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http://math.stackexchange.com/questions/139575/can-someone-explain-cremer-mcleans-astonishing-result-in-auction-theory/139576
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# Can someone explain Cremer-Mclean's astonishing result in auction theory?
In mechanism design/auction theory, there is a famous result by Cremer and Mclean that if agents'/bidders' valuations are even slightly correlated, then all the surplus can be extracted by the principal/auctioneer. This is what makes auction theory without i.i.d. valuations an unpopular research topic even though it's interesting from a practical point of view.
My question is: Can someone explain to me why the Cremer-Mclean result is true?
The original paper is Econometrica 1998, "Full extraction of the surplus in Bayesian and dominant strategy auctions" by Cremer and Mclean.
There is an explanation beginning on page 151 in Vijay Krishna's Auction Theory textbook.
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Could you include links, please. – Turukawa Nov 15 '11 at 22:23
They won't be opensource links anyway, but I'll add them nonetheless. – Zermelo Nov 15 '11 at 22:52
## 2 Answers
Suppose you and I are consultants for competing oil companies. We independently estimate the value of some new drilling site, and recommend our employers make bids based upon our estimates. Since some amount of error is involved, but our estimates are correlated (we are experts, after all!), if I can observe the bid your company makes before my company makes its bid, I can get a sense of whether or not I'm over- or under-estimating the value of the new site.
The problem is that we each only have one estimate. If the seller adopts an auction style that permits me to see your bid (i.e., not sealed-bid), I get additionally information about the distribution of probable values for the new drill site.
We're interested in discovering something about the random variable $t$, the distribution of estimates of the value of the new site. Suppose I have the correct estimate (the true value of the new site, $T$) by accident, and call my estimate $t_i = T$. Suppose further your estimate $t_j$ is higher than $T$. If I observe your bid, I might conclude that the true value actually lies somewhere between $t_i$ and $t_j$, so I tell my boss to bid $(t_i+t_j)/2$. This sort of "watering down" is desirable in order to avoid the "winner's curse". This is where "Bayesian" comes into the jargon of the paper. I obtain new information, your estimate of the value, and use it to update by estimation of the value. After the auction, my company starts drilling, only to discover the true value of production is $T<(t_i+t_j)/2$, so we've overbid.
So, thinking about surplus extraction, the seller of the new field wants to encourage high bidding, or to discourage shading. Discouraging shading is another way of saying "encouraging revelation of your true valuation $t$." This is what Krishna calls "truth-telling". Suppose you value the asset at $t_j$, and you bid $b_j$. Your goal is to maximize $t_j-b_j$, or to bid as low as possible. Ideally, you'd bid $\epsilon$ above the second highest valuation, $t_{j-1}$. This isn't truth-telling.
If you don't know the true value, as in this example, the seller can use other bidders' estimates to reduce your shading. You wouldn't ever bid more than your estimate of the value for the asset, so $b_j \leq t_j$. The seller wants to force you to reveal your estimate, in other words to make you bid your value, or make $b_j=t_j$.
I haven't worked through the full Bayesian-Nash equilibrium, so I can't explain why the seller is successfully able to push $b_j$ toward $t_j$, but I hope this helps!
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Hmmm, thanks that was actually quite helpful. The reason why I called it "astonishing" was that in a private values auction truthful revelation is possible only in the second price auction, and even there the all of the surplus is not extracted (since the second highest bid is paid). How all the surplus can be extracted when there is correlation, no matter how slight, is a mystery to me. But your post gets me thinking about how the auctioneer might go about preventing shading. Let me know if you work it out though! :) – Zermelo Nov 16 '11 at 6:26
@Zermelo I'll definitely give this some more thought and edit my answer later. My story isn't robust to $i$ observing a first bid from $j$ lower than $i$'s estimate. – jrhorn424 Nov 16 '11 at 6:40
Let me try to explain in rough terms why the Cremer-Mclean's result is true. Lets assume that thare are $n$ bidders and $B$ possible values for the item for each bidder. For each bidder $i$ we compute what is his expected gain $g(i,v(i))$ in a second price auction conditioned on his value $v(i)$. We can also compute the expected value $V(j;i,v(i))$ of the item for bidder $j$ conditioned on the value for $i$ being $v(i)$.
We want to express $g(i)$ as a linear combination
(*) $a(1)V(1,i,v(i))+a(2)V(2,i,v(i))+...+a(B)V(n,i,v(i))$, (The sum is over all integers from 1 to $n$ except $i$)
So that the coefficients do not depend on the value $v(i)$!
This is possible if the dependencies among the agents bids are of sufficiently general type. (Namely, certain matrix described this system of equations has a full rank.)
Now, you can charge bidder $i$ entry price which is $a(1)v(1)+a(2)v(2)+...+a(n)v(n)$, (The sum is again over all integers from $1$ to $n$ except $i$)
This way the expected entry price is precisely the expected gain given the bidder's bid but it is computed as a linear combinations of the other bids.
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http://mathoverflow.net/questions/32111/compactness-and-covering-spaces/32112
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## Compactness and Covering Spaces
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Let p : Y -> X be an n-sheeted covering map, where X and Y are topological spaces. If X is compact, prove that Y is compact.
I realize that this seems like a very simple problem, but I want to stress the lack of assumptions on X and Y. For example, this is very easy to prove if we can assume that X and Y are metrizable, for sequential compactness is then equivalent to compactness and it is easy to lift sequential compactness from X to Y.
I asked three people in person this question and all of them immediately made the assumption that X and Y are metrizable, so I feel like I should put in this warning here that they are not.
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Are you assuming X and Y are also Hausdorff? If so, then I can't see what goes wrong with the natural approach: take an open cover of Y, push it down to an open cover of $X$ (because $p$ is surjective it will be open) take a finite subcover downstairs and lift it up with multiplicity $n$ to a finite subcover upstairs. What have I missed? – Yemon Choi Jul 16 2010 at 4:31
@Yemon: What does "lift it up with multiplicity n" mean? How do you choose sets from the original cover to cover your new cover? (And I apologize for that sentence.) – Tyler Lawson Jul 16 2010 at 4:50
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If I recall correctly, you don't need Hausdorff. – Dylan Moreland Jul 16 2010 at 4:57
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@Tyler: good point, I was being over hasty. Seeing as my topology is rusty: by an n-sheeted covering, do we mean that (a) p is a quotient map of topological spaces; (b) each point $x\in X$ has an open neighbourhood $U$ suchthat $p^{-1}U)$ is the disjoint union of $n$ open sets, each of which is mapped homeomorphically onto $U$? – Yemon Choi Jul 16 2010 at 5:12
I think a fix of the proof above goes as follows: Let $U_\gamma$ be a cover of $Y$. Choose a refinement $V_\alpha$ of this cover so that each $V_\alpha$ is small enough to be homeomorphic to its image under the covering map. It suffices to show that $V_\alpha$ has a finite subcover. Since the sets $p(V_\alpha)$ form an open cover of $X$, they have a finite subcover, $p(V_\alpha)_\beta$, each of which has $n$ lifts. The lifts of this subcover provide a finite subcover of $Y$. – BMann Jul 16 2010 at 15:48
## 3 Answers
A direct argument without the use of nets:
Let $\mathcal{C}$ be an open cover of $Y$. For each $p \in X$, choose an open set $p \in U \subseteq X$ such that $Y$ is trivial over $U$, and such that each lift of $U$ is contained in some element of $\mathcal{C}$. This is an open cover $\mathcal{D}$ of $X$, which has a finite subcover $\mathcal{D}'$ since $X$ is compact. The lift of $\mathcal{D}'$ to $Y$ is also a finite cover, as well as a cover that refines $\mathcal{C}$. Thus $\mathcal{C}$ must have a finite subcover. (The fact that $Y$ is a finite cover is used twice, first to make each $U$, second to lift $\mathcal{D}'$.)
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Can you enlighten as to what is the closest statement that may be true for infinite sheeted covers? I understand that for a cover downstairs to lift to a cover upstairs one needs the number of sheets to be finite but why does the definition of $U$ depend on finiteness of the number of sheets? Take a locally trivializing neighbourhood of $p$ and look at some intersection of its pre-image with some open set in $C$ and then project it down. Doesn't that give the kind of $U$ one needs? May be I am missing something obvious! – Anirbit Jul 16 2010 at 9:57
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@Anirbit: infinite intersections of open sets are not open. Because in the end you want a finite subcover from $\mathcal{C}$, so on each "sheet" you need to pick an open set in $\mathcal{C}$ containing $U$. – Willie Wong Jul 16 2010 at 15:20
Ah..yes. I was mistaken to think that taking intersection on only 1 sheet should be enough. To define the kind of $U$ that Greg wanted one needs to take intersection on all the sheets and that will mess up. – Anirbit Jul 16 2010 at 17:35
### 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.
Well, the obvious argument that any sequence has a convergent subsequence that your three friends used for the metrizeable case generalizes easily to show that any net has a convergent subnet in the general case.
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Nice! maybe this could be used as an example for mathoverflow.net/questions/32035/… – Yemon Choi Jul 16 2010 at 5:15
Dear Eric, here is a Bourbaki-style proof.
Recall that a continuous map $f: Y\to X$ is called proper by Bourbaki if, for all spaces $Z$, the map $f\times 1_Z: Y \times Z\to X \times Z$ is closed. For example the trivial finite covering $X\times \{ 1,\ldots n \}\to X$ is proper.
Now, your $X$ is covered by opens $X_\iota \subset X$ such that the restricted/corestricted maps $f_{X_\iota }:f^{-1} (X_\iota) \to X_\iota$ are trivial finite coverings, hence are proper by the example above. We deduce that the original covering $f:Y\to X$ is proper: this follows easily from the definition of "proper" and (if a reference is needed) is proved in Bourbaki's General Topology, Chapter 1, §10, Proposition 3.
But a proper map has the property that the inverse image of a quasi-compact subset of the target (in our case all of $X$) is quasi-compact (ibid., Proposition 6). Hence $Y$ is quasi-compact if $X$ is.
NB I have used Bourbaki's definition "universally closed" for proper. As I said, this implies that inverse images of quasi-compact subsets are quasi-compact.This last property is often taken as the definition of proper. For locally compact spaces, both definitions coincide.
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http://mathhelpforum.com/calculus/120525-spread-disease.html
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# Thread:
1. ## Spread of a disease
Under certain circumstances, the rate at which a disease spreads through a population of P individuals is proportional to the product of the number I of people infected and the number not infected
a) Write and equation for dI/dt
b) Invert to get dt/dI
c) Find t as a function of I, assuming time is 0 at the moment when half the population is infected
d) Find I as a function of t
Hint for part c: What happens when you combine 1/I+1/(P-1) into a single fraction.
Any help that you can offer is appreciated!
2. Originally Posted by Alice96
Under certain circumstances, the rate at which a disease spreads through a population of P individuals is proportional to the product of the number I of people infected and the number not infected
a) Write and equation for dI/dt
b) Invert to get dt/dI
c) Find t as a function of I, assuming time is 0 at the moment when half the population is infected
d) Find I as a function of t
Hint for part c: What happens when you combine 1/I+1/(P-1) into a single fraction.
Any help that you can offer is appreciated!
a) $\frac{dI}{dt}=I(P - I)$
b) $\frac{dt}{dI}=\frac{1}{I(P - I)}$
c) $\frac{dt}{dI}=\frac{1}{I(P - I)}=\frac{1}{PI} + \frac{1}{P(P-I)}$, so $dt = \frac{1}{PI} ~dI+ \frac{1}{P(P-I)}~dI$. solve this differential equation.
d) solve part (a)
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http://physics.stackexchange.com/questions/44185/what-is-the-characteristic-length-of-a-cylinder?answertab=votes
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# What is the characteristic length of a cylinder
I have a cold cylinder that is submerged in hot water and I need to find the convective heat transfer coefficient. I can do the whole process but I am stuck finding the characteristic length. I found that the characteristic length of an object is $$L_{c}=\frac{A}{P}$$
Now I assume that heat transfer area in this case includes the top and botom of the vertical cylinder. So does my characteristic length become $$L_{c}=\frac{2\pi r^{2}+\pi DH}{2\pi r}$$
Or do I neglect the surface area of the top and bottom to have $$L_{c}=\frac{\pi DH}{2\pi r}=\frac{\pi DH}{\pi D}=H$$
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What is P in the first equation? – Jason Davies Nov 13 '12 at 23:02
P must be perimeter? – Johannes Dec 14 '12 at 13:01
## 2 Answers
There is no unique definition for characteristic length. One can define the length scale as the square root of the surface area, as the third root of the volume, as the volume-to-surface ratio, etc. It all depends what length scale you want identify. For your problem (convective heat transport of an object submerged in a liquid at elevated temperature) the characteristic length is the vertical size of the object. This is because the convection will take place in the vertical direction (cold water flowing down, hot water rising).
If your object is a cylinder in vertical orientation, the convective length scale is the height H of the cylinder. If your object is a cylinder placed horizontally the characteristic length is determined by the diameter D of the cylinder.
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The characteristic length is commonly defined as the volume of the body divided by the surface area of the body, i.e.
$L_c = \frac{V_{body}}{A_{surface}}$
The surface area should include the top and bottom of the cylinder.
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does the equation $L_{c}=\frac{A_{s}}{P}$ not hold true? – Greg Harrington Nov 13 '12 at 23:13
Again, what is P? – Jason Davies Nov 14 '12 at 8:45
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http://stats.stackexchange.com/questions/43720/minimum-population-size-for-chi-square-test
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# Minimum population size for chi-square test?
I'm analyzing data from an experiment in which two independent groups were exposed to an experimental setup without and with treatment.
I am testing whether treatment changed the second group's behaviour by performing a chi-square test that compares group 2 (the observed) vs group 1 (the expected). The result indicates there is a significant change in behaviour X² p-value < 0.00014.
Now, I am trying to test subgroups to understand better the change, i.e looking at gender, age, and other self reported metrics.
My question is, given that group 2 N=40 if I look at age for instance I find people in their 20s and their 60s show significant change but other age groups don't. However people in their 20s N=12 and people in their 60s N=5. Is there a heuristic/rule that says there is a minimum number of people needed to consider a result significant? for instance anything below N=5 cannot be considered significant or anything below N=20% of the population?
EDIT: Just to clarify, I am doing a chi-square test of independence (between group 1&2) not a chi-square goodness of fit test.
EDIT 2: With this edit I consider the question closed. None of the answers/comments gave me a definitive solution, which I believe says more about the question than the answers. I was hoping for a definitive answer along the lines you need at least 5 ppl or 20% of your sample. It seems the answer is less direct as it is sensitive to many factors.
Thanks.
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I don't understand what you mean by "expected" -- I thought that group 1 was also collected data, and not theoretical distribution? – January Nov 16 '12 at 8:52
Group one is not exposed to treatment but yes to environment so their behaviour is what is expected when the treatment is not present in a given environment. Once you introduce the treatment then you compare the behaviour in the same environment pre and post treatment. The resulting change is the observed effect of treatment in that environment. This is standard procedure in HCI/social sciences research. – G Garcia Nov 16 '12 at 11:26
You are going to confuse a lot of people if you use "expected" in that way, as it means something else when talking about chi-square tests. – Peter Flom Nov 16 '12 at 11:54
Why are you using chi-square at all? You have some dependent variable, apparently dichotomous (although you haven't said) and several independent variables (treatment, age, gender etc). That calls for regression of some sort, probably logistic regression if I am right about the DV. – Peter Flom Nov 16 '12 at 11:58
I'm doing chi-square because I need to look at each variable in isolation not as a group, in which case I would probably do an ANOVA. – G Garcia Nov 16 '12 at 14:31
## 1 Answer
For small sample sizes, use Fisher's exact test, because the $\chi^2$ test sampling statistics has only approximately the $\chi^2$ distribution, and this approximation is problematic for small sample sizes.
While lower sample size decreases the power of the test, the p-values (and not the sample size) are indicators of the statistical significance. A significant p-value stays significant whatever the sample size; the sample size has been taken care of through the calculation of the test statistic.
However, someone might claim that a small sample size is more likely to be biased. This is not necessarily true, but I think there might exist a correlation between the study sample size and whether the data was collected in an unbiased way as it should.
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My understanding is that Fisher tests for the independence of variables on a contingency table. However I want to compare the observed against the expected not between two observed measures. Fisher could tell me how independent the variables are, but that whould not shed any light on whether the treatment had any effect when compared to previous behaviour. Am I correct on my interpretation? I know p-value is the indicator of significance, my question is if p-value is significant but sample is a small proportion of the population, does that invalidate the significance? – G Garcia Nov 16 '12 at 8:10
You are not doing a goodness-of-fit here, I thought you are comparing two populations. I read your question again, and I think that I am confusing about what you are doing. Can you please elaborate on what you are comparing? Regarding the sample size: see updated answer. – January Nov 16 '12 at 8:46
I am comparing group1 (my null case, i.e no treatment) vs group2 (my group exposed to treatment). So comparing two contingency tables as opposed to columns within one. – G Garcia Nov 16 '12 at 9:37
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http://mathoverflow.net/questions/108953?sort=votes
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## Motivic generalisation of Neron-Ogg-Shaferevich criterion
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Given a variety $X$ over $\mathbb{Q}$ with good reduction at $p$, proper smooth base change tells us that its $l$-adic cohomology groups are unramified at $p$ (and I'd guess some $p$-adic Hodge theory tells us its p-adic cohomology is crystalline).
My question is to what extent it's possible to find a converse to this statement. More precisely, I have yet to see a counterexample to the following "conjecture" (though I still suspect it's wrong).
"Conjecture": Let $K$ be a number field, $p$ and $l$ primes, and $V$ a geometric (say, coming from the variety $Y$) $l$-adic representation of $G_K$ that is unramified/crystalline at $\mathfrak{p}|p$. Then there exists a smooth proper variety $X$ such that $X$ has good reduction at $\mathfrak{p}$ and $V$ can be cut out of the cohomology of $X$.
From googling around, the things I know so far are (at least for $l \not= p$):
• If $Y$ is an abelian variety, the classical Neron-Ogg-Shafarevich condition means that $Y$ itself is a witness to the conjecture.
• We can take torsors for abelian varieties with no $K$-rational points, and these can have the same representations, but fail to have good reduction (in this paper http://arxiv.org/abs/math/0605326 of Dalawat).
• There exist curves which have bad reduction, but whose Jacobians have good reduction.
If anyone knows any more about this story I'd be interested to hear. Ultimately I guess it would be nice to have a definition for when a motive is unramified/has good reduction, and cohomologically this surely has to mean unramified/crystalline, but it would be nice if this could always be realised "geometrically".
Thanks, Tom.
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## 2 Answers
(could be a comment but too long...)
That's quite a natural question. I am not sure it is possible to prove that the "conjecture" you state is true with the current technology (and to be sure I have no idea how to prove it), but my intuition would differ from yours in that I believe the conjecture to be true. Let me give a very rough "argument" why. Let us assume for the sake of the argument that your Galois representation satisfies a self-duality condition so that it is expected to be attached to an automorphic representation for an unitary group. Then because of your hypothesis that your Galois representation is unramified, the automorphic representation should be unramified at places $\mathfrak{p}$ above $p$, that is have invariant by maximal compact hyperspecial subgroups $K_\mathfrak p$. The Shimura variety for the unitary group with hyper special maximal level structure is conjectured (and actually, known) to have good reduction (Milne conjecture). Assume also that the representation $\pi_\infty$ is such that the Galois representation attached to $\pi$ appears in the étale cohomology of that Shimura variety (like for a modular eigenform of weight 2, whose representation appears in the cohomology of the modular curve and not of a Kuga-Sato variety over it). Then your Galois rep. appears in the cohomology of a variety with good reduction.
Admittedly this argument is not very convincing, because I have made very strong assumption on the Galois representation. However, as those assumptions seem (to me) quite orthogonal to the problem discussed, I believe this is a decent evidence in favor of the conjecture.
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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.
Unfortunately I don't know much about motives in genereal, but this might be relevant to your question. One result of my thesis, that I am currently writing, is to prove Neron-Ogg-Shafarevich for 1-motives. The proof is not particularly difficult and it ultimately reduces to the corresponding results for the components of the 1-motive. I will describe below what good reduction means in this particular case.
A 1-motive $M = [u\colon Y\to G]$ over a scheme $S$ consists of a group scheme $Y$, which is locally etale isomorphic to $\mathbb{Z}^r$, a group scheme $G$ which is an extension of an abelian scheme $A$ by a torus $T$ and a homomorphism $u\colon Y\to G$. If $S$ is the spectrum of a field $K$, this means that $Y$ is a free finitely-generated $\mathbb{Z}$-module with a continuous action of the absolute Galois group $\Gamma_K$ and that $u$ is a $\Gamma_K$-equivariant homomorphism $u\colon Y\to G(\bar K)$.
If $R$ is a complete discrete valuation ring with a fraction field $K$ we say that a 1-motive $M$ over $K$ has good reduction if there exists a 1-motive $\widetilde{M}$ over $R$ whose generic fiber is isomorphic to $M$. This is equivalent to the following:
• $G$ has good reduction $\widetilde{G}$ over $R$, which is equivalent to saying that both $A$ and $T$ have good reduction;
• The action of $\Gamma_K$ on $Y$ is unramified;
• The image of $u(Y)$ is contained in the set of those points in $G(K')$ which can be reduced, where $K'/K$ is some finite field extension. Equivalently, $u(Y)$ is contained in the maximal compact subgroup of $G(K')$;
With this definition, the criterion of Neron-Ogg-Shafarevich is as follows: Let $l,p$ be primes, with $l\neq p$. A 1-motive $M/\mathbb{Q}$ has good reduction mod p if and only if the Tate module $T_l(M)$ is unramified at $p$. For general number fields replace $p$ by a prime ideal.
If you want to learn more about reduction of 1-motives you can look at M. Raynaud's paper 1-Motifs et Monodromie Géométrique.
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http://math.stackexchange.com/questions/88300/if-fx-is-continuous-on-a-b-and-m-max-fx-is-m-lim-limits-n/88309
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If $f(x)$ is continuous on $[a,b]$ and $M=\max \; |f(x)|$, is $M=\lim \limits_{n\to\infty} \left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n}$?
Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=\max\{|f(x)| \; :\; x \in [a,b]\}$. Is it true that: $$M= \lim_{n\to\infty}\left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n} ?$$
Thanks!
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Did you forget $\lim\limits_{n\to\infty}$ somewhere? – Ilya Dec 4 '11 at 17:46
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In other words, you are asking if $\| f \|_n \to \| f \|_\infty$ for a continuous $f$. – Srivatsan Dec 4 '11 at 19:09
1 Answer
Put $S_{\delta}:=\{x\in\left[a,b\right], |f(x)|\geq M-\delta\}$ for any $\delta>0$. Then we have for all $n$ $$M\cdot (b-a)^{\frac 1n}\geq\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq \left(\int_{S_{\delta}}|f(x)|^ndx\right)^{\frac 1n}\geq (M-\delta)\left(\lambda(S_{\delta})\right)^{\frac 1n}.$$ Since $f$ is continuous, the measure of $S_{\delta}$ is positive and taking the $\liminf$ and $\limsup$, we can see that $$M\geq \limsup_n \left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq M-\delta\quad \mbox{and }\quad M\geq \liminf_n\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}\geq M-\delta$$ for all $\delta>0$, so $\lim_{n\to\infty}\left(\int_a^b|f(x)|^ndx\right)^{\frac 1n}=M$.
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Also, it will be nice to point out where continuity is being used here: to argue that $S_\delta \cap [a,b]$ is of positive measure. – Srivatsan Dec 4 '11 at 18:16
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You may replace $S_\delta \cap [a,b]$ by $S_\delta$ everywhere. – Did Dec 4 '11 at 22:21
Why do you absolutely need to take the lim inf and lim sup? All the limits exists. – Patrick Da Silva Jan 5 '12 at 10:42
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@PatrickDaSilva We don't know a priori that the limit $\lim_n\left(\int_a^b|f(x)|^n\right)^{\frac 1n}$ exists. – Davide Giraudo Jan 5 '12 at 12:36
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out – Patrick Da Silva Jan 5 '12 at 23:13
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http://mathoverflow.net/revisions/98869/list
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## Return to Question
3 added 63 characters in body
I am writing a paper for game theorists where I use (countable) amenable groups to do some things. So I am writing up a preliminary section about countable amenable groups whose main purpose is to convince the reader that invariant means are the generalization of the uniform measure.
Well, the first obvious thing to say is that Every every finite group is amenable and the uniform measure is its unique invariant mean. But this sounds like only a merely technical reason. Well, I have at least one deeper reason. Let $G$ be a group, fix $W\subseteq G$ and consider the two-person game where the two players choose $x,y\in G$, respectively, and player 1 wins if $xy\in W$. For generic $W$ we can consider this game as a game without relevant information and therefore, a game theoretical reformulation of the Maximum Entropy Principle, would predict that the players will play casually (whatever it means). Indeed, if $G$ is finite, a Nash equilibrium is given by the uniform measure. The point is that if $G$ is countable amenable, then this game admits Nash equilibria and they are exactly suitable invariant means.
I think this is already a pretty convincing motivation, but I would like to include some more reason. For instance,
I am not expert in ergodic theorycertainly open to any other proposal. In particular, but I know that amenable groups and invariant means are used quite a lot in ergodic theory. Moreover, this This field sounds like something able to produce an example of the kind I am looking for. So I am not expert in ergodic theory and so I am wondering whether there is some result that can be used as a support of the interpretation of invariant means as the generalization of the uniform measure. I hope there are many and, in case, I would like to know one which is easy to state and to understand by somebody that may have no background in ergodic theory, but may have, potentially, a background in classical probability theory. Summarizing,
Question. Are there ergodic theoretical results that really support the interpretation of invariant means as the generalization of invariant measure for groups?
Thanks in advance,
Valerio
2 edited title
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# Invariant means as the generalization of uniform measure
I am writing a paper for game theorists where I use (countable) amenable groups to do some things. So I am writing up a preliminary section about countable amenable groups whose main purpose is to convince the reader that invariant means are the generalization of the uniform measure.
Well, the first obvious thing to say is that Every finite group is amenable and the uniform measure is its unique invariant mean. But this sounds like only a technical reason. Well, I have at least one deeper reason. Let $G$ be a group, fix $W\subseteq G$ and consider the two-person game where the two players choose $x,y\in G$, respectively, and player 1 wins if $xy\in W$. For generic $W$ we can consider this game as a game without relevant information and therefore, a game theoretical reformulation of the Maximum Entropy Principle, would predict that the players will play casually (whatever it means). Indeed, if $G$ is finite, a Nash equilibrium is given by the uniform measure. The point is that if $G$ is countable amenable, then this game admits Nash equilibria and they are exactly suitable invariant means.
I think this is already a pretty convincing motivation, but I would like to include some more reason. For instance, I am not expert in ergodic theory, but I know that amenable groups and invariant means are used quite a lot. Moreover, this field sounds like something able to produce an example of the kind I am looking for. So I am wondering whether there is some result that can be used as a support of the interpretation of invariant means as the generalization of the uniform measure. I hope there are many and, in case, I would like one which is easy to state and to understand by somebody that may have no background in ergodic theory, but may have, potentially, a background in classical probability theory. Summarizing,
Question. Are there ergodic theoretical results that really support the interpretation of invariant means as the generalization of invariant measure for groups?
Thanks in advance,
Valerio
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http://math.stackexchange.com/questions/319137/determine-k-if-using-a-factor-of-x2
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# Determine $k$ if using a factor of $(x+2)$
$f(x)=2x^3-5x^2=kx-20$ Help is appreciated.
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Welcome to MSE! It helps to share what you have tried, and where you are stuck, so people can take it from there. – gnometorule Mar 3 at 3:46
I see you have some answers, but I cannot understand really what your question is. – Peter Tamaroff Mar 3 at 4:10
## 3 Answers
$$\begin{align} f(x) & = 2x^3 - 5x^2 = kx - 20 \\ \\ & = 2x^3 - 5x^2 - kx + 20 = 0 \tag{1}\\ \\ &= (x+2)(2x^2 -9x + 10)=0\tag{2}\\ \\ \end{align}$$
Note that $(-18 +10)x = -kx \implies k = 8.$
Determining what the remaining factors must be in $(2)$: we know $2x^2$ must lead, and it must end in $10$ to obtain $2x^3$ and the constant $20$. We must also then have a term of $-9x$ because we need for $4x^2 - 9x^2 = - 5x^2$. And we see that we can argue that $k = 8$.
You can try doing this using polynomial division, dividing $(1)$ by $(x + 2)$, or you can use trial an error to determine what the remaining term in the second factor must be. To make this work, you'll see $k$ must equal 8.
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Thank you . That clears up things a little. – Robert Mar 3 at 3:43
You're welcome, Robert! – amWhy Mar 3 at 4:11
You're also Welcome, Amy. hAVE A NICE SLEEP – Babak S. Mar 3 at 5:40
Hint $\$ By the Factor Theorem, $\rm\: x\!+\!2\:$ is a factor of $\rm\:g(x)\iff g(-2) = 0.\:$ Applying this to your polynomial $\rm\ g(x) = 2x^3-5x^2-kx+20,\$ the criterion is: $\rm\ g(-2) = 2k-16 = 0\iff k =\: \ldots$
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If $x+2$ is a factor of $2x^3-5x^2-kx+20$, then $-2$ is a zero of that polynomial. So, $(-2)^3-5(-2)^2-k(-2)+20=0$. Now you just have to do the arithmetic to find $k$.
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http://math.stackexchange.com/questions/75031/solving-quadratic-diophantine-equations
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# Solving quadratic diophantine equations
I hope it's not inappropriate asking this here. I stumbled upon this site recently while researching a Project Euler problem, now I figure I'd use it to ask about a recurring theme in these problems: quadratic Diophantine equations.
I've recently boiled down another Project Euler problem (I won't say which, it should be unrecognizable from the original problem and should probably be kept that way) to the following Diophantine equation:
$$5n^2+2n+1=y^2$$
I've been trying to use http://www.alpertron.com.ar/METHODS.HTM as a reference, but I seem to get lost in a sea of constants. And the steps that program on the bottom of that page takes don't seem to match what he says to do. I'd rather be able to understand the steps I'm taking anyway rather than just copying a method.
I'm interested in all positive integer values of n and I'm more or less given a solution exists with n=2. How would I go about finding the rest of the solutions? And how would I solve these kinds of equations in general? If that last part is too complex a question to be handled here, is there any other resource that might help? As far as my current level of math, I have a degree in engineering (and helped a math major with some courses I never took myself) and I've already worked on Project Euler problems involving Pell's equations and continued fraction expansions of square roots.
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## 1 Answer
Start by completing the square on the left: $$5\big( n+\tfrac15)^2 + \tfrac45 = y^2$$ Multiply by 5 to clear denominators: $$(5n+1)^2 + 4 = 5y^2.$$ Therefore you're looking for solutions to the Pell equation $$x^2 - 5y^2 = -4$$ that happen to satisfy $x\equiv1\pmod5$ (so that $n=(x-1)/5$).
Given that you know how to solve Pell's equation, you should be able to find a way to generate all integer solutions $(x_k,y_k)$; some of these values $x_k$ will be $1\pmod 5$, and the set of such $k$ should be an arithmetic progression.
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I guess I should be more specific. I've had problems dealing with Pell's equations of the form x^2-ny^2=1. I've had to calculate solutions for such equations by checking the convergents of the square root of n. I've had other equations where I've had to do similar substitutions as the above, though I wasn't aware about the arithmetic progression thing and checked each answer to make sure it was still an integer after resubstition. I do not know why the convergents work and assume the answer would be complex. I also don't know how to generalize to solve x^2-ny^2=k. – Mike Oct 23 '11 at 10:29
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There's one obvious solution $(a,b)$ to $x^2-5y^2=-4$. Prove that if $(x,y)$ is any solution then $(u,v)$ is another given by $(u+v\sqrt5)/2=((3+\sqrt5)/2)(x+y\sqrt5)/2$. It follows that $((3+\sqrt5)/2)^r(a+b\sqrt5)/2$ gives solutions for all $r$. If you choose $r$ cleverly, you'll even get $x\equiv1\pmod5$. If that's not enough, the libraries are full of intro number theory texts that discuss quadratic diophantine equations. – Gerry Myerson Oct 23 '11 at 11:47
How do you do mathematical notation here like exponents and radicals? Ah well, I was able to prove that statement. First I went to prove that if a+bz=c+dz, where a,b,c, and d are rational and z is irrational, then a=c and b=d. I rewrite the equation as a-c=(d-b)z. The left side is rational. The right size is rational if and only if d-b=0. So that step is proven. I cleared the parentheses in your equation to yield u=(3x+5y)/2 and v=(x+3y)/2. I then wrote u+v(root 5) in terms of x and y, then multiplied by the original equation you gave and simplified, giving u^2-5v^2=x^2-5y^2 – Mike Oct 23 '11 at 23:13
I guess I'll need to check the libraries for that and maybe something on game theory. In any case, I used u as x(n+1) and checked to see if that equation with (x0,y0)=(1,1) would generate all the solutions I needed. It appears the answers I needed were given for even values of r, so I redid the recurrences to take that into account. I was given what the 10th value should be for the problem, which matched the 10th iteration. So while I don't know how you arrived at that formula, it did generate all solutions I needed. Another problem solved. Thanks. – Mike Oct 24 '11 at 2:36
If you have $x^2 - 5^2 = -4$, then $(x / 2)^2 - 5 (y / 2)^2 = -1$. That I recognize as Pell's equation. – vonbrand Mar 25 at 1:15
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http://www.physicsforums.com/showthread.php?t=663388
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Physics Forums
Please help with calculation involve solid angle.
This is only an example from Kraus Antenna 3rd edition page 404. The question is really a math problem involves calculation of ratio of solid angles. Just ignore the antenna part. this is directly from the book:
Example 12-1.1 Mars temperature
The incremental antenna temperature for the planet Mars measured with the U.S. Naval Research Lab 15-m radio telescope antenna at 31.5mm wavelength was found to be 0.24K(Mayer-1). Mars subtended an angle of 0.005 deg at the time of the measurement. The antenna HPBW=0.116 deg. Find the average temperature of the Mars at 31.5mm wavelength.
The answer from the book:
$$T_s=\frac {\Omega_A}{\Omega_S}\Delta{T_A}\;\;\approx\;\frac {0.116^2}{\frac{\pi}{4}0.005^2}(0.24)=164K$$
I don't understand where the $\frac {\pi}{4}$ in the denominator comes from. This is just a simple problem where the ratio of the area of two disk one with Θ=0.116/2 and the other with Θ=0.005/2. I try to use the following to calculate:
$$\Omega=\int_0^{2\pi} d\phi\;\int_0^{\theta/2}\sin {\theta} d \theta$$
So
$$\frac{\Omega_A}{\Omega_S}\;=\;\frac{\int_0^{2\pi} d\phi\;\int_0^{0.116/2}\sin {\theta} d \theta}{\int_0^{2\pi} d\phi\;\int_0^{0.005/2}\sin {\theta} d \theta}\;=\;\frac{2\pi\int_0^{0.116/2}\sin {\theta} d \theta}{2\pi\int_0^{0.005/2}\sin {\theta} d \theta}$$
$$\frac{\Omega_A}{\Omega_S}\;=\;\frac{(1- 0.999999487)}{(1-0.999999999)}=538.2$$
But using the answer from the book:
$$\frac{0.116^2}{\frac{\pi}{4}0.005^2}\;=\;685.3$$
I just don't get the answer from the book. Can it be the limitation of my calculator to get cos 0.0025 deg? Can anyone use their calculator to verify the number? please help.
