the transformation properties of $\eta(z)$, in particular that $\eta(-1/z)=(-iz)^{1/2}\eta(z)$, it's possible to see that hitting any $\eta$-product with the appropriate Fricke involution takes it to another $\eta$-product, perhaps after scaling. Thus, hitting an $\eta$-product identity with the Fricke involution produces another identity, which is moreover dual to the original. At least the first example you give has this property, i.e., it follows from the Fricke involution.
As to the question of unique duals, you've already seen that things get wonky if you allow for self-dual identities. But even if you require the identities to not be self-dual, things are still bad (although perhaps unsatisfyingly so). Let $I_1$ and $I_2$ be two linearly independent, not dual to each other, and not self-dual identities. Let $I_1^\prime$ and $I_2^\prime$ denote their duals, respectively (if you want, just take them to be images under the Fricke involution). Consider now the identity $I=I_1+I_2$. By your definition of dual, any of the identities $I^\prime=aI_1^\prime+bI_2^\prime$ will be dual to $I$, and so the dual-space is at least two-dimensional. Modifying this in the obvious way suggests that the dimension of the dual-space can be arbitrarily large, with the requisite $I_1, \dots, I_n$ being of the form $I_1^aI_2^b$ varying over $a+b=n-1$, say.
I think the right way to phrase the uniqueness question is by asking whether there is a set of primitive $\eta$-product identities from which all other identities can be obtained, with the property that every dual of an identity is generated by Fricke involutions of the constituent primitive elements of the original identity. In particular, if an identity is primitive, is its dual-space one-dimensional? I don't know the answer to this question.
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### 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 would guess that this is just due to the fact that taking the dual preserves the space (i.e. if P_1 and P_2 are in the same space, the duals Q_1 and Q_2 will be as well, or at least some same space, maybe the level goes up a bit, I can't remember) of the eta quotient, and that this space is finite dimensional.
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I suppose that's about the question of unique duals? It's not that simple. Even once we have established that a dual always exists, a simple "infinite descend" argument may not necessarily work. See my edit above. – spanferkel Mar 25 2012 at 21:31
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http://crypto.stackexchange.com/questions/6052/why-work-in-a-subgroup-qrn-of-an-rsa-group-z-n?answertab=oldest
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Why work in a subgroup QR(n) of an RSA group $Z^*_n$?
I sometimes read in papers that a (sub-)group generator $g$ is taken from $\mathrm{QR}(n)$ instead of $\mathbb{Z}^*_n$, where $n = p \cdot q$ and $p$ and $q$ are prime. Is there a reason for this? What properties does $\mathrm{QR}(n)$ have? Is it an especially big subgroup?
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1 Answer
Well, the reason that a specific cryptographical object needs to work in a specific subgroup probably has to do with the details of that object, and the cryptographical properties it needs from the subgroup.
One obvious possibility is that they need to avoid leaking information via the Jacobi symbol; that is an easily computed function that maps values in $Z^*_n$ into the values in the set $\{1, 0, -1\}$; it is interesting because the RSA operation preserves it; $Jacobi(x) = Jacobi(RSA(x))$ (where $RSA$ is either the public or the private operation, and $Jacobi$ implicitly relies on $n$).
Depending on what you're doing, this property may allow people to track which inputs refer to which outputs (for example, if you're shuffling values by encrypting them, and then randomly permuting them). One way of making sure this sort of thing doesn't lead to a problem is the specify that all input values are in $QR(n)$; all these values have Jacobi symbol 1, and so there's no leakage.
It's also easy to create random values in $QR(n)$; pick a random vale $r$ relatively prime to n, and compute $r^2 \bmod n$; that's a random value in $QR(n)$
As for the size of $QR(n)$, well, that'd be $(p-1)(q-1)/4 \approx N/4$; obviously, that's a fairly sizable group.
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thanks a lot, this helps... – user4811 Jan 21 at 19:33
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http://math.stackexchange.com/questions/153238/finding-a-point-on-the-line-joining-two-points-in-cartesian-coordinate-system?answertab=votes
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# Finding a point on the line joining two points in Cartesian coordinate system
Given a point A with coordinates (a,b) ; a point B with coordinates (c,d) , I want to find a point C's coordinate (x,y) ,with C lying on the line joining A and B and C is at a distance Dist from point A. This is the representation
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## 1 Answer
You will want to use Linear Interpolation. You can interpolate both the x and y coordinate independently. Let $\alpha = Dist/\sqrt{(c-a)^2+(d-b)^2}$. Then the interpolated point is given by:
$$x = a + \alpha(c-a)$$
and
$$y = b + \alpha(d-b)$$
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http://mathhelpforum.com/calculus/207725-partial-fractions.html
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1Thanks
• 1 Post By hollywood
# Thread:
1. ## Partial fractions
Please explain the step in the green box(#5), I remember doing this in algebra 2 but not this specifically.
Attached Thumbnails
2. ## Re: Partial fractions
Which bit don't you understand?
Incidentally substituting x=i (as well as x=0 and x=1) is quite handy too if you're happy with complex numbers. It gives you C and D and ten using the firs two equations you can also get A and B.
3. ## Re: Partial fractions
Specifically equating coefficients with like terms after "foil"ing, I see this in other problems and I'm not sure why its there or why it makes sense exactly.
Also, I am comfortable with imaginaries, but doesnt that needlessly make it more complicated? or from your perspective do you find it simpler by combining the i's, but what happens whenyou go back to reals when not all i's become an even power of i?
4. ## Re: Partial fractions
Using x=i gives you $1=4(Ci+D)$.
On the left you have $1+0i$
On the right you have $4D+4Ci$
Equating real and imaginary parts you see that 4D=1 and 4C=0.
I may get back to your other points later..
5. ## Re: Partial fractions
Starting with:
$1=(Ax+B)(x^2+1)+(Cx+D)(x^2+5)$
multiply and add the polynomials to get:
$1=(Ax^3+Bx^2+Ax+B)+(Cx^3+Dx^2+5Cx+5D)$
$1=(A+C)x^3+(B+D)x^2+(A+5C)x+(B+5D)$
You need these to be equal as polynomials, so in other words, all the coefficients must be equal:
$0=A+C$
$0=B+D$
$0=A+5C$
$1=B+5D$
Usually when you see equations like this, you're trying to figure out which values of x make the equation true, but here you want 1, which is $0x^3+0x^2+0x+1$, to be the same polynomial as $(A+C)x^3+(B+D)x^2+(A+5C)x+(B+5D)$, so that they are equal for all values of x.
The rest is using these four equations and the two equations from the previous part to solve for A, B, C, and D.
Step 4 is actually unnecessary, by the way; the four equations in step 5 are all you need to solve for the four unknowns A, B, C, and D.
- Hollywood
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http://math.stackexchange.com/questions/138090/solving-quadratic-residue/138100
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# Solving quadratic residue
Suppose we have $25x^2 + 70x + 37 \equiv 0 \pmod{13}$. Since it doesn't factor we obviously have to subtract/add $(ax + b)$ from both sides of the congruence. However I'm getting different answers. What is the proper way to approach solving for all residues that solve this congruence?
Thanks!
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If you have a root mod 13, the quadratic will factor mod 13. – Mark Bennet Apr 28 '12 at 16:43
## 2 Answers
The prime $13$ is very small. A useful strategy may be to try everything. Or not quite everything, since we know that a quadratic congruence modulo a prime has at most $2$ solutions. If we reduce our coefficients modulo $13$ to make calculations easier, we can fairly quickly find the solutions $x\equiv 8\pmod{13}$, $x\equiv 10\pmod{13}$.
Although for small primes "trial and error" is efficient, we will examine an approach through general theory. Consider the congruence $ax^2+bx+c\equiv 0\pmod{p}$, where $p$ is prime, and $a\not\equiv 0\pmod p$. Multiply through by $4a$. We get the equivalent congruence $$4a^2x^2+4abx+4ac\equiv 0\pmod{p}.$$ The purpose of multiplying through by $4a$ is to make completing the square easy. The above congruence can be rewritten as $$(2ax+b)^2-b^2+4ac\equiv 0\pmod{p},$$ or equivalently $$(2ax+b)^2\equiv b^2-4ac\pmod p.$$ Now we turn to our particular case. The supplied coefficients are largish. It is useful to note that $25\equiv -1$, $70\equiv 5$, and $37\equiv -2$, all modulo $13$. Our congruence is equivalent to $$(-2x+5)^2\equiv 17\equiv 4 \pmod{13}.$$ So we need to solve $y^2\equiv 4 \pmod{13}$, $-2x+5\equiv y \pmod{13}$.