Thanks
Alan
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions: Science Advisor I'll take a guess at this. It seems like the same idea as "geometric form factors" in radiative heat transfer. The telescope is "seeing" a piece of space that you can take as a flat disk (diameter 0.116 degrees) But the surface of Mars that is radiatiing is not a flat disk, it is a hemisphere. The "brightness" as seen by the telescope will be maximum in the center and less at the edges. That will give you another factor of ##\sin\theta## or ##\cos\theta## in the integral for Mars. Probably the "correction factor" of ##\pi/4## is a well known result, since most astronomical objects are approximately spherical.
Thanks for you time. I don't mean to say you are not right as it might well can be. On the first pass, I don't think that's what the book meant, the reason is, this example is right at the beginning of the chapter and the book did not mention anything about the concaveness of the star. Also the answer of the book: $$T_s=\frac {\Omega_A}{\Omega_S}\Delta{T_A}$$ Which does not contain any compensation for the concaveness of the planet. So far, this book has been pretty straight forward and concise. I type out the exact words from the book in my post and nothing about this. I scan through the later part of the chapter and there's nothing about this either. Basically it only talked about the ratio of the solid angle for object smaller than the half power angle. Again thank you for your time. I manage to go on line and use a cosine calculator that don't flag error of 0.0025 deg and gave me 2 more digits in the decimal place than my calculator, the answer is even more off!! So I don't think the small angle is the cause of the problem.
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Please help with calculation involve solid angle.
Quote by AlephZero But the surface of Mars that is radiatiing is not a flat disk, it is a hemisphere. The "brightness" as seen by the telescope will be maximum in the center and less at the edges.
Are you sure about that? It doesn't seem right to me.
yungman, is it possible that HPBW is to be interpreted as a square, not a circle?
Quote by haruspex Are you sure about that? It doesn't seem right to me. yungman, is it possible that HPBW is to be interpreted as a square, not a circle?
It kind of make sense as the surface is concave and you have the max reflection at the center of the planet where the surface is true normal to the propagation. But as I said, the equation don't reflect this, all it gave was the ratio of the two solid angle where there is no reason for the π/4 to appear.
That I am pretty sure, the main lope a very directional antenna is like a long balloon or sausage!!! But still don't explain the pi/4 even if you look at it as square.
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Quote by yungman It kind of make sense as the surface is concave and you have the max reflection at the center of the planet where the surface is true normal to the propagation.
Seems to me the model is to suppose the surface is uniform temperature and is radiating on that basis. I believe a sphere at uniform intrinsic brightness would appear appear as a uniformly bright disc whatever position it is viewed from, but I'm prepared to be proved wrong about that.
That I am pretty sure, the main lope a very directional antenna is like a long balloon or sausage!!!
Yes, but what I'm suggesting is that when an antenna is quoted as having a diameter of such-and-such, that is to be interpreted as the square root of the 'effective' area, whatever shape and response pattern the actual area is. I've no idea whether this is the case - just trying to make sense of the equation
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Quote by haruspex Are you sure about that? It doesn't seem right to me.
I'm not sure either, whcih is why I started my post with "I'll take a guess at this".
The physics question here is whether each point on the surface of Mars radiates energy in all directions like a "point source", or whether some effect (e.g. electrical conductivity of the Martian material) means sonething different happens. It's not self-evident to me that EM radiation with wavelength 31.5mm would behave the same way as visible light (the wavelengths are different by a factor of 10^6) but I don't know whether it does or it doesn't.
Quote by haruspex Seems to me the model is to suppose the surface is uniform temperature and is radiating on that basis. I believe a sphere at uniform intrinsic brightness would appear appear as a uniformly bright disc whatever position it is viewed from, but I'm prepared to be proved wrong about that. Yes, but what I'm suggesting is that when an antenna is quoted as having a diameter of such-and-such, that is to be interpreted as the square root of the 'effective' area, whatever shape and response pattern the actual area is. I've no idea whether this is the case - just trying to make sense of the equation
Thanks for the answer. In the whole, they use round shape like a cone to represent the main lobe. I don't believe the book mean square. Besides, using square don't explain ∏/4 either.
The question is whether I am doing it right with my approach assuming the Mars has uniform temperature on the surface. I really believe the book is wrong to put the ∏/4 in. The square of the degree will be good already as the ratio is the same using Sr or degree.
I just try taking my result (538) divide by ∏/4, the answer is 685 which is the same as the book!!!
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Quote by yungman Besides, using square don't explain ∏/4 either.
Yes it does. If the antenna 'diameter', A, is interpreted as the side of a square then its area is A2. Mars' diameter, M, gives an area πM2/4.
Yes, is the only way to for it to make sense. I just don't know of any antenna that have a square cross section beam. Mostly has round or at least roundish cross section. I guess we never know what the author think. Anyway, thanks for the help.
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Quote by yungman I just don't know of any antenna that have a square cross section beam.
I'm sure they don't, but it's quite possible that the apertures are conventionally quoted as though they are square. In other words, it is not saying it is actually width D, merely that the effective area is D2.
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http://mathoverflow.net/questions/39312?sort=votes
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## Smallest non-isomorphic strongly regular graphs
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Motivation: I want to see how the 3-dimensional Weisfeiler-Lehman algorithm (see Logical complexity of graphs, p. 14) distinguishes between two non-isomorphic strongly regular graphs srg(v,k,λ,μ) in a specific example.
Question: What are the smallest non-isomorphic strongly regular graphs with the same v,k,λ,μ?
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## 1 Answer
This page http://www.maths.gla.ac.uk/~es/srgraphs.html lists some strongly regular graphs on few vertices, and gives two (16,6,2,2) graphs (which I didn't check but I presume they're non-isomorphic). I imagine they're the smallest possible but I haven't checked: http://www.maths.gla.ac.uk/~es/16.vertices
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The two (16,6,2,2) graphs are the Shrikhande graph and the line graph of $K_{4,4}$. The Shrikhande graph may be obtained by forming a 5x5 grid of squares, adding a diagonal in the same direction to each square, and gluing opposite edges of the square grid to form a torus. – David Eppstein Sep 19 2010 at 19:10
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I forgot to add: they are obviously nonisomorphic because the neighborhood of a vertex in the Shrikhande graph is a 6-cycle, whereas the neighborhood in the line graph of $K_{4,4}$ is a pair of 3-cycles. – David Eppstein Sep 19 2010 at 19:13
Thanks for the clarification, David. Do you know if they are indeed the smallest, as they seem to be? – Andrew D. King Sep 19 2010 at 19:40
3
Yes, e.g. in oai.cwi.nl/oai/asset/1817/1817A.pdf Brouwer and van Lint write that this is the only nonisomorphic pair with fewer than 25 vertices. – David Eppstein Sep 19 2010 at 20:04
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Peter Cameron discussed these graphs in a blog post recently: cameroncounts.wordpress.com/2010/08/26/… – Emil Sep 19 2010 at 21:33
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http://crypto.stackexchange.com/questions/5955/how-can-we-determine-if-two-discrete-logarithms-are-equal/5957
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How can we determine if two discrete logarithms are equal?
Let $p$ be a prime number, and let $g_{1},g_{2},...,g_{n}$ be $n$ generator of $Z^{*}_{p}$. We have a list $y_{1},y_{2},\dotsc,y_{n}$ of elements in $Z^{*}_{p}$ such that for every $i\in \lbrace1,2,\dotsc,n \rbrace$ we have $y_{i}=g_{i}^{x_{i}} \bmod p$ for some number $x_{i}$ (but we don't know $x_{i}$). I am trying to find an algorithm to determine all pairs $(y_{i},y_{j})$, $i\neq j$ such that $x_{i}=x_{j}$.
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If you know what powers of one generator results in the other generators then the problem is easy. Please clarify your question to indicate whether this is the case or not. – Barack Obama Jan 11 at 2:06
1 Answer
This problem is equivalent to the decisional Diffie-Hellman problem, and hence your problem is intractable (assuming, of course, that the group is well chosen).
Here's how we can use an Oracle that can solve the above problem to solve the DDH problem:
• In the DDH problem, we're given values $g, g^x, g^y, g^z$, and we're asked whether $xy = z$.
• We call the Oracle with the following instance: $n=2$, $g_1 = g$, $g_2 = g^x$, $y_1 = g^y$, $y_2 = g^z$.
• The Oracle will return that $x_1 = x_2$ is a matching pair iff, for that same $w$, $g_1^w = y_1$ and $g_2^w = y_2$, that is, if $g^w = g^y$ and $g^{wx} = g^z$, which is to say, if $g^{xy} = g^z$
It's also obvious how to solve your problem with an Oracle that solves the DDH problem (using $n (n-1)/2$ calls to the Oracle).
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I believe there are some groups in which the decisional DH problem is easy, but the computational DH problem is hard. These are useful for some unusual protocols. – CodesInChaos Jan 9 at 19:51
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@CodesInChaos: yes, there are such groups; however, the group $Z^{*}_{p}$ is not one of them. – poncho Jan 9 at 19:55
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http://mathoverflow.net/questions/31629/advantages-of-working-with-cw-complexes-spaces-over-kan-complexes-simplicial-sets/31638
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## Advantages of working with CW complexes/spaces over Kan complexes/simplicial sets?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Many topologists express a clear preference for working with CW complexes instead of simplicial sets.
One of the reasons is that the cellular chain complex of a CW complex is often easier to work with than a simplicial chain complex. However, simplicial sets have many nice features that spaces do not. The category of simplicial sets has a proper and combinatorial (in the sense of Jeff Smith) model structure and is a presheaf topos, which makes the objects behave very much like sets. Surely these make up for the problems with specifying combinatorial data?
The question: Why do many topologists and homotopy theorists prefer to work with spaces and CW complexes over simplicial sets and Kan complexes? What are some other advantages that CW complexes enjoy over Kan complexes?
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2
I'm pretty surprised to hear that you've found a substantial number of homotopy theorists willing to express a clear preference for CW complexes over simplicial sets. Both are very useful, for different purposes. I see no reason to prefer one to the other in general, although certainly there are specific situations in which one is easy to work with and the other would be very difficult or annoying to use. – Dan Ramras Jul 12 2010 at 23:24
8
It suffices to specify the degree of the attaching maps (only) if you are interested in computing homology groups, but a CW structure needs the full (homotopy class) of attaching maps. For example, CP^2 and S^2 \vee S^4 have isomorphic cellular chain complexes but are not homotopy equivalent as can be seen through their cohomology rings. The attaching map for CP^2 is the Hopf map S^3 --> S^2, which does not have a sensible degree. Indeed, one fact which "everyone should know" is that the cellular chain complex loses information needed to compute cohomology ring structure. – Dev Sinha Jul 12 2010 at 23:31
I've never worked with CW complexes. All of my experience with homotopy theory is with simplicial sets and model categories. That's why I'm asking this question =). – Harry Gindi Jul 12 2010 at 23:46
13
It's hard to talk about manifolds, classifying spaces, the Pontryagin-Thom construction, $G$-equivariant homotopy theory where $G$ is compact Lie, etc. without making reference to topological spaces. – Sam Isaacson Jul 12 2010 at 23:56
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By the way, there's an amazing theorem by Mike Mandell that roughly says that as an $E_\infty$-algebra, $C^\ast(X; F_p)$ retains all the homotopical information about $X$ if $X$ is $p$-complete, nilpotent, connected, and of finite type. The rational version of this statement is due to Quillen. But as Dev pointed out, there's no obvious way to get this multiplicative structure when you work with cellular chains. – Sam Isaacson Jul 13 2010 at 1:09
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## 2 Answers
I think there are many times that simplicial sets are preferable (e.g for classifying spaces the simplicial construction is often advantageous), but to answer the stated question:
• CW complexes connect more immediately to manifold theory (Morse functions give CW structures; a finite CW complex is homotopy equivalent to a manifold by embedding it in some Euclidean space and "fattening it up").
• CW structures can be simpler and more explicit in "small" cases. For example, I do not know an explicit simplicial set whose realization is $CP^2$ (though perhaps I could work one out using a simplicial model for the Hopf map.)
• CW complexes can be analyzed using manifold theory. For example, maps from manifolds to $n$-dimensional CW complexes such as attaching maps can be understood in part by taking a "smooth" approximation and looking at preimages of points in each cell (Goodwillie uses this kind of technique to generalize the Blakers-Massey theorem).
But why should one have to choose "once and for all" between building things from sets vs. from vector spaces, anyways?
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2
Good points! Just to register a counterpoint, I think it's at least as easy associating a simplicial set to a manifold as associating a CW-complex to one: choose a Riemannian metric, then take the Cech nerve of a covering by geodesically convex open sets. This isn't a Kan complex, though, which reminds us of another convenience of CW-complexes: more constructions are automatically homotopy-invariant. – Dustin Clausen Jul 13 2010 at 16:10
### 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.
My gut reaction is always to work with CW complexes because, being a topologist, I like to work with spaces. Simplicial sets, as nice as they may be, are definitely not spaces.
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1
Why the negative vote? – Mariano Suárez-Alvarez Jul 13 2010 at 0:57
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I don't know why this was downvoted. There are many types of topologists in the world. While the more algebraically-minded sometimes prefer simplicial objects, the more geometrically-minded (I could myself among this group) feel more comfortable working with actual spaces, and thus often prefer CW complexes. Sometimes it is a matter of technology (eg if your space is a manifold it would be perverse to replace it w/ a simplicial set, as you would lose the ability to talk about the tangent/stable normal bundles, embedded submanifolds, etc), but often it is just a matter of taste. – Andy Putman Jul 13 2010 at 0:58
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I'm not sure I agree with the statement that simplicial sets are definitely not spaces. Of course, set theoretically, that's true. But would you say that simplicial complexes are not spaces? One can describe simplicial complexes as sets equipped with a "downward closed" collection of subsets, and yet topologists certainly think of simplicial complexes as spaces. Simplicial sets are just another, more flexible, way to describe spaces combinatorially. Some homotopy theorists may prefer not to think of simplicial sets as geometric objects, but that's hardly a universal point of view. – Dan Ramras Jul 13 2010 at 3:23
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CW complexes aren't exactly spaces either for that matter - they're spaces with a decomposition... Sometimes we might confuse the notions of "CW complexes" and "spaces homotopy equivalent to CW complexes," and thus not as readily recognize their respective drawbacks. – Dev Sinha Jul 13 2010 at 6:26
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Dan, as a topologist who's only recently learned not to run screaming from the room when simplicial sets come up, I'd argue (from a naive point of view) that simplicial complexes are spaces more than simplicial sets are. This is probably because I tend to (happily) confuse complexes with their realizations. But back to the original question, CW complexes come ready-made with topologies that simplicial sets need an extra step to get to. I'd also argue (honestly) that the literature is better for a student to learn about CW complexes than simplicial sets. – Greg Friedman Jul 13 2010 at 10:07
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http://quant.stackexchange.com/questions/1833/duality-between-constant-rebalanced-portfolio-crp-and-corresponding-derivative/1922
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# Duality between constant rebalanced portfolio (CRP) and corresponding derivative
One of the greatest achievements of modern option pricing theory is finding corresponding dynamical trading strategies in linear instruments with which you can replicate and by that price derivative instruments; of course the most prominent example being Black Scholes and dynamic hedging.
I wonder concerning another trading strategy, the constant rebalanced portfolio (CRP), what a corresponding derivative would look like:
• How would the payoff diagram and resulting return distribution look like?
• How to price it (in a BS world for a start)?
• How to statically hedge it with plain vanilla options?
I think this will be especially interesting since it is a well known fact that (under appropriate assumptions) a CRP uses the volatility component to shift the drift upwards (unlike most other trading strategies which leave the drift unchanged but only form the return distribution differently). Therefore volatility is not only an element of risk but of chance too. I want to fully understand this phenomenon, also known as volatility pumping, by using the machinery of option theory.
Do you have any literature concerning this topic (I found none) or any ideas and hints where to start?
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1
Fascinating question! I'd be very interested to know the answer as well, and will hopefully try to find an answer myself (if no one else does first) when I have some time. Do you have a source for the "well known fact" regarding volatility pumping? Also, I doubt you could achieve this with a static options position, since options expire and a CRP is meant to be held indefinitely. – Tal Fishman Sep 5 '11 at 14:53
– vonjd Sep 5 '11 at 16:12
– vonjd Sep 5 '11 at 18:27
– vonjd Sep 5 '11 at 18:40
## 2 Answers
Actually there are more than just ideas and hints concerning this topic. There is an intuitive model and solution to your question already using machinery of option theory. But don't worry, it's not a surprise that you didn't find any useful literature in your search because the proposed solution actually comes from a very different topic.
In addition to the references recommendation, here are some analogies and highlights that hopefully will bridge the proposed solution and your interests:
• Note that in volatility pumping, which is the easiest version of constant rebalanced portfolio (CRP), we always invest less than 100% capitals in risky assets. What if now, I propose to invest more than 100% capitals in risky assets? Will you buy it? :)
• For whatever reasons, many investors buy it! And that's why we have a fast growing leveraged exchange-traded funds (Leveraged ETFs) market. Yeah, >100% capital investment is 'leveraged'. As the business grow, there thus arise greater demands for better understanding and modeling of these leveraged services/products.
• There you go! The answer you are looking for, for a 'de-leveraged' trading strategy, actually lies in the research of modeling leveraged assets. All you need to do is to plug in your favored leverage (x1/2 for volatility pumping) and double think about the new formula and consequences (mostly reversed). Here is a list of the recommended references:
[1] Leveraged ETFs: All You Wanted to Know but Were Afraid to Ask
[2] A Dynamic Model for Leveraged Funds
[3] The Dynamics of Leveraged and Inverse Exchange-Traded Funds
[4] Path-Dependence Properties of Leveraged Exchange-Traded Funds: Compounding, Volatility and Option Pricing
• There is nothing magic after you read it: your volatility pumping (upward drift) will become volatility 'dragging' (downward) for leveraged ETFs. If you can fully understand dragging in leveraged ETFs (which is exactly the goal these references try to help you), you gain a complete understanding of pumping in CRP at the same time. Even better, these references indeed use the machinery of option theory :)
• You can also prove to yourself that x1/2 leverage will maximize your volatility pumping (not in paper discussion, but it's very obvious once you understand it).
Unfortunately, there are not much discussions regarding derivatives on leveraged assets in these references and others except for the last paper. But honestly once you master the underlying dynamics, solution for derivatives pricing isn't too far away :)
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Oh? The bounty reward was given away while I am posting my answer? I remember there are still few hours to go before the bounty deadline? :) Too bad I just back from a vocation! – 楊祝昇 Sep 15 '11 at 20:32
Do any of these papers discuss replicating a leveraged ETF with options? If so, then it may be applicable to this problem, but if not, then I'm not sure you've offered anything besides a useful framework (not irrelevant, but, like my answer, not a complete answer). BTW, if vonjd feels your answer is best you can still get 25 points with an up-vote and check mark. If Louis also feels your answer is much better than mine, I'd be willing to re-offer the bounty and award it to you. – Tal Fishman Sep 15 '11 at 20:44
Hello Tal, please keep your bounty. I am teasing myself :) As for your question, in my opinion, once you have sound underlying dynamics, replicating is trivial. To myself, my answer is complete enough, i.e. the information I suggest here is enough to solve vonjd's questions. – 楊祝昇 Sep 15 '11 at 21:21
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Thank you Chu-Sheng: I am accepting this answer since it opens up a whole new line of research relevant to the topic + it connects different quant topics impressively. – vonjd Sep 16 '11 at 6:15
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Thank you for accepting the answer, vonjd :) And thanks for your inspiring question. I always believe it's the smart questions and curiosity that lead us to great answers and ideas! I wish I could answer you more but I really favor short and inspiring answer in public. However, I am more than happy to discuss further in private :) Please feel free to contact me (Google/Linked) if you are interested in more discussions. – 楊祝昇 Sep 16 '11 at 18:35
While not really an answer, here are my thoughts on the problem.
For starters, I would approach the problem as one of whether a portfolio which is constantly rebalanced over some horizon can be replicated using vanilla options. Obviously the portfolio will have to be rolled over when the options reach expiration. Then the rest, such as payoff diagram and price in a B-S world can be easily derived.
Suppose you have a \$100 portfolio to be invested equally in two assets, X and Y, both of which have liquid options over a closely-spaced range of strikes. Let's say both assets start out at a price of \$50, so that initially you want to hold 1 share of each. In a traditional CRP, as the asset prices evolve, you will want to sell some of whichever asset outperforms and buy the underperformer. With options, however, an alternative strategy is to simply try to keep the exposure to each asset fixed at \$50. In order for the P&L from the options portfolio to match the hypothetical CRP, though, one must then "rebalance" by buying and selling the entire replicating portfolio. For example, if both assets increase by 1%, then the replicating portfolio should show a P&L of \$1. However, since the exposures will remain \$50/\$50, and additional \\$1 unit of the options portfolio must be purchased (essentially reinvesting the P&L).
Proceeding under these assumptions, the replicating portfolio of options for asset X must satisfy \begin{equation} \sum_i\Delta_{t,i}^X=\frac{p_0^X}{p_t^X} \end{equation} where the sum of $i$ is over all option contracts held for asset X, and $p_t^X$ is the price of X at time $t$. A similar equation holds for asset Y, which is henceforth interchangeable with X (so I drop the superscript).
In order to keep the exposures constant, the sum of deltas must not depend on volatility, interest rates, or other extraneous factors besides price. I'm not sure what the ultimate dependence between sum of deltas and price will be, so perhaps some additional dynamic trading will be necessary because of this factor. What remains now is to solve for the portfolio of options with this property. Perhaps with this first step someone else can take up the mantle, but I believe this will take me more time than I have to spend on this problem at the moment.
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This is really interesting - Thank you! I hope that somebody will take up the mantle - or that you will find the time to continue your work. Give me a shout if you would be interested in publishing a (white) paper on this, perhaps we can work together here... I would very much appreciate that! Thank you again. – vonjd Sep 16 '11 at 6:17
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http://math.stackexchange.com/questions/280876/definition-of-a-peculiar-quotient-group-of-isometries
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# Definition of a peculiar quotient group of isometries
I just wrote a text file to sum up my ideas about the Riemann Hypothesis. This text is a draft, and I don't expect people here to say if this approach is interesting or not (but if by chance you think so, please let me know). My question is about the definition of a peculiar quotient group, which might not be correct in definition 6.
Can someone tell me if it's ok or not?
Thank you in advance.
Here comes the text:
Main ideas towards a proof of GRH
Definition 1
Let $A$ be be a subclass of the Selberg class containing $1$, closed under products, and such that every element of $A$ can be factored in a unique fashion in a product of primitive elements of $S$, these primitive elements belonging to $A$. Such a subclass of $S$ will be called a Galois class of L-functions.
Definition 2
Let $A$ be a Galois class of L-functions. $T$ is an automorphim of $A$ iff the following properties simultaneously hold true:
1) $T$ is a bijective map from $A$ to itself
2) $T$ maps a primitive element of $A$ to a primitive element of $A$
3) for all $F$ in $A$, the degree of $F$ and the degree of $T(F)$ are the same
4) for all $F$, $G$ in $A$, $T(F.G)=T(F).T(G)$
Definition 3
$\tilde{\phi}:F\mapsto \phi\circ F\circ \phi^{-1}$ is said to be an isometric automorphism of $A$ if $\tilde{\phi}$ is an automorphism of $A$ and $\phi$ is an isometry of the complex plane preserving any rectangle centered in $1/2$.
Proposition 1
$\tilde{\phi}(F)=F$ implies $\phi(0)=0$.
Definition 4
A group of equivalence classes of isometries is said to be nice iff each equivalence class contains exactly one representant $\phi$ such that $\phi(0)=0$.
Let $F\in A$ and let $G_F$ be the group of all equivalence classes of isometries preserving the set of non trivial zeros of $F$, where the equivalence relation $R_F$ is defined by : $f R_F g$ iff for every non-trivial zero $s$ of $F$, $f(s)=g(s)$.
Let $G'_F$ be the group of isometric automorphisms of $A$ preserving $F$.
Proposition 2
$G_F$ and $G'_F$ are isomorphic iff $G_F$ is nice.
Definition 5
Let $A$ and $B$ two Galois classes of L-functions such that $A$ is a sub Galois class of L-functions of $B$. Let's define the L-type (Isometric) Structure group of $B$ over $A$ as the group of (isometric) automorphisms of $B$ preserving $A$ pointwise. The L-type structure group of $B$ over $A$ will be denoted as LStr(B/A) and the L-type Isometric Structure Group of $B$ over $A$ LIStr(B/A).
Definition 6
Let $GZ(A)$ be the set $\bigcup_{F\in A}\{Z(F):=\{s,F(s)=0, 0<\Re(s)<1\}\}$. The group of all isometries preserving $GZ(A)$ will be denoted as $PG_A$. Now let's consider the relation $R'_A$ defined by $f R'_A g,$, where $f$, $g$ are elements of $PG_A$, iff for all $F$, $H$ in $A$, $f R_F g$ iff $f R_H g$. The quotient group $PG_A/R'_A$ will be denoted as $SG_A$. Let's now consider the group of elements of $SG_B$ preserving $GZ(A)$ pointwise. This last group will be called "Z-type Isometric Structure Goup of $B$ over $A$" and denoted as ZIStr(B/A).
Conjecture 1
Let $A$ be a Galois class of L-functions, $F$ an element of $A$. $G_F$ is nice iff for all $H$ in $A$, $G_H$ is nice.
Conjecture 2 (Main Conjecture)
Let $A$ and $B$ be two Galois classes of L-functions such that $A$ is a sub-Galois class of L-functions of $B$. Then LIStr(B/A) and ZIStr(B/A) are isomorphic.
Idea of proof: prove that LIStr(B/A) and ZIStr(B/A) are both isomorphic to $Gal(F_{B}/F_{A})$, where $F_B$ is the field generated by the union of all images of $\mathbb{R}-\{1\}$ by the elements of $B$, and $F_{A}$ is defined in a similar way. If $B$ is the maximal Galois class of L-functions and $A=\{\zeta^{n}, n\geq 0\}$, one can expect to have a proof of the Riemann Hypothesis.
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You should maybe think about changing the title to something more specific. I for one came here thinking this question is about the definition of 'quotient groups' and nothing too advanced ;) – Rand al'Thor Jan 17 at 18:10
Thank you for your comment. I changed the title. – Sylvain Julien Jan 17 at 18:33
Not exactly an answer to the initial question, but one can show that $F_A=\mathbb{R}$ if and only if all the elements of $A$ are self-dual, otherwise $F_{A}=\mathbb{C}$. Hope this can be of interest... – Sylvain Julien Jan 21 at 19:22
## 1 Answer
$SG_{A}=PG_{A}/[Id]$, where $[Id]$ is the equivalence class of the neutral element of $PG_{A}$ for the relation $R'_{A}$. So that $SG_{A}$ is well defined.
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http://math.stackexchange.com/questions/tagged/banach-spaces+real-analysis
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1answer
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### (Real Analysis) Integration of two functions
Note ($\Omega,A,\mu)$ is a finite additive space. Let $f\in \bar S(A, \mathbb{R})$ and $y\in Y$ where $Y$ is Banach space. Prove $\int _\Omega yf d\mu = y \int _\Omega f d\mu$. Also, for any $X\in A$, ...
3answers
82 views
### Completeness proof of $\ell^p$
Say $\{x_n\}$ is Cauchy in $\ell^p$ and $x$ is its pointwise limit. To argue that $x \in \ell^p$ would the following be correct: Let $\varepsilon > 0$ and let $N$ be s.t. $n,m > N$ ...
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53 views
### A question about convergence in $L^p$. [duplicate]
Let $E$ be measurable and $1 \le p \le \infty$. Suppose $\{f_n\}_{n \in \mathbb{N}}$ all measurable and $\{f_n\}_{n \in \mathbb{N}} \to f$ pointwise a.e. $E$. For $p$ as above, I want to show that: ...
1answer
43 views
### Clarification of Reed and Simon proof of the open mapping theorem
I was reading Reed and Simon Methods of Mathematical Physics Volume 1 and have a question about a small their proof of the open mapping theorem for Banach spaces. Let $T:X\rightarrow Y$ be a bounded ...
0answers
72 views
### Show that the space $l^2= \{ a \in \mathbb{R}^\mathbb{N} | \sum |a_k|^{2} < \infty \}$ is a Banach space
If we have $l^2= \{ a \in \mathbb{R}^{\mathbb{N}} | \sum_{k=0}^{\infty} |a_k|^{2} < \infty \}$ and $||a||_2 = (\sum_{k=0}^{\infty} |a_k|^2 )^{1/2}$ 1 proposition: $l^2$ is a vector ...
2answers
127 views
### Prove that $(B, \|-\|_{\infty})$ complete. B the set of bounded real valued functions on [0,1] which are pointwise limit of continuous functions.
Question: Prove that $(B, \|-\|_{\infty})$ is complete. B the set of bounded real valued functions on [0,1] which are pointwise limit of continuous functions on [0,1]. Context: Old exam problem I'm ...
0answers
26 views
### how to show this operator is in fact convolution with a specific kernal
Firstly, the operator T is defined as taking the fourier inverse transform of the function $(1-|\zeta|)1_{[-1,1]}(\zeta)\hat{f(\zeta)}$. ie.$\hat{Tf}$=$(1-|\zeta|)1_{[-1,1]}(\zeta)\hat{f(\zeta)}$. I ...
3answers
95 views
### Does $C_0(X)$ completely determine $X$?
Let $X$ and $Y$ be compact metric spaces. Let $C_0(X)$ and $C_0(Y)$ be the Banach spaces of continuous real-valued functions over $X$ and $Y$, respectively. If $F : X \rightarrow Y$ is a ...
1answer
134 views
### Weak convergence and weak$^*$ convergence question
Let $X$ be a Banach space and $X^*$ be its dual space. Let $\phi_n\in X^\ast$ and for all $x\in X$ we have $\phi_n(x)\to c\in\mathbb{C}$ as $n\to\infty$. I want to show that the sequence $\phi_n$ has ...
1answer
96 views
### prove a subset of squence space lp closed in strong topology
Let $l^p$ be the space of $p$-summable sequences. von Neumann constructed a subset of $l^p$ space $$S=\{X_{mn}: m,n≥1\}$$ where $X_{mn}\in l^p$ are defined by $X_{mn}(m)=1, X_{mn}(n)=m$ and ...
1answer
72 views
### Proving $L^2$ convergence (application of dominated convergence?)
For any $f\in L^2(\mathbb{R}^d)$ prove \begin{align}\left\lVert \int_{\mathbb{R}^d} e^{i |x-y|^2}f(y) dy-\int_{\mathbb{R}^d}e^{i |x-y|^2} e^{-|y|^2/a}f(y) dy \right\rVert_{L^2} \rightarrow 0\ \ \ ...
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46 views
### Cantor intersection theorem HW [duplicate]
Possible Duplicate: Nested sequences of balls in a Banach space I am asked to show the intersection of a decreasing sequence of nonempty closed balls in a Banach space is nonempty. It ...
2answers
199 views
### fail of cantor intersection property on closed , bounded , convex sets of integrable functions
This is from my recent homework. I am asked to find a descending nested sequence of closed , bounded , nonempty convex sets $\{D_n\}$ in $L^1(\mathbb{R})$ such that the intersection is empty , where ...
3answers
121 views
### References on relationships between different $L^p$ spaces
I am looking for detailed references containing proofs of inclusion relationships between different $L^p$ spaces and multiple counterexamples of functions in one but not the others.
1answer
123 views
### $C^1 [0,1]$ with different norm
If the space $C^1 [0,1]$ is equiped with norm $\Vert \cdot\Vert_1$,where $$\Vert f\Vert_1=|f(0)|+\Vert f'\Vert _{C}=|f(0)|+\sup_{t\in [0,1]}|f'(t)|$$ for any $f\in C^1 [0,1]$, is this space Banach? ...
1answer
96 views
### Map bounded if composition is bounded
Let $X,Y,Z$ Banach spaces and $A:X\rightarrow Y$ and $B:Y\rightarrow Z$ linear maps with $B$ bounded and injective and $BA$ bounded. Prove that $A$ is bounded as well. If I knew that $B(Y)$ is ...
2answers
185 views
### Boundary of $L^1$ space
Is there any rigorous or heuristic notion of boundary of $L^1$ that is studied? I mean something loosely like the collection of functions or distributions defined by \left\{f\notin L^1: f_n\to ...
2answers
147 views
### Shifting a function is continuous
I'm slightly puzzled by the following: if $g(t)$ is a function in $L^q(X)$ then we can show that $g(t-x)$ is continuous function of $t$, i.e. for $\varepsilon > 0$ we can find $\delta$ such that ...
1answer
353 views
### Space of Complex Measures is Banach (proof?)
How can we prove that the space of Complex Measures is Complete? with the norm of Total Variation. I have stuck on the last part of the proof where I have to prove that the limit function of a Cauchy ...
1answer
75 views
### differentiability classes
I am beginning to study multidimensional calculus using Spivak's Calculus on Manifolds, and so far I understand the purpose of considering the classes $C^n$ to be twofold. First, being in $C^n$ is a ...
1answer
114 views
### Range of bounded operator is of first category
Let $T$ be a bounded operator from a Banach Space $X$ to a normed space $Y$ such that $T$ is not onto, but $R(T)\subset Y$ is dense. Prove that $R(T)$ is of first category and not no-where dense. ...
1answer
213 views
### Every multiplicative linear functional on $\ell^{\infty}$ is the limit along an ultrafilter.
It is well-known that for any ultrafilter $\mathscr{u}$ in $\mathbb{N}$, the map\begin{equation}a\mapsto \lim_{\mathscr{u}}a\end{equation} is a multiplicative linear functional, where ...
1answer
157 views
### Cancellation law for Minkowski sums
Let $(X,\|\cdot\|)$ be a Banach space and $A,B,C\subset X$ closed bounded non-empty convex subsets. Let $+$ denote the Minkowski symbol for addition. Does the $+$ satisfy: A+C\subset B+C\implies ...
2answers
102 views
### Composition of continuous and closed operators is closed
Let $X,Y,Z$ Banach spaces, $\text{dom}(S)\subset Y$, let $T:X\rightarrow Y$ be linear and continuous and let $S:\text{dom}(S)\rightarrow Z$ be linear and closed. Show that the composition $ST$ is ...
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86 views
### Question about proof of completeness of $L^p$
In my notes we prove completeness of $L^p$ by showing that if $\sum\|f_k\|_p < \infty$ then $\sum f_k$ converges in $L^p$. (That's a lemma we prove a bit earlier, namely that $(V, \|\cdot\|)$ is ...
1answer
346 views
### Inclusion of $L^p$ spaces
Let $X \subset L^1(\mathbb{R})$ a closed linear subspace satisfying \begin{align} X\subset \bigcup_{p>1} L^p(\mathbb{R})\end{align} Show that $X\subset L^{p_0}(\mathbb{R})$ for some ...
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73 views
### What is the dual of the space L-infinity ($L^\infty$)? [duplicate]
Possible Duplicate: The Duals of $l^\infty$ and $L^{\infty}$ In learning real analysis, I do understand that the dual of $L^\infty$ cannot be $L^1$ because the latter is separable, whereas ...
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136 views
### What is the precise definition of predual
How does one define "predual" and the surrounding notions? More specifically: Why must there be only one predual of $X$ when $X$ is a Banach space? What is the correct notion of similarity here ...