We got a little lucky, the solutions of $y^2\equiv 4\pmod{13}$ are $y\equiv \pm2\pmod{13}$. Now solve the congruences $-2x+5\equiv 2\pmod{13}$, $-2x+5\equiv -2\pmod{13}$.
Remark: For large primes $p$, one can use exactly the same strategy to arrive at the system $y^2\equiv b^2-4ac\pmod{p}$, $2ax+b\equiv y \pmod{p}$. The only place where there is computational difficulty is in determining whether the congruence $y\equiv b^2-4ac\pmod{p}$ has a solution, and if it does, finding one.
Note also that, with suitable interpretation, what we did amounts to deriving the Quadratic Formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. Of course, square root has to be interpreted modulo $p$, as a solution of the congruence $y^2\equiv b^2-4ac\pmod{p}$. And division by $2a$ should be thought of as multiplication by the multiplicative inverse, modulo $p$, of $2a$.
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Hint $\rm\: mod\ 13\!:\ 0 \equiv25\:x^2 + 70\:x + 37 \equiv -(x^2 -5\:x+2).\:$ Applying the quadratic formula
$$\rm x\equiv \dfrac{5\pm\sqrt{17}}2\equiv \dfrac{18\pm\sqrt{4}}2\equiv\{10,8\}\ \ \ since\ \ \ 5\equiv 18,\ 17\equiv 4$$
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http://mathoverflow.net/questions/19213?sort=newest
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Changing the orientation of a Landweber exact cohomology theory
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Let the ring R be a MU`*`-module via a ring homomorphism φ and suppose it satisfies the condition of the Landweber exact functor theorem such that we obtain a cohomology theory `$R^*(-) := R \otimes_{MU_*} MU^*(-)$`. If ω denotes the complex orientation class in $\widetilde{MU}^2(\mathbb{C}P^\infty)$, then R`*` is oriented by the class $\omega_R := 1 \otimes \omega$.
Any other complex orientation of R`*` is obtainable by homogeneous power series θ with leading term x over R: θ(ω). These power series are in 1-1 correspondence with multiplicative natural transformations `$t_\theta\colon MU^*(-) \to R^*(-)$`.
Question: Which tθ restrict to ring homomorphisms which satisfy the Landweber criterion on coefficients? For which theories is this true for any θ?
The place to start seems to be by noting that if the formal group law associated to R`*` (with the orientation given by ωR) is F, then tθ classifies the FGL `$F^\theta(x,y) := \theta\big(F(\theta^{-1}(x),\theta^{-1}(y))\big)$` over R. Further, the p-series are related by `$[p]_{F^\theta}(x) = \theta\big([p]_{F}(\theta^{-1}(x))\big)$`, so it would suffice to show that the sequence of coefficients in the right degrees stay regular under this conjugation by θ.
This seems to be true for any θ as long as `$[p]_F(x)$` is of the form `$\sum_{n \geq1} a_n x^{p^n}$` modulo p. In general, it is of the form `$\sum_{k\geq1} a_kx^{kp^m}$`, where m can be taken to be the height of the FGL (Ravenel's Green Book), but I don't see why it should be true in the general case.
I am sure this has been treated by someone, but have yet to see it on print. If anyone has seen question discussed somewhere, please let me know.
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1 Answer
The property of being Landweber exact is independent of the orientation. In terms of Landweber's criterion, this is generally phrased as saying that the element vn is invariant modulo the ideal (p,v1,...,vn-1), and so any change-of-orientation (which induces a strict isomorphism on the formal group law) does not change the property of vn being or not being a zero divisor after modding out the previous terms.
This follows from Lemma A2.2.6 in Ravenel's green book, which implies that any endomorphism of the formal group law over an Fp-algebra R is of the form g(xph) for some h and some power series g. In particular, the p-series [p]F(x) over R/(p,v1,...,vn-1) has this property, and so the leading coefficient vn is invariant under strict isomorphisms.
It should be noted that vn is not invariant before taking this quotient, but that has no effect on whether these elements form a regular sequence in R.