1answer
183 views
### Duality of $L^p$ and $L^q$
If $X$ is an arbitrary measure space, I already know with proof that $L^p(X)$ and $L^q(X)$ are mutually duals as Banach spaces, when $1<p$ and $p$, $q$ are dual indices. I also know a different ...
2answers
663 views
### Prove that the normed space $L^{\infty}$ equipped with $\lVert\cdot\rVert_{\infty}$ is complete. [duplicate]
Possible Duplicate: Understanding proof of completeness of $L^{\infty}$ Most of the materials I have in Real Analysis consider this statement as a trivial one: "The normed space ...
3answers
482 views
### Gâteaux derivative
Let $X$ be a Banach space and $\Omega \subset X$ be open. The functional $f$ has a Gâteaux derivative $g \in X'$ at $u \in \Omega$ if, $\forall h\in X,$ \lim_{t \rightarrow 0}[f(u+th)-f(u)- \langle ...
0answers
228 views
### A problem with $\ell_p$-norm
Let $1<p<\infty$ be fixed. Suppose $L=\{(x_1,\dots,x_n):x_i\ge 0, \sum_i x_i=1, \sum_i a_i x_i=b\}$ for some real numbers $a_i$ and $b$. I am wondering whether the following would be true. ...
3answers
100 views
### There exists a unique function $u\in C^0[-a,a]$ which satisfies this property
The problem: Let $a>0$ and let $g\in C^0([-a,a])$. Prove that there exists a unique function $u\in C^0([-a,a])$ such that $$u(x)=\frac x2u\left(\frac x2\right)+g(x),$$ for all $x\in[-a,a]$. ...
1answer
61 views
### Construct locally lipschitz map from a bounded one
Let $X$ be a Banach space and $BC(X)$ the space of all bounded closed subsets in $X$. It can be shown that $(BC(X),d_H)$ is a complete metric space (see this page for a definition of $d_H$). If ...
1answer
74 views
### Prove that the space is not complete
Let $X$ be a separable space with infinite dimension, let $(\cdot,\cdot)$ and $\|\cdot \|$ be the scalar product and the norm of $X$, and $\{e_n\}_n$ be an orthonormal basis of $X$. We define ...
0answers
45 views
### existance of the interpolation space
Let $X\subset L_1+L_2$ and let $Y$ be interpolation space between $L_2$ and $L_{\infty}.$ Given $U:X\longrightarrow Y$. My question is the following: Is there exists space $Z\subset Y$, such that ...
2answers
279 views
### Totally bounded space
Suppose $M=\left \{ f\in L^1([0,1])\, |\, 0<f(x)<\frac 1{\sqrt x} \text{almost everywhere on} \, (0,1) \right\}$. Is it true or not, that $M$ is totally bounded?
1answer
114 views
### Is this set corresponding to a bounded linear operator necessarily open?
Let $\Lambda : X \to X$ be a bounded linear operator on a Banach space $X$. My question is whether the set $$\{\lambda \in \mathbb C: \lambda I - \Lambda \quad\text{is surjective} \}$$ is ...
2answers
136 views
### How to show that the sum of $L^p$ spaces is Banach.
Let $p<q$ be positive integers (with the allowance that $q$ may be $\infty$). How can we show that the sum of $L^p$ and $L^q$ is a Banach space under the norm \$\|f\|=\inf\{\|g\|_p+\|h\|_q: ...
1answer
509 views
### If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic
If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic Maybe I would have to use the Rademachers.
1answer
193 views
### Swapping a limit and a $\sup$
In this proof of the completeness of $(C(K), \| \cdot \|_\infty)$ they use the following inequality: \sup_{x \in K} \lim_{m \to \infty} | f_n(x) - f_m(x) | \leq \liminf_{m \to \infty}\, \sup_{x ...
0answers
148 views
### An explicit functional in $(l^\infty)^*$ not induced by an element of $l^1$? [duplicate]
Possible Duplicate: Nonnegative linear functionals over $l^\infty$ Setup: Let $l^\infty$ be the set of bounded sequences (with terms in $\mathbb{R}$), and let $l^1$ be the set of sequences ...
4answers
301 views
### Does the completeness of a normed vector space only depend on its topology?
Let $V \space$ be a vector space over $\mathbb{R}$, and $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ norms over $V$, which generate the same topology. Is it always true that if $v_n$ is a Cauchy ...
1answer
183 views
### Cauchy, Bolzano-Weierstrass, Convergence
Why is it that if for every bounded sequence we can find a convergent subsequence (in a normed vector space) then every Cauchy sequence converges (in this normed space)? Thanks.
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http://stats.stackexchange.com/questions/1478/testing-implementation-of-anderson-darling-test-for-uniform-rv/1510
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# Testing implementation of Anderson-Darling test for uniform RV
I am trying to write unit tests for a whole mess of statistics code. Some of the unit tests take the form: generate a sample following a null hypothesis, use code to get a p-value under that null, repeat hundreds of times, then look at all the p-values: if they are reasonably uniform, then the code passes. I usually check if the ratio of p-values < $\alpha$ is near $\alpha$ for $\alpha = 0.01, 0.05, 0.1$. But I am usually also interested in whether the p-values output deviate from uniformity. I usually test this with the Anderson-Darling test.
Here is where I have a circularity problem: how can I unit test my Anderson-Darling code? I can easily feed it uniformly generated variables, and get a p-value, repeat hundreds of times, but then I just have a bunch of p-values. I can q-q plot them, but I'm more interested in an automatic unit test I can run. What are some basic sanity checks I can implement automatically? there is the naive check of ratio of p-values < $\alpha$ noted above. I can also implement a Kolmogorov-Smirnov test. What else can I easily check for?
(I realize this question may seem hopelessly pedantic or naive or subject to infinite regress...)
edit some additional ideas:
1. test the code on $\frac{i}{n}$ for $i = 1,2,...,n$, for different values of $n$. Presumably I can compute, by hand, the p-value for the A-D test in this case.
2. compute $n$ p-values by feeding many uniform samples to the code $n$ times, then regress the order statistics of the p-values, $p_{(i)}$ vs $i/n$, to get $p_{(i)} = \beta_1 i/n + \beta_0$ and test the null $\beta_1 = 1, \beta_0 = 0$. Presumably I can simplify this test by hand in such a way that inspection reveals it to be correct.
3. make sure the code is invariant with respect to permutation of the input. (duh)
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Could you give a couple of examples of what your code does? – csgillespie Aug 10 '10 at 9:42
briefly `pv = AD_test_uni(xs)` takes a sample vector `xs`, which are restricted to the range $[0,1]$ and returns a p-value, `pv` under the null that `xs` are drawn from the Uniform distribution. I can then use this elsewhere to test other tests which spit out a p-value. – shabbychef Aug 10 '10 at 16:28
## 2 Answers
You could test your Anderson-Darling code using data that is generated from an external library. However, you then run into the issue of how to test/trust the external library. At some point you have to trust that well established libraries are error free and that their output can be relied on.
Once you have the Anderson-Dalring code tested against data generated from an external library the circularity will be broken and you can rely on your own code if it passes the Anderson-Darling tests. The same will hold for the K-S (I presume Kolmogorov–Smirnov) test.
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There is indeed an infinite regress problem here. However, it is certainly not the case that all external libraries are error-free (I have found such errors in the past). Because all of my tests ultimately rest on this A-D test, I thought it might be worthy of extra scrutiny. – shabbychef Aug 10 '10 at 16:44
Perhaps, you can use multiple external libraries whose code base does depend on each other. It is unlikely that independent libraries have the same errors which should eliminate some of your concerns. – user28 Aug 10 '10 at 16:49
I can q-q plot them, but I'm more interested in an automatic unit test I can run.
If a visual inspection of a q-q plot would suffice, then you could calculate an entropy measure such as the Gini coefficient and accept the test after allowing for some tolerance for deviation from the 45 degree line. For more on comparing distributions in general, you could look to the reldist package in R for ideas.
Though not directly applicable, I think you'll also find the method presented by Cook, Gelman & Rubin interesting:
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http://en.wikipedia.org/wiki/MSbar_scheme
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# Minimal subtraction scheme
(Redirected from MSbar scheme)
In quantum field theory, the minimal subtraction scheme, or MS scheme, is a particular renormalization scheme used to absorb the infinities that arise in perturbative calculations beyond leading order, introduced independently by 't Hooft (1973) and Weinberg (1973). The MS scheme consists of absorbing only the divergent part of the radiative corrections into the counterterms.
In the similar and more widely used modified minimal subtraction, or MS-bar scheme ($\overline{MS}$), one absorbs the divergent part plus a universal constant (which always arises along with the divergence in Feynman diagram calculations) into the counterterms.
## References
• 't Hooft, G. (1973). "Dimensional regularization and the renormalization group". 61: 455. Bibcode:1973NuPhB..61..455T. doi:10.1016/0550-3213(73)90376-3.
• Collins, J.C. (1984). Renormalization. Cambridge Monographs on Mathematical Physics. Cambridge University Press. ISBN 978-0-521-24261-5. MR778558.
• Weinberg, S. (1973). "New Approach to the Renormalization Group". 8 (10): 3497–3509. Bibcode:1973PhRvD...8.3497W. doi:10.1103/PhysRevD.8.3497.
This particle physics-related article is a stub. You can help Wikipedia by expanding it.
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http://mathoverflow.net/questions/41795?sort=oldest
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## Self homeomorphisms of $S^2\times S^2$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Every matrix $A\in SL_2(\mathbb{Z})$ induces a self homeomorphism of $S^1\times S^1=\mathbb{R}^2/\mathbb{Z}^2$. For different matrices these homeomorphisms are not homotopic, as the induced map on $\pi_1(S_1\times S_1)$ is given by $A$ (w.r.t the induced basis).
So I am wondering, whether a similar construction also works for other spheres than $S^1$. So is there any non-obvious self-homeomorphism of $S^2\times S^2$ (I know i can use degree $-1$ maps in any choice of coordinates / flip the coordinates /compose such maps).
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There are some more coming from the fact that the Grassmannian of oriented 2-planes in 4-dimensions is diffeomorphic to $S^2$x$S^2$. Thus $S^2$x$S^2$ is also $SO(4)/(SO(2)\times SO(2))$. – jc Oct 11 2010 at 15:38
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The good question might be: "what can be said about the homotopy type of Diff($S^2\times S^2$)?" – André Henriques Oct 11 2010 at 15:56
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@Andre. That's a fine question, but sadly we don't have a single computation of $\pi_i (Diff(X))$ for a closed 4-manifold $X$ and an integer $i \geq 0$. – Tim Perutz Oct 11 2010 at 17:31
Which finite groups can act freely on a product of two 2-spheres? Are there any besides the obvious Klein 4-group? – Robert Bell Oct 11 2010 at 18:47
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rbell - No. Any such action would map to $\text{Aut}(H^*(S^2 \times S^2))$, which is the 4-group. The kernel would act trivially on homology and would therefore have fixed points by the Lefschetz Fixed-Point Theorem. – Greg Kuperberg Oct 11 2010 at 18:58
show 2 more comments
## 2 Answers
The matrix $\text{SL}(2,\mathbb{Z})$ acts on $H^n(S^n \times S^n)$; one interpretation of your question is whether this action lifts to $\text{Diff}(S^n \times S^n)$. There is a simple reason that it doesn't when $n$ is even: The intersection form on $H^n(S^n \times S^n)$ is symmetric rather than anti-symmetric, and any diffeomorphism has to preserve this form. This more or less nails down everything, in the weaker category of homotopy self-equivalences; in this setting you can't do anything other than exchange the spheres or apply degree $-1$ maps. (But the full homotopy structure of the $\text{Diff}(S^n \times S^n)$ could be much more complicated.)
When $n$ is odd things are much more complicated. If $n=3$ or $n=7$, you can use multiplication in the unit quaternions or the unit octonions to lift the matrix $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ and its transpose. These matrices generate $\text{SL}(2,\mathbb{Z})$. On the other hand, for any other value of $n$, there is no diffeomorphism, nor even any homotopy equivalence, that realizes this matrix. Because, if you composed such a map with projection onto one of the factors, it would turn $S^n$ into an H-space.
I don't know of a way to prove more than that just by citation for some other large, odd value of $n$. In other words, I know that you can't get all of $\text{SL}(2,\mathbb{Z})$, but I don't know how big of a group you can get.
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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 small comment extending Greg's comments, given an element of $[g] \in \pi_n Diff(S^n)$ you can construct a diffeomorphism
$f : S^n \times S^n \to S^n \times S^n$ by sending the pair $(x,y)$ to $(x,g_x(y))$, so the corresponding matrices would be upper-triangular as in Greg's examples.
$Diff(S^n)$ has the homotopy-type of $O_{n+1} \times Diff(D^n)$.
Greg's examples come from situations where there are non-trivial elements of $\pi_n O_{n+1}$ that you "know" because they're coming from splittings of the fibrations $SO_n \to SO_{n+1} \to S^n$, but you could also get non-trivial elements from finding elements in $\pi_n Diff(D^n)$. My understanding is there's little to nothing known about these homotopy groups. The rational homotopy $\pi_j Diff(D^n)\otimes \mathbb Q$ of $Diff(D^n)$ is only understood in the "Igusa Stable Range" of $j \leq \min\{ \frac{n-4}{3}, \frac{n-7}{2} \}$ so it's not useful for the kind of constructions you'd like.
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http://math.stackexchange.com/questions/199967/how-can-i-compute-a-posterior-distribution-using-bayes
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# how can I compute a posterior distribution using Bayes?
This may be a silly question, but I cannot figure out a convincing (to myself) answer to it. Suppose that you want to buy a new car. Let $v$ be the value you attach to the car. Before visiting the dealer, you cannot tell for sure how much you value the car, i.e., you're uncertain about the true valuation of the car. However, you receive a observe the realization of a random variable $S\sim U[0,1]$, that tells you something about $v$: with probability $p$ the signal tells you the truth ($s=v$) and with probability $(1-p)$ the signal is noise, $s=\epsilon$, where $\epsilon$ is independent of $v$, and $\epsilon \sim U[0,1]$. This means that the expected value of $v$ given $s$ and $p$ is $\omega=ps+(1-p)\frac{1}{2}$.
Now, so long as $p>0$, the posterior of $\omega$ given $p$ and $s$ is: $$\mathbb{P}[\omega \leq x]=\mathbb{P}\left[s \leq \frac{x - (1-p)1/2}{p} \right]=\frac{x - (1-p)1/2}{p}$$ because $s\sim U[0,1]$. My question is, how can I obtain the posterior when $p=0$? Any help to understand this is really appreciated it!
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What do you mean by "because $s\sim U[0,1]$"? Is this new information being introduced, or something you're trying to infer from the information given in the first paragraph? How can you specify a distribution for $s$ when in one case $s$ is equal to $v$, which presumably is not random? Also, why is $\omega=ps+(1-p)\frac{1}{2}$ the expected value of $v$? Shouldn't it be that $\omega=pv+(1-p)\frac{1}{2}$ is the expected value of $s$? – joriki Sep 20 '12 at 22:23
@joriki Sorry I forgot to mention that $s$ is uniform $[0,1]$. I added an edit into my the question. $v$ is random because you don't know it. What you do know is the realization of a random variable $S$ (the signal) which tells you the truth with probability $p$. Hence, when you compute the expected value of your valuation, you do it conditional on what you have observed, i.e., your sigal $s$ and the probability $p$. – Cristian Sep 20 '12 at 22:29
I find this presentation of the problem somewhat confusing. Let me check whether I understand it correctly by rephrasing it. The value $v$ of the car is random, and it's uniformly distributed between $0$ and $1$. There is also another value $\epsilon$, which is also uniformly distributed between $0$ and $1$. The signal $s$ has the value $v$ with probability $p$ and the value $\epsilon$ with probability $1-p$. You know $p$, and you observe $s$, and $\omega$ is the expected value of $v$ given that information. Is that equivalent to what you're saying? – joriki Sep 20 '12 at 22:55
@joriki yup. What you say is equivalent to the problem I presented when you add independence between $\epsilon$ and $s$. – Cristian Sep 21 '12 at 1:03
You mean between $\epsilon$ and $v$? – joriki Sep 21 '12 at 10:58
## 1 Answer
Your result for $p\gt0$ is only correct if $\displaystyle\frac{x - (1-p)/2}{p}\in[0,1]$, which need not be the case.
If $p=0$, then $\omega=\frac12$, so $\mathbb{P}[\omega \leq x]$ is $0$ for $x\lt\frac12$ and $1$ for $x\ge\frac12$.
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http://mathoverflow.net/questions/121097/spectrum-of-a-laplacianized-matrix/121101
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## Spectrum of a Laplacianized matrix
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose that $A$ is a positive matrix and that we let $R$ be the diagonal matrix of $A$'s row-sums. What can be said about the spectrum of $R-A$? I am particularly interested in the largest eigenvalue of $R-A$.
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## 1 Answer
It is just a point of view. But it is more long for writing it as a comment.
If $A$ be the adjacency matrix of graph $G$, then $R-A$ is its Laplacian matrix and there are some good bounds for the radius (or maximum eigenvalues of this matrix) of matrix. For example, if we denote the maximum eigenvalue of this matrix by $\mu$, we have: $$\mu\leq \sqrt(2)\times max(d(v)^2+\sum_{uv\in E(G)}{d(u)})^\frac{1}{2}$$ where $v$ changes in the vertices of the graph $G$. In most cases, we can extend such relations to the positive definite matrices. For more such bounds you can see the paper:
Bounds on the (Laplacian) spectral radius of graphs, written by $Lingsheng$ $Shi$ and published in Linear Algebra and Its Application.
But in general, as you want, I think there is not good bound.
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Shahrooz, I was not aware of the bound you mention. Do you know similar bounds for the normalized Laplacian of a graph? – Delio Mugnolo Feb 8 at 8:43
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In general no, but these references maybe good for further study: 1. "THE NORMALIZED LAPLACIAN MATRIX AND GENERAL RANDIC INDEX OF GRAPHS", a thesis by "Michael Scott Cavers" 2. "Generalizing some results to the normalized Laplacian" by "Steve Butler" – Shahrooz Feb 8 at 10:24
thanks a lot, shahrooz. – Delio Mugnolo Feb 8 at 11:13
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http://mathhelpforum.com/number-theory/17157-greatest-product-sum-print.html
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# Greatest product from a sum
Printable View
• July 23rd 2007, 11:33 PM
DivideBy0
Greatest product from a sum
The sum of n positive integers is 19. What is the maximum possible product of these n numbers?
Can you also please explain why this is so?
• July 24th 2007, 11:30 PM
CaptainBlack
Quote:
Originally Posted by DivideBy0
The sum of n positive integers is 19. What is the maximum possible product of these n numbers?
Can you also please explain why this is so?
Well as nobody else has answered this I will tell you what I know though its not a complete solution.
The arithmetic-geomentric mean inequality tells us that if there are $N$ numbers involved, and that these are $n_i,\ i=1, .. N$, then:
$<br /> \left[ \frac{19}{N} \right]^N \ge \prod_1^N n_i<br />$
Regarding the lefthand side of this inequality as a function of N this has a maximum $\approx 1085.4$ when $N=7$. So we have that the maximum posible value of the product of an integer partition of $19$ is less than or equal $1085$.
The partition $3,3,3,3,3,2,2$ has product $972$ and I would not be supprised if this were the maximum, but have not done enough to convince even myself that this is the case.
RonL
• July 25th 2007, 02:06 AM
CaptainBlack
Quote:
Originally Posted by CaptainBlack
Well as nobody else has answered this I will tell you what I know though its not a complete solution.
The arithmetic-geomentric mean inequality tells us that if there are $N$ numbers involved, and that these are $n_i,\ i=1, .. N$, then:
$<br /> \left[ \frac{19}{N} \right]^N \ge \prod_1^N n_i<br />$
Regarding the lefthand side of this inequality as a function of N this has a maximum $\approx 1085.4$ when $N=7$. So we have that the maximum posible value of the product of an integer partition of $19$ is less than or equal $1085$.
The partition $3,3,3,3,3,2,2$ has product $972$ and I would not be supprised if this were the maximum, but have not done enough to convince even myself that this is the case.
RonL
Now as
$<br /> \left[ \frac{19}{9} \right]^9 \approx 832.9<br />$
and
$<br /> \left[ \frac{19}{5} \right]^5 \approx 792.4<br />$
The above example of a 7-partition with a product of $972$ shows that the partition that gives the maximum product must be a 6, 7 or 8-partition.
(Exhaustive search shows that $972$ is indeed the maximum product and it can be achived with either a 6 or a 7-partition.
RonL
• July 25th 2007, 03:25 AM
CaptainBlack
Quote:
Originally Posted by CaptainBlack
The partition $3,3,3,3,3,2,2$ has product $972$ and I would not be supprised if this were the maximum, but have not done enough to convince even myself that this is the case.
The reason for guessing that this partition is a good candidate for the maximum product is that the arithmetic-geometric mean inequality is an equality when all the numbers are equal. We cant achive that in this case as 19 is prime, but we can look at partitions where the elements are as near equal as possible, and that is what I did here.
The 6-partition of 19 which achives the maximum product can also be found im this way.
RonL
• July 27th 2007, 10:38 AM
ray_sitf
What about [19/(19/e)] to the power of (19/e)?
This gives 1085.405992...
I know the OP wanted integers, but this suggests the following procedure.
If the number leaves a remainder of 2 when divided by three, then the max. answer is 3*3*3*...*2.
If the number leaves a remainder of 1 when divided by three, then the max. answer is 3*3*3*...*2*2.
If there is no remainder on division by three, then it's 3*3*3*...*3.
E.g. for 23 it's 3^7 * 2, and for 25 it's 3^7 * 2 * 2
• July 27th 2007, 10:52 AM
CaptainBlack
Quote:
Originally Posted by ray_sitf
What about [19/(19/e)] to the power of (19/e)?
This gives 1085.405992...
I know the OP wanted integers, but this suggests the following procedure.
I know what you are talking about, but I doubt many others will
RonL
• July 27th 2007, 11:09 AM
CaptainBlack
Quote:
Originally Posted by ray_sitf
I know the OP wanted integers, but this suggests the following procedure.
If the number leaves a remainder of 2 when divided by three, then the max. answer is 3*3*3*...*2.
If the number leaves a remainder of 1 when divided by three, then the max. answer is 3*3*3*...*2*2.
If there is no remainder on division by three, then it's 3*3*3*...*3.
E.g. for 23 it's 3^7 * 2, and for 25 it's 3^7 * 2 * 2
Very likely, but you will need to prove it.
RonL
• July 27th 2007, 11:59 PM
DivideBy0
You're correct Captainblack, the answer is 972.
I had a hard time understanding the inequality, but I guess as long as I take it for granted I should be fine.
Also, is it an actual fact that 3 is the best integer and $e$ is the best number to use for something like this?
• July 28th 2007, 06:14 AM
CaptainBlack
Quote:
Originally Posted by DivideBy0
You're correct Captainblack, the answer is 972.
I had a hard time understanding the inequality, but I guess as long as I take it for granted I should be fine.
Also, is it an actual fact that 3 is the best integer and $e$ is the best number to use for something like this?
The logic says that the nearest integer to N/e is a good choice for the
number of terms in the sum and product, and that lots of 3's and a few 2's
will get you close to the maximum for the product, but its not definitive.
For instance you can get as large a product with 6 terms and with 7 when
the sum is 19.
RonL
All times are GMT -8. The time now is 11:05 AM.
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http://mathoverflow.net/revisions/6841/list
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## Return to Answer
Post Undeleted by D. Savitt
2 deleted 255 characters in body; added 8 characters in body
Here's a strategy that I've been toying with. It seems unlikely to me that it works, but perhaps it'll inspire someone to have a better idea. Fix an embedding of $\overline{\mathbb{Q}}$ into $\mathbb{C}$, thereby fixing Try to arrange a complex conjugation $c$. Let $L$ be the fixed field of $\overline{\mathbb{Q}}$ under a Sylow $3$-subgroup of the absolute Galois group of $\mathbb{Q}$, and let $K$ be the fixed field of $L$ under $c$. Then the whose Galois group of $K$ is an extension of has trivial pro-$5$ Sylow subgroup but nontrivial pro-$p$ Sylow subgroup for $\mathbb{Z}/2\mathbb{Z}$ by a big pro-$3$ group. p=2,3$. Let$f$be the product of an irreducible quadratic and an irreducible cubic over$K$. Since there are no irreducible quintics over$K$, for each$a \in K$, the polynomial$f-a\$ either has a root, or factors as an irreducible quadratic times an irreducible cubic. We'd be done if we can arrange for the latter to occur only finitely many times. But this seems like a stretch....
Post Deleted by D. Savitt
Post Undeleted by D. Savitt
Post Deleted by D. Savitt
1
Here's a strategy that I've been toying with. It seems unlikely to me that it works, but perhaps it'll inspire someone to have a better idea. Fix an embedding of $\overline{\mathbb{Q}}$ into $\mathbb{C}$, thereby fixing a complex conjugation $c$. Let $L$ be the fixed field of $\overline{\mathbb{Q}}$ under a Sylow $3$-subgroup of the absolute Galois group of $\mathbb{Q}$, and let $K$ be the fixed field of $L$ under $c$. Then the Galois group of $K$ is an extension of $\mathbb{Z}/2\mathbb{Z}$ by a big pro-$3$ group. Let $f$ be the product of an irreducible quadratic and an irreducible cubic over $K$. Since there are no irreducible quintics over $K$, for each $a \in K$, the polynomial $f-a$ either has a root, or factors as an irreducible quadratic times an irreducible cubic. We'd be done if we can arrange for the latter to occur only finitely many times. But this seems like a stretch....
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http://math.stackexchange.com/questions/287058/how-to-choose-x-evenly-distributed-points-from-within-an-n-ball
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How to choose $x$ evenly distributed points from within an n-ball
I would like to know how to choose $x$ evenly distributed points from within an n-ball. I think a formal way of defining this is that we want to choose $x$ points from within the n-ball such that we maximize the closest distance between any two points. As a result, it seems, all points should be evenly spaced and many should be located at the surface. What is an algorithm to generate such a set?
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Here is an algorithm for studying the 2D sphere to get a feel for potential theory techniques, and here is a stackoverflow thread discussing the general problem. – Eugene Shvarts Jan 26 at 1:42
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This question is almost an exact duplicate, except it's for points on the sphere instead of the ball. The answers there may still be helpful. – Rahul Narain Jan 26 at 1:46
Ok, so far what I'm getting from those questions is that only approximate solutions exist for this type of problem. – Matt Munson Jan 26 at 2:03
1 Answer
Suppose the closest distance between two points is $d$. This implies that $x$ spheres of radius $d/2$ centered at your points can fit inside a sphere of radius $1+d/2$ without overlapping. Equivalently, $x$ spheres of radius $d/(2+d)$ fit in a unit sphere, and maximizing $d$ is equivalent to maximizing $d/(2+d)$.
So your problem amounts to finding the densest packing of $x$ spheres in a sphere. There are some (approximate) precomputed solutions for $x\le51$ in three dimensions, but there is probably no closed-form solution in general. I guess this is not an answer to your question of what is an algorithm to generate such sets, but searching for "sphere packing algorithms" may find you some useful references.
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Cool. But what is the values of $x$ for a given value of $d$? And where did you get $1+d/2$? – Matt Munson Jan 26 at 3:18
@Matt: Finding the largest $x$ for a given $d$ is just as hard as finding the largest $d$ for a given $x$; as I said there is no known explicit formula. You get $1+d/2$ as follows: If a point lies within distance $1$ of the origin, then a sphere of radius $d/2$ around it may not be contained in the unit sphere around the origin, but it will be contained in a sphere of radius $1+d/2$ around the origin. – Rahul Narain Jan 26 at 4:45
Ok, I see. And its 1 because we start from the unit circle, but it could actually be any positive number. Nice. – Matt Munson Jan 26 at 9:15
@Matt: Right, I somehow assumed you meant the unit $n$-ball in your question. It's just a matter of scaling in the end. – Rahul Narain Jan 26 at 9:19
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http://math.stackexchange.com/questions/133689/an-algebra-of-nilpotent-linear-transformations-is-triangularizable
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# An algebra of nilpotent linear transformations is triangularizable
How to prove "An algebra of nilpotent linear transformations is triangularizable" using linear algebra only?
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What do you mean by triangularizable? That each element can be represented by a triangular element? – Manos Apr 19 '12 at 1:23
@Manos: Yes, it is. – Sunni Apr 19 '12 at 1:31
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Do you know the result that every linear operator $T$ on a vector space $V$ (real or complex) is nilpotent iff there is a basis $\mathcal{B}$ for $V$ such that $T$ in that basis is upper triangular? – BenjaLim Apr 19 '12 at 1:53
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@BenjaminLim: Dear Benjamin, I believe that Sunni wants to simultaneously triangularize the matrices in the algebra, as in my answer. Regards, – Matt E Apr 19 '12 at 4:04
@MattE Ah ok thanks :D – BenjaLim Apr 19 '12 at 6:46
## 2 Answers
The following argument is a simplified version of an earlier argument posted here:
Let $A$ be the algebra in question, and let $V$ be the finite-dimensional vector space on which $A$ acts. Suppose that $V \neq 0$ (as we may, since otherwise $A = 0$ and there is nothing to prove). I claim that $A V$ is a proper subspace of $V$. Indeed, suppose not, i.e. suppose that $V = AV$, and choose a minimal generating set $v_1,\ldots, v_n$ for $V$ over $A$. Then in particular we may find $a_{i} \in A$ such that $v_1 = \sum_i a_{i} v_i$, and hence so that $$(1-a_{1})v_1 = \sum_{i> 1} a_{i}v_i.$$ Since $a_{1}$ is nilpotent, say $a_{1}^N = 0$, the operator $1 - a_{1}$ is invertible, with inverse $1 + a_{1} + \cdots + a_{1}^{N-1}$. Acting this on both sides of the displayed identity, we find that $v_1 = \sum_{i>1} b_{i} v_i$ for appropriately chosen $b_{i} \in A$, and so in fact our original generating set was not minimal ($v_1$ was superfluous), a contradiction.
Thus $AV$ is a proper subspace of $V$. Similary $A(AV)$ is a proper subspace of $AV$, and continuing, we find a strictly decreasing sequence of $A$-invariant subspaces $$V \supset AV \supset A^2V \supset \ldots \supset A^NV \supset \cdots$$ which must eventually reach $0$ (just for dimension reasons). Choosing a basis for $V$ compatible with this descending sequence, we obtain a triangularization of the action of $A$ on all of $V$. QED
This argument uses nothing more than the notion of dimension and basic algebra, and so in principal is just linear algebra, but of course it is more sophisticated than the term "linear algebra" might suggest. You could probably make it less sophisticated on a line-by-line basis at the expense of adding (many) more lines. My feeling is that the result you are asking about is sufficiently non-trivial as a statement of algebra that any proof is going to be either somewhat sophisticated (using a view-point of abstract algebra, even if it only uses facts from commutative algebra) or else quite painful to write down. But maybe I'm wrong!
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This reminds me a lot of the non-Cayley-Hamilton proof of Nakayama. – Dylan Moreland Apr 19 '12 at 4:43
@Dylan: Dear Dylan, That's what I had in mind when I wrote it. Cheers, – Matt E Apr 19 '12 at 4:43
– BenjaLim Apr 19 '12 at 6:48
@MattE I know how if we have a nilpotent operator $T$ on some (real or complex) vector space, then looking at the ascending chain $0 \subset \ker T \subset \ker T^2, etc$ which must eventually terminate, choosing an appropriate basis from $\ker T$, $\ker T^2$, etc gives us a basis for which $T$ in that basis is upper triangular. However now we are doing this for an entire collection of nilpotent operators, I am not so clear how all this can be done at once. – BenjaLim Apr 19 '12 at 6:52
@Benjamin: Dear Benjanmin, Just to fix ideas, imagine that $V$ were $5$ dimensional, that $AV$ were three dimensional, that $A^2V$ were two-dimensional, and that $A^3V = 0$. Choose a basis $v_1,\ldots,v_5$ for $V$ so that $v_1$ and $v_2$ form a basis for $A^2 V$, and $v_1,\ldots , v_3$ is a basis for $AV$. Then each element of $A$ will have a matrix of the form $$\begin{pmatrix} 0 & 0 & *& *& *\\ 0&0&*&*&*\\ 0 & 0& 0& * & * \\ 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \end{pmatrix}.$$ This should help illustrate the general assertion. Regards, – Matt E Apr 19 '12 at 12:18
For the record, the non-«linear algebra» proof of this is:
Consider the local artinian algebra $B=k1_V\oplus A\subseteq\mathrm{End}(V)$. Its radical is $A$ with $1$-dimensional semisimple quotient $B/A\cong k$, so the unique simple module is of dimension $1$ over the base field. The vector space $V$ is tautologically a $B$-module which has finite length, so it has a composition series and each subquotient, being simple, is of dimension $1$. Any ordered basis of $V$ such that each of its prefixes is a basis of one of the layers of this composition series triangularizes $A$.