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Thanks for your quick comment! A2.2.6 was indeed the result I was referring to above. You remarking that it is an endomorphism over R/(p,v<sub>1</sub>,...,v<sub>n-1</sub>) (and not just R/(p)) really did the trick. – Johan Mar 24 2010 at 18:31
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http://mathoverflow.net/questions/18058/
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## On statements provably independent of ZF + V=L
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Are there any known statements that are provably independent of $ZF + V=L$? A similar question was asked here but focusing on "interesting" statements and all examples of statements given in that thread are not provably indepedent of $ZF + V=L$, they all raise the consistency strength bar. For example, the claim that "there exists an inaccessible" is independent of $ZF + V=L$, is really just an assumption. Because of Gödel's second incompleteness theorem, we cannot prove this. It is well possible that $ZFC$ proves "there is no inaccessible". The same holds for $Con(ZF)$ or "there is a set model of ZF". Those are assumed to be independent of $ZF + V=L$, but this cannot be proved without large cardinal assumptions.
So my question is: Is there any known (not necessarily "interesting" or "natural") statement $\phi$ and an elementary proof of $Con(ZF) => Con(ZF + V=L + \phi) \wedge Con(ZF + V=L + \neg\phi)$? Or is there at least a metamathematical argument that such statements should exists? (Contrast this with the situation of $ZFC$ and $CH$!)
And if not: Might $ZF + V=L$ be complete in a weak sense: There is no statement provably independent of it?
What is known about this?
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Does the answer to your question not depend on the exact system in which you allow the proof of the independence to happen? – Mariano Suárez-Alvarez Mar 13 2010 at 14:50
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If ZF turns out to be inconsistent then no statement is provably independent of ZF+V=L because every statement is provable assuming it :P . Hence no known statements are provably independent of ZF+V=L ;-) – Kevin Buzzard Mar 13 2010 at 15:46
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But Kevin, the OP asked for a proof of the implication Con(ZF) implies Con(ZF+ psi) + Con(ZF + not-psi). If Con(ZF) fails, then this implication is vacuously true. – Joel David Hamkins Mar 13 2010 at 17:24
@Joel: Serves me right for not reading the question. I feel I should delete my comment but perhaps I'll leave it and just say here that of course you're right. Let me run another one past you though while you're here though: how about letting psi be "Con(ZF+V=L)"? – Kevin Buzzard Mar 13 2010 at 17:52
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Yes, that statement is what Goedel uses in the 2nd Incompleteness theorem, and it is provably equivalent to "I am not provable". The situation in general is complicated by the need for omega-consistency in Goedel's version of the Incompletness theorem, to get both sides of the indpendence. Thus, one should use the Rosser sentence instead. For example, some consistent theories T do prove their own inconsistency. e.g. ZFC+not-Con(ZFC), and so one cannot always say that Con(T) is independent of T, even when T is consistent, although Con(T) is not probable. – Joel David Hamkins Mar 13 2010 at 18:01
show 1 more comment
## 1 Answer
The Incompleteness theorem provides exactly the requested independence. (But did I sense in your question that perhaps you thought otherwise?)
The Goedel Incompleteness theorem says that if T is any consistent theory interpreting elementary arithmetic with a computable list of axioms, then T is incomplete. Goedel provided a statement σ, such as the "I am not provable" statement, which is provably equivalent to Con(T), or if you prefer, the Rosser sentence, "there is no proof of me without a shorter proof of my negation", such that T does not prove σ and T does not prove ¬σ.
This establishes Con(T) implies Con(T + σ) and Con(T + ¬σ), as you requested. [Edit: one really needs to use the Rosser sentence to make the full conclusion here.]
In particular, this applies to the theory T = ZFC+ V=L, since this theory interprets arithmetic and has a computable list of axioms. Thus, this theory, if consistent, is incomplete, using the corresponding sentences σ above. Since it is also known (by another theorem of Goedel) that Con(ZF) is equivalent to Con(ZFC + V=L). This establishes the requrested implication:
• Con(ZF) implies Con(ZFC + V=L + σ) and Con(ZFC + V=L + ¬σ)
The Incompleteness theorem can be proved in a very weak theory, much weaker than ZFC or even PA, and this implication is similarly provable in a very weak theory (PA suffices).
One cannot provably omit the assumption Con(ZF) of the implication, since the conclusion itself implies that assumption. That is, the existence of an independent statement over a theory directly implies the consistency of the theory. So since we cannot prove the consistency of the theory outright, we cannot prove the existence of any independent statements. But in your question, you only asked for relative consistency (as you should to avoid this triviality), and this is precisely the quesstion that the Incompleteness theorem answers.
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Thank you! But i'm still somewhat confused: I don't think that Con(T) implies Con(T + Con(T)), otherwise the theory T + Con(T) could prove it's own consistency. So one cannot take Con(ZF + V=L) as an example for a statement provably independent of ZF + V=L. Or am i missing something? – gerald-thaler Mar 13 2010 at 18:20
In this general case, one must use the Rosser sentence. The issue you mention is exactly the difference between the Goedel Incompleteness theorem and the Goedel-Rosser Incompleteness theorem. The statement sigma you want is: "For any proof of me, there is a shorter proof of my negation". – Joel David Hamkins Mar 13 2010 at 18:41
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http://math.stackexchange.com/questions/96007/integration-involving-product-of-inverse-error-and-polynomial-functions
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# Integration involving product of inverse error and polynomial functions
I am stuck with the following integral: $$\int_{x \ge 1} \frac{18 \exp \left(-\frac12 \left(\phi^{-1}[1 - \frac{1}{x^3}]\phi^{-1}[1 - \frac{1}{(z-x)^3}] \right) \right)}{\sqrt{3}x^3 (z-x)^4} dx$$ where $z$ is a constant and $\phi^{-1}$ is the inverse of the standard normal cdf which can also be expressed in terms of the error function: $$\phi^{-1}[p] = \sqrt{2} \operatorname{erf}^{-1}(2p-1)\qquad\text{for}\quad p \in (0,1)$$
Can this be solved analytically or numerically?
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Analytically, unlikely. Numerically... difficult, but doable I think. – J. M. Jan 3 '12 at 7:31
Where does it come from? What does cdf mean? – draks ... Apr 5 '12 at 10:48
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http://mathoverflow.net/questions/48123/regarding-a-feature-of-multivariate-real-function
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## Regarding a Feature of Multivariate Real Function
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Any real function can be expressed as a function of the sum of two monotonic real functions?
More precisely, for real function p(x, y), there exist continuous real functions P(x), h(x,y), g(x) such that:
$p(x,y)=P(h(x,y)+g(x))$
Where $P(x), h(x,y), g(x)$ are arbitrary satisfying $\frac {d(g(x))}{dx}>0$, $\frac {\partial h(x,y)}{\partial y}>0$
This is equivalent to mine another question “Solving Functional Equation”. By letting $h(x,y)=ln(w(x,y)), g(x)=-ln(u(x)), p(x,y)=\frac {u(f(x,y))}{u(x)}$, we have:
$p(x,y)=\frac {u(f(x,y))}{u(x)}=F[ln(w(x,y))-ln(u(x))]=F(ln \frac {w(x,y)}{u(x)})=\Psi (\frac {w(x,y)}{u(x)})$
Thank you very much!
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You seem to be missing a continuity hypothesis on $p$. – Joel David Hamkins Dec 3 2010 at 11:57
I'm voting to close as the question is too localised. But a quick counterexample: Let $p(0,y)$ be some function such that $p(0,y) = 0$. Since $h$ is continuous and monotonic in $y$, there exists some $z_0 \in \mathbb{R}$ and a small $\delta$ such that $P(z) = 0$ whenever $|z-z_0| = \delta$, and such that there is some $(0,y_0)$ such that $h(0,y_0) + g(0) = z_0$. Now using continuity of $g, h$, the function $P(h(x,y) + g(x))$ must vanish on a neighborhood of $(0,y_0)$. So a function like $p(x,y) = xy$ cannot be written in the form you want. – Willie Wong Dec 3 2010 at 12:00
By setting $P(x)=e^x, g(x)=ln(x), h(x,y)=ln(y)$, we have: $p(x,y)=xy=e^{ln(xy)}=e^{ln(x)+ln(y)}=P(g(x)+h(x,y))$ Where: $\frac {\partial lnx}{\partial x}=\frac {1}{x}>0$, $\frac {\partial lny}{\partial y}=\frac {1}{y}>0$ for $x>0, y>0$. – Wang Tao Dec 4 2010 at 2:40
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http://math.stackexchange.com/questions/156094/construct-a-new-relation-from-several-relations/156099
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# Construct a new relation from several relations
Let $f_0,\dots,f_{n-1}$ are relations, where $f_i$ is a relation of arity $m_i$ for every $i=0,\dots,n-1$.