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http://mathoverflow.net/users/17907?tab=recent
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# René Pannekoek
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## Registered User
Name René Pannekoek
Member for 1 year
Seen 16 mins ago
Website
Location Leiden, The Netherlands
Age
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| May9 | comment | r-torsion points on elliptic curve on finite fieldNo; let $E$ be $y^2=x^3+x$ over $\mathbf{F}_q$ with $q \equiv 3\pmod{4}$. Then we have $2 \mid q−1$ and $\# E(\mathbf{F}_q)[2] > 1$, but since $−1 \notin \mathbf{F}_q^{\ast 2}$ we have $\# E(\mathbf{F}_q)[2] = 2 < 4$. |
| Apr20 | comment | non-singular cubics are not rationalThe funny thing is that the arithmetic genus of a singular (e.g. nodal) cubic curve is also 1, at least if it is assumed to be geometrically integral. And such curves are actually rational. But in the non-singular case, the arithmetic and geometric genera agree, and the latter is a birational invariant (unlike the former). |
| Apr12 | comment | Counter-example to faithfully flat descentYou can use \rightrightarrows. |
| Apr7 | comment | Tricks to produce examples of hypersurfaces with index greater than $1$Thanks! I realized that you can indeed take any curve and 'pinch' some effective zero-cycle into a rational point. |
| Apr6 | comment | Tricks to produce examples of hypersurfaces with index greater than $1$Right. But since the second part of my answer was incorrect, I have deleted it. The first part read: "A non-trivial Severi-Brauer variety of dimension $2$ over $K$ has index $3$ and is birational to a smooth cubic surface over $K$: just blow up a zero-cycle of degree $6$." |
| Mar28 | awarded | ● Nice Question |
| Mar23 | comment | Rational points on a sphere in $\mathbb{R}^d$That certainly looks interesting; I can't explain the patterns off the top of my head. Can you make an animation where you vary the height bound? I'd be interested to see it if you can. |
| Mar22 | comment | Rational points on a sphere in $\mathbb{R}^d$I don't understand why Peter Michor's comment got two upvotes. If a quadric hypersurface $X \subset \mathbf{A}^{n+1}$ defined over $\mathbf{Q}$ has a point $P \in X(\mathbf{Q})$, then projecting away from $P$ gives a birational map $X \stackrel{\sim}{\dashrightarrow} \mathbf{A}^n$ that is defined over $\mathbf{Q}$. Restricting this birational map gives an isomorphism between open subsets of $X$ and $\mathbf{A}^n$ that is defined over $\mathbf{Q}$. In particular, the rational point son $X$ are dense in the real locus of $X$ iff the same holds for $\mathbf{A}^n$, which is trivially the case. |
| Mar4 | comment | Do isogenies with rational kernels tend to be surjective?Dear Chris. You correctly observed that, since the $(x_0y_0 + y_0 - x_0^2)d^i$ come from rational points $(x_0,y_0)$ on $E_d$, they lie in the $\widehat{\eta}$-Selmer group. This does not imply they are 5-th powers in $\mathbf{Q}_v$ for all good $v$ (if so, the $(x_0y_0 + y_0 - x_0^2)d^i$ would be 5-th powers in $\mathbf{Q}$ by Grunwald-Wang). It does say that choice of $d$ imposes restrictions on the values that $x_0,y_0$ may take. On the other hand, I don't see why these restrictions force $(x_0y_0 + y_0 - x_0^2)d^i$ to be 5-th powers more often than if $x_0,y_0$ were random. Best, René. |
| Mar1 | comment | Do isogenies with rational kernels tend to be surjective?Hmm, I am not sure I agree. Why do you think that any (let alone all) of the $(x_0y_0 + y_0 - x_0^2)d^i$ should be fifth powers almost everywhere locally? I do not see how that follows from anything. Here is a heuristic reason to believe that the pairs $(x_0,y_0)$ one gets are in fact quite random. The curves $E_d$ trace out a pencil in $\mathbf{P}^2$ if you let $d$ vary. In particular you will get all pairs $(x_0,y_0)$ as points on some $E_d$. Moreover if you believe that most elliptic curves have rank $0$ or $1$, then most $(x_0,y_0)$ will lie on an $E_d$ that has rank $1$. |
| Mar1 | comment | Do isogenies with rational kernels tend to be surjective?Moreover, I think the condition is necessary as well as sufficient. This remains true if, in the statement above, one replaces "some point of infinite order" by "a generator of $E_d(\mathbf{Q})$ modulo torsion". |
| Mar1 | comment | Do isogenies with rational kernels tend to be surjective?If you follow the approach via Galois cohomology, you get the following: "Assume that $d \in \mathbf{Q}^{\ast}$ is not a $5$-th power, and that the abelian group $E_d(\mathbf{Q})$ is of rank $1$. If for some point $(x_0,y_0) \in E_d(\mathbf{Q})$ of infinite order, and each integer $0 \leq i \leq 4$, we have that the non-zero rational number $(x_0 y_0 + y_0 - x_0^2) d^i$ is not a $5$-th power in $\mathbf{Q}^{\ast}$, then $\eta$ is surjective on $\mathbf{Q}$-points." Looks like it might be a condition that is generically satisfied, but it's hard to say. |
| Feb28 | comment | Do isogenies with rational kernels tend to be surjective?Perhaps you could do the same experiment for the Legendre family $E_\lambda:y^2=x(x−1)(x−\lambda)$ with the role of $P$ played by the $2$-torsion point $(0,0)$. This gives $2$-isogenies $\eta$ for each $E_\lambda$. Barring mistakes on my part, the surjectivity of $\eta$ on $\mathbf{Q}$-points can be rephrased (a small condition on $\lambda$ aside) in terms of the image of an element in $E_\lambda(\mathbf{Q})$ of infinite order under the coboundary map associated to $\widehat{\eta}$ (dual of $\eta$). It seems much easier to make this explicit for $2$-isogenies than for $5$-isogenies. |
| Feb27 | comment | Do isogenies with rational kernels tend to be surjective? |
| Feb14 | comment | Properties of quotient varietyAs Dmitri points out, it is not so much the singularities of Y as the branch locus of π that you have to worry about. Outside of that, π will be étale and therefore the inverse image of the part of C that lies outside of the branch locus will be non-singular. In the case where π:X→Y is the map from an abelian variety X to its quotient by -1, the non-étale locus is given by π(X[2]), which is a union of $2^{2 \operatorname{dim}(X)}$ points. Finding out what the inverse image of C looks like above any of the points in π(X[2]) should be a perfectly local problem. |
| Feb13 | revised | E an elliptic curve over Z[1/N], how many p such that E(Z/p^2) = (Z/p)^2?deleted 31 characters in body; edited title |
| Feb11 | revised | Do torsors give a long exact sequence of cohomology?added 1040 characters in body |
| Feb7 | comment | How many curves in a family possess a rational point?I guess the answer is trivially yes, because the hypotheses are never fulfilled. If $B(\mathbb{Z})$ contains an element $b$, then $X_b$ must be a proper smooth curve over $\operatorname{Spec}(\mathbb{Z})$ of genus $>1$, and such curves don't exist. |
| Feb4 | comment | Do torsors give a long exact sequence of cohomology?Thank you, Will, this is indeed part of the motivation for my question. I should have included it in the post. |
| Feb4 | comment | Do torsors give a long exact sequence of cohomology?Thank you, this is very interesting! Actually however, I was thinking along the lines of Will's comment: if X,Y are k-group schemes and f is a homomorphism, then we are able to extend the exact sequence: if G is abelian we have a long exact sequence in the right sense of the word, whereas in general with the help of non-abelian cohomology we can get at least 2 more terms. I was hoping that one could do this in more generality. |
| Feb4 | revised | Do torsors give a long exact sequence of cohomology?deleted 74 characters in body |
| Feb4 | revised | Do torsors give a long exact sequence of cohomology?deleted 28 characters in body; deleted 63 characters in body |
| Feb4 | revised | Do torsors give a long exact sequence of cohomology?deleted 5 characters in body |
| Feb4 | revised | Do torsors give a long exact sequence of cohomology?G has to be finite-type, so removed parentheses |
| Feb4 | revised | Do torsors give a long exact sequence of cohomology?some elaboration |
| Feb4 | asked | Do torsors give a long exact sequence of cohomology? |
| Feb3 | comment | Are rational varieties simply connected?I believe that the Godeaux surface is a quotient of your quintic surface, not the quintic itself. |
| Jan27 | revised | Elliptic fibration of K3 surfaceadded 43 characters in body |
| Jan27 | answered | Elliptic fibration of K3 surface |
| Dec14 | comment | How do you compute the primes of bad reduction?Dear François: Thank you very much. I must say though I wonder: in terms of complexity, one might be worse off with the Gröbner basis calculation than if you would upper-bound $S$ using my method and then for each $p \in S$ would check if $X_{\mathbf{F}_p}$ is smooth. I can't prove this, of course, after all you do have to factor an integer $N$ over whose size you don't really seem to have much control |
| Dec13 | revised | How do you compute the primes of bad reduction?added 84 characters in body |
| Dec13 | comment | How do you compute the primes of bad reduction?That makes a lot of sense, thanks! Your second remark agrees with what I thought myself, I just thought it made the notation simpler if I restricted to $k=1$. |
| Dec13 | asked | How do you compute the primes of bad reduction? |
| Dec2 | comment | Solved cubic Thue equation@Richard: You should convince people who read your article that solving that equation is a routine matter - given the bounds established by Baker, say. Indeed, all that Mathematica uses is the existence of these bounds on $x,y$, which are part of the (by now) standard theory. The fact that you learned about this only recently doesn't force you to belabor the point; on the contrary, if you spend too many words on this your readers might get the false impression that there's something unusual going on here (which there ain't). |
| Dec2 | comment | How does “modern” number theory contribute to further understanding of $\mathbb{N}$?@Johnny: even such a fundamental statement as "every elliptic curve over Q has finitely generated Mordell-Weil rank", which doesn't refer to algebraic number theory in the least, needs algebraic number theory to prove it. But you see the same principle in much more basic examples as well: try finding all integer solutions to $x^2+4=y^3$ without using the Gaussian integers. (I'm not saying it can't be done, but it's completely routine once you're using some basic ANT.) |
| Dec1 | comment | Solved cubic Thue equationMay I be so bold as to ask how you know these are the only ones? |
| Nov30 | awarded | ● Enthusiast |
| Nov26 | revised | The boundedness of the rank of twists of a fixed curve.added 39 characters in body |
| Nov26 | revised | The boundedness of the rank of twists of a fixed curve.added 177 characters in body |
| Nov26 | revised | The boundedness of the rank of twists of a fixed curve.added 409 characters in body |
| Nov26 | revised | The boundedness of the rank of twists of a fixed curve.added 82 characters in body; added 11 characters in body |
| Nov26 | answered | The boundedness of the rank of twists of a fixed curve. |
| Nov26 | comment | Does finite+reduced fibers+connected fibers imply isomorphism?Fiber above the cusp is not reduced. |
| Nov25 | comment | Surjectivity of reduction maps of elliptic curves over QSo far, all your counterexamples seem to come from isogenies of degree $3$. Any reason to expect isogenies $\phi$ of degree $2$ not to satisfy $\mathbf{Q}(\phi^{-1}E(\mathbf{Q})) = \mathbf{Q}$? |
| Nov24 | answered | reference for (co)homology theories |
| Nov24 | revised | Surjectivity of reduction maps of elliptic curves over Qadded 6 characters in body |
| Nov23 | comment | Elliptic Curves and Torsion PointsDear Joe: Out of curiosity, can you prove your last statement without appealing to Mazur? I like the result, but at the same time it seems too simple a statement for me to require something so deep as Mazur's theorem... |
| Nov21 | comment | Are ranks of Jacobians over number fields unbounded?@Noam Elkies: A belated remark. You not only want the projections to be non-constant, but you'll want them to be "independent". For instance, simply choosing $C$ to be the diagonal $E \subset E^r$ clearly won't do the job. So I'm guessing the condition must be that the images of the $r$ pull-back maps $H^0(E,\Omega_E) \rightarrow H^0(C,\Omega_C)$ must span an $r$-dimensional subspace. Alternatively, someone suggested to me that one could look at curves $C$ in $\prod_{i=1}^r E_i$, where the $E_i$ are non-isogenous elliptic curves of positive rank, with non-trivial projections to each factor. |
| Nov21 | revised | Surjectivity of reduction maps of elliptic curves over Qadded 340 characters in body; edited title |
| Nov21 | comment | Surjectivity of reduction maps of elliptic curves over QHi Felipe, I don't mind getting all these answers, they've been very helpful :) I've checked out the reference to Gupta-Murty, which does indeed prove what you say. Rank $\ge 6$ is large though, so I'll check for improvements later. (Although if Maarten Derickx's calculation proves correct, I guess some assumption on the size of $E(\mathbf{Q})$ on top of rank $>0$ is necessary.) I don't understand your solution to 3., do you mean to start by picking an isogeny that maps to the reduced curve? If so, I fail to see how you can get anything from that unless the isogeny lifts to $\mathbf{Q}$. |
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http://mathoverflow.net/questions/87303/best-algorithm-software-for-solving-a-planar-transportation-problem
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## Best algorithm/software for solving a planar transportation problem ?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for software (open-source or otherwise) or an implementable algorithm for solving a continuous transportation problem. The input consists of a pointset in a planar rectangle, and we need to relocate these points within the rectangle to decrease the peak density below a given threshold (feasibility can be assumed), while minimizing total displacement.
The distances inside the rectangle are Manhattan/taxicab, although efficient solutions for the Euclidean distance can also be helpful. Total displacement is interpreted in the $L_1$ sense, but efficient solutions for the $L_2$ case can also be helpful. The peak density can be evaluated with respect to a uniform grid (is there another practical way ?)
My students implemented a geometric algorithm (without having any background in transportation) that works great in our application, but we don't know how far the results are from optimal. Just in case, our application also imposes rectangular "exclusion zones", where no points can be placed (more generally, we can assume a "bounding probability distribution").
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Have you tried it on examples where you know what the optimal placement is? Gerhard "Ask Me About System Design" Paseman, 2012.02.04 – Gerhard Paseman Feb 5 2012 at 5:07
Yes, and it is suboptimal. Recall that 1D transportation has a simple/closed-form solution, implemented for a pointset input by sorting. In 2D, we alternate optimal transportation in X,Y directions, position cutlines at median locations, then recursively divide and conquer. Now consider a pointset distributed normally in 2D. The 1D solution can be applied radially, but our algorithm produces something blocky. A more important question for us is how the suboptimality affects our application (which can "transport" millions of points 40-50 times per run)- this is difficult to answer w/o a solver – Igor Markov Feb 6 2012 at 8:15
## 2 Answers
How much have you looked into the theory of optimal transport? It's very popular for image warping/registration.
There's codes available to compute the $l1$-optimal transport distance (also referred to as "Earth mover's distance") here: http://ai.stanford.edu/~rubner/emd/default.htm and here: http://www.cs.huji.ac.il/~ofirpele/FastEMD/code/
The $l2$ optimal mass transport problem is quite difficult but can be solved: http://www.springerlink.com/index/40PGJBKDC9V0UH94.pdf
Once it's possible to compute the cost to get between two distributions of points I guess you'll have to optimize to see which distribution is closest to the one you have. Maybe something like: $$\min_\rho d(\rho_0, \rho)\;subject\;to\;\rho \leq c, \rho \geq 0,\int \rho = 1$$ where $\rho$ is a probability distribution describing the density of points and $c$ is your threshold. Maybe you can do this with a Lagrange multiplier and gradient descent?
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Looking into it very closely now. While we are now leaning toward $l_2$-optimal transport because it tends to preserve the relative ordering of "the grains of sand" being transported. Also, the uniqueness of $l_2$-optimal solutions is useful. But we'll definitely take a look at the software you suggested. – Igor Markov Mar 30 2012 at 21:36
Sure. The book by Villani (who got a fields medal last time around) "Topics in Optimal Transportation" has more information in this field than I can give you. – rcompton Mar 30 2012 at 22:36
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You can use the code packages rcompton mentioned to compute transportation with any kind of cost function (called also ground distance). If you have "exclusion zones" you can just set the cost there to be a very high number there (you can for example, compute a solution that is not optimal and set it as a cost). If you have many edges with this maximum distance, it also has the advantage of this code running faster: http://www.cs.huji.ac.il/~ofirpele/FastEMD/code/
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http://math.stackexchange.com/questions/142142/about-the-order-of-groups
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# About the Order of Groups
Is there any theorem that states the all the finite groups of order n are the same? or some sort of theorem that refers to the order of two finite groups? If anyone can post a reference to this topic will be great... 10x in advance
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This is boldly not true for $n=4$. – Asaf Karagila May 7 '12 at 8:22
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Take any noncommutative group, compare it with the cyclic group of the same order. – anon May 7 '12 at 8:23
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"some sort of theorem that refers to the order of two finite groups"? I'm pretty sure there's more than one such theorem... – Zev Chonoles♦ May 7 '12 at 8:24
Side remark: On Wikipedia for "finite field" I found: "Any two finite fields with the same number of elements are isomorphic.". But that's about fields. – Gerenuk May 7 '12 at 8:39
@Gerenuk: your comment about finite fields is correct. However, there are very few finite fields: there is a field of order $n$ if and only if $n=p^m$ for some prime $p$, $m\in\mathbb{N}$. – user1729 May 7 '12 at 9:29
## 4 Answers
No, this is unimaginably not true. In fact, there is a theorem which says that almost the opposite of what you have just said:
Theorem: Let $n\in\mathbb{N}$. Then, there exists only one group (up to isomorphism) of order $n$ if and only if $n=p_1\cdots p_m$ for distinct primes $p_i$, such that $p_i\not\equiv 1\text{ mod }p_j$ for any $i,j$.
This is a standard set of exercises in most algebra textbooks--ask if you'd like an explicit reference.
(I prove this on my blog, here)
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which is equivalent to $gcd(n,\varphi(n))=1$, where $\varphi$ is the Euler totient. – Nicky Hekster May 7 '12 at 8:32
– lhf May 7 '12 at 11:56
– lhf May 7 '12 at 12:01
Not all groups of order n are the same.
$\mathbb{Z}_6$ and $S_3$ are both of order 6. However $S_3$ has 3 subgroups of order 2, where $\mathbb{Z}_6$ has only one subgroup of order 2.
Therefore they cannot be isomorphic.
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Consider the group $C_2\times C_2$ and $C_4$. Both of order $4$ but the latter has an element of order $4$ while the former does not
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I use $C_n$ to denote the cyclic group of $n$ elements. – Asaf Karagila May 7 '12 at 8:26
You might want to read the paper of Besche, Eick and O'Brien http://www.math.auckland.ac.nz/~obrien/research/2000.pdf which contains a table of the number of groups of order $n<2001$.
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http://physics.stackexchange.com/questions/53381/statistical-mechanic?answertab=active
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# Statistical Mechanic
One can define entropy as $$S=k\log{\omega(E)},$$ where $\omega(E)$ is the numbers of states with energy equal $E$; and the canonical partition function for a set of N particles is defined as$$Z_N=\sum_{\phi}e^{-\beta E[\phi]}=e^{-\beta F(\beta,N)},$$ where the sum run on states $\phi$ and the free energy is defined as $F(\beta,N)=U-TS.$ The mean value of the internal energy and the entropy i learned that would be $$\langle U\rangle=\frac{\partial(\beta F)}{\partial\beta}$$ $$\langle S\rangle=\beta^2\frac{\partial(F)}{\partial\beta}.$$ For definition, the mean value of any physical observable is $$\langle O\rangle=Z_N^{-1}\sum_{\phi}O[\phi]e^{-\beta E[\phi]}.$$ I'm quite sure that there will be a problem of amount $k$, if one verify the definition of $\langle S\rangle$. Am i wrong?
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If you mean missing factors of $k$, there are no missing factors of $k$ in your formulae. They're just fine. – Luboš Motl Feb 8 at 15:32
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What makes you think there is a problem with a $k$? At a glance your formulae look correct. Maybe if you could show us where you are getting hung up in your calculation we could help. – Michael Brown Feb 8 at 15:47
$F=-\frac{\log{Z_N}}{\beta}$, so $$\beta^2\frac{\partial F}{\partial\beta}=-\beta^2\frac{Z_N^{-1}\frac{\partial Z_N}{\partial\beta}\beta-\log{Z_N}}{\beta^2}=-\beta(-\langle U\rangle+F)=\beta TS=\frac{S}{k}$$ – ivax Feb 8 at 20:12
– Qmechanic♦ Feb 8 at 20:55
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http://mathoverflow.net/questions/102900/free-modules-over-integers/102901
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## Free modules over integers
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
After the course of linear algebra I'm more familiar with vector spaces rather than modules so my question may seem to be silly but I think it's quite natural for someone who thinks of modules as 'vector spaces over a ring': which of the following is free module (free means: having a basis, all of them are over ring $\mathbb{Z}$):
a) $\mathbb{Z}^{\infty}$-all sequences of integers,
b) $\mathbb{Z}^{\mathbb{R}}$-all functions $f: \mathbb{R} \to \mathbb{Z}$,
c) the set of all functions $f: \mathbb{R} \to \mathbb{Z}$ with at most countable support?
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It is well known non of them is free. This is not at a research level. Actually, it is enough to check that a) is not free, as it is a direct summand of b) and c). – Fernando Muro Jul 22 at 22:23
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This raises an interesting question about "research level". The proofs are not difficult and are many decades old, yet I know from personal experience that there are very capable research mathematicians who not only didn't know that (a) isn't free but, when I mentioned the result, were sure I had made a mistake. – Andreas Blass Jul 22 at 22:27
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For me, a research question is a question arising in your research that you, an experienced mathematician (at least to some extent), have tried to solve on your own, looking up in the literature, etc. This question can be solved in one google search. – Fernando Muro Jul 22 at 23:17
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A module over the integers is precisely an abelian group. So, for example, the volumes by László Fuchs will tell you this story and of course much more. @Fernando Muro...what did you google?...one try for me landed in the world of photoshop freeware LOL. – David Feldman Jul 23 at 0:59
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This question and the accompanying answers deal with a very similar topic: mathoverflow.net/questions/10239 – Emerton Jul 23 at 5:40
show 1 more comment
## 1 Answer
None of these are free. For (a), there is a theorem of Specker ("Additive Gruppen von Folgen ganzer Zahlen" Portugaliae Math. 1950) that covers this group and lots of its subgroups (the so-called monotone subgroups). I believe that for this particular group, the result may already be in a 1937 paper of Baer. Since the groups in (b) and (c) have subgroups isomorphic to the group in (a), and since subgroups of free abelian groups are free, it follows that (b) and (c) aren't free either.
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Here reh.math.uni-duesseldorf.de/~schroeer/… is a short self-contained proof that an infinite direct product of the integers is non-free. – SJR Jul 23 at 7:31
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http://physics.stackexchange.com/questions/4170/is-the-z-boson-one-entity-or-are-there-as-many-entities-as-decay-pairs-but-they
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is the Z boson one entity or are there as many entities as decay pairs, but they are equivalent and lumped together
just wondering if it is a distinction without a difference - it seeming a bit weird that one thing can decay into different things.
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Ok, I am givin a bounty to someone able to answer setup and answer this question in reference to bootstrap theory (see en.wikipedia.org/wiki/Bootstrap_model en.wikipedia.org/wiki/S-matrix_theory). – arivero Feb 18 '11 at 16:53
I mean, I plan to offer a bounty :-D... It seems I can not have two open bounties. – arivero Feb 18 '11 at 16:55
4 Answers
No more weird than that a single particle can take different paths through different slits with different amplitudes. In this case it's mutually exclusive different decay channels. And yes, on occasion there can be interference between them. An excited atom can also decay through different channels to different final states in the same way.
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I can offer a different way to think about it. Many different particles are electric charged, right? Do you think this is strange?
Well, the same happens with the weak nuclear force. Many different particles are charged under this force and, therefore, interact via a Z boson.
The only thing is that the Z, different from the photon, has a non-zero mass. So you start thinking about it more particle-like than the photon, because when it is mediating an interaction it can come as close to the mass shell as one want. But that's not always a very good way to interpret it.
Anyhow... the idea is that many things are charged under the weak force. And there is nothing odd about it.
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There is only one Z boson. If there were more, it would change the decay rate for Weak interaction. If there were a different Z boson for electrons and muons, for example, then you wouldn't have a Z boson linking electron lines to muon lines, and this is possible.
The number of different types of particles are meaningful in quantum field theory, because the total amplitude is the sum of the different ways, and if there are different particles which contribute, it gets added up. There is a qualitative difference between "same particle" and "different particle with same properties" in QM.
This has nothing to do with Bootstrap or field theory. There is only one Z.
-
awarded the +100 for contributing after the revival of the question. I am not sure of which is the right interpretation of OP expression "lumped together". – arivero Nov 4 '11 at 11:49
A closer similar weirdness is that a particle can have different charges, eg colour and electric, and then interact via the exchange of different particles. Worse, a charge for a non abelian force (such as colour, or also electroweak theory itself) immediately has different particles for the same charge.
So there is not problem here. It is perfectly possible for an elementary Z to decay into different pairs, and in fact it does with the probabilities predicted in the standard model.
-
Still, there is a puzzling -to me, not mainstream of course- numerical coincidence, and it is that the reduced total decay width, ie the quantity $Γ/M^3$, is the same for the Z0 than for $\pi^0$ and a lot of total electomagnetic decay widths of neutrals. – arivero Jan 30 '11 at 0:56
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http://mathforum.org/mathimages/index.php?title=The_Golden_Ratio&diff=35107&oldid=33491
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# The Golden Ratio
### From Math Images
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| | |ImageName=The Golden Ratio | | |ImageName=The Golden Ratio |
| - | |Image=Goldenrectangleappwarp.jpg | + | |Image=Goldenratio.jpg |
| - | |ImageIntro=The '''golden number,''' often denoted by lowercase Greek letter "phi", is <br /><math>{\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...</math>. <br /> | + | |ImageIntro=The '''golden number,''' often denoted by lowercase Greek letter "phi", is <br /> |
| - | The term '''golden ratio''' refers to any ratio which has the value phi. The image to the right is a warped representation of dividing and subdividing a rectangle into the golden ratio. The result is [[Field:Fractals|fractal-like]]. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number. | + | :<math> \frac{a+b}{a} = \frac{a}{b} \equiv \varphi,</math> |
| - | |ImageDescElem=The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. <br /> | + | where the Greek letter [[Phi (letter)|phi]] (<math>\varphi</math>) represents the golden ratio. Its value is: |
| - | Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids. <br /> | + | |
| - | However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle. | + | |
| | | | |
| | | + | <math>{\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...</math>. <br /> |
| | | + | The term '''golden ratio''' refers to any ratio which has the value phi. The image to the right illustrates dividing and subdividing a rectangle into the golden ratio. This page explores how the Golden Ratio can be observed and found in the arts, mathematics, and nature. |
| | | + | |ImageDescElem===The Golden Ratio as an Irrational Number== |
| | | + | ==The Golden Ratio in the Arts== |
| | | | |
| - | ===Misconceptions about the Golden Ratio=== | + | ===Parthenon=== |
| - | Many rumors and misconceptions surround the golden ratio. There have been many claims that the golden ratio appears in art and architecture. In reality, many of these claims involve warped images and large margins of error. One claim is that the Great Pyramids exhibit the golden ratio in their construction. This belief is illustrated to the right. | + | :[[Image:ParthenonDIAG.gif]] |
| - | [[Image:Finalpyramid1.jpg|Markowsky has determined the above dimensions to be incorrect.|thumb|400px|left]] | + | There is an abundance of artists who consciously used the Golden Ratio from as long ago as 400 BCE. Once such example is the Greek sculptor Phidias, who built the Parthenon. The exterior dimensions of the Parthenon form the golden rectangle, and the golden rectangle can also be found in the space between the columns. |
| - | [[Image:Monalisa01.jpg|Does the Mona Lisa exhibit the golden ratio?|thumb|400px|right]] | + | |
| | | | |
| - | In his paper, ''Misconceptions about the Golden Ratio,'' George Markowsky disputes this claim, arguing that the dimensions assumed in the picture are not anywhere close to being correct. Another belief is that a series of [[The Golden Ratio#Jump2|golden rectangles]] appears in the ''Mona Lisa''. | + | ===Vitruvian Man=== |
| - | However, the placing of the golden rectangles seems arbitrary. Markowsky also disputes the belief that the human body exhibits the golden ratio. To read more, [http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf click here!] | + | :[[Image:Vitruvian man mixed.jpeg|350px]] |
| | | + | Another instance in which the Golden ratio appears is in Leonardo Da Vinci’s drawing of the Vitruvian Man. Da Vinci’s picture of man’s body fits the approximation of the golden ratio very closely. This picture is considered to a depiction of a perfectly proportioned human body. The Golden Ratio in this picture is the distance from the naval to the top of his head, divided by the distance from the soles of the feet to the top of the head. |
| | | + | ===Music=== |
| | | + | Some people have argued that the Golden Ratio in music produces aesthetically pleasing sounds. Major sixth and the minor sixth chords are considered to be the most pleasing intervals, and subsequently, they are the intervals related to the Golden Ratio. The standard tuning tone used is an A and it vibrates at 440 vibrations per second. A major sixth interval below that would be a C, which has a frequency of 264 vibrations per second. The ratio of the two frequencies reduces to 5/3, which is the ratio of two Fibonacci numbers. Similarly, a minor sixth can be obtained from a high C, which has a frequency of 528 vibrations per second, and an E, which is 330 vibrations per second. This ratio reduces to 8/5, which is also a ratio of two Fibonacci numbers. |
| | | + | :[[Image:Chromaticscales.gif|350px]] |
| | | | |
| - | ====''What do you think?''==== | + | The Golden Ratio manifests in the idea that music is proportionally balanced. However, it is an area of debate, because some scholars claim that these appearances are purely coincidental, or a result of number juggling from aficionados. However, many contemporary artists have intentionally used the Golden Ratio in their pieces. Composer, mathematician and teacher Joseph Schillinger believed that music could be based entirely on mathematical formulation. He developed a System of Musical Composition in which successive notes in a melody were followed by Fibonacci intervals when counted in half steps. Half steps are the smallest intervals possible, and it is the closest note that can be played higher or lower. |
| - | George Markowsky argues that, like the ''Mona Lisa,'' the Parthenon does not exhibit a series of golden rectangles (discussed below). Do you think the Parthenon was designed with the golden ratio in mind or is the image below simply a stretch of the imagination? | + | For example, if C were the first note in a composition, the following note would be half a step higher, because one is a Fibonacci number. So, the second note would be a D flat. Schillinger would alternate between moving up and down in intervals. Thus, the third note would be two half steps (because two is the next Fibonacci number), down from the D flat, which would be B flat. The next Fibonacci number after two is three, so the next note would be three half steps higher than the last. Three half steps higher than B flat is E flat. Schillinger believed that these notes convey the same sense of harmony as the phyllotactic ratios found in leaves. |
| - | :[[Image:Golden ratio parthenon.jpg|300px]] | + | |ImageDesc===Fibonacci Numbers== |
| - | <ref>[http://lotsasplainin.blogspot.com/2008/01/wednesday-math-vol-8-phi-golden-ratio.html "Parthenon"], Retrieved on 16 May 2012.</ref> | + | One of the greatest breakthroughs regarding the Golden Ratio came when its relation to Fibonacci numbers, also known as the Fibonacci sequence, was discovered. Fibonacci numbers can also be found in the arts, and nature. The Fibonacci sequence is the series of numbers, |
| | | + | 0,1,1,2,3,5,8,13,21,34,55,89,144,233… |
| | | + | The next term in the Fibonacci sequence, starting from the third, is determined by adding the previous two terms together. The Fibonacci sequence is related to the Golden Ratio because as the sequence grows, the ratio of consecutive terms gradually approaches the Golden Ratio. For example here are the ratios of the successive numbers in the Fibonacci sequence: <br/> |
| | | + | 1/1=1.000000 <br/> |
| | | + | 2/1=2.000000 <br/> |
| | | + | 3/2=1.500000 <br/> |
| | | + | 5/3=1.666666 <br/> |
| | | + | 8/5=1.600000 <br/> |
| | | + | 13/8=1.625000 <br/> |
| | | + | 21/13=1.615385 <br/> |
| | | + | 34/21=1.619048 <br/> |
| | | + | 55/34=1.617647 <br/> |
| | | + | 89/55=1.618182 <br/> |
| | | + | 144/89=1.717978 <br/> |
| | | + | 233/144=1.618056 <br/> |
| | | + | 377/233=1.618026 <br/> |
| | | + | 610/377=1.618037 <br/> |
| | | + | 987/610=1.618033 <br/> |
| | | | |
| - | ==A Geometric Representation== | | |
| | | | |
| - | ===The Golden Ratio in a Line Segment=== | + | ==The Golden Ratio in Nature== |
| - | [[Image:Golden_segment.jpg|400px]][[Image:Animation2.gif]] | + | |
| | | | |
| - | | + | ===Spirals & Phyllotaxis=== |
| - | The golden number can be defined using a line segment divided into two sections of lengths a and b. If a and b are appropriately chosen, the ratio of a to b is the same as the ratio of a + b to a and both ratios are equal to φ. The line segment above (left) exhibits the golden proportion. The line segments above (right) are also examples of the golden ratio. In each case, | + | [[Image:Sunflower head.jpeg|px200]] |
| | | | |
| - | <math>\frac{{\color{Red}\mathrm{red}}+\color{Blue}\mathrm{blue}}{{\color{Blue}\mathrm{blue}} }= \frac{{\color{Blue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} }= \varphi . </math> | + | Spirals are abundant concerning phyllotaxis, which describes the way leaves are arranged on a plant stem. In 92 percent of Norway spruce cones, the spirals were found to appear in rows five and eight rows. They appeared in rows of four and seven in six percent, and four and six in four percent. In addition, the number of right-handed spirals appears to be equal to the number of left-handed spirals. The arrangements of the spirals in these spruce cones are found as the following pairs of rows: 2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89, and 89/144. These numbers are all numbers that belong to the Fibonacci Series. |
| | | + | [[Image:Pinecone3.gif]] |
| | | | |
| - | <div id="Jump2"></div> | + | A similar phenomenon can be observed in sunflower heads. A sunflower head has both clockwise and counterclockwise spirals. The numbers of the spirals in a sunflower usually depend on the size of the sunflower. However, the ratios of the spirals that usually occur are 89/55,144/89, and even 233/144. Once again, all of the numbers in these ratios are consecutive Fibonacci numbers. |
| - | ===The Golden Rectangle=== | + | |
| - | A '''golden rectangle''' is any rectangle whose sides are proportioned in the golden ratio. When the sides lengths are proportioned in the golden ratio the rectangle is said to possess the '''golden proportions.''' A golden rectangle has sides of length <math>\varphi \times r</math> and <math>1 \times r</math> where <math>r</math> can be any constant. Remarkably, when a square is cut off of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle below. | + | |
| - | :[[Image:Coloredfinalrectangle1.jpg]] | + | |
| | | | |
| - | | + | |AuthorName=Joyce Han |
| - | | + | |
| - | ===Triangles=== | + | |
| - | [[Image:1byrrectangle1.jpg|500px]][[Image:Pentagon_final.jpg|300px]] | + | |
| - | | + | |
| - | The golden number, φ, is used to construct the '''golden triangle,''' an isoceles triangle that has legs of length <math>\varphi \times r</math> and base length of <math>1 \times r</math> where <math>r</math> can be any constant. It is above and to the left. Similarly, the '''golden gnomon''' has base <math>\varphi \times r</math> and legs of length <math>1 \times r</math>. It is shown above and to the right. These triangles can be used to form regular pentagons (pictured above) and <balloon title="A pentagram is a five pointed star made with 5 straight strokes">pentagrams.</balloon> | + | |
| - | | + | |
| - | The pentgram below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio. | + | |
| - | :[[Image:Star1.jpg]] | + | |
| - | :::<math>\frac{{\color{SkyBlue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} } = \frac{{\color{Red}\mathrm{red}} }{{\color{Green}\mathrm{green}} } = \frac{{\color{Green}\mathrm{green}} }{{\color{Magenta}\mathrm{pink}} } = \varphi . </math> | + | |
| - | | + | |
| - | These triangles can be used to form [[Field:Fractals| fractals]] and are one of the only ways to tile a plane using '''pentagonal symmetry'''. Pentagonal symmetry is best explained through example. Below, we have two fractal examples of pentagonal symmetry. Images that exhibit pentagonal symmetry have five symmetry axes. This means that we can draw five lines from the image's center, and all resulting divisions are identical. | + | |
| - | | + | |
| - | :[[Image:Penta1.jpg|400px]] | + | |
| - | :[[Image:Pent111.jpg|400px]] | + | |
| - | |ImageDesc==Mathematical Representations of the Golden Ratio= | + | |
| - | | + | |
| - | | + | |
| - | ==An Algebraic Derivation of Phi== | + | |
| - | | + | |
| - | | + | |
| - | {{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText=How can we derive the value of φ from its characteristics as a ratio? We may algebraically solve for the ratio (φ) by observing that ratio satisfies the following property by definition: | + | |
| - | :<math>\frac{b}{a} = \frac{a+b}{b} = \varphi</math>|FullText= | + | |