How to construct an $(m_0+\dots+m_{n-1})$-ary relation from them?
For my previous erroneous attempt to do this see here: Name for product of relations
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If all on same set, can define new relation as holding if all the component relations hold, or if at least one holds, or any other desired Boolean combination. – André Nicolas Jun 9 '12 at 15:09
So, you want to combine several sets of n-tuples into a single set? There is, of course, no standard way to do this. Do you have an application in mind? – Dan Christensen Jun 15 '12 at 15:24
## 1 Answer
Oh, I found:
$$\left\{ \operatorname{uncurry}z \hspace{0.5em} | \hspace{0.5em} z \in \prod f \right\}$$
where uncurrying is defined in Wikipedia: http://en.wikipedia.org/wiki/Currying by the formula $\operatorname{uncurry}(f)(x;y)=(fx)y$.
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This would not seem to be the answer to your question. This is "a technique of transforming a function that takes multiple arguments (or an n-tuple of arguments) in such a way that it can be called as a chain of functions each with a single argument." – Dan Christensen Jun 15 '12 at 15:41
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http://math.stackexchange.com/questions/259047/solving-a-trig-equation
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# Solving a trig equation?
How would I solve the following equation? $$1+ \sin (x)=2 \cos(x)$$ I am having difficult with it.
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## 5 Answers
We have $1 + \sin(x) = 2 \cos(x)$. Recall that $1 - \sin^2(x) = \cos^2(x)$. Hence, we get that $$(1 + \sin(x))(1- \sin(x)) = \cos^2(x)$$ i.e. $$2 \cos(x) (1 - \sin(x)) = \cos^2(x)$$ If $\cos(x) \neq 0$, then we get that $$1 - \sin(x) = \dfrac{\cos(x)}2$$ Hence, have \begin{align} 1 + \sin(x) & = 2\cos(x)\\ 1 - \sin(x) & = \dfrac{\cos(x)}2 \end{align} This gives us that $2 = \dfrac52 \cos(x) \implies \cos(x) = \dfrac45$. This gives us $\sin(x) = \dfrac35$. Hence, we get one possible solution as $$x = 2n \pi + \theta$$ where $\sin(\theta) = \dfrac35$ with $0 < \theta < \dfrac{\pi}2$.
If $\cos(x) = 0$, then we need $\sin(x) = -1$. Hence, $x = 2n\pi + \dfrac{3 \pi}2$.
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,according to my method, $x=\arccos \frac45=\arcsin\frac45$ is another solution. – lab bhattacharjee Dec 15 '12 at 5:19
@labbhattacharjee Try plugging your answer in... $1 + \frac{4}{5} \neq 2 \cdot \frac{4}{5}$, so it looks like something went wrong. – andybenji Dec 15 '12 at 5:51
@labbhattacharjee Thanks. Yes. the other solution is $x = \arcsin(3/5) = \arccos(4/5)$. – user17762 Dec 15 '12 at 6:03
@andybenji, sorry there was terrible typo--it should be $\arcsin\frac35=\arccos\frac45$ as Marvis has noted. – lab bhattacharjee Dec 15 '12 at 6:58
Hint: Square both sides, and express $\cos^2 x$ in terms of $\sin^2 x$.
You will get a quadratic equation in $\sin^2 x$. After finding the solutions of this equation, make sure to substitute into the original equation to check whether they are indeed roots of the original equation. The squaring process could produce "spurious" roots. (It happens not to, but checking is cheap.)
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I have a quick question I squared both sides and got. 1+2sin(x)+sin^2(x)=4cos^2(x) so I made 4cos^2(x) into 4-sin^2(x) Is this correct. – Fernando Martinez Dec 15 '12 at 4:32
It would be $4-4\sin^2 x$. The equation then becomes $5\sin^2 x+2\sin x-3=0$. You can then use the Quadratic Formula or factor as $(5\sin x-3)(\sin x+1)$. – André Nicolas Dec 15 '12 at 4:48
Oh thanks very much. – Fernando Martinez Dec 15 '12 at 4:53
This may be the hard way, but if you let $q=\tan(x/2)$, then $\sin x=2q/(1+q^2)$, and $\cos x=(1-q^2)/(1+q^2)$.