| - | Let <math> r </math> denote the ratio : | + | |
| - | :<math>r=\frac{a}{b}=\frac{a+b}{a}</math>. | + | |
| - | | + | |
| - | So | + | |
| - | :<math>r=\frac{a+b}{a}=1+\frac{b}{a}</math> which can be rewritten as | + | |
| - | | + | |
| - | :<math>1+\cfrac{1}{a/b}=1+\frac{1}{r}</math> thus, | + | |
| - | | + | |
| - | :<math>r=1+\frac{1}{r}</math> | + | |
| - | | + | |
| - | Multiplying both sides by <math>r</math>, we get | + | |
| - | | + | |
| - | :<math> {r}^2=r+1</math> | + | |
| - | | + | |
| - | which can be written as: | + | |
| - | :<math>r^2 - r - 1 = 0 </math>. | + | |
| - | | + | |
| - | Applying the <balloon title="load:myContent">quadratic formula</balloon> | + | |
| - | <span id="myContent" style="display:none"> | + | |
| - | An equation, <math>\frac{-b \pm \sqrt {b^2-4ac}}{2a}</math>, which produces the solutions for equations of form <math>ax^2+bx+c=0</math> | + | |
| - | </span>, we get <math>r = \frac{1 \pm \sqrt{5}} {2}</math>. | + | |
| - | | + | |
| - | The ratio must be positive because we can not have negative line segments or side lengths. Because the ratio has to be a positive value, | + | |
| - | | + | |
| - | :<math>r=\frac{1 + \sqrt{5}}{2} = 1.61803399... =\varphi</math>. | + | |
| - | |NumChars=500}} | + | |
| - | | + | |
| - | | + | |
| - | | + | |
| - | | + | |
| - | ==Continued Fraction Representation and [[Fibonacci sequence|Fibonacci Sequences]]== | + | |
| - | The golden ratio can also be written as what is called a '''continued fraction,'''a fraction of infinite length whose denominator is a quantity plus a fraction, which latter fraction has a similar denominator, and so on. This is done by using <balloon title="Recursion is the method of substituting an equation into itself">recursion</balloon>. | + | |
| - | | + | |
| - | {{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText= |FullText=We have already solved for φ using the following equation: | + | |
| - | | + | |
| - | <math>{\varphi}^2-{\varphi}-1=0</math>. | + | |
| - | | + | |
| - | We can add one to both sides of the equation to get | + | |
| - | | + | |
| - | <math>{\varphi}^2-{\varphi}=1</math>. | + | |
| - | | + | |
| - | Factoring this gives | + | |
| - | | + | |
| - | <math> \varphi(\varphi-1)=1 </math>. | + | |
| - | | + | |
| - | Dividing by <math>\varphi</math> gives us | + | |
| - | | + | |
| - | <math>\varphi -1= \cfrac{1}{\varphi }</math>. | + | |
| - | | + | |
| - | Adding 1 to both sides gives | + | |
| - | | + | |
| - | <math>\varphi =1+ \cfrac{1}{\varphi }</math>. | + | |
| - | | + | |
| - | Substitute in the entire right side of the equation for <math> \varphi </math> in the bottom of the fraction. | + | |
| - | | + | |
| - | <math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }</math> | + | |
| - | | + | |
| - | Substituting in again, | + | |
| - | | + | |
| - | <math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}}</math> | + | |
| - | | + | |
| - | | + | |
| - | | + | |
| - | <math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}</math> | + | |
| - | | + | |
| - | Continuing this substation forever, the last infinite form is a continued fraction | + | |
| - | | + | |
| - | If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction, replacing <math>\varphi</math> by 1, we produce the ratios between consecutive terms in the [[Fibonacci sequence]]. | + | |
| - | | + | |
| - | <math>\varphi \approx 1 + \cfrac{1}{1} = 2</math> | + | |
| - | | + | |
| - | <math>\varphi \approx 1 + \cfrac{1}{1+\cfrac{1}{1}} = 3/2</math> | + | |
| - | | + | |
| - | <math>\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1} } } = 5/3</math> | + | |
| - | | + | |
| - | <math>\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5</math> | + | |
| - | | + | |
| - | Thus we discover that the golden ratio is approximated in the Fibonacci sequence. | + | |
| - | | + | |
| - | <math>1,1,2,3,5,8,13,21,34,55,89,144...\,</math> | + | |
| - | <table style="position:relative;left:50px"> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>1/1</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>2/1</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>2</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>3/2</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1.5</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>8/5</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1.6</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>13/8</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1.625</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>21/13</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1.61538462...</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>34/21</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1.61904762...</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>55/34</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1.61764706...</math></td> | + | |
| - | </tr> | + | |
| - | <tr> | + | |
| - | <td align="right"><math>89/55</math></td> | + | |
| - | <td align="center"><math>=</math></td> | + | |
| - | <td align="left"><math>1.61818182...</math></td> | + | |
| - | </tr> | + | |
| - | </table> | + | |
| - | | + | |
| - | | + | |
| - | <!--Alan, here's your old stuff just in case I messed up the numbers or something. --Maria | + | |
| - | | + | |
| - | <math>2/1 = 2, 3/2 = 1.5, 8/5 = 1.6, 13/8 = 1.625, 21/13 = 1.61538462...,\,</math> | + | |
| - | <math>34/21 = 1.61904762..., 55/34=1.61764706..., 89/55 = 1.61818182..., ...\,</math> | + | |
| - | --> | + | |
| - | <math>\varphi = 1.61803399...\,</math> | + | |
| - | | + | |
| - | As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio. Many real-world applications of the golden ratio are related to the Fibonacci sequence. For more real-world applications of the golden ratio [[Fibonacci Numbers|click here!]] | + | |
| - | | + | |
| - | In fact, we can prove that the ratio between terms in the Fibonacci sequence approaches the golden ratio by using mathematical [[Induction]]. | + | |
| - | | + | |
| - | {{Switch|link1=Click to show proof|link2=Click to hide proof|1= |2= | + | |
| - | | + | |
| - | Since we have already shown that | + | |
| - | | + | |
| - | <math> \varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}} </math>, | + | |
| - | | + | |
| - | we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above. | + | |
| - | | + | |
| - | First, let | + | |
| - | :<math> x_1=1</math>, | + | |
| - | :<math> x_2=1+\frac{1}{1}=1+\frac{1}{x_1} </math>, | + | |
| - | :<math> x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2} </math> and so on so that | + | |
| - | :<math> x_n=1+\frac{1}{x_{n-1}} </math>. | + | |
| - | | + | |
| - | These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as | + | |
| - | :<math> s_n=s_{n-1}+s_{n-2} </math> where <math>s_1=1</math>,<math>s_2=1</math>,<math>s_3=2</math>,<math>s_4=3</math> etc. | + | |
| - | <br> | + | |
| - | | + | |
| - | We want to show that | + | |
| - | :<math> x_n=\frac{s_{n+1}}{s_n} </math> for all n. | + | |
| - | | + | |
| - | First, we establish our [[Induction|base case]]. We see that | + | |
| - | :<math> x_1=1=\frac{1}{1}=\frac{s_2}{s_1} </math>, and so the relationship holds for the base case. | + | |
| - | | + | |
| - | Now we assume that | + | |
| - | :<math> x_k=\frac{s_{k+1}}{s_{k}} </math> for some <math> 1 \leq k < n </math> (This step is the [[Induction|inductive hypothesis]]). We will show that this implies that | + | |
| - | :<math> x_{k+1}=\frac{s_{(k+1)+1}}{s_{k+1}}=\frac{s_{k+2}}{s_{k+1}} </math>. | + | |
| - | | + | |
| - | <br><br> | + | |
| - | | + | |
| - | By our assumptions about ''x<sub>1</sub>'',''x<sub>2</sub>''...''x<sub>n</sub>'', we have | + | |
| - | | + | |
| - | :<math> x_{k+1}=1+\frac{1}{x_k} </math>. | + | |
| - | | + | |
| - | By our inductive hypothesis, this is equivalent to | + | |
| - | | + | |
| - | :<math>x_{k+1}=1+\frac{1}{\frac{s_{k+1}}{s_{k}}}</math>. | + | |
| - | | + | |
| - | Now we only need to complete some simple algebra to see | + | |
| - | | + | |
| - | :<math> x_{k+1}=1+\frac{s_k}{s_{k+1}} </math> | + | |
| - | | + | |
| - | :<math> x_{k+1}=\frac{s_{k+1}+s_k}{s_{k+1}} </math> | + | |
| - | | + | |
| - | Noting the definition of <math>s_n=s_{n-1}+s_{n-2}</math>, we see that we have | + | |
| - | | + | |
| - | <math> x_{k+1}=\frac{f_{k+2}}{f_{k+1}} </math> | + | |
| - | | + | |
| - | So by the principle of mathematical induction, we have shown that the terms in our continued fraction are represented by ratios of consecutive Fibonacci numbers. | + | |
| - | | + | |
| - | The exact continued fraction is | + | |
| - | :<math> x_{\infty} = \lim_{n\rightarrow \infty}\frac{f_{n+1}}{f_n} =\varphi </math>. | + | |
| - | | + | |
| - | }}|NumChars=75}} | + | |
| - | | + | |
| - | | + | |
| - | | + | |
| - | ==Proof of the Golden Ratio's Irrationality== | + | |
| - | | + | |
| - | {{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText= |FullText= | + | |
| - | Remarkably, the Golden Ratio is <balloon title="A number is irrational if it cannot be expressed as the fraction between two integers.">irrational</balloon>, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers. | + | |
| - | We will use the method of <balloon title="A method of proving a statement true by assuming that it's false and showing this assumption would logically lead to a statement that is already known to be untrue.">contradiction</balloon> to prove that the golden ratio is irrational. | + | |
| - | | + | |
| - | Suppose <math>\varphi </math> is rational. Then it can be written as fraction in lowest terms <math> \varphi = b/a</math>, where a and b are integers. | + | |
| - | | + | |
| - | Our goal is to find a different fraction that is equal to <math> \varphi </math> and is in lower terms. This will be our contradiction that will show that <math> \varphi </math> is irrational. | + | |
| - | | + | |
| - | First note that the definition of <math> \varphi = \frac{b}{a}=\frac{a+b}{b} </math> implies that <math> b > a </math> since clearly <math> b+a>b </math> and the two fractions must be equal. | + | |
| - | | + | |
| - | <br> | + | |
| - | | + | |
| - | Now, since we know | + | |
| - | | + | |
| - | :<math> \frac{b}{a}=\frac{a+b}{b} </math> | + | |
| - | | + | |
| - | we see that <math> b^2=a(a+b) </math> by cross multiplication. Foiling this expression gives us <math> b^2=a^2+ab </math>. | + | |
| - | | + | |
| - | Rearranging this gives us <math> b^2-ab=a^2 </math>, which is the same as :<math> b(b-a)=a^2 </math>. | + | |
| - | | + | |
| - | Dividing both sides of the equation by ''a(b-a)'' gives us | + | |
| - | | + | |
| - | :<math> \frac{b}{a}=\frac{a}{b-a} </math>. | + | |
| - | | + | |
| - | Since <math> \varphi=\frac{b}{a} </math>, this means <math> \varphi=\frac{a}{b-a} </math>. | + | |
| - | | + | |
| - | Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since <math> a<b </math>, we know that <math> \frac{a}{b-a} </math> must be in lower terms than <math> \frac{b}{a} </math>. | + | |
| - | | + | |
| - | Since we have found a fraction of integers that is equal to <math> \varphi </math>, but is in lower terms than <math> \frac{b}{a} </math>, we have a contradiction: <math> \frac{b}{a} </math> cannot be a fraction of integers in lowest terms. Therefore <math> \varphi </math> cannot be expressed as a fraction of integers and is irrational. | + | |
| - | |NumChars=75}} | + | |
| - | | + | |
| - | | + | |
| - | | + | |
| - | :*Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 2-19. | + | |
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| - | | + | |
| - | |other=Algebra, Geometry | + | |
| - | |AuthorName=azavez1 | + | |
| - | |SiteName=The Math Forum | + | |
| - | |SiteURL=http://mathforum.org/mathimages/index.php/Image:180px-Pentagram-phi.svg.png | + | |
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## Current revision
The Golden Ratio
Fields: Algebra and Geometry
Image Created By: Joyce Han
The Golden Ratio
The golden number, often denoted by lowercase Greek letter "phi", is
$\frac{a+b}{a} = \frac{a}{b} \equiv \varphi,$
where the Greek letter phi ($\varphi$) represents the golden ratio. Its value is:
${\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...$.
The term golden ratio refers to any ratio which has the value phi. The image to the right illustrates dividing and subdividing a rectangle into the golden ratio. This page explores how the Golden Ratio can be observed and found in the arts, mathematics, and nature.
# Basic Description
==The Golden Ratio as an Irrational Number==
## The Golden Ratio in the Arts
### Parthenon
There is an abundance of artists who consciously used the Golden Ratio from as long ago as 400 BCE. Once such example is the Greek sculptor Phidias, who built the Parthenon. The exterior dimensions of the Parthenon form the golden rectangle, and the golden rectangle can also be found in the space between the columns.
### Vitruvian Man
Another instance in which the Golden ratio appears is in Leonardo Da Vinci’s drawing of the Vitruvian Man. Da Vinci’s picture of man’s body fits the approximation of the golden ratio very closely. This picture is considered to a depiction of a perfectly proportioned human body. The Golden Ratio in this picture is the distance from the naval to the top of his head, divided by the distance from the soles of the feet to the top of the head.
### Music
Some people have argued that the Golden Ratio in music produces aesthetically pleasing sounds. Major sixth and the minor sixth chords are considered to be the most pleasing intervals, and subsequently, they are the intervals related to the Golden Ratio. The standard tuning tone used is an A and it vibrates at 440 vibrations per second. A major sixth interval below that would be a C, which has a frequency of 264 vibrations per second. The ratio of the two frequencies reduces to 5/3, which is the ratio of two Fibonacci numbers. Similarly, a minor sixth can be obtained from a high C, which has a frequency of 528 vibrations per second, and an E, which is 330 vibrations per second. This ratio reduces to 8/5, which is also a ratio of two Fibonacci numbers.
The Golden Ratio manifests in the idea that music is proportionally balanced. However, it is an area of debate, because some scholars claim that these appearances are purely coincidental, or a result of number juggling from aficionados. However, many contemporary artists have intentionally used the Golden Ratio in their pieces. Composer, mathematician and teacher Joseph Schillinger believed that music could be based entirely on mathematical formulation. He developed a System of Musical Composition in which successive notes in a melody were followed by Fibonacci intervals when counted in half steps. Half steps are the smallest intervals possible, and it is the closest note that can be played higher or lower. For example, if C were the first note in a composition, the following note would be half a step higher, because one is a Fibonacci number. So, the second note would be a D flat. Schillinger would alternate between moving up and down in intervals. Thus, the third note would be two half steps (because two is the next Fibonacci number), down from the D flat, which would be B flat. The next Fibonacci number after two is three, so the next note would be three half steps higher than the last. Three half steps higher than B flat is E flat. Schillinger believed that these notes convey the same sense of harmony as the phyllotactic ratios found in leaves.
# A More Mathematical Explanation
[Click to view A More Mathematical Explanation]
## Fibonacci Numbers
One of the greatest breakthroughs regarding the Golden Ratio came when its rela [...]
[Click to hide A More Mathematical Explanation]
## Fibonacci Numbers
One of the greatest breakthroughs regarding the Golden Ratio came when its relation to Fibonacci numbers, also known as the Fibonacci sequence, was discovered. Fibonacci numbers can also be found in the arts, and nature. The Fibonacci sequence is the series of numbers, 0,1,1,2,3,5,8,13,21,34,55,89,144,233… The next term in the Fibonacci sequence, starting from the third, is determined by adding the previous two terms together. The Fibonacci sequence is related to the Golden Ratio because as the sequence grows, the ratio of consecutive terms gradually approaches the Golden Ratio. For example here are the ratios of the successive numbers in the Fibonacci sequence:
1/1=1.000000
2/1=2.000000
3/2=1.500000
5/3=1.666666
8/5=1.600000
13/8=1.625000
21/13=1.615385
34/21=1.619048
55/34=1.617647
89/55=1.618182
144/89=1.717978
233/144=1.618056
377/233=1.618026
610/377=1.618037
987/610=1.618033
## The Golden Ratio in Nature
### Spirals & Phyllotaxis
Spirals are abundant concerning phyllotaxis, which describes the way leaves are arranged on a plant stem. In 92 percent of Norway spruce cones, the spirals were found to appear in rows five and eight rows. They appeared in rows of four and seven in six percent, and four and six in four percent. In addition, the number of right-handed spirals appears to be equal to the number of left-handed spirals. The arrangements of the spirals in these spruce cones are found as the following pairs of rows: 2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89, and 89/144. These numbers are all numbers that belong to the Fibonacci Series.
```
```
A similar phenomenon can be observed in sunflower heads. A sunflower head has both clockwise and counterclockwise spirals. The numbers of the spirals in a sunflower usually depend on the size of the sunflower. However, the ratios of the spirals that usually occur are 89/55,144/89, and even 233/144. Once again, all of the numbers in these ratios are consecutive Fibonacci numbers.
# Teaching Materials
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Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://mathhelpforum.com/number-theory/98565-any-number-can-divides-infinity.html
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# Thread:
1. ## any number can divides infinity???
Definition: An integer $b$ is said to be divisible by an integer $a \neq 0$, in symbols $a|b$, if there exists some integer $c$ such that $b=ac$.
If b is infinity $\infty$, then what are the possible a?
I think $a$ can be any number. Is it true?
2. ## according to me
yes. (b/a) should not be undefined.
even infinity/infinity is valid though its a inderminate form as infinity can not be compared with other infinity unless we know its generating function.
3. Your definition says that $a, b \mbox{ and } c$ are integers.
Can you reasonably say that $\infty$ is an integer, (which would be necessary to include it within your definition) ?
4. Originally Posted by BobP
Your definition says that $a, b \mbox{ and } c$ are integers.
Can you reasonably say that $\infty$ is an integer, (which would be necessary to include it within your definition) ?
well if one is talking about integers then infinity will be an integer
if one is talking about natural numbers then infintity will be an natural number for sure.
it depends upon the function used.
note:
if there are two person A and B. they are talking about natural numbers, both of them then think about infinity then you will never be able to find if
(infinity suggested by A)/(infinity suggested by B)>1 or <1 this will be totally indeterminate.
5. Originally Posted by nikhil
yes. (b/a) should not be undefined.
even infinity/infinity is valid though its a inderminate form as infinity can not be compared with other infinity unless we know its generating function.
What do generating functions have to do with this? Perhaps you mean rate of convergence and such.
well if one is talking about integers then infinity will be an integer
if one is talking about natural numbers then infintity will be an natural number for sure.
it depends upon the function used.
note:
if there are two person A and B. they are talking about natural numbers, both of them then think about infinity then you will never be able to find if
(infinity suggested by A)/(infinity suggested by B)>1 or <1 this will be totally indeterminate.
You have it all wrong. "Infinity" is not a natural number, not a real number, not a complex number. For instance, take the Peano axioms as a foundation for arithmetic; then if $\infty \in \mathbb{N}$ we must have $\infty+1 \in \mathbb{N}$ and $(\infty+1)+1 \in \mathbb{N}$.. But you will agree that both of these are equal to $\infty$ hence they cannot be natural numbers because the successor function $n \mapsto n+1$ is injective (it's one of the axioms).
In other words, if you define $\infty$ as a natural number you end up with things such as $0=2$. The cancellation laws for addition are easily derived from the Peano axioms and they do not hold for "infinity".
6. Assuming that b can be infinity, and a is any integer not equal to zero, consider what integer c multiplied to a would equal infinity? If you multiply any two integers, no matter how "large", you'll still get an integer which might be extremely "large" but still not infinity.
So, to answer your question, a can't be any number ...
Regarding infinity:
Wikipedia: "Infinity (symbolically represented by ∞) refers to several distinct concepts – usually linked to the idea of "without end"".
i.e. Infinity is not a number, it's an idea of being "endless".
7. Originally Posted by Bingk
Assuming that b can be infinity, and a is any integer not equal to zero, consider what integer c multiplied to a would equal infinity? If you multiply any two integers, no matter how "large", you'll still get an integer which might be extremely "large" but still not infinity.
So, to answer your question, a can't be any number ...
Regarding infinity:
Wikipedia: "Infinity (symbolically represented by ∞) refers to several distinct concepts – usually linked to the idea of "without end"".
i.e. Infinity is not a number, it's an idea of being "endless".
On that note, thread closed.
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http://mathhelpforum.com/algebra/75675-binomial-expansion.html
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# Thread:
1. ## Binomial expansion
Two positive integers , p and q are connected by p=q+1 . By using the binomial expansion , show that the expression $p^{2n}-2nq-1$ can be divided exactly by $q^2$ for all positive integers of n .
2. Originally Posted by thereddevils
Two positive integers , p and q are connected by p=q+1 . By using the binomial expansion , show that the expression $p^{2n}-2nq-1$ can be divided exactly by $q^2$ for all positive integers of n .
Using
$(1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$
$(1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$
3. Originally Posted by danny arrigo
Using
$(1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$
$(1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$
Thanks Danny , but i still don see how it works . Need a little more explaination .
Thanks again.
4. Originally Posted by danny arrigo
Using
$(1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$
$(1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$
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http://physics.stackexchange.com/questions/24822/percolation-in-a-2d-ising-model?answertab=oldest
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# Percolation in a 2D Ising model
For a 2D Ising model with zero applied field, it seems logical to me that the phases above and below T_c will have different percolation behaviour. I would expect that percolation occurs (and hence there is a spanning cluster) below T_c and that it does not occur (no spanning clusters) above T_c. I'm looking for conformation that this is true, but haven't found anything so far. Is my conjecture correct?
-
1
– Vijay Murthy May 4 '12 at 9:33
Thanks, though my suspicions were that it is related to site percolation. Specifically, I would expect that the spin configuration of the high temperature Ising model can be mapped to the configuration of sites for p \approx 1/2 bond percolation. I suppose that the bond-correlated percolation mapping also works at finite temperature. – Matthew Matic May 4 '12 at 9:42
@VijayMurthy it would be great if you post that as an answer (and just add a few words about the argument given in the paper). – David Zaslavsky♦ May 4 '12 at 15:50
## 1 Answer
You conjecture is correct. One can relate the 2d Ising model with the bond correlated percolation model. The details are in the paper Percolation, clusters, and phase transitions in spin models.
The basic idea is to consider interacting (nearest neighbor) spins as forming a bond with a certain probability. One can then show that the partition function of the Ising model is related to the generating function of the bond-correlated percolation model.
The above paper demonstrates that the bond-correlated percolation model has the same critical temperature and critical exponents as the 2d Ising model. However, the values of $T_c$ and the critical exponents seem to be dependent on exactly how one defines a bond. See section III.A.1 in Universality classes in nonequilibrium lattice systems (or arxiv version).
Nonetheless your intuitive picture that there would be spanning clusters below $T_c$ and no such clusters above $T_c$ remains valid.
EDIT 21 May 2012
I found a pedagogical paper that discusses this issue.
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http://math.stackexchange.com/questions/192189/integrating-with-infinite-radius-semicircle
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# integrating with infinite-radius semicircle
Sorry, I'm really confused on this one.... My math physics class is having us evaluate closed contours, where the contour is on the real line and at the infinities we turn towards the upper half of the space (i.e. for $z=a+ib$ we have $b>0$), anyway, the semicircle is essentially in the upper half of the plane. And we are trying to evaluate this integral... I don't really understand this concept (i'm a mathematician in a classroom of physicists, so I'm really frustrated as they seem to have no rigor)
Second issue, there seems to be contour integrals that are evaluated on poles and then the professor seems to "bypass singular points" and this is done by getting arbitrarily close to the pole and then integrating radially around it (but why integrate clockwise or counterclockwise around this point?).... why? It seems that the professor always chooses one direction arbitrarily and evaluates! I don't know what's going on, I've just read an entire book on complex analysis (M W Wong) and he never seems to cover this topic. Any general suggestions?
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if you bypass a singularity it doesn't seem to be the same integral? – Squirtle Sep 7 '12 at 2:25
this is pretty open-ended – Jonathan Sep 7 '12 at 2:37
Are you sure? Any book on complex analysis certainly covers contour integrals.. – Euler....IS_ALIVE Sep 7 '12 at 2:50
Well, there're lots of complex analysis books. Go into your maths library and begin diving into the section of complex analysis searching for books with real integration with the help of complex integration. Some authors that is worth checking are Ahlfors, Ullrich, Freitag (I love this one!), Remmert, Silverman... – DonAntonio Sep 7 '12 at 2:54
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Perhaps it would help if you provided a concrete example. From what you have given, it sounds as if there is some confusion about contour integration (either on the professor's part or yours). Both answerers are trying to help, but the question is a bit vague; it is hard to tell exactly what the professor is doing. I have never seen the book you mention, so I don't know how well it covers contour integration. – robjohn♦ Sep 7 '12 at 9:26
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## 2 Answers
It sounds as if the professor is talking about contour integration. If complex analysis is a prerequisite for this course, then that is most likely what it is.
An example of an integral with a real singularity and a removable singularity is the principal value integral $$\mathrm{PV}\int_{-\infty}^\infty\frac{e^{ix}}{x}\mathrm{d}x =\lim_{R\to\infty}\left(\int_{-R}^{-1/R}\frac{e^{ix}}{x}\mathrm{d}x+\int_{1/R}^{R}\frac{e^{ix}}{x}\mathrm{d}x\right)\tag{1}$$ Since $e^{ix}=\cos(x)+i\sin(x)$, $(1)$ covers a principal value integral with a real singularity and an integral with a removable singularity: $$\mathrm{PV}\int_{-\infty}^\infty\frac{\cos(x)}{x}\mathrm{d}x\qquad\text{and}\qquad\int_{-\infty}^\infty\frac{\sin(x)}{x}\mathrm{d}x\tag{2}$$ The integral on the left hand side of $(1)$ can be handled with the following contour:
$\hspace{3.2cm}$
The integral in $(1)$ equals the integral of $\dfrac{e^{iz}}{z}$ over the two red pieces of the contour above as the large curve (radius $R$) gets larger and the small curve (radius $1/R$) gets smaller. To use contour integration, we close the contour with the two circular curves.
Since there is no singularity of inside the contour, the integral over the entire contour is $0$.
The integral over the large curve where $y<\sqrt{R}$ tends to $0$ since the absolute value of the integrand is $\le1/R$ over two pieces of the contour whose length is essentially $\sqrt{R}$. The integral over the large curve where $y\ge\sqrt{R}$ also tends to $0$ since the absolute value of the integral is less than $e^{-\sqrt{R}}/R$ over a contour whose length is less than $\pi R$. Thus, the integral over the large curve tends to $0$.
Near the origin, we have $$\frac{e^{iz}}{z}=\frac1z+i-\frac{z}{2}-i\frac{z^2}{6}+\dots\tag{3}$$ Aside from $\frac1z$, the series in $(3)$ is bounded and so its integral over the small curve tends to $0$ since the length of the curve is $\pi/R$. The integral of $\frac1z$ over a counter-clockwise circular arc centered at the origin is $i$ times the angle of the arc. Thus, the integral over the small curve is $-\pi i$ (clockwise circular arc with angle $\pi$).
Since the integral over the entire contour is $0$ and the integral over the two curves tends to a total of $-\pi i$, the integral over the two red pieces must tend to $\pi i$. Taking real and imaginary parts yields $$\mathrm{PV}\int_{-\infty}^\infty\frac{\cos(x)}{x}\mathrm{d}x=0\qquad\text{and}\qquad\int_{-\infty}^\infty\frac{\sin(x)}{x}\mathrm{d}x=\pi\tag{4}$$ Alternatively, we can also use the following contour:
$\hspace{3.2cm}$
Everything is as above except that the contour contains the singularity at $0$ and the small circular curve is counter-clockwise.
The integral over the entire contour is $2\pi i$ since the residue of the singularity is $1$.
The integral over the large curve tends to $0$ as above.
The integral over the small curve is $\pi i$ (counter-clockwise circular arc with angle $\pi$).
Therefore, the integral over the two red pieces must again tend to $\pi i$.
Thus, the integral can be handled circling the singularity either way.
Perhaps something like this is what is going on in class.
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I think you are being very rude to the people that are taking time to answer your question. If you are the mathematician that knows what you say you do then you should understand how to answer your own question. This is why people are responding by telling you what contour integration is. If you even read this particular answer properly you will see that he answers your question. He evaluates a countour integral very similar to the one you have but circles the singularity both ways and shows that it doesn't matter. – fretty Sep 7 '12 at 9:30
@fretty: to be fair, I added everything past the first paragraph after the comment. – robjohn♦ Sep 7 '12 at 9:33
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Ah right, I didn't know this. However this still does not excuse the ignorance and rudeness of the OP. – fretty Sep 7 '12 at 9:34
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Why is this from afar very nice and detailed looking answer downvoted? Does it contain any mistakes or wrong things I dont see ...? – Dilaton Sep 7 '12 at 11:07
@Dilaton: it was downvoted when only the first paragraph had been written and accidentally posted before the rest. – robjohn♦ Sep 7 '12 at 11:19
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The method of integration like this is rigorous if you understand what's going on. For example, the infinite semicircle integration you're talking about comes from parametrizing the family of semicircles by radius, then take the limit as radius goes to infinity. Most of the time the integral on the semicircle part converges to zero as radius goes to infinity.
Avoiding singularities is valid because as long as you don't have a singularity inside your curve, you can deform the curve however you want and the integral will not change. Having a branch cut might be a bit more tricky as it requires you to think about continuity of your variable as you progress along the contour. Anyway, all concepts are well justified, rigorously. Wikipedia page on contour integration is actually pretty good.
P.S. I find it surprising that a complex analysis book does not cover this. According to Amazon, the table of contents of the said book looks like it covers contour integration, but I have no idea how much is covered. (The price is so high for a book with 160 pages!)
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@dustanalysis Please, save useless words and write down an explicit example. We can't be in your head, and after reading your rude comments I would not like to be in your head at all :-) Calm down and write some mathematical example you do not understand. – Siminore Sep 7 '12 at 9:41
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http://math.stackexchange.com/questions/285030/random-walk-with-zero-drift/285045
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# Random walk with zero drift
Let $X_{k+1} = X_k+\xi_k$ be a random walk on $\Bbb R$ starting from $0$ and such that $\mathsf E\xi_0 = 0$, $\mathsf{Var}[\xi_0]>0$. Is that true that $$-\infty = \liminf_nX_n<\limsup_nX_n=\infty\quad \mathsf P\text{-a.s.}?$$
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## 1 Answer
If the variance is finite, the law of the iterated logarithm tells you that for i.i.d random variables $\xi_k$ with mean $0$, the limsup of their sum grows like the square root of $n\log(\log(n))$. A similar argument will show that the liminf grows like the square root of $-n\log(\log(n))$.
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http://mathhelpforum.com/advanced-statistics/117131-conditions-process-martingale.html
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# Thread:
1. ## Conditions for a process to be a martingale
Hi !
I'll give the introduction but it's not the main problem.
We have E a countable space of states. Q is an irreducible transition matrix. Let $(\Omega,\mathcal{F},(X_n)_n,\mathbb{P}_x)$ ( $\mathbb{P}_x$ denotes the initial probability) be the canonical Markov chain with the transition matrix Q.
Let $f ~:~ E\to [0,\infty)$ such that $Qf=f$
Prove that $(f(X_n))_n$ is a martingale.
So I have no problem showing the equality of the conditional expectation, and showing that it's measurable wrt the filtration.
But I don't know how to show that $f(X_n)$ is integrable...
So my questions :
Just to be sure : does it mean to show that $\mathbb{E}_x(f(X_n))$ is finite ?
A friend said that there were martingales with an infinite expectation... How so ?
Is something positive always integrable ?
Would the hypothesis that $(X_n)_n$ is a Markov chain be useful ? Or the hypothesis that Q is irreducible ?
Erm... That's all for now.
Thanks
2. Hi, Moo,
You can indeed define conditional expectation of positive random variables even if they're not integrable: have a look p.147 of this wonderful reference
In your case, the reason why $E[f(X_n)]$ is finite is obtained by induction: $E[f(X_0)]=f(x)<\infty$ and
$E[f(X_{n+1})]=\sum_y f(y)P(X_{n+1}=y)=\sum_y \sum_z f(y)P(X_n=z)Q(z,y)$ $= \sum_z P(X_n=z)\sum_y f(y) Q(z,y)$ $=\sum_z P(X_n=z) f(z)=E[f(X_n)]$.
But because of the first remark, you don't really have to do this beforehand, it may come as a consequence of the martingale property; it may depend on what you're supposed to know...
3. Yep, I know this reference, already downloaded the pdf
Actually, the reason why we have to say that it's integrable is to justify the use of the expectation over f(Xn).
So we can't first calculate the conditional expectation and then that it equals a constant.
But actually, the integrability is useful if the random variable is not positive. If it is positive, then it's always correct to take its integral.
That's what M. TA said ^^'
4. Originally Posted by Moo
Yep, I know this reference, already downloaded the pdf
Actually, the reason why we have to say that it's integrable is to justify the use of the expectation over f(Xn).
So we can't first calculate the conditional expectation and then that it equals a constant.
But actually, the integrability is useful if the random variable is not positive. If it is positive, then it's always correct to take its integral.
That's what M. TA said ^^'
I feel like this is exactly what I wrote (and the justification comes from what Le Gall says on the page I mentioned)... Anyway, we agree, that's the most important.
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http://physics.stackexchange.com/questions/7778/how-can-you-tell-if-a-critical-energy-density-is-actually-a-black-hole
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# How can you tell if a critical energy density is actually a black hole?
Here's a question inspired by Edward's answer to this question.
It's my understanding that the average energy density of a black hole in its rest frame is $\rho_\text{BH}(A)$, a function of surface area. I calculated $3c^2/2GA$ for a Schwarzschild black hole, but that's presumably not applicable here since I'm talking about an extended energy distribution. Anyway, suppose you are in a space filled with some sort of energy, matter, or whatever, which produces a potentially time-dependent stress-energy tensor. And further suppose that there is some finite, spherical region of surface area $A$ in this space, over which you measure the average energy density to be $\rho_\text{BH}(A)$, the net charge to be zero, and the total angular momentum of the matter within the region to be zero. (I'm assuming there is a measurement procedure available which can be carried out without entering the region, if it matters.)
Now, Edward's argument in the other question shows that there are at least two ways to produce that (average) energy density:
1. The energy density arises from a fluid or other material at rest, such that the region is a black hole and there are no timelike geodesics exiting it
2. The energy density has been "augmented" by a Lorentz boost from some other frame, implying that there are timelike geodesics exiting the region
Is it possible, in general, to distinguish between case 1 and case 2 by only looking at the other components of the stress energy tensor, without actually calculating the geodesics? If so, how? What would be the "signature" of a black hole in the components of $T^{\mu\nu}$ (or rather, their averages over the region in question)?
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– Edward Mar 30 '11 at 3:41
@Edward: thanks, I figured there was a decent chance that I was asking something which can't be answered. I'll see what people have to say, though. – David Zaslavsky♦ Mar 30 '11 at 4:28
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Sorry, David, but the energy density $T_{00}$ in the Schwarzschild black hole solution is zero everywhere except for the singularity where it's ill-defined. Your disagreement with this simple fact makes your question confusing, to say the least. Are you actually asking about the stars that are going to collapse into black holes sometime in the future? – Luboš Motl Mar 30 '11 at 8:02
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@Luboš: Yes, I know the stress-energy tensor itself is singular, but I was talking about the average energy density, which (as I tried to make clear) could be defined as the mass of the black hole divided by the volume of a Euclidean sphere with the same surface area as the event horizon. I would be very surprised to hear that either mass or surface area is ill-defined for a black hole. I've tried to formulate the question in a way that doesn't rely on measuring the value of $T^{00}$ at an individual point, only on its average over a region. – David Zaslavsky♦ Mar 30 '11 at 18:12
the asymptotic structure of the spacetime will define you a covariant asymptotic 4-momentum vector, which will have magnitude $M$, so the mass is indeed well-defined. But the volume enclosed by the black hole area is not an invariant. You can get a bit tricky with this, as the $r=0$ singularity has a similar structure to the $r=0$ singularity of a point charge, so you, in some sense, can think of $\rho$ as a delta function, but I dont' think this generalizes to even the Kerr metric, so it's probably not useful. – Jerry Schirmer Mar 31 '11 at 3:30
## 3 Answers
David,
As you can see from Lubos's comment there is a serious flaw in the formulation of this question, so I will use this answer simply to explain some aspects of the topic. First the flaw:
the energy density of a Schwarzchild black hole
The Schwarzchild solution is a solution of the Vacuum Einstein equations ie $R^{u,v}=0$ and so $T^{u,v}=0$ ie the Stress Energy Tensor is zero throughout the Schwarzchild solution (removing the origin from this manifold avoids the undefinedness there). With zero stress-energy there are no fluids or local energy densities to measure or examine.
What is meant in the question (as it arises from the earlier Stack questions) has some relationships with the Hoop conjecture as Edward points out, but just for clarification I shall add more.
Let us not consider any more an Einstein vacuum and assume that some form of matter (or radiation - I will just say matter below) must have been present "originally". This matter is part of a non-vacuum solution of the Einstein equations, and so there will be a corresponding non-zero Stress-Energy Tensor whose matter is destined to become the Black Hole. So now the question begins to make sense.
So the question is really about what determines whether a given non-vacuum solution of Einstein's equations forms a Black Hole and whether this fact is measurable locally. The sentence that asks this is:
What would be the "signature" of the Black Hole in the components of $T^{u,v}$?\$
I dont believe that this answer is known, partly because the space of all solutions of Einstein's equations is not yet known. If one considers the points made below, one might also conclude that GR alone was inadequate to predict the BH formation - matter properties are central too.
There are the classical Hawking-Penrose theorems which give a topological-geometric answer to this question by positing the existence of "closed trapped surfaces", along with certain properties of $T^{u,v}$. In that sense there is an answer, but it doesnt tell us when the closed trapped surfaces will form (generically).
Black Holes arose as physically plausible solutions to Einstein's equations because of the early work of Oppenheimer et al. Here there are two metrics combined to form the matter leading to a Black Hole:
Friedmann Dust (interior) + Schwarzchild (exterior)
(a clever trick to consider "massless dust" allows one to use only the Friedmann Dust solution). These two solutions need to be "glued together" to form the surface of the star. The $T^{u,v}$ (in the comoving frame and its translation into other frames) for this was given in Edward's earlier answer, and is non-zero in the star interior.
What causes another layer of complication in discussing the formation of Black Holes and Event Horizons is the teleological nature of their formation in General Relativity. This arises because "time" is just a parameter in the theory and the formation of the Black Hole is determined by the overall solution (thus in a time independent way). Now it has been concluded that for stellar objects of mass > TOV limit a Black Hole will form. But as remarked above translating this condition into a condition on the Stress-Energy Tensor alone may not be possible.
Physically one might expect that there is some local condition despite all these issues, such as the Hoop conjecture which includes the object's mass in its formulation. There are several subtleties connected with this unproven conjecture and one problem here is that "mass" is not a local property in GR (because mass = energy and the gravitational field contributes energy too, not just the Stress-Energy Tensor - hence we again may need the full solution of $G^{u,v}=T^{u,v}$.)
I shall add this link from Willie Wong for anyone interested in the latest on the Hoop Conjecture.
Finally my answer to the linked question might be of interest.
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Nice answer :) I will have to mull this over a bit. Incidentally, my $\rho_\text{BH}(A)$ is in fact the same thing you labeled as $D_R$ in your answer to the other question. – David Zaslavsky♦ Mar 31 '11 at 0:25
Also, note that there are exact solutions of Einstein's equation that correspond to naked singularities. For an easy example, if you have a charged black hole with $Q>M$, there is no event horizon at all, but you still have a mass $M$ measured by an observer at asymptotic infinity. So, these solutions are a class of solutions that have "infinite" density at the singularity, but at the same time, aren't black holes.
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Right. But then again you have the cosmic censorship conjecture which says that naked singularities cannot form. And even if they do, the existence of a singularity in any form is an indication of a breakdown of the existing theory. In any theory the values of various fields and observables where the theory develops singularities are de facto limiting values. – user346 Mar 30 '11 at 23:54
@Jerry: Interesting, I didn't think about that... thanks for pointing it out. So I guess the question is a moot point for black holes with sufficient charge or angular momentum, but I had in mind a situation where $Q = 0$ and $J = 0$ so I will update the question to make that clear, in case it changes anything. +1 for your insight ;) – David Zaslavsky♦ Mar 30 '11 at 23:55
@Deepak: but the cosmic censorship conjecture has been proven false on at least a the complement of a dense subset of initially physically reasonable initial states of GR. Regardless, this shows that the notion of a 'universaal black hole energy density' isn't going to work out. – Jerry Schirmer Mar 30 '11 at 23:56
Also, there are more exotic naked singularities corresponding to sheets of or lines of mass for which no horizon forms, even when you get infinite mass. You can get these for (very contrived) physically reasonable (but not asymptotically flat) initial conditions. – Jerry Schirmer Mar 30 '11 at 23:59
@Jerry, you say proven false on a complement of a dense subset ... wouldn't the complement of a dense subset be a non-dense subset of the solution space? Can you explain that statement in somewhat less technical language? – user346 Mar 31 '11 at 0:13
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It really is not a density. The critical spatial quantity is the Schwarzschild radius $R~=~2GM/c^2$, where the mass $M$ is contained in a volume with a radius $R$. The critical factor is not the amount of matter per unit volume, or density. A super massive black hole has a mass of 10 billion solar masses, and a radius 10 billion times the solar mass radius of 1.5 km. The volume in standard coordinates is $3.4\times 10^{30}km^3$. The sun has a volume of $1.4\times 10^{18}km^3$. !0 billion suns have a volume of $1.4\times 10^{28}km^3$. So 10 billion suns could be packed into the volume of a 10 billion solar mass black hole with room to spare.
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Right, and the critical average density for a black hole of that size would be 0.7 kg/m^3, which is less than the Sun's density by a factor of nearly 2000. I don't really see what you're getting at... – David Zaslavsky♦ Mar 31 '11 at 0:02
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Absolutely @David, and this is what suggests that the line of inquiry based on volume densities might be incorrect. Regardless of the density of the interior bulk - which may be large or very small as you point out - one common characteristic of all black holes is that their bounding surface (the "horizon") satisfies $S=A/4$. Therefore IMHO it is the entropy density per unit area on the boundary of a region with any given field configuration which will ultimately determine whether or not that region will undergo grav. collapse. This also sidesteps all issues about the definition of energy. – user346 Mar 31 '11 at 0:17
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http://physics.stackexchange.com/questions/8940/how-can-gravitational-forces-influence-time
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# How can gravitational forces influence time?
How does it work that gravitational forces can affect time and what usable applications could arise from this?
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+1 good question, can we get Einstein in here? – Mark Schultheiss Apr 20 '11 at 14:48
You may be interested in the classic experiment that verified the existence of the effect: en.wikipedia.org/wiki/Pound-Rebka_experiment It was also one of the effects involved in the memorable Hafele-Keating experiment: en.wikipedia.org/wiki/Hafele-Keating_experiment – Ben Crowell Apr 6 at 19:08
## 4 Answers
Gravity is currently understood to be curvature of spacetime, so mathematically it's unambiguous why gravity should affect time. But to leave it at that would be cheating, no? So I'll try to give an explanation of why you'd intuitively expect gravity to affect time. This is called the Einstein Tower thought experiment. If you've seen the phenomenon of time dilation in special relativity before, you'll notice that the way time behaves depends a lot on how light behaves. It's not crucial you know this but the spirit of the following explanation is the same.
Let's assume for a moment that gravity doesn't affect light in any way. Suppose you create a photon with energy $E$ on the ground and fire it up to the top of a tower where this photon is converted into an equivalent mass $m=E/c^2$. This mass will fall back to the ground and once it reaches the ground it will have gained extra energy $mgh$ because it fell through the gravitational field. If you started with energy $E$, you end up with energy $$E'=E+mgh=E(1+gh/c^2)$$ So every time you do this, you create energy! Surely something is wrong -- it must've been our assumption that gravity doesn't affect light. The most obvious fix then is to assume that the photon lost energy as it climbed up the gravitational field and that its energy at the top is $$E_{top}=E(1+gh/c^2)^{-1}$$ This is (a first approximation to) the phenomenon of gravitational red shift where photons lose energy as they climb out of a gravitational field.
Now, if you remember that the photon frequency is related to its energy by $E=h\nu$, you'll see that the frequency (~$1/T$ which is the "internal clock" of a photon) obeys the same relation as above. This is indicative of gravity affecting time -- in fact it's immediately obvious that any clock based on the frequency of light will run at a slower rate higher up in the gravitational field.
In reality, the result is far more general, though you'll need the apparatus of general relativity to see this. See http://en.wikipedia.org/wiki/Gravitational_time_dilation for more.
There's probably no practical application you can milk out of gravitational time dilation (yet) since the effect is weak. But you definitely need to account for these effects in engineering projects such as GPS satellites.
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+1! That's a very nice argument you got there. – Olaf Apr 21 '11 at 12:44
Since the OP phrased the question in terms of "gravitational forces," it may be worth pointing out explicitly that the effect has absolutely no dependence on gravitational force. It depends only on a difference in gravitational potential. – Ben Crowell Apr 6 at 19:10
Since your question is from the How Things Work site I guess you're looking for some basic understanding rather than lots of equations. What follows is in that spirit ...
Your question doesn't really have an answer, because it isn't meaningful, or at the least it's misleading, to say that gravitational forces affect time. Mainly that's because, assuming you believe in General Relativity, gravity isn't really a force at all. At least it's not a force in the same way that there's an electric force between two charges or a strong force between two subatomic particles.
General Relativity is based on the idea that mass (and energy and even pressure) cause spacetime to curve. The path that objects moving through spacetime follow is affected by this curvature so objects moving through curved space don't move in straight lines. If you've watched any popular science programmes you've probably seen the analogy of balls rolling on rubber sheets. The key point is that the freely moving object thinks it's moving in a straight line and it doesn't feel any gravitational force.
For example, suppose you leap off a high cliff. As long as nothing is in your way you'll move in a locally straight line and you won't feel any force at all (ignoring wind resistance). That is, you won't experience any gravitational force. The trouble is that the presence of the Earth curves spacetime in your vicinity, and as a result your locally straight line will hit the surface of the earth. Assuming you land with no injury, you'll now find you feel a force holding you to the ground that you (and Newton!) would describe as a gravitational force. But the apparent force is just because you and the Earth are trying to move along different locally straight lines. There isn't really any force there.
So my point is that it isn't meaningful to ask how a gravitational force affects time, because there isn't any such thing as a gravitational force. What does affect time, or more precisely spacetime, is the presence of matter, and that produces the appearance of a gravitation force. Your question should really be "How can (the curvature of) time affect gravity?" i.e. exactly the reverse of what you asked!
So now you're going to ask "How does matter affect time", and that's a good question that doesn't have an easy answer. The Einstein equation tells you how matter affects spacetime, and you can feed in numbers and get the right answers. What it doesn't tell you is why matter affects spacetime. We tend to leave the "why" questions to the philosophers :-)
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The Schwarzschild metric is $$ds^2~=~(1~-~2GM/rc^2)(cdt)^2~-~(1~-~2GM/rc^2)^{-1}dr^2~-~r^2d\Omega^2$$ This has all the constants and the like here to illustrate how the time part of the metric has $O(c^2)$ larger dependency than the spatial part, so $|g_{00}~-~1|~>>$ $|g_{rr}~-~1|$. So we may approximate this with $$ds^2~\simeq~(1~-~2GM/rc^2)(cdt)^2~-~dr^2~-~r^2d\Omega^2$$ So we may then consider a stationary situation with $dr~=~0$ so the propertime is $$d\tau~=~\sqrt{1~-~2GM/rc^2}dt~\simeq~(1~-~GM/rc^2)dt,$$ which for near Earth surface gravity the $1~-~GM/r~\simeq~1~+~gh/c^2$, $r~=~R~+~h$. This recovers the dbrane result. For a moving body $dr~=~vdt$ we may the compute a gamma factor $$ds^2~\simeq~(1~-~2GM/rc^2~-~(v/c)^2)dt^2$$ which can be used to derive modified Lorentz transformations of spacetime. These then may be used to compute time dilation results for satellites in Earth orbit.
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I changed the last equation. In looking at it I see I made a small mistake. – Lawrence B. Crowell Apr 21 '11 at 22:46
The easy answer is that in day to day life, it can't. And so not any really usable applications.
The hard answer is that in special circumstances it can (see Special relativity), or more accurately the speed you travel at can affect how you experience time. To get a really good understanding of why would require a university degree in both maths and theoretical physics, which I don't have and neither would I have chance to explain in a couple of paragraphs.
Put simply, the closer to the speed of light you travel, the slower you experience time, so that if you could theoretically reach the speed of light (which you can't without an infinite amount of energy, unless you have zero mass, which is a mind bender in its self) time would appear to stop (from your point of view at least).
Consider this (sorry its quoted directly from wikipedia but its illustrates it quite well): Since one can not travel faster than light, one might conclude that a human can never travel further from Earth than 40 light years, if the traveler is active between the age of 20 and 60. One would easily think that a traveller would never be able to reach more than the very few solar systems which exist within the limit of 20-40 light years from the earth. But that would be a mistaken conclusion. Because of time dilation, he can travel thousands of light years during his 40 active years. If the spaceship accelerates at a constant 1G, he will after a little less than a year (mathematically) reach almost the speed of light, but time dilation will increase his life span to thousands of years, seen from the reference system of the Solar System, but his subjective lifespan will not thereby change. If he returns to Earth he will land thousands of years into its future. Even if he should accelerate for a longer period, his speed will not be seen as higher than the speed of light by observers on Earth, and he will not measure his speed as being higher than the speed of light. This is because he will see a length contraction of the universe in his direction of travel. And during the journey, people on Earth will experience much more time than he does. So, although his (ordinary) speed cannot exceed c, his four-velocity (distance as seen by Earth divided by his proper (i.e. subjective) time) can be much greater than c. This is similar to the fact that a muon can travel much further than c times its half-life (when at rest), if it is traveling close to c.
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"the closer to the speed of light you travel, the slower you experience time" could be misleading. A person moving near the the speed of light will age slower relative to someone who is moving slower, but each person will experience time the same way. 40 years would still feel like 40 years to both people. As described in the Wikipedia article, the traveller returning to earth would find that much more time has passed on earth, even though the traveller himself has only experienced 40 years of life as a high-speed astronaut. – e.James Apr 20 '11 at 17:46
That's a mind bender. Not a scientific reference by any means, but the Stargate episode "A Matter of Time" illustrates this theory if you're interested in a less technical explanation. – Michael Apr 20 '11 at 20:34
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Note: special relativity specifically does not deal with gravity: the special means it applies in inertial reference frames. To answer this question, one must appeal to general relativity. – dmckee♦ Apr 20 '11 at 22:44
## protected by Qmechanic♦Apr 6 at 18:45
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http://physics.stackexchange.com/questions/tagged/dimensional-analysis+density
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### Gaussian integration and dimension argument
I made a mistake recently regarding the Gaussian density, by putting the determinant of the variance to the power $\frac{d}{2}$. Would the following argumentation be valid to highlight it should be to ...
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### Center Of Mass Troubles
I understand the concept of Center Of Mass(com), but I am having a difficult time interpreting the equation of the simplified case of one-dimension. The book I am reading defines the position of the ...
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# Tagged Questions
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### Understanding units and the units of the derivative operator
Suppose that $f$ is a function from unit $A$ to $B$, then what is the unit of $f'(x)$?. We can do $f'(x)\Delta x$ to get an estimate of $f(x + \Delta x)$. Since the latter has unit $B$, so has the ...
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http://math.stackexchange.com/questions/133999/can-one-construct-a-non-measurable-set-without-axiom-of-choice
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# Can one construct a non-measurable set without Axiom of choice?
Is axiom of choice required to show the existence of non-measurable sets? Is there a Lebesgue non-measurable set that can be constructed without axiom of choice?
Related question on MO says it is consistent: http://mathoverflow.net/questions/73902/axiom-of-choice-and-non-measurable-set
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2
Solovay proved that there is a model of Set Theory without Choice where every subset of the real numbers is measurable. In that model, the Axiom of Dependent Choice is true. So you need at least something stronger than DC in order to establish the existence of non-measurable sets. – Arturo Magidin Apr 19 '12 at 17:00
2
– Qiaochu Yuan Apr 19 '12 at 17:01
That's very interesting. I had thought that the existence of ultrafilters was equivalent to AC. Thanks! – Neal Apr 19 '12 at 17:03
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@Neal: AC is equivalent to the statement "Every lattice has an ultrafilter." It is strictly stronger than "There is a non-principal ultrafilter over $\mathbb{N}$." This is perhaps the source of your confusion. – Cameron Buie Apr 19 '12 at 17:12
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@Cameron: I actually believe that the main source of confusion is the fact that people are usually ignorant to how much choice is really needed, and they simply use Zorn's lemma to do things. This gives the impression that many things require full choice while some require none. – Asaf Karagila Apr 19 '12 at 17:14
show 4 more comments
## 1 Answer
The answer is, you cannot.
It is consistent with ZF that the real numbers are a countable union of countable sets, this implies that every set of reals is Borel and therefore measurable. Of course, in such model it is nearly impossible to develop the analysis we know.
However it is consistent relative to an inaccessible cardinal that there is a model of ZF+DC where all the sets of real numbers are Lebesgue measurable, and DC allows us to do most of classical analysis too.
Non-measurable sets can be generated by free ultrafilters over $\mathbb N$ too, which as remarked is a strictly weaker assumption that the axiom of choice. If there are $\aleph_1$ many real numbers and DC holds then there is an non-measurable set as well, which implies that ZF+DC($\aleph_1$) also implies the existence of non-measurable sets of real numbers - however this is not enough to imply the existence of free ultrafilters over the natural numbers!
Several other ways to generate non-measurable sets of real numbers:
1. The axiom of choice for families of pairs;
2. Hahn-Banach theorem;
3. The existence of a Hamel basis for $\mathbb R$ over $\mathbb Q$.
There are several other ways as well, but none are quite close to the full power of the axiom of choice.
One important remark is that we can ensure that the axiom of choice holds for the real numbers as usual, but breaks in many many severe ways much much further in the universe (that is counterexamples will be sets generated much later than the real numbers in the von Neumann hierarchy). This means that the axiom of choice is severely negated - but the real numbers still behave as we know them.
The above constructions and to further read about ways to construct non-measurable sets cf. Horst Herrlich, Axiom of Choice, Lecture Notes in Mathematics v. 1876, Springer-Verlag (2006).
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Can you provide a source of the Hahn-Banach way to generate a non-measurable? – leo Apr 19 '12 at 18:41
@leo: The HB theorem is enough to prove the Banach-Tarski paradox. – Asaf Karagila Apr 19 '12 at 18:48
Thanks Asaf :-) – leo Apr 19 '12 at 19:02
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http://mathoverflow.net/questions/66613?sort=oldest
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## A toy model for the t-section problem
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Let $S(x)$ be the area of the yellow curvilinear triangle. I'd like to find a graph for which $S(x)=H(x)$ where $H$ is some prescribed function (small, smooth, vanishing near the endpoints to any order you wish, etc.). Is it always possible or there are some non-obvious hidden restrictions?
The question comes from the infamous t-section problem (if you know the areas of all sections of a symmetric convex body by the hyperplanes at some fixed small distance $t$ from the origin (so small that all sections are non-empty), can you recover the body?). The problem is open even on the plane. I do not say that this toy question is directly relevant here but an answer to it will certainly make a few things clearer for me.
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What happens when you try the obvious, which to me is adding dx to x, and then figuring the incremental shape of the curve based on the difference in the area of the triangles? Or is 45 not 45 degrees but an angle that depends on x? Gerhard "Ask Me About System Design" Paseman, 2010.05.31 – Gerhard Paseman Jun 1 2011 at 2:16
This strikes me as something that deserves a 'calculus of variations' tag. Gerhard "Ask Me About System Design" Paseman, 2011.05.31 – Gerhard Paseman Jun 1 2011 at 2:32
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The answer to the first question is "nothing good". You get a delayed differential equation, which is highly unstable. As to the tag, I'll add it :). – fedja Jun 1 2011 at 4:32
## 1 Answer
There are restrictions.
At most points, $S'(x)$ is the length of the right leg minus the length of the left leg of the curvilinear triangle, perhaps with exceptions on a null set where there are tangencies. If $S(0)=S(1)=0$ then the lengths of these legs are at most $\sqrt{2}(1-x)$ and $\sqrt{2}x$. For almost all $0\le x \le 1$, $S'(x)$ satisfies $-\sqrt{2}\le -\sqrt{2} x \lt S'(x) \lt \sqrt(2) (1-x) \le \sqrt{2}$. This is an extra condition on $H$ which rules out some smooth small functions which have large derivatives near some points, such as $10^6 \exp(-1/(x (1-x))^2)$ for $0\lt x \lt 1$, which has a derivative of $1.132$ at $x=0.436$ although the value of the function is small.
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Thanks a lot! I upvoted it but, unfortunately, it still fits under the "trivial restriction" category. To be more precise, you can assume $H$ to be small in any function space you desire (say, $C^k$ with any $k$). What I mean is that "roughly speaking" $S(x)=\frac f(x)^2$ and cannot change too much when $x$ goes by $f(x)$ to the right or to the left where $f$ is the function whose graph gives $S$, so if you have hard time with the square root or the rate of change is too big, you have trouble for sure. But what if you have $H(x)=10^{-100}[x(1-x)]^{10}$, say? – fedja Jun 5 2011 at 17:40
I wasn't sure if this was supposed to be ruled out by the condition that $H$ is small. One of the things I considered was whether you meant that if $H$ is smooth and vanishes to all orders, then there is some function so that $S = cH$ for some $c \gt 0$. – Douglas Zare Jun 5 2011 at 18:03
Yes, that would be another good way to formalize it (provided that you do not start playing with coming very close to $0$ and then lifting off infinitely many times). The bad thing is that I don't know myself what effect I'm looking for. I rather know a long list of trivial effects that I do not care much about. The particular function I wrote would have none of them though, so if you can show that it is impossible, it'll tell me something new. – fedja Jun 6 2011 at 1:35
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http://mathoverflow.net/questions/78232/is-there-a-combinatorial-interpretation-of-this-triangle-sequence-is-there-a-si
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## Is there a combinatorial interpretation of this triangle sequence? Is there a “simpler” formula?
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Hi.
`$$a_{n,k} = \sum_{1 = m_1 < m_2 < \cdots < m_k = n}\ \prod_{j=2}^k S(m_j, m_{j-1})$$.
where the $S(n, k)$ are Stirling numbers of the second kind. The triangle begins:
1,
0, 1,
0, 1, 3,
0, 1, 13, 18,
0, 1, 50, 205, 180,
...
Something very much like this is listed on oeis.org as "Triangle of coefficients from fractional iteration of $e^x - 1$" (and it is doing such fractional iteration that piques my interest in this sequence, and I got the formula while fiddling around with the one mentioned in my earlier question here: http://mathoverflow.net/questions/64450/is-there-a-theory-about-these-kinds-of-recurrence-equations-is-this-a-known-form). The two look to be identical except for possible indexing differences and the presence/absence of leading zeroes, but I don't have a proof (yet?).
Now, my questions are: does my $a_{n,k}$ have a combinatorial interpretation of some sort, and if so, what is it? Is there a "simpler" formula for it (i.e. one that does not involving summing over up to an exponential or worse amount of terms, even if it is not symbolically shorter)? And finally, is there a combinatorial interpretation of a product of Stirling numbers $S(m_2, m_1) S(m_3, m_2) \cdots S(m_k, m_{k-1})$ where $m_1 < m_2 < \cdots < m_k$, and if so, what is it?
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## 2 Answers
There is the obvious combinatorial interpretation that follows from the definition of Stirling numbers: $a_{n,k}$ counts the number of ways you can take $n$ elements and partition them into some identical boxes, take those boxes and partition them into some identical boxes and so on $k$ times, in the end you use only one box. I'm not sure if this can help prove anything about the sequence combinatorially, though.
Edit: Actually there is a neat way to think about this interpretation in terms of trees. Let's call a rooted tree monotone if there are more vertices at depth $h+1$ than at $h$, for all $h$, and for which all leaves are at the same distance from the root. Then by the previous paragraph, $a_{n,k}$ counts the number of monotone rooted trees of depth $k$ with $n$ leaves. What follows below can also be phrased in terms of the corresponding combinatorial species, so if you think in terms of such trees you can prove these statements purely combinatorially.
Perhaps this can clarify the computational part of the question. Let $$F=\text{diag}(0!,1!,2!,\dots)$$ and let $$D(x)=(1,x,x^2,\dots)$$ also denote the Stirling matrix by $S=(p_{ij})_{i,j\geq 0}$, where $p_{ij}=S(i,j)$ if $i\geq j$ and $p_{ij}=0$ otherwise.
I might have my indexing off here but by looking at the (exponential) generating function of Stirling numbers the following is easy to check $$D(x)F^{-1}SF=D(e^x-1)$$ So that $$D(x)F^{-1}S^kF=D((e^x-1)^{(k)}),$$ where $f^{(k)}$ is $f$ composed $n$ times with itself. And the entries of the first column of $S^k$ are precisely the $a_{n,k}$ from which your statement that these are coefficients of the functional iterates of $e^x-1$, follows.
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Thanks for the response. But when I mentioned about a "simpler formula" I was thinking more along the lines of the double-sum formula for Bernoulli numbers, the single-sum formula for Stirling numbers of the 2nd kind, sum formulas for the Stirling numbers of the 1st kind (the one that sums over the 2nd-kind Stirling numbers, that is), etc. I.e. involving a simple linear-indexed sum over n or n^2, etc. terms. Maybe with a product involved as well, but the total terms summed/producted should be like n, n^2, n^3, etc. (or otherwise polynomial), not a matrix formula like that one. – mike3 Oct 25 2011 at 23:25
### 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 just saw this question. If I'm not missing something, I think I have a quick answer to your question asking for a combinatorial interpretation for $a_{n,k}$.
Edit: I just realized that my answer is closely related to Gjergji's answer -- I hadn't read his closely enough before. But what I would add to his interpretation is that one may think of his recursive process as giving chains of a particular length in the partition lattice (as detailed below). So you might tap into the literature on the partition lattice.
This number seems to count the chains $\hat{0} = u_k < u_{k-1} < \cdots < u_1 < \hat{1}$ in the partition lattice $\Pi_n$, namely the poset of set partitions of {1,...,n} ordered by reverse refinement (i.e. with $u \le v$ iff $v$ is obtained from $u$ by merging blocks). I've indexed in reverse to align with your notation. Notice that going up a cover relation reduces the number of blocks by exactly one, so $S(n,n-j)$ counts poset elements of rank $j$. Once you choose $u_{k-1}$ of rank $j$, you might as well think of each block in $u_{k-1}$ as a letter, so that counting ways to go up from your chosen $u_{k-1}$ which has rank $j$ to any element $u_{k-2}$ of rank $j'$ satisfying $u_{k-1} < u_{k-2}$, you have $S(n-j,n-j')$ choices for this $u_{k-2}$, etc. Thus, your summands $1 = m_1 < \cdots < m_k = n$ are a choice of which ranks in the partition lattice to use for your chain, and then your products are progressively calculating the ways to choose the next element in a chain having the appropriate rank and above the lower elements of the chain, given the choices so far of lower elements in the chain. In particular, $S(m_j,m_{j-1})$ seems to count the choices for $u_{j-1}$ of rank $m_{j-1}$ once you have chosen the chain elements $u_k,\dots ,u_j$ that are all below it in the chain.
Hopefully I didn't mess up any indexing, but I easily could have. There is a large literature regarding the partition lattice, so maybe it's worth seeing what's been done with chain enumeration there.
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Regarding the connection you mention to iterating $e^x-1$, you might figure out an explanation from looking at chapter 5 in Enumerative Combinatorics, Vol 2, section 5.1 (on exponentiating generating functions and a connection to set partitions). Now I'd better get off MO and get back to my own work. Good luck! – Patricia Hersh Aug 20 at 23:07
I just discovered that my answer comes full circle back to the answer my Ph.D. advisor gave you to a different question. Would you like me to delete my answer so you can focus more on trying to get fresh ideas (if you're still thinking about this question)? One comment is that the flag $h$-numbers are dimensions of $S_n$-representations, which is one way to see where your $(n-1)!$ is coming from -- a Lie character tensored with sign character. There's been a lot of study of these representations, which you could find by searching on "rank-selected homology" and "partition lattice". – Patricia Hersh Aug 21 at 17:02
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http://mathoverflow.net/questions/40518/subsets-of-products-of-trees/40546
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## subsets of products of trees
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A subset of a geodesic metric space is called convex if for every two points in the subset one of the geodesics connecting these points lies in the subset. Is it true that every convex subset of a product of two trees with $l_1$-metric is a median space, that is for every three points A,B,C in the subset there exists a point D in the subset such that D is on (some) geodesics connecting A and B, B and C, A and C?
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It is not true even for $\mathbb R\times\mathbb R$ with $\ell_1$-metric... – Anton Petrunin Sep 29 2010 at 19:50
Consider the set $\{(x,y):\, xy=0, x\geq 0,y\geq 0\}$ (right angle). – Fedor Petrov Sep 29 2010 at 20:06
1
Anton and Fedor: Yes, of course. Sorry for being so sloppy. I have modified the question. I only need to know if every triangle ABC in the subset contains a "center". It would follow from strong convexity, but, say, a geodesic is not necessarily strongly convex, but satisfies the median property. – Mark Sapir Sep 29 2010 at 20:12
## 2 Answers
Yes. Let $A=(A_1,A_2)$, $B=(B_1,B_2)$ and $C=(C_1,C_2)$. For the triangle $A_1B_1C_1$ in the first tree, there is a "center" $M_1$ such that the (unique) geodesics $[A_1B_1]$, $[A_1C_1]$ and $[B_1C_1]$ contain $M_1$. Similarly, there is a "center" $M_2$ for the triangle $A_2B_2C_2$ in the second tree. Let us show that $(M_1,M_2)$ belongs to any convex set $S$ containing $A$, $B$ and $C$.
Suppose $A_1\ne M_1$ and $A_2\ne M_2$. Then consider a geodesic $A(t)=(A_1(t),A_2(t))$ connecting $A$ to $B$ in $S$. Its coordinate projections are geodesics in the two trees. Let $A_1(t)$ hit $M_1$ before $A_2(t)$ hits $M_2$ and this first hit happens at $t=t_0$. Replace $A$ by $A'=A(t_0)$. It suffices to prove the assertion for $A'BC$ in place of $ABC$ (note that the centers of the new coordinate triangles are the same points $M_1$ and $M_2$). What we gained is that now one of $A_1$ and $A_2$ coincides with the corresponding center $M_1$ or $M_2$.
Repeat this procedure for $B$ and $C$. Now, for each of $A$, $B$ and $C$, one of the coordinate projections is at $M_1$ or $M_2$. So at least two of these projections are at the same point $M_1$ or $M_2$. Assume w.l.o.g. that $A_1=B_1=M_1$. Consider a geodesic from $A$ to $B$ in $S$. Its first coordinate projection is constant $M_1$, and the second one is a geodesic from $A_2$ to $B_2$ that must go through $M_2$ eventually.
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@Sergei: Thanks! – Mark Sapir Sep 29 2010 at 22:20
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Hi. I simply wanted to add to the above that this is, I think, specific to a product of two trees, it is not true for at least three. Take R^3 with the l^1 metric and inside it the plane x+y+z=1. It is convex (because Euclidean segments are geodesics also for l^1), not strongly convex, and not median: it contains the points (1,0,0), (0,1,0), (0,0,1) but not their median point (0,0,0).
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http://mathhelpforum.com/advanced-statistics/165979-covariance-normal-vector-x-y-distribution-z-x-2y-1-a.html
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# Thread:
1. ## covariance, a normal vector (X,Y) and the distribution of Z = X - 2Y + 1
hi, I am preparing for a final in my least favorite class and could use a little help with this preparation problem.
Initially X and Y were normal with given mean and variance, and it asked for the distribution of the same Z = h(X,Y). I know that's a linear combination and was able to calculate the distribution no problem.
Now I am a little stuck on the next part,
Assume now that (X,Y) is a normal vector with covariance Cov(X,Y) = 4. What is the distribution of Z = X - 2Y + 1? Compute P( Z > 5).
I can get the second part if I am able to find the distribution of Z, which is where I get stumped.
Does the information in the beginning about the normal dist. of X, Y help? All I can see in my notes about this is
Cov(X,Y) = E(XY) - ux*uy where ux is the mean of x, same for y
thanks for any help at all!
2. Z = X - 2Y + 1
$V(aX+bY +c)=a^2V(X)+b^2V(Y)+2abCOV(X,Y)$
so $V(Z)=V(X)+4V(Y)-4COV(X,Y)$
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http://mathematica.stackexchange.com/questions/21454/how-to-define-functions-for-a-parameter-dependent-recursive-sequence
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# How to define functions for a parameter-dependent recursive sequence
I am trying to generate a two variable recursive sequence
For instance on Mathematica, I did
````z[1] := {1, 1}
B := {{t, 1}, {-1, t}}
z[n_, t_] := B.z[n - 1, t]
````
When I tried to see `z[2]`, I get a bunch of gibberish.
$n$ is the variable that is iterating, $t$ is just a parameter that I can throw in whenever I want. My goal is to generate a table of terms based on various iterations of $n$. From that, my sequence would be defined in terms of only $t$, then I would like test out various values of $t$ to generate a numerical sequence.
There are two problems I am having: how to define the functions properly, and how to implement the recursion.
I use Mathematica 8.0.4
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(1) Replace delayed set `:=` by set `=` in the first two lines. (2) When you run this, look at the error messages. Think about how MMA is going to tell when to stop the recursion. (Hint: `z[1]` is not the same as `z[1,t]`.) BTW, you might be interested in `MatrixExp`. – whuber Mar 16 at 3:22
I tried using z[1,t] before, but there was no difference – jak Mar 16 at 3:47
That's because you need to use a pattern in the argument, as in `z[1, t_] := {1, 1}`. Incidentally, you would appreciate the information you get in a search for memoizing. See mathematica.stackexchange.com/questions/2639/… for instance. Your code is still a little whacky, because `B` implicitly has the literal symbol `t` within it. It will work with the change I suggested, but if you supply anything other than `t` for the second argument of `z`, be prepared for surprises. – whuber Mar 16 at 3:51
## 2 Answers
I see two ways to do the kind of thing desired. You could make the parameter an argument to every function, or have the parameter be a global symbol and not an argument to any function.
Parameters as arguments
Put an argument in each function representing the parameter:
````z[1, t_] := {1, 1};
B[t_] := {{t, 1}, {-1, t}};
z[n_Integer, t_] /; n > 0 := B[t].z[n - 1, t];
{z[2, t], z[2, s], z[2, 100]}
(* {{1 + t, -1 + t}, {1 + s, -1 + s}, {101, 99}} *)
````
Parameters as global variables
Make sure `t` is undefined when `z` and `B` are defined. (I clear any definitions of `z` and `B` as well.)
````Clear[t];
Clear[B, z]; (* clear previous definitions *)
z[1] = {1, 1};
B = {{t, 1}, {-1, t}};
z[n_Integer] /; n > 0 := B.z[n - 1];
{z[2], z[2] /. t -> s, z[2] /. t -> 100}
(* {{1 + t, -1 + t}, {1 + s, -1 + s}, {101, 99}} *)
````
Memoizing
You can often improve performance of recursively defined function if you use a technique called "memoization." An good discussion can be found What does the construct f[x_] := f[x] = ... mean?, cited in a comment by @whuber. The Mathematica documentation has a good introductory tutorial, "Functions That Remember Values They Have Found", too. For example, in the example using `t` as a global variable, the definition of `z` would be
````z[n_Integer] /; n > 0 := z[n] = B.z[n - 1]
````
See the links for further discussion.