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Let me take a stab at this. I had to double check a few identities. First, we use the identities $\cos(\frac\pi 2-x)=\sin x$ and $\sin(\frac\pi 2-x)=\cos x$.
$$1+\cos(\frac\pi 2-x)=2\sin(\frac\pi 2-x)$$ $$\frac{1+\cos(\frac\pi 2-x)}{\sin(\frac\pi 2-x)}=2$$
Next, we use the half-angle identity $\tan\frac\theta 2=\frac{\sin\theta}{1+\cos\theta}$.
$$\cot(\frac\pi 4-\frac x2)=2$$ $$\tan(\frac\pi 4-\frac x2)=\frac12$$
From here, take the inverse tangent and the rest is algebra.
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$\cos\left(\frac\pi4-\frac x2\right)$ can also be $0$ – lab bhattacharjee Dec 15 '12 at 9:38
@labbhattacharjee I see how I might have lost some roots by not making sure $\cos x$, or $\sin(\frac\pi 2-x)$ if you prefer, was 0 before dividing by it. Did I lose roots in any other step somehow? – Mike Dec 15 '12 at 18:22
We can try to avoid squaring to avoid tests for extraneous root. $$2\cos x-\sin x=1$$
Putting $r\cos\theta=2,r\sin\theta=1$ where $r>0$
Squaring and adding we get $r^2=2^2+1^2\implies r=\sqrt5$
On division, $\tan \theta=\frac12$ where $0<\theta<\frac\pi2$ as $\sin\theta,\cos\theta>0$
we get $$\cos\theta\cos x-\sin\theta\sin x=\sin\theta$$(cancelling $r$ in either sides )
or, $\cos(x+\theta)=\cos(\frac\pi2-\theta)$
So, $$x+\theta=2n\pi\pm\left(\frac\pi2-\theta\right)$$ where $n$ is any integer.
Taking $'-'$ sign, $$x+\theta=2n\pi-\left(\frac\pi2-\theta\right)\implies x=2n\pi-\frac\pi2$$
Taking $'+'$ sign, $$x+\theta=2n\pi+\left(\frac\pi2-\theta\right)\implies x=2n\pi+\frac\pi2-2\theta=2n\pi+\frac\pi2-2\arctan \frac12$$
Now, we know $\cos2y=\frac{1-\tan^2y}{1+\tan^2y},\sin2y=\frac{2\tan y}{1+\tan^2y},\tan2y=\frac{2\tan y}{1-\tan^2y}$
If $\arctan \frac12=y,\tan y=\frac 12$
$$\cos2y=\frac35,\sin2y=\frac45,\tan2y=\frac43$$
So, $$2y=2\arctan \frac12=\arccos\frac35=\arcsin\frac45=\arctan \frac43$$
So, $$x=2n\pi+\frac\pi2-\arccos\frac35=2n\pi+\arcsin\frac35$$ as $\arcsin z+\arccos z=\frac12$ for $-1\le z\le1$
Similarly, $$x=2n\pi+\arccos\frac45=2n\pi+\arccot\frac43$$
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It is not clear we should avoid squaring, since checking is cheap. A good reason to avoid squaring is if degree is increased beyond the comfortable. But degree $2$ is comfortable. – André Nicolas Dec 15 '12 at 5:48
@AndréNicolas, squaring is easing the calculation a lot for $A\cos x+B\sin x=C$ w/o introducing much intricacies. There is no extraneous root as such,right?. We just need to identify the quadrant of $x$ by getting the sign of $\cos x$ from the given equation, if we have the quadratic equation in $\sin x$ or vice versa. Is $\arccot$ not a valid latex? – lab bhattacharjee Dec 15 '12 at 9:44
@labbhattacharjee Apparently not. Try "\mathrm{arccot}" – Thomas Dec 15 '12 at 10:32
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http://physics.stackexchange.com/questions/16664/how-do-i-rearrange-de-dt-to-find-an-electrons-half-life-due-to-synchrotron-ra
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# How do I rearrange dE/dt to find an electron's 'half-life' due to synchrotron radiation?
I know that $-\frac{\mathrm{d}E}{\mathrm{d}t} \propto E^2$ for an electron losing energy to synchrotron radiation, but I can't find how to arrange this to give the time it would take for the electron to lose half of its original energy. How would I go about working that out?
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The half-life description only works for exponential decay. As others have shown, this leads to a power-law decay, as 1/t. So you can ask for the asymptotic half-life in log-t variable, but not in t. – Ron Maimon Nov 8 '11 at 6:48
But the OP does not speak of half-life! His problem has meaning per se. – Vladimir Kalitvianski Nov 9 '11 at 15:21
## 2 Answers
Not to worry, it's fairly easy: right now you have a differential equation which can be written
$$\frac{\mathrm{d}E}{\mathrm{d}t} = -CE^2$$
for some constant $C$. You need to solve that differential equation for $E(t)$. (If you're wondering how to do that, you can find more information at the math site.) Then you can determine the electron's energy at the initial time $t_0$ as $E(t_0)$, and find the time at which its energy becomes half of that:
$$E(t) = \frac{E(t_0)}{2}$$
and solve for $t$.
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$$\frac{dE}{dt} = -CE^2$$
$$d\left (\frac{1}{E}\right ) = Cdt$$
$$\frac{1}{E_2}-\frac{1}{E_1}= C(t_2-t_1)$$
$$E_2=E_1/2$$
$$\frac{1}{E_1} = C\Delta t$$ $$\Delta t = \frac{1}{CE_1}$$
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http://mathoverflow.net/revisions/120352/list
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## Return to Answer
2 added 13 characters in body
You can just prove it yourself directly in local holomorphic coordinates. Indeed, the $\overline{\partial}$ Laplacian on functions is equal to $\Delta_{\overline{\partial}}f=g^{i\overline{j}}\partial_i \partial_{\overline{j}}f$. Apply this to $|\partial f|^2=g^{k\overline{\ell}}\partial_k f \partial_{\overline{\ell}}f$ (the gradient of the complex length squared of $f$, \partial f=(df)^{(1,0)}$, which is equals$1/2$of the usual$|\nabla f|^2$), using if you want local holomorphic normal coordinates for$g\$ at a point, and you will immediately get
$$\Delta_{\overline{\partial}}|\partial f|^2=|\nabla_i \nabla_j f|^2+|\nabla_i \nabla_{\overline{j}} f|^2+2\mathrm{Re}\langle \partial f, \partial\Delta_{\overline{\partial}}f\rangle+R^{i\overline{j}}\partial_i f\partial_{\overline{j}}f,$$ where $R^{i\overline{j}}$ is the Ricci curvature of $g$ with the indices raised.
If $g$ is not Kähler, and you define the complex Laplacian by the same formula $g^{i\overline{j}}\partial_i \partial_{\overline{j}}f,$ then a similar result holds, with the Ricci curvature now being one of the several Ricci curvatures of the Chern connection of $g$, and with several new terms involving the torsion of $g$ and its covariant derivative. The calculation is again completely strightforward, using local holomorphic coordinates (not normal anymore!), and using the definitions of covariant derivative and curvature of the Chern connection of $g$.
1
You can just prove it yourself directly in local holomorphic coordinates. Indeed, the $\overline{\partial}$ Laplacian on functions is equal to $\Delta_{\overline{\partial}}f=g^{i\overline{j}}\partial_i \partial_{\overline{j}}f$. Apply this to $|\partial f|^2=g^{k\overline{\ell}}\partial_k f \partial_{\overline{\ell}}f$ (the gradient of the complex of $f$, which is $1/2$ of the usual $|\nabla f|^2$), using if you want local holomorphic normal coordinates for $g$ at a point, and you will immediately get
$$\Delta_{\overline{\partial}}|\partial f|^2=|\nabla_i \nabla_j f|^2+|\nabla_i \nabla_{\overline{j}} f|^2+2\mathrm{Re}\langle \partial f, \partial\Delta_{\overline{\partial}}f\rangle+R^{i\overline{j}}\partial_i f\partial_{\overline{j}}f,$$ where $R^{i\overline{j}}$ is the Ricci curvature of $g$ with the indices raised.
If $g$ is not Kähler, and you define the complex Laplacian by the same formula $g^{i\overline{j}}\partial_i \partial_{\overline{j}}f,$ then a similar result holds, with the Ricci curvature now being one of the several Ricci curvatures of the Chern connection of $g$, and with several new terms involving the torsion of $g$ and its covariant derivative. The calculation is again completely strightforward, using local holomorphic coordinates (not normal anymore!), and using the definitions of covariant derivative and curvature of the Chern connection of $g$.
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