Discussion
Each approach can do what you want, and each is probably used often. The biggest consideration is that if the parameter `t` is a global variable, then generally you want to avoid it being set (no `t = 3` for example). It might cause unexpected behavior. Thus I have a slight preference for using the argument approach
Why `_Integer.../; n > 0` instead of `_`?
If the definition of `z` begins `z[n_, t_] :=...`, then for an undefined symbol `n`, `z[n, t]` matches the definition and is expanded. This leads to `z[n-1,t]`, then `z[n-2,t]` etc. Since `z[1,t]` is never reached, the recursion goes until the limit is hit. The `n_Integer` restricts the definition to matching only when `n` is an integer. The `Condition` `/; n > 0` further restricts the definition to positive integers. This guarantees that recursive definition will be applied only to positive `n`.
What went wrong in the original definitions?
What is wrong with the first line has to do with how it matches the third line. The recursive definition of `z` has two arguments and so it will never lead to this pattern:
````z[1] := {1, 1}
````
In the second, `t` is a global symbol. If it has a value, it will be substituted whenever `B` is evaluated, because `B` is defined using `SetDelayed`.
````B := {{t, 1}, {-1, t}}
````
Another difficulty arises in the different way `z` and `B` are defined.
````z[n_, t_] := B.z[n - 1, t]
````
The `t` in the definition of `z` is not a variable per se, just the name of the second argument. Technically, it is the `Pattern` `Blank[]` that matches whatever is passed as the second argument. It has nothing to do with the global `t` in `B`. For instance, `z[2, s]` will expand to
````{{t, 1}, {-1, t}}.z[1, s]
````
where `s` would be whatever was passed to `z` and does not have to be the same as `t`. (Of course, the recursion will go on until `$RecursionLimit` is hit, because of the mismatch with the first definition.)
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Awesoem it's working just as I expected. – jak Mar 29 at 1:26
There are two issues here: getting the syntax of the functions correct and then doing the iteration. wHuber shows in the comments a way to correct the syntax of the functions (using pattern matching in the argument of the calling function). You can find this kind of approach applied to calculate the Fibonnaci sequence which essentially uses the `SetDelayed` command `:=` to do the iteration. While this is quite slick, it is also tricky to get right and does not generalize well.
Mathematica also has several functions devoted to solving iterations. `RSolve` is one, `LinearRecurrence` is another. What I would like to do here is to show a structured way of iterating that can be generalized easily just by changing the function definitions.
Let's define two functions. `b` defines your time varying matrix. `f` defines the function you want to iterate: in this case, each iteration increases the `n` and calculates the matrix multiply:
````b[n_] := {{n, 1}, {-1, n}};
f[{n_, z_}] := {n + 1, b[n].z};
````
Then the iteration can be done by NestList:
````NestList[f, {1, {1, 1}}, 3]
````
In this case, your answer comes out looking like:
````{{1, {1, 1}}, {2, {2, 0}}, {3, {4, -2}}, {4, {10, -10}}}
````
where each entry is {n, {z1,z2}}, that is, the iteration number followed by the state z at that time. Of course you can replace the `3` by any number to specify how many iterations to carry out and you can replace the initial state `{1,1}` with whatever values are convenient.
The above us purely numerical. In the final part of the question, the OP asks to keep the `t` variable symbolic. This can be done by changing the definition of `f` to
```` f[{n_, z_}] := {n + 1, b[t].z};
````
Now, running the same `NestList` gives:
````{{1, {1, 1}}, {2, {1 + t, -1 + t}},
{3, {-1 + t + t (1 + t), -1 - t + (-1 + t) t}},
{4, {-1 - t + (-1 + t) t + t (-1 + t + t (1 + t)),
1 - t - t (1 + t) + t (-1 - t + (-1 + t) t)}}}
````
which is of course, growing rapidly in length.
-
HUh I could've sworn someone closed my question. ANyways. I tried yours, it worked to some degree. I modified it to another recursion, but it didn't work. The NestList doesn't give me the iterations. I'll show you later of what i mean – jak Mar 25 at 21:01
lang-mma
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http://math.stackexchange.com/questions/tagged/pattern-recognition+ring-theory
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Tagged Questions
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Multiplication structure for finite abelian rings of order $p^2$.
Consider an odd prime $p$ and a positive integer $a$ as close to 2 as possible that is not a quadratic residue $mod$ $p$. If we extend the ring $mod$ $p$ with the element $b = a^{\frac{1}{2}}$ then ...
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http://math.stackexchange.com/questions/71257/can-asymptotes-be-imaginary/71320
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# Can asymptotes be imaginary?
I'm currently learning about asymptotes and I'm trying to work out the following example:
$$f(x)=\frac{1}{x^2+1}$$
To find the vertical asymptotes, the book I'm following says that after factoring completely, you should set each factor of the denominator to $0$ and:
Every solution you get that does not make the numerator 0 will give you a vertical asymptote of the function.
According to that, I do:
$$\begin{align} x^2+1=0 \\ x^2=-1 \\ x = \pm\sqrt{-1} \\ x = \pm i \end{align}$$
Now, neither $i$ nor $-i$ make the numerator $0$, so does that mean that $x=i$ and $x=-i$ are vertical asymptotes to the function $f$? And if so, what does that really mean?
Wolfram|Alpha doesn't mention any vertical asymptotes, only the horizontal one at $y=0$.
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4
Do note that $x^2+1$ is never zero when $x$ is real. Complex asymptotes are usually called poles in this context (of a ratio between two polynomials). – Asaf Karagila Oct 9 '11 at 21:54
## 4 Answers
To illustrate what 3296 was getting at, consider these two plots:
These are plots of the real and imaginary parts of the function $\dfrac1{z^2+1}$, where $z=x+i y$. The poles of the function at $z=\pm i$ are easily seen in the plots. Since we're limited to seeing (a two-dimensional projection of) three dimensions, we are forced here to illustrate the poles by plotting the real and imaginary parts of the function separately.
Now, look at a "slice" of the real part:
You see at once the plot of the function you are accustomed to on the real line. The poles do not lie in the slice, and this corresponds to you seeing no vertical asymptotes in the plots of your function on the real line.
Incidentally, this function is the usual example for demonstrating the so-called "Runge phenomenon": any attempt to approximate this function with a polynomial fails due to the poles in the complex plane, even if you are only considering real values of the argument.
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The "asymptote" concept is mostly used only in real analysis, so when one asks about asymptotes it's generally implied that we're considering the function only on the real axis.
In the complex case we'd say that the function has poles at $i$ and $-i$, which corresponds more or less to vertical asymptotes (but in fact have much nicer and subtler properties). You shouldn't need to be concerned about such things until you get to complex analysis.
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Since presumably your domain is the real numbers, the fact that $\pm i$ causes the denominator to be 0 is of no consequence to you. That is, if you run over all the real numbers and plug them in one-by-one to the function, you're never going to see $x = i$ or $x = -i$ because they are not real numbers. Since you're never going to select those two "bad" $x$'s, you're never going to divide by 0, and so you're never going to have a place where asymptotes might arise.
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Let me attempt to answer this in a more direct way that avoids allusions to complex analysis "in the future."
When you graph this function on a 2D piece of paper, you are graphing a real input versus a real output. You get a one-dimensional curve in two-dimensional space. An asymptotes is a meaningful property of a one-dimensional curve embedded in a larger space.
If you were to graph this function as a complex function, regarding the complex numbers as living on the (two-dimensional) complex plane, you'd have two input axes and two output axes, for a total of four dimensions. The graph of the function would be a two-dimensional surface in four-dimensional space. Obviously this situation is going to be substantially more complicated, so it's a safe assumption that you're only going to be asked to deal with the simpler one-dimensional problem above.
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http://wikipedia.sfstate.us/Nonlinearity
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edit
# Nonlinear system
(Redirected from Nonlinearity)
Not to be confused with Non-linear editing system.
This article describes the use of the term nonlinearity in mathematics. For other meanings, see nonlinearity (disambiguation).
In mathematics, a nonlinear system is one that does not satisfy the superposition principle, or one whose output is not directly proportional to its input; a linear system fulfills these conditions. In other words, a nonlinear system is any problem where the equation(s) to be solved cannot be written as a linear combination of the unknown variables or functions that appear in it (them). It does not matter if nonlinear known functions appear in the equations; in particular, a differential equation is linear if it is linear in the unknown function and its derivatives, even if non linear known functions appear as coefficients.
Nonlinear problems are of interest to engineers, physicists and mathematicians because most physical systems are inherently nonlinear in nature. Nonlinear equations are difficult to solve and give rise to interesting phenomena such as chaos.1 Some aspects of the weather (although not the climate) are seen to be chaotic, where simple changes in one part of the system produce complex effects throughout. A nonlinear system is not random.
## Definition
In mathematics, a linear function (or map) $f(x)$ is one which satisfies both of the following properties:
• additivity (Superposition), $\textstyle f(x + y)\ = f(x)\ + f(y);$
• homogeneity, $\textstyle f(\alpha x)\ = \alpha f(x).$
(Additivity implies homogeneity for any rational α, and, for continuous functions, for any real α. For a complex α, homogeneity does not follow from additivity; for example, an antilinear map is additive but not homogeneous.) The conditions of additivity and homogeneity are often combined in the superposition principle
$f(\alpha x + \beta y) = \alpha f(x) + \beta f(y) \,$
An equation written as
$f(x) = C\,$
is called linear if $f(x)$ is a linear map (as defined above) and nonlinear otherwise. The equation is called homogeneous if $C = 0$.
The definition $f(x) = C$ is very general in that $x$ can be any sensible mathematical object (number, vector, function, etc.), and the function $f(x)$ can literally be any mapping, including integration or differentiation with associated constraints (such as boundary values). If $f(x)$ contains differentiation of $x$, the result will be a differential equation.
## Nonlinear algebraic equations
Main article: Algebraic equation
Main article: Systems of polynomial equations
Nonlinear algebraic equations, which are also called polynomial equations, are defined by equating polynomials to zero. For example,
$x^2 + x - 1 = 0\,.$
For a single polynomial equation, root-finding algorithms can be used to find solutions to the equation (i.e., sets of values for the variables that satisfy the equation). However, systems of algebraic equations are more complicated; their study is one motivation for the field of algebraic geometry, a difficult branch of modern mathematics. It is even difficult to decide if a given algebraic system has complex solutions (see Hilbert's Nullstellensatz). Nevertheless, in the case of the systems with a finite number of complex solutions, these systems of polynomial equations are now well understood and efficient methods exist for solving them.2
## Nonlinear recurrence relations
A nonlinear recurrence relation defines successive terms of a sequence as a nonlinear function of preceding terms. Examples of nonlinear recurrence relations are the logistic map and the relations that define the various Hofstadter sequences.
## Nonlinear differential equations
A system of differential equations is said to be nonlinear if it is not a linear system. Problems involving nonlinear differential equations are extremely diverse, and methods of solution or analysis are problem dependent. Examples of nonlinear differential equations are the Navier–Stokes equations in fluid dynamics, the Lotka–Volterra equations in biology.
One of the greatest difficulties of nonlinear problems is that it is not generally possible to combine known solutions into new solutions. In linear problems, for example, a family of linearly independent solutions can be used to construct general solutions through the superposition principle. A good example of this is one-dimensional heat transport with Dirichlet boundary conditions, the solution of which can be written as a time-dependent linear combination of sinusoids of differing frequencies; this makes solutions very flexible. It is often possible to find several very specific solutions to nonlinear equations, however the lack of a superposition principle prevents the construction of new solutions.
### Ordinary differential equations
First order ordinary differential equations are often exactly solvable by separation of variables, especially for autonomous equations. For example, the nonlinear equation
$\frac{\operatorname{d} u}{\operatorname{d} x} = -u^2\,$
will easily yield u = (x + C)−1 as a general solution. The equation is nonlinear because it may be written as
$\frac{\operatorname{d} u}{\operatorname{d} x} + u^2=0\,$
and the left-hand side of the equation is not a linear function of u and its derivatives. Note that if the u2 term were replaced with u, the problem would be linear (the exponential decay problem).
Second and higher order ordinary differential equations (more generally, systems of nonlinear equations) rarely yield closed form solutions, though implicit solutions and solutions involving nonelementary integrals are encountered.
Common methods for the qualitative analysis of nonlinear ordinary differential equations include:
• Examination of any conserved quantities, especially in Hamiltonian systems.
• Examination of dissipative quantities (see Lyapunov function) analogous to conserved quantities.
• Linearization via Taylor expansion.
• Change of variables into something easier to study.
• Bifurcation theory.
• Perturbation methods (can be applied to algebraic equations too).
### Partial differential equations
The most common basic approach to studying nonlinear partial differential equations is to change the variables (or otherwise transform the problem) so that the resulting problem is simpler (possibly even linear). Sometimes, the equation may be transformed into one or more ordinary differential equations, as seen in separation of variables, which is always useful whether or not the resulting ordinary differential equation(s) is solvable.
Another common (though less mathematic) tactic, often seen in fluid and heat mechanics, is to use scale analysis to simplify a general, natural equation in a certain specific boundary value problem. For example, the (very) nonlinear Navier-Stokes equations can be simplified into one linear partial differential equation in the case of transient, laminar, one dimensional flow in a circular pipe; the scale analysis provides conditions under which the flow is laminar and one dimensional and also yields the simplified equation.
Other methods include examining the characteristics and using the methods outlined above for ordinary differential equations.
### Pendula
Main article: Pendulum (mathematics)
Illustration of a pendulum
Linearizations of a pendulum
A classic, extensively studied nonlinear problem is the dynamics of a pendulum under influence of gravity. Using Lagrangian mechanics, it may be shown3 that the motion of a pendulum can be described by the dimensionless nonlinear equation
$\frac{d^2 \theta}{d t^2} + \sin(\theta) = 0\,$
where gravity points "downwards" and $\theta$ is the angle the pendulum forms with its rest position, as shown in the figure at right. One approach to "solving" this equation is to use $d\theta/dt$ as an integrating factor, which would eventually yield
$\int \frac{d \theta}{\sqrt{C_0 + 2 \cos(\theta)}} = t + C_1\,$
which is an implicit solution involving an elliptic integral. This "solution" generally does not have many uses because most of the nature of the solution is hidden in the nonelementary integral (nonelementary even if $C_0 = 0$).
Another way to approach the problem is to linearize any nonlinearities (the sine function term in this case) at the various points of interest through Taylor expansions. For example, the linearization at $\theta = 0$, called the small angle approximation, is
$\frac{d^2 \theta}{d t^2} + \theta = 0\,$
since $\sin(\theta) \approx \theta$ for $\theta \approx 0$. This is a simple harmonic oscillator corresponding to oscillations of the pendulum near the bottom of its path. Another linearization would be at $\theta = \pi$, corresponding to the pendulum being straight up:
$\frac{d^2 \theta}{d t^2} + \pi - \theta = 0\,$
since $\sin(\theta) \approx \pi - \theta$ for $\theta \approx \pi$. The solution to this problem involves hyperbolic sinusoids, and note that unlike the small angle approximation, this approximation is unstable, meaning that $|\theta|$ will usually grow without limit, though bounded solutions are possible. This corresponds to the difficulty of balancing a pendulum upright, it is literally an unstable state.
One more interesting linearization is possible around $\theta = \pi/2$, around which $\sin(\theta) \approx 1$:
$\frac{d^2 \theta}{d t^2} + 1 = 0.$
This corresponds to a free fall problem. A very useful qualitative picture of the pendulum's dynamics may be obtained by piecing together such linearizations, as seen in the figure at right. Other techniques may be used to find (exact) phase portraits and approximate periods.
## Types of nonlinear behaviors
• Classical chaos – the behavior of a system cannot be predicted.
• Multistability – alternating between two or more exclusive states.
• Aperiodic oscillations – functions that do not repeat values after some period (otherwise known as chaotic oscillations or chaos).
• Amplitude death – any oscillations present in the system cease due to some kind of interaction with other system or feedback by the same system.
• Solitons – self-reinforcing solitary waves
## Examples of nonlinear equations
See also the list of nonlinear partial differential equations
## Software for solving nonlinear system
• interalg: solver from OpenOpt / FuncDesigner frameworks for searching either any or all solutions of nonlinear algebraic equations system
• A collection of non-linear models and demo applets (in Monash University's Virtual Lab)
• FyDiK Software for simulations of nonlinear dynamical systems
## References
1. Lazard, D. (2009). "Thirty years of Polynomial System Solving, and now?". Journal of Symbolic Computation 44 (3): 222–231. doi:10.1016/j.jsc.2008.03.004.
## Further reading
• Diederich Hinrichsen and Anthony J. Pritchard (2005). Mathematical Systems Theory I - Modelling, State Space Analysis, Stability and Robustness. Springer Verlag. ISBN [[Special:BookSources/09783540441250|09783540441250[[Category:Articles with invalid ISBNs]]]] Check `|isbn=` value (help).
• Jordan, D. W.; Smith, P. (2007). Nonlinear Ordinary Differential Equations (fourth ed.). Oxford Univeresity Press. ISBN 978-0-19-920824-1.
• Khalil, Hassan K. (2001). Nonlinear Systems. Prentice Hall. ISBN 0-13-067389-7.
• Kreyszig, Erwin (1998). Advanced Engineering Mathematics. Wiley. ISBN 0-471-15496-2.
• Sontag, Eduardo (1998). Mathematical Control Theory: Deterministic Finite Dimensional Systems. Second Edition. Springer. ISBN 0-387-98489-5.
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http://mathoverflow.net/questions/20791?sort=votes
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## isomorphism of abelian varieties
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $A, B, C$ and $D$ be abelian varieties (over $\mathbb{C}$) such that $A \times B \cong C \times D$, and $A \cong C$. From the irreducibility of abelian varieties, we can say that $B$ and $D$ are isogeneous. But do we actually have $B \cong D$?
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I think that rather than "irreducibility" you mean "complete reducibility" or "Poincare's complete reducibility theorem". As for the question itself: I seem to recall that it is famously false for supersingular abelian varieties, i.e., in positive characteristic. Over C, I might guess that it's true, but that's just a guess. – Pete L. Clark Apr 9 2010 at 0:21
I meant the complete reducibility theorem. Thanks. – Tuan Apr 9 2010 at 3:23
It is false also in characteristic $0$ though there it is true for elliptic curves (see Angelo's reply). It is a question of finding an example of non-cancellation for projective modules over a suitable ring and then mirror it for abelian varieties. See Poonen, "The Grothendieck ring of varieties is not a domain" for an example. (Bjorn's example is somewhat involved as he wants an example over $\mathbb Q$. An example over $\mathbb C$ is easier to construct.) – Torsten Ekedahl Apr 9 2010 at 4:04
@Torsten: I think the example in my paper was of something slightly different, namely A x A = B x B with A and B not isomorphic. – Bjorn Poonen Apr 10 2010 at 1:02
@Bjorn: You are right of course, I misremembered. (In my defence, an example of non-cancellation also gives examples of zero-divisors in the Grothendieck ring, the tricky thing is to get an example over $\mathbb Q$.) – Torsten Ekedahl Apr 11 2010 at 18:39
show 1 more comment
## 1 Answer
This is false even for elliptic curves over $\mathbb{C}$. This was proved by T. Shioda in "Some remarks on abelian varieties" J. Fac. Sci. Univ. Tokyo Sect. IA Math. 24 (1977), no. 1, 11-21, http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6164/1/jfs240102.pdf.
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Thank you. That was very helpful! – Tuan Apr 9 2010 at 4:31
This is a very interesting paper. I look forward to reading it carefully. – Pete L. Clark Apr 9 2010 at 5:04
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http://math.stackexchange.com/questions/271205/whats-the-reason-why-this-sequence-of-function-doesnt-converge-uniformly-to-f
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# What's the reason why this sequence of function doesn't converge uniformly to $f$?
Consider $f_n = \sqrt[n]{x}$ on $[0,1]$
So it converges to the step function $f = 0$ if $x = 0$ and $f=1$ otherwise
I could see why it doesn't converge if i draw an epsilon rectangle over one part since for each $n$, $f_n$ lies completely outside of the function.
If I draw an epsilon rectangle over the whole $f$, I don't see why this isn't uniform convergence
EDIT: I got this from Spivak, so keep it at that level please...
Added question: if $f_n\nrightarrow f$ for some $x \in \mathbb{R}$, can I conclude that it is not uniformly convergent over $\mathbb{R}$?
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What do you know about the limit of a uniformly convergent sequence of continuous functions? – hardmath Jan 5 at 23:17
Oh then the limit $f$ must also be continuous, if the set is closed and bounded right? But in my notebook, I wrote down that pointwise convergence does not imply regular convergence. – sizz Jan 5 at 23:25
@sizz: You are right to say that pointwise convergence does not imply uniform convergence. The problem that you have described proves exactly this fact. However, uniform convergence implies pointwise convergence, as I have shown below. – Haskell Curry Jan 6 at 2:56
What about pointwise divergence? Does that imply $f_n$ cannot uniformly converge? – sizz Jan 6 at 4:12
@Sizz: I also answered this question below. You see, uniform convergence implies pointwise convergence. If you have pointwise divergence, then you cannot have uniform convergence, otherwise you would have pointwise convergence --- a contradiction. :) – Haskell Curry Jan 6 at 4:21
show 4 more comments
## 3 Answers
I like @hardmath's approach in the comments above. But here is an approach using the definition of uniform convergence.
Assume $\{f_n\}$ converges uniformly to $f$. Pick $\varepsilon < 1/2$. We can find an $N$ for which $\forall x \in [0, 1] : |f_N(x) - f(x)| < \varepsilon$. Clearly, $f_N(1) = 1$ and $f_N(0) = 0$. Since $f_N$ is continuous, we can find $x \in (0, 1)$ for which $f_N(x) = 1/2$ (by the intermediate value theorem). This means that $|f_N(x) - 1| = 1/2 > \varepsilon$. This is a contradiction and $\{f_n\}$ doesn't converge uniformly to $f$.
Here is a plot of $f_{10}(x) = \sqrt[10]{x}$:
Since $f_n$ is continuous, there will always be values of $x > 0$ for which $f_n(x)$ is too far away from $1$. Uniform convergence requires that after a certain $N$, $f_n(x)$ must be within a small distance $\varepsilon$ from $f(x)$ for all $x$ . This fails for the sequence we have.
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Why is $|f_N(x) - 1| = 1/2$? – sizz Jan 5 at 23:52
@sizz Because we picked $x$ so that $f_N(x) = 1/2$. – Ayman Hourieh Jan 5 at 23:53
Would it be too much to ask to draw me a picture? I am have trouble visualing $|f_N(x) - f(x)| < \epsilon$ part – sizz Jan 5 at 23:55
@sizz I edited my question with a plot and more text to help you visualize it. – Ayman Hourieh Jan 6 at 0:10
We will first prove two results.
Theorem 1 Let $(f_{n})_{n \in \mathbb{N}}$ be a sequence of continuous functions on $[0,1]$ that converges uniformly to some function $f$ on $[0,1]$. Then $f$ must be continuous on $[0,1]$.
Proof: Let $\epsilon > 0$. Then there exists an $N \in \mathbb{N}$ such that for all integers $n \geq N$, we have $$\forall x \in [0,1]: \quad |{f_{n}}(x) - f(x)| < \frac{\epsilon}{3}.$$ To prove that $f$ is continuous, pick an arbitrary $x_{0} \in [0,1]$. As $f_{N}$ is continuous by assumption, there exists a $\delta > 0$ such that $|{f_{N}}(x_{0}) - {f_{N}}(x)| < \dfrac{\epsilon}{3}$ for all $x \in (x_{0} - \delta,x_{0} + \delta) \cap [0,1]$. Hence, by the Triangle Inequality, we see that for all $x \in (x_{0} - \delta,x_{0} + \delta) \cap [0,1]$, the following relations hold: \begin{align} |f(x_{0}) - f(x)| &= |[f(x_{0}) - {f_{N}}(x_{0})] + [{f_{N}}(x_{0}) - {f_{N}}(x)] + [{f_{N}}(x) - f(x)]| \\ &\leq |f(x_{0}) - {f_{N}}(x_{0})| + |{f_{N}}(x_{0}) - {f_{N}}(x)| + |{f_{N}}(x) - f(x)| \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\ &= \epsilon. \end{align} As $x_{0}$ and $\epsilon$ are arbitrary, we conclude that $f$ is indeed continuous on $[0,1]$. $\quad \spadesuit$
Theorem 2 Let $(f_{n})_{n \in \mathbb{N}}$ be a sequence of (not-necessarily-continuous) functions on $[0,1]$ that converges uniformly to some function $f$ on $[0,1]$. Then $(f_{n})_{n \in \mathbb{N}}$ converges pointwise to $f$.
Proof: This follows directly from the definition of uniform convergence. For any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all integers $n \geq N$, we have $$\forall x \in [0,1]: \quad |{f_{n}}(x) - f(x)| < \epsilon.$$ Therefore, for all $x \in [0,1]$, we get $\displaystyle \lim_{n \to \infty} {f_{n}}(x) = f(x)$. $\quad \spadesuit$
This corollary is in response to the OP's latest question.
Corollary Let $(f_{n})_{n \in \mathbb{N}}$ be a sequence of (not-necessarily-continuous) functions on $[0,1]$ and $f$ a function on $[0,1]$ also. If ${f_{n}}(x) \nrightarrow f(x)$ for some $x \in [0,1]$, then $(f_{n})_{n \in \mathbb{N}}$ does not converge uniformly to $f$.
Proof: By Theorem 2, uniform convergence implies pointwise convergence; if pointwise convergence fails, then uniform convergence fails. $\quad \spadesuit$
Assume now, for the sake of contradiction, that the sequence $(f_{n})_{n \in \mathbb{N}} := (\sqrt[n]{\bullet})_{n \in \mathbb{N}}$ of continuous functions on $[0,1]$ converges uniformly to some function $f$ on $[0,1]$. By Theorem 1, $f$ is continuous on $[0,1]$. By Theorem 2, $f$ can be computed as the pointwise limit of $(f_{n})_{n \in \mathbb{N}}$. Hence, \begin{equation} f(x) = \left\{ \begin{array}{ll} 0 & \text{if $x = 0$}; \\ 1 & \text{if $x \in (0,1]$}. \end{array} \right. \end{equation} However, $f$ is clearly not continuous at $0$, thus contradicting Theorem 1.
Conclusion: $(f_{n})_{n \in \mathbb{N}}$ does not converge uniformly to any function on $[0,1]$. However, it does converge pointwise to the piecewise-defined function $f$ described above.
The main point here is that uniform convergence and pointwise convergence are two different concepts. Uniform convergence implies pointwise convergence, but not vice-versa.
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Is that really the correct pointwise limit? – mrf Jan 5 at 23:44
Thanks! I kept thinking it was $x^{n}$. – Haskell Curry Jan 5 at 23:45
@sizz : It does look as if you don't know the difference between pointwise convergence and uniform convergence.
If for every $x\in\mathbb R$, $\lim_{n\to\infty} f_n(x)=f(x)$, that is pointwise convergence of $f_n$ to $f$. That's what you've got here (although in order for the statement to be true just as you've stated it, you ought to have $\sqrt[n]{|x|}$).
Now look at the supremum over all $x\in\mathbb R$ of the distance between $f_n(x)$ and $f(x)$. If that goes to $0$ as $n\to\infty$, then that is uniform convergence of $f_n$ to $f$. That doesn't happen here. You have $$f(x) = \begin{cases} 1 & \text{if }x\ne 0, \\ 0 & \text{if }x=0 \end{cases}$$ and $f_n(x)=\sqrt[n]{|x|}$. As $x\to0$, you have the distance between $f_n(x)$ and $f(x)$ approaching $1$. So the "uniform distance" between $f_n$ and $f$ is $1$. And that doesn't go to $0$ as $n\to\infty$. So you don't have uniform convergence.
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http://crypto.stackexchange.com/questions/5345/breaking-rsa-given-a-special-kind-of-oracle-that-decrypts-related-ciphertexts-f
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# Breaking RSA, given a special kind of oracle that decrypts related ciphertexts for us
Let $c=E^{RSA}_{e}(w)$ be the ciphertext belonging to the plaintext $w$ if an $RSA$ system is used. Assume that the public exponent $e$ satisfies $e \le 10$. Furthermore, assume there is an oracle that, on input $r>0$, responds with the value $c_{r}= E^{RSA}_{e} (w + r)$.
Can the plaintext be decrypted efficiently, given this oracle?
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You should probably do your own homework problems. They are assigned for a good reason, you know (hint: it's not to test your Google skills). – D.W. Nov 13 '12 at 6:39
## 2 Answers
Hint 1: what's the binomial expansion of $E_e^{RSA}(w+r)$? That is, how can that be expressed as a polynomial in variables $w$ and $r$? What degree are those polynomials, in terms of $w$?
Hint 2: suppose we ask for $E_e^{RSA}(w+r)$ for $0 < r < e$ (and we also know $E_e^{RSA}(w)$); what can we do with the corresponding $e$ polynomials?
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okay, thank you very very much – Sam Nov 12 '12 at 9:38
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@Kemo: I suppose this answer helped you to find the proof, as you accepted it. Could you add your own answer with a more complete proof, for all of us who don't have this as a homework, but are nevertheless interested? – Paŭlo Ebermann♦ Nov 12 '12 at 19:06
Here is a simple example for the algorithm given in the last two hints:
Let $n=55,e=3$ and let $w$ be the plaintext and let
$E_{e}^{RSA}(w)=25,E_{e}^{RSA}(w+1)=21,E_{e}^{RSA}(w+2)=33$
but:
$E_{e}^{RSA}(w)=25\;\;\;\;\;\; \Rightarrow w^{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\equiv 25\;mod\;55\;\;............(1)$ $E_{e}^{RSA}(w+1)=21 \Rightarrow w^{3}+3w^{2}+3w+1\;\,\equiv 21\;mod\;55\;\;............(2)$ $E_{e}^{RSA}(w+2)=33 \Rightarrow w^{3}+6w^{2}+12w+8\equiv 33\;mod\;55\;\;............(3)$
Now we just solve this system of congruence equations and then we choose the solution which lives in $\mathcal{Z}_{n}$,
We multiply the first congruence equation with $(-1)$ and add it to the second and third:
$3w^{2}+3w+1\;\,\equiv -4\;mod\;55 \Rightarrow 3w^{2}+3w\;\,\equiv -5\;mod\;55 .....(4)$ $6w^{2}+12w+8\equiv 8\;mod\;55\;\Rightarrow\;\; 6w^{2}+12w\equiv 0\;\;\;\;mod\;55 .....(5)$
We multiply the $4^{th}$ congruence equation with $(-2)$ and add it to the $5^{th}$ then we get:
$6w\equiv10\;mod\;55$
Which is a linear congruence equation and can be efficient solved: $gcd(6,55)=1|10$ so it has a solution and using the extended euclidean algorithm we get:
$6(-9)+55(1)=1 \Rightarrow 6(-90)+55(10)=10 \Rightarrow 6(-90)\equiv 10\; mod\;55$ so the general solution is $w\equiv -90\;mod\;55$ and the solution which lives in $Z_{55}$ is $w=20$.
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http://mathoverflow.net/questions/38701?sort=newest
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Which group does not satisfy the Tits alternative?
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A group is said to satisfy the Tits alternative if every finitely generated subgroup of $G$ is either virtually solvable or contains a nonabelian free subgroup.
Tits proved this for linear groups, and a MathSciNet search gives 38 papers with "Tits alternative" in the title (and 154 papers quoting Tits's original paper), so certainly a lot of groups do enjoy this property.
What then is an example of a group which does not satisfy the Tits alternative?
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7 Answers
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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.
In the generalized sense of measurable group theory, every infinite group satisfies the Tits alternative. Indeed, every group is either amenable and hence orbit equivalent (isomorphic in the category of groups with randomorphisms) to ${\mathbb Z}$ or non-amenable and hence contains ${\mathbb F}_2$ as a random subgroup. The second result is recent and due to Damien Gaboriau and Russell Lyons (see here).
The notion of randomorphism is due to Nicolas Monod, see his ICM talk from 2006 (see here).
EDIT: Answering Henry's comment: $H$ is a random subgroup of $G$ if there is a $H$-equivariant probability measure on the space of maps $\lbrace f: H \to G \mid f(e)=e \rbrace$ endowed with the action $(h.f)(k)= f(kh)f(h)^{-1}$; supported on injective maps. Clearly, every injective homomorphism yields an atomic randomorphism, but there are others.
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Could you say a little bit more about what 'contains $\mathbb{F}_2$ as a random subgroup' means? – HW Sep 14 2010 at 18:54
Tarski monsters are counterexamples to almost anything.
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The paper of Hartley
A conjecture of Bachmuth and Mochizuki on automorphisms of soluble groups. Canad. J. Math. 28 (1976), no. 6, 1302--1310
provides many counterexamples.
Let me quote from MathSciNet review:
"J. Tits [J. Algebra 20 (1972), 250--270; MR0286898 (44 #4105)] showed that a finitely generated linear group either contains a soluble subgroup of finite index or else contains a nonabelian free subgroup. S. Bachmuth and H. Y. Mochizuki [Bull. Amer. Math. Soc. 81 (1975), 420--422; MR0364452 (51 #706)] conjectured that a finitely generated group of automorphisms of a finitely generated soluble group $G$ satisfies the same conclusion. The conjecture is known to be true when $G$ is nilpotent-by-abelian. This paper shows that the conjecture is false in general and a family of counterexamples is constructed. In particular there are such examples where $G$ has derived length three".
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There are also the Burnside's groups $B(m,n)$ for $n\ge 665$ odd: they are of exponential growth and have the law $x^n=1$ so that they cannot contain any free subgroup on two generators. The fact that they are not solvable follows by the theorem of Rosenblatt:
"A f.g. solvable group is of exponential growth if and only if it contains a free sub-semigroup on two generators."
You can find details on paragraphs VII.C.27/28 of Pierre de la Harpe's book "Topics in Geometric Group Theory" (Chicago Lectures in Mathematics, 2000)
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It seems it could be made simplier, without referring to the theorem of Rosenblatt. The Burnside groups are free groups in the variety of groups determined by the law x^n = 1, so would they be virtually solvable, every infinite group satisfying the law x^n = 1 would be solvable. The latter is obviously not true. Similarly, a free group in any variety of group containing infinite non-solvable groups is not virtually solvable (and, obviously, does not contain a free subgroup). – Pasha Zusmanovich Sep 18 2010 at 8:28
I am taking this back: "obviously" in my previous comment is blatantly wrong (if one replaces solvability by nilpotency, this is what Burnside's problems about, far from being "obvious"). – Pasha Zusmanovich Sep 18 2010 at 9:40
Thompson's group F is also an example: it isn't virtually solvable (actually, its commutator subgroup is simple) and does not contain a free subgroup, according to a theorem of Brin and Squier ("Groups of piecewise linear homeomorphisms of the real line.", Invent. Math. 79 (1985))).
You can find a survey about this group (and two cousins of his) written by Cannon, Floyd and Parry on Brin's webpage at http://www.math.binghamton.edu/matt/thompson/cfp.pdf
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I'm a bit confused by the first parenthetical remark, which seems to suggest that simplicity of the commutator subgroup implies virtual solvability. Certainly this is not literally true -- e.g. $S_n$ for $n \geq 5$. But even supposing that the commutator subgroup $G'$ is infinite, I'm still not quite seeing it: off the top of my head, I would think that you need to know also that $G'$ has finite index in $G$. Please help... – Pete L. Clark Sep 14 2010 at 18:51
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What I meant is : 1. a subgroup of a virtually solvable group is virtually solvable. So F cannot be virtually solvable if F' isn't. 2. An infinite simple group has no finite-index subgroup at all (it is easy to show that every finite-index subgroup contains a finite-index normal subgroup). So, if it were virtually solvable, it would be solvable, but that's absurd. – Maxime Bourrigan Sep 14 2010 at 20:09
@MB: Thanks for the help. For some reason I was vaguely doubtful of 1., but now that you assert it I see how to prove it (a subgroup of a solvable group is solvable!). – Pete L. Clark Sep 15 2010 at 2:02
Though this is not the only example of this kind, I think you might like to study the Grigorchuk group. This Wiki page has lots of information, so there is little point of repeating it. Enjoy! -- IP
P.S. I am especially partial to the arXiv preprint mentioned on the bottom of that article, though, apparently, Wikipedia does not realize that it was published awhile ago... :)
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As far as I remember the following is true: the Grigorchuk group has intermidia word growth. Now linear groups have either polynomial or exponential word growth and the Tits alternative played an important role in the original proof of this. So the Grigorchuk group is even more than not satisfying the Tits alternative. – Yiftach Barnea Sep 14 2010 at 17:04
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http://physics.stackexchange.com/questions/4749/is-there-any-physics-that-cannot-be-expressed-in-terms-of-lagrange-equations
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# Is there any physics that cannot be expressed in terms of Lagrange equations?
A lot of physics, such as classical mechanics, General Relativity, Quantum Mechanics etc can be expressed in terms of Lagrangian Mechanics and Hamiltonian Principles. But sometimes I just can't help wonder whether is it ever possible (in the future, maybe) to discover a physical law that can't be expressed in terms of Lagrangian Equations?
Or to put it in other words, can we list down for all the physical laws that can be expressed in terms of Lagrangian equations, what are the mathematical characteristics of them( such as it must not contain derivatives higher than 2, all the solutions must be linear etc)?
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All solutions must be linear? Why? – QGR Feb 7 '11 at 14:12
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– user566 Feb 7 '11 at 14:19
@QGR, I'm not saying that all solutions to physics law must obey linearity!! It's just an example I use to illustrate what I mean by "mathematical characteristics". – Graviton Feb 7 '11 at 14:22
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This is pretty much a duplicate of the question cited by Moshe, vote to close as a duplicate. – pho Feb 7 '11 at 15:26
@Morshe, not too similar! I'm also asking about what are the mathematical characteristics physical laws must obey if they are expressible in terms of Lagrangian mechanics. – Graviton Feb 8 '11 at 1:52
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## 4 Answers
Hamilton's dynamics occurs on a phase space with an equal number of configuration and momentum variables $\{q_i,~p_i\}$, for $i~=~1,\dots n.$ The dynamics according to the symplectic two form ${\underline{\Omega}}~=~\Omega_{ab}dq^a\wedge dp^b$ is a Hamiltonian vector field $$\frac{d\chi_a}{dt}~=~\Omega_{ab}\partial_b H,$$ with in the configuration and momentum variables $\chi_a~=~\{q_a,~p_a\}$ gives $${\dot q}_a~=~\frac{\partial H}{\partial p_a},~{\dot p}_a~=~-\frac{\partial H}{\partial q_a}$$ and the vector $\chi_a$ follows a unique trajectory in phase space, where that trajectory is often called a Hamiltonian flow.
For a system the bare action is $pdq$ ignoring sums. The Hamiltonian is found with imposition of Lagrangians as functions over configuration variables. This is defined then on half of the phase space, called configuration space. It is also a constraint, essentially a Lagrange multiplier. The cotangent bundle $T^*M$ on the configuration space $M$ defines the phase space. Once this is constructed a symplectic manifold is defined. Therefore Lagrangian dynamics on configuration space, or equivalently the cotangent bundle defines a symplectic manifold. This does not mean a symplectic manifold defines a cotangent bundle. The reason is that symplectic or canonical transformations mix the distinction between configuration and momentum variables.
As a result there are people who study bracket structures which have non-Lagrangian content. The RR sector on type IIB string is non-Lagrangian. The differential structure is tied to the Calabi-Yau three-fold, which defines a different dynamics.
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If you have a classical theory specified by some partial differential equations, you can automatically come up with a Lagrangian by introducing a Lagrange multiplier for each PDE.
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Not too sure what you mean here. Classical theory, as opposed to, quantum theory? – Graviton Feb 7 '11 at 14:23
@Graviton: doesn't matter. You can use the classical Lagrangian to quantize your theory. Whether that works in a particular case is a different question though. – Marek Feb 7 '11 at 16:33
You cannot model friction very well with Lagrangian Mechanics.
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And why would that be? – Marek Feb 7 '11 at 16:32
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It is true that in introductory classes to classical mechanics usually only systems with energy conservation are considered. But it is possible to extend the framework to systems with dissipation, see e.g. Brogliato et. alt., “Dissipative Systems, Analysis and Control”, Springer, 2nd edition. – Tim van Beek Feb 7 '11 at 16:39
– ja72 Feb 7 '11 at 17:10
– kharybdis Feb 7 '11 at 17:51
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Why is this downvoted? This is exactly correct--- odd order differential equations, disspiative ones, cannot come out of an explicitly time independent Lagrangian, because they cannot conserve a time independent symplectic volume--- their time evolution takes any motion in phase space to a single point generically! – Ron Maimon Oct 19 '11 at 17:21
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This question is rather generic in several different respects. For example it is not clear that even Schrodinger's equation is an equation that can be "expressed in Hamiltonian principles". Yes we have $\delta/\delta t \Psi=H \Psi$ but is this an expression in "Hamiltonian Principles"? How different must the equation become to not meet this requirement - assuming it does now, perhaps adding a non-linear term?
Furthermore "Hamilton's Principle" does not itself apply in the Quantum context, although the action paths that it introduces are used in Feynman Path Integrals. Hamilton's Principle being a classical concept.
Another generality is in the range of Physics. The whole area of Thermodynamics comes to mind. Now there are phase space explanations, but is that "Hamiltonian Principles?"
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http://stats.stackexchange.com/questions/2691/making-sense-of-principal-component-analysis-eigenvectors-eigenvalues/2706
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# Making sense of principal component analysis, eigenvectors & eigenvalues
In today's pattern recognition class my professor talked about PCA, eigenvectors & eigenvalues.
I got the mathematics of it. If I'm asked to find eigenvalues etc. I'll do it correctly like a machine. But I didn't understand it. I didn't get the purpose of it. I didn't get the feel of it. I strongly believe in
you do not really understand something unless you can explain it to your grandmother -- Albert Einstein
Well, I can't explain these concepts to a layman or grandma.
1. Why PCA, eigenvectors & eigenvalues? What was the need for these concepts?
2. How would you explain these to a layman?
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Good question. I agree with the quote as well. I believe there are many people in statistics and mathematics who are highly intelligent, and can get very deep into their work, but don't deeply understand what they are working on. Or they do, but are incapable of explaining it to others.I go out of my way to provide answers here in plain English, and ask questions demanding plan English answers. – Neil McGuigan Sep 15 '10 at 21:43
– G. Jay Kerns Sep 16 '10 at 2:18
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– whuber♦ Sep 16 '10 at 5:03
Similar to explanation by Zuur et al in Analyzing ecological data where they talk about projecting your hand on an overhead projector. You keep rotating your hand so that the projection on the wall looks pretty similar to what you think a hand should look like. – Roman Luštrik Sep 16 '10 at 9:00
– vonjd Oct 26 '10 at 6:25
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## 18 Answers
This manuscript really helped me grok PCA. I think it's still too complex for explaining to your grandmother, but it's not bad. You should skip first few bits on calculating eigens, etc. Jump down to the example in chapter 3 and look at the graphs.
I have some examples where I worked through some toy examples so I could understand PCA vs. OLS linear regression. I'll try to dig those up and post them as well.
edit: You didn't really ask about the difference between Ordinary Least Squares (OLS) and PCA but since I dug up my notes I did a blog post about it. The very short version is OLS of y ~ x minimizes error perpendicular to the independent axis like this (yellow lines are examples of two errors):
If you were to regress x ~ y (as opposed to y ~ x in the first example) it would minimize error like this:
and PCA effectively minimizes error orthogonal to the model itself, like so:
More importantly, as others have said, in a situation where you have a WHOLE BUNCH of independent variables, PCA helps you figure out which ones matter the most. The examples above just help visualize what the first principal component looks like in a really simple case.
In my blog post I have the R code for creating the above graphs and for calculating the first principal component. It might be worth playing with to build your intuition around PCA. I tend to now really own something until I write code that reproduces it.
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+1 for really caring to answer. – claws Sep 16 '10 at 18:55
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The tutorial was really great. Could you suggest any further tutorials as a follow-up? – Edward Sep 21 '10 at 9:54
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Good call on the Lindsay I Smith manuscript - just read it today; very helpful. – Stedy Oct 23 '10 at 5:58
So is PCA equivalent to Total Least Squares if it optimizes orthogonal distances from points to the fit line? – Marcin Apr 16 '11 at 14:25
@Marcin - this is correct. You can re-phrase PCA as finding the best rank $m$ estimate ($1\leq m\leq p$) of the original $p$ variables ($\hat{x}_{ij}\;\;\;\; i=1,\dots,n\;\;\;j=1,\dots,p$), with an objective function of $\sum_{i=1}^{n}\sum_{j=1}^{p}(x_{ij}-\hat{x}_{ij})^{2}$. Choosing the number of PCs is equivalent to choosing the rank of the predictions. – probabilityislogic Sep 5 '11 at 7:24
Let's do (2) first. PCA fits an ellipsoid to the data. An ellipsoid is a multidimensional generalization of distorted spherical shapes like eggs, cigars, and pancakes. These are all neatly described by the directions and lengths of their principal (semi-)axes, such as the axis of the cigar or egg or the plane of the pancake. No matter how the ellipsoid is turned, the eigenvectors point in those principal directions and the eigenvalues give you the lengths. The smallest eigenvalues correspond to the thinnest directions having the least variation, so ignoring them (which collapses them flat) loses relatively little information: that's PCA.
(1) Apart from simplification (above), we have needs for pithy description, visualization, and insight. Being able to reduce dimensions is a good thing: it makes it easier to describe the data and, if we're lucky to reduce them to three or less, lets us draw a picture. Sometimes we can even find useful ways to interpret the combinations of data represented by the coordinates in the picture, which can afford insight into the joint behavior of the variables.
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To add to this, when you have (near-)equal semiaxes (i.e. the ellipsoid has a (near-)circular slice), it indicates that the two pieces of data corresponding to those axes have (near-)dependency; one can talk about principal axes for an ellipse, but circles only have one radius. :) – J. M. Sep 16 '10 at 9:43
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I would be more cautious here, J.M. First, just to clarify, by "near-dependency" you must mean "nearly independent." This would be true for a multinormal variate, but in many cases PCA is performed with data that are markedly non-normal. Indeed, the clustering analyses that follow some PCA calculations can be viewed as one way to assess a strong form of non-normality. Mathematically, circles do have principal axes, but they are just not uniquely determined: you can choose any orthogonal pair of radii as their principal axes. – whuber♦ Sep 16 '10 at 14:11
Yes, sorry, I suppose "the principal axes of a circle are indeterminate" would have been a better way of putting it. – J. M. Sep 16 '10 at 16:13
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+1 for the geometric explanation. – vqv Dec 20 '10 at 4:49
Hmm, here goes for a completely non-mathematical take on PCA...
Imagine you have just opened a cider shop. You have 50 varieties of cider and you want to work out how to allocate them onto shelves, so that similar-tasting ciders are put on the same shelf. There are lots of different tastes and textures in cider - sweetness, tartness, bitterness, yeastiness, fruitiness, clarity, fizziness etc etc. So what you need to do to put the bottles into categories is answer two questions:
1) What qualities are most important for identifying groups of ciders? e.g. does classifying based on sweetness make it easier to cluster your ciders into similar-tasting groups than classifying based on fruitiness?
2) Can we reduce our list of variables by combining some of them? e.g. is there actually a variable that is some combination of "yeastiness and clarity and fizziness" and which makes a really good scale for classifying varieties?
This is essentially what PCA does. Principal components are variables that usefully explain variation in a data set - in this case, that usefully differentiate between groups. Each principal component is one of your original explanatory variables, or a combination of some of your original explanatory variables.
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What about the eigenvectors & eigenvalues? – Ηλίας Oct 14 '10 at 8:29
Okay: the Eigenvalue associated with each principal component tells you how much variation in the data set it explains (in my example, how clearly it separates your bottles into groups). They are usually expressed as a percentage of the total variation in the data set. As for the Eigenvectors, well, that's where as claws said I follow the output of an analysis like a machine ;) In my head, they are related to how you rotate Vince's mobile to its 'best' orientation, but this might not be the right way to think of them. – Freya Harrison Oct 27 '10 at 13:07
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Eigenvectors are just the linear combinations of the original variables (in the simple or rotated factor space); they described how variables "contribute" to each factor axis. Basically, think of PCA as as way to construct new axes that point to the directions of maximal variance (in the original variable space), as expressed by the eigenvalue, and how variables contributions are weighted or linearly transformed in this new space. – chl♦ Nov 23 '10 at 21:46
Alright, I'll give this a try. A few months back I dug through a good amount of literature to find an intuitive explanation I could explain to a non-statistician. I found the derivations that use Lagrange multipliers the most intuitive.
Let's say we have high dimension data - say 30 measurements made on an insect. The bugs have different genotypes and slightly different physical features in some of these dimensions, but with such high dimension data it's hard to tell which insects belong to which group.
PCA is a technique to reduce dimension by:
1. Taking linear combinations of the original variables.
2. Each linear combination explains the most variance in the data it can.
3. Each linear combination is uncorrelated with the others
Or, in mathematical terms:
1. For $Y_j = a_j' x$ (linear combination for jth component)
2. For $k > j$, $V(Y_k) < V(Y_j)$ (first components explain more variation)
3. $a_k' a_j = 0$ (orthogonality)
Finding linear combinations that satisfy these constraints leads us to eigenvalues. Why?
I recommend checking out the book An Introduction to Multivariate Data Analysis for the full derivation (p. 50), but the basic idea is successive optimizations problems (maximizing variance) constrained such that a'a = 1 for coefficients a (to prevent the case when variance could be infinite) and constrained to make sure the coefficients are orthogonal.
This leads to optimization with Lagrange multipliers, which in turn reveals why eigenvalues are used. I am too lazy to type it out (sorry!) but, this PDF goes through the proof pretty well from this point.
I would never try to explain this to my grandmother, but if I had to talk generally about dimension reduction techniques, I'd point to this trivial projection example (not PCA). Suppose you have a Calder mobile that is very complex. Some points in 3-d space close to each other, others aren't. If we hung this mobile from the ceiling and shined light on it from one angle, we get a projection onto a lower dimension plane (a 2-d wall). Now, if this mobile is mainly wide in one direction, but skinny in the other direction, we can rotate it to get projections that differ in usefulness. Intuitively, a skinny shape in one dimension projected on a wall is less useful - all the shadows overlap and don't give us much information. However, if we rotate it so the light shines on the wide side, we get a better picture of the reduced dimension data - points are more spread out. This is often what we want. I think my grandmother could understand that :-)
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That's very layman ;-) – mbq♦ Sep 15 '10 at 21:24
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It's a little mathy, but the best way to understand something is to derive it. – Vince Sep 16 '10 at 1:08
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You have an exceptionally well-educated grandmother :-). – whuber♦ Sep 16 '10 at 1:44
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i like the explanation with the light shining on a 3-d structure – Neil McGuigan Jun 7 '11 at 18:40
OK, a totally non-math answer:
If you have a bunch of variables on a bunch of subjects and you want to reduce it to a smaller number of variables on those same subjects, while losing as little information as possible, then PCA is one tool to do this.
It differs from factor analysis, although they often give similar results, in that FA tries to recover a small number of latent variables from a larger number of observed variables that are believed to be related to the latent variables.
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Hey Peter! Good to see you here. This is a really good, simple, no math answer. – JD Long Sep 16 '10 at 19:40
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+1 for mentioning FA, which no one else seems to discuss, and which some people's explanations seem to blend towards. – gung Jan 31 '12 at 3:52
I'd answer in "layman's terms" by saying that PCA aims to fit straight lines to the data points (everyone knows what a straight line is). We call these straight lines "principal components". There are as many principal components as there are variables. The first principal component is the best straight line you can fit to the data. The second principal component is the best straight line you can fit to the errors from the first principal component. The third principal component is the best straight line you can fit to the errors from the first and second principal components, etc., etc.
If someone asks what you mean by "best" or "errors", then this tells you they are not a "layman", so can go into a bit more technical details such as perpendicular errors, don't know where the error is in x- or y- direction, more than 2 or 3 dimensions, etc. Further if you avoid making reference to OLS regression (which the "layman" probably won't understand either) the explanation is easier.
The eigenvectors and eigenvalues are not needed concepts per se, rather they happened to be mathematical concepts that already existed. When you solve the mathematical problem of PCA, it ends up being equivalent to finding the eigenvalues and eigenvectors of the covariance matrix.
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+1, this is truly in "layman's terms", and I know you could derive it very rigorously if you wanted to! – gung Jan 31 '12 at 3:53
Trying to be non-technical... Imagine you have a multivariate data, a multidimensional cloud of points. When you compute covariance matrix of those you actually (a) center the cloud, i.e. put the origin it the multidimensional mean, the coordinate system axes now cross in the centre of the cloud, (b) encrypt the information about the shape of the cloud and how it is oriented in the space by means of variance-covariance entries. So, all important info about the data as a whole - i.e. except information of each individual data point - is stored in the covariance matrix.
Then you do eigen-decomposition of that martrix and obtain the list of eigenvalues and the corresponding number of eigenvectors. Now, the 1st principal component is the new, latent variable which can be displayed as the axis going through the origin and oriented along the direction of the maximal variance (thickness) of the cloud. The variance along this axis, i.e. the variance of the coordinates of all points on it, is the first eigenvalue, and the orientation of the axis in space referenced to the original axes (the variables) is defined by the 1st eigenvector: its entries are the cosines between it and those original axes. The aforementioned coordinates of data points on the 1st component are the 1st pr. component values, or component scores; they are computed as the product of (centered) data matrix and the eigenvector.
"After" the 1st pr. component got measured it is, to say, "removed" from the cloud with all the variance it accounted for, and the dimensionality of the cloud drops by one. Next, everything is repeated with the second eigenvalue and the second eigenvector - the 2nd pr. component is being recorded, and then "removed". Etc.
So, once again: eigenvectors are direction cosines for principal components, while eigenvalues are the magnitude (the variance) in the principal components. Sum of all eigenvalues is equal to the sum of variances which are on the diagonal of the variance-covariance matrix. If you transfer the "magnitudinal" information stored in eigenvalues over to eigenvectors to add it to the "orientational" information stored therein you get what is called principal component loadings; these loadings - because they carry both types of information - are the covariances between the original variables and the principal components.
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I might be a bad person to answer this because I'm the proverbial grandmother who has had the concept explained to me and not much more, but here goes:
Suppose you have a population. A large portion of the population is dropping dead of heart attacks. You are trying to figure out what causes the heart attacks.
You have two pieces of data: height and weight.
Now, it's clear that there's SOME relationship between weight and heart attacks, but the correlation isn't really strong. There are some heavy people who have a lot of heart attacks, but some don't.
Now, you do a PCA, and it tells you that weight divided by height ('body mass') is a much more likely predictor of heart attacks then either weight or height, because, lo and behold, the "reality" is that it's body mass that causes the heart attacks.
Essentially, you do PCA because you are measuring a bunch of things and you don't really know if those are really the principle components or if there's some deeper underlying component that you didn't measure.
[Please feel free to edit this if it's completely off base. I really don't understand the concept any more deeply than this].
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– Shane Sep 16 '10 at 2:11
From someone who has used PCA a lot (and tried to explain it to a few people as well) here's an example from my own field of neuroscience.
When we're recording from a person's scalp we do it with 64 electrodes. So, in effect we have 64 numbers in a list that represent the voltage given off by the scalp. Now since we record with microsecond precision, if we have a 1-hour experiment (often they are 4 hours) then that gives us 10e6 * 60^3 == 216,000,000,000 time points at which a voltage was recorded at each electrode so that now we have a 216,000,000,000 x 64 matrix. Since a major assumption of PCA is that your variables are correlated, it is a great technique to reduce this ridiculous amount of data to an amount that is tractable. As has been said numerous times already, the eigenvalues represent the amount of variance explained by the variables (columns). In this case an eigenvalue represents the variance in the voltage at a particular point in time contributed by a particular electrode. So now we can say, "Oh, well electrode `x` at time point `y` is what we should focus on for further analysis because that is where the most change is happening". Hope this helps. Loving those regression plots!
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I can give you my own explanation/proof of the PCA, which I think is really simple and elegant, and doesn't require anything except basic knowledge of linear algebra. It came out pretty lengthy, because I wanted to write in simple accessible language.
Suppose we have some $M$ samples of data from an $n$-dimensional space. Now we want to project this data on a few lines in the $n$-dimensional space, in a way that retains as much variance as possible (that means, the variance of the projected data should be as big compared to the variance of original data as possible).
Now, let's observe that if we translate (move) all the points by some vector $\beta$, the variance will remain the same, since moving all points by $\beta$ will move their arithmetic mean by $\beta$ as well, and variance is lineary proportional to $\sum_{i=1}^M ||x_i - \mu||_2$. Hence we translate all the points by $-\mu$, so that their arithmetic mean is $0$, for computational comfort. Let's denote the translated points as $x_i' = x_i - \mu$. Let's also observe, that the variance can be now expressed simply as $\sum_{i=1}^M ||x_i'||_2$.
Now the choice of the line. We can describe any line as set of points that satisfy the equation $x = \alpha * v + w$, for some vectors $v,w$. Note that if we move the line by some vector $\gamma$ orthogonal to $v$, then all the projections on the line will also be moved by $\gamma$, hence the mean of the projections will be moved by $\gamma$, hence the variance of the projections will remain unchanged. That means we can move the line parallel to itself, and not change the variance of projections on this line. Again for convenience purposes let's limit ourselves to only the lines passing through the zero points (this means lines described by $x = \alpha *v$).
Alright, now suppose we have a vector $v$ that describes the direction of a line that is a possible candidate for the line we search for. We need to calculate variance of the projections on the line $\alpha * v$. What we will need are projection points and their mean. From linear algebra we know that in this simple case the projection of $x_i'$ on $\alpha * v$ is $\frac{<x_i, v>}{||v||_2}$. Let's from now on limit ourselves to only unit vectors v. That means we can write the length of projection of point $x_i'$ on $v$ simply as $<x_i', v>$.
In some of the previous answers someone said that PCA minimizes the sum of squares of distances from the chosen line. We can now see it's true, because sum of squares of projections plus sum of squares of distances from the chosen line is equal to the sum of squares of distances from point $0$. By maximizing the sum of squares of projections, we minimize the sum of squares of distances and vice versa, but this was just a thoughtful digression, back to the proof now.
As for the mean of the projections, let's observe that $v$ is part of some orthogonal base of our space, and that if we project our data points on every vector of that base, their sum will cancel out (it's like that because projecting on the vectors from the base is like writing the data points in the new orthogonal base). So the sum of all the projections on the vector $v$ (let's call the sum $S_v$) and the sum of projections on other points from the base (let's call it $S_o$) is 0, because it's the mean of the data points. But $S_v$ is orthogonal to $S_o$! That means $S_o = S_v = 0$.
So the mean of our projections is $0$? Well, that's convenient, because that means the variance is just the sum of squares of lengths of projections, or in symbols $\sum_{i=1}^M (x_i' \cdot v)^2 = \sum_{i=1}^M v^T \cdot x_i'^T \cdot x_i' \cdot v = v^T \cdot (\sum_{i=1}^M x_i'^T \cdot x_i) \cdot v$.
Well well, suddenly the covariance matrix popped out. Let's denote it simply by $X$. It means we are now looking for a unit vector $v$, that maximizes $v^T \cdot X \cdot v$, for a semi-positive definite matrix $X$.
Now, let's take the eigenvectors and eigenvalues of matrix $X$, and denote them by $e_1, e_2, \dots , e_n$ and $\lambda_1 , \dots, \lambda_n$ respectively, such that $\lambda_1 \geq \lambda_2 , \geq \lambda_3 \dots$. If the values $\lambda$ do not duplicate, eigenvectors form an orthonormal base. If they do, we choose the eigenvectors in a way that they form an orthonormal base.
Now let's calculate $v^T \cdot X \cdot v$ for eigenvector $e_i$. We have $e_i^T \cdot X \cdot e_i = e_i^T \cdot (\lambda_i * e_i) = \lambda_i (||e_i||_2)^2 = \lambda_i$.
Pretty good, this gives us $\lambda_1$ for $e_1$. Now let's take an arbitrary vector $v$. Since eigenvectors form an orthonormal base, we can write $v = \sum_{i=1}^n e_i * <v, e_i>$, and we have $\sum_{i=1}^n <v, e_i>^2 = 1$. Let's denote $\beta_i = <v, e_i>$.
Now let's count $v^T \cdot X \cdot v$. We rewrite $v$ as a linear combination of $e_i$, and get : $(\sum_{i=1}^n \beta_i * e_i)^T \cdot X \cdot (\sum_{i=1}^n \beta_i *e_i) = (\sum_{i=1}^n \beta_i * e_i) \cdot (\sum_{i=1}^n \lambda_i * \beta_i * e_i) = \sum_{i=1}^n \lambda_i *(\beta_i)^2 * (||e_i||_2)^2$.
The last equation comes from the fact thet eigenvectors where chosen to be pairwaise orthogonal, so their dot product's are zero. Now, because all eigenvectors are also unit, we can write $v^T \cdot X \cdot v = \sum_{i=1}^n \lambda_i * \beta_i^2$, where $\beta_i ^2$ are all positive, and sum to $1$.
That means, that the variance of the projections is a weighted mean of eigenvalues. Certainly, it is always less then the biggest eigenvalue, which is why it should be our choice of the first vector.
Now suppose we want another vector. We should chose it from a space orthogonal to the already chosen one, that means the subspace $lin(e_2, e_3, \dots , e_n)$. By analogical inference we arrive at the conclusion, that the best vector to project on is $e_2$. And so on, and so on...
Btw. it should be now clear, why the variance retained can be expresed by $\frac{\sum_{i=1}^k \lambda_i}{\sum_{i=1}^n \lambda_i}$.
Hope this helps and I didn't make any big mistakes.
We should also justify the greedy choice of vectors. When we want too choose $k$ vectors to project onto, it might not be the best idea to first choose the best vector, then the best from what reamains and so on. I'd like to argue that in this case it is justified and makes no difference. Lets denote the $k$ vector we wish to project onto by $v_1, \dots , v_k$. Also, let's assume the vectors are pairwaise orthogonal. As we already know, the total variance of the projections on those vectors can be expressed by : $\sum_{j=1}^k \sum_{i=1}^n \lambda_i * \beta_{ij}^2 = \sum_{i=1}^n \lambda_i * \gamma_i$ where $\gamma_i = \sum_{j=1}^k \beta_{ij}^2$.
Now, let's write e_i in some orthonormal base that includes $v_1, \dots , v_k$. Let's denote the rest of the base as $u_1, \dots, u_{n-k}$. We can see that $e_i = \sum_{j=1}^k \beta_{ij} v_j + \sum_{j=1}^{n-k} \theta_j * <e_i, u_j>$. Because $||e_i||_2 = 1$, we have $\sum_{j=1}^k \beta_{ij}^2 + \sum_{j=1}^{n-k} \theta_j^2 = 1$, and hence $\gamma_i \leq 1$ for all $i$.
Now we have a similar case to one vector only, we now know that the total variance of projections is $\sum_{i=1}^n \lambda_i * \gamma_i$ with $\gamma_i \leq 1$ and $\sum_{i=1}^n \gamma_i = k$. This is yet another weighted mean, and is certainly no more than $\sum_{i=1}^k \lambda_i$ which corresponds to projecting on $k$ eigenvectors corresponding to biggest eigenvalues.
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yes very simple:) – berkay Dec 15 '12 at 23:34
awesome and elegant – Vass Mar 11 at 12:07
Eigens are mathematical concept used for implementing PCA, so they are out of discussion. PCA is based on the idea that the multivariate data is so hard to interpret not because the reality is complex, but because we have measured wrong variables, and the reality is trivial; namely there are only linear relations involved.
To this end we use some math-magick to expose this structure, and voilà.
This is pure wishful thinking, but it often works because of the Taylor expansion.
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I view PCA as a geometric tool. If you are given a bunch of points in 3-space which are pretty much all on a straight line, and you want to figure out the equation of that line, you get it via PCA (take the first component). If you have a bunch of points in 3-space which are mostly planar, and want to discover the equation of that plane, do it via PCA (take the least significant component vector and that should be normal to the plane).
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Some time back I tried to understand this PCA algorithm and I wanted to make a note about eigen vectors and eigen values. That document stated that the purpose of EVs is to convert a model of the large sized model to a very small sized model.
For example, instead of constructing first the full sized bridge and then carrying out experiments and tests on it, it is possible to use EVs to create a very small sized bridge where all the factors/quantities will be reduced by the same margin and moreover the actual result of tests and stress related tests carried out on it can be calculated and enlarged appropriately as needed for the original model. In a way EVs help to create abstracts of the original.
To me, this explaination had profound meaning to what I was trying to do! Hope it helps you too!
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The way I understand principle components is this: Data with multiple variables (height, weight, age, temperature, wavelength, percent survival, etc) can be presented in three dimensions to plot relatedness.
Now if you wanted to somehow make sense of "3D data", you might want to know which 2D planes (cross-sections) of this 3D data contain the most information for a given suite of variables. These 2D planes are the principle components, which contain a proportion of each variable.
Think of principal components as variables themselves, with composite characteristics from the original variables (this new variable could be described as being part weight, part height, part age, etc). When you plot one principle component (X) against another (Y), what you're doing is building a 2D map that can geometrically describe correlations between original variables. Now the useful part: since each subject (observation) being compared is associated with values for each variable, the subjects (observations) are also found somewhere on this X Y map. Their location is based on the relative contributions of each underlying variable (i.e. one observation may be heavily affected by age and temperature, while another one may be more affected by height and weight). This map graphically shows us the similarities and differences between subjects and explains these similarities/differences in terms of which variables are characterizing them the most.
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Although there are myriad examples given to provide an intuitive understanding of PCA, that fact can almost make it more difficult to grasp at the outset, at least it was for me.
"What was the one thing about PCA that all these different examples from different disciplines have in common??"
What helped me intuitively understand were a couple of math parallels I found, since it's apparent the maths is the easy part for you, although this doesn't help explain it to your grandmother...
Think of a regularization problem, trying to get
$$|| XB - Y || = 0$$
Or in English, break down your data $Y$ into two other matrices which will somehow shed light on the data? If those two matrices work well, then the error between them and $Y$ shouldn't be too much.
PCA gives you a useful factorizaton of $Y$, for all the reasons other people have said. It breaks the matrix of data you have, $Y$, down into two other useful matrices. In this case, $X$ would be a matrix where the columns are first $k$ PCs you kept, and $B$ is a matrix giving you a recipe to reconstruct the columns of matrix $Y$ using the columns of $X$. $B$ is the first $k$ rows of $S$, and all of $V$ transpose. The eigenvalues on the diagonal of $S$ basically weights which PCs are most important. That is how the math explicitly tells you which PCs are the most important: they are each weighted by their eigenvalues. Then, the matrix $V^\mathrm{T}$ tells the PCs how to combine.
I think people gave many intuitive examples, so I just wanted to share that. Seeing that helped me understand how it works. There are a world of interesting algorithms and methods which do similar things as PCA. Sparse coding is a subfield of machine learning which is all about factoring matrix $A$ into two other useful and interesting ones that reflect patterns in $A$.
Anyhow have fun!
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Basically PCA finds new variables which are linear combinations of the original variables such that in the new space, the data has fewer dimensions. Think of a data set consisting of the points in 3 dimensions on the surface of a flat plate held up at an angle. In the original x, y, z axes you need 3 dimensions to represent the data, but with the right linear transformation, you only need 2.
Basically what @Joel said, but only linear combinations of the input variables.
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Why so eigenvalues/eigenvectors ?
When doing PCA, you want to compute some orthogonal basis by maximizing the projected variance on each basis vector.
Having computed previous basis vectors, you want the next one to be:
• orthogonal to the previous
• norm 1
• maximizing projected variance, i.e with maximal covariance norm
This is a constrained optimization problem, and the Lagrange multipliers (here's for the geometric intuition, see wikipedia page) tell you that the gradients of the objective (projected variance) and the constraint (unit norm) should be "parallel" at the optimium.
This is the same as saying that the next basis vector should be an eigenvector of the covariance matrix. The best choice at each step is to pick the one with the largest eigenvalue among the remaining ones.
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2
Definitely not an explanation to a layman - orthogonal basis vectors? maximising projection variance? constrained optimisation problem? Lagrange multiplier? These are highly "jargonised" terms. Show a layman who understands what these mean and i'll show you a mathematician/statistician – probabilityislogic Sep 4 '11 at 22:55
Here is a math answer: the first principal component is the longest dimension of the data. Look at it and ask: where is the data widest? That's the first component. The next component is the perpendicular. So a cigar of data has a length and a width. It makes sense for anything that is sort of oblong.
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3
Unfortunately, the correctness of this answer depends on how the vague expression "longest" is interpreted. Many natural and relevant interpretations, such as the diameter, would be wrong. – whuber♦ Mar 20 at 21:56
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http://quant.stackexchange.com/questions/7202/what-do-eigenvalues-eigenvectors-of-the-yield-forward-rates-covariance-matrices/7204
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# What do eigenvalues/eigenvectors of the yield/forward rates covariance matrices mean?
I have 5 bonds (with maturities 1,2,3,4,5 years) which I calculated the yield curve for 10 days. I also calculated the forward rates from the yield rates. Now I've been told to calculate the covariance of the yield rates and the forward rates, as well as their eigenvalues/eigenvectors.
I'm assuming the covariance of the yield rates tell me how the bonds with different maturities move along with each other over time? But I'm not sure what the forward rates tell me.
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## 1 Answer
The PCA analysis does not really tell you what the bonds do but it tells you how the rates move together. The variations of $n$ rates (i.e. 1 y, 2y, ...) are split up in (at first) abstract factors like $$\Delta R_i = \sum_{j=1}^n e_{i,j} f_j$$ where $\Delta R_i$ is the change in the rate $i$ and $f_j$ is factor $j$ and $e_{i,j}$ is the (factor loading=) influence of factor $j$ to the rate $i$. The factors coming from PCA are uncorrelated and ordered by the size of their variance (largest first). Then it turns out that usually all rates have $e_{i,1}$, the influence of the first factor, with the same sign. This means that a change in this factor is in the same direction for all rates. The $e_{i,2}$ have a different sign for short terms as opposed to longer terms. Thus the second factor influences short and long rates differently - this is interpreted as steepening/flattening factor. For the third one often sees a curvature pattern (same sign for short and long and another sign for the middle terms).
Mathematical detail: the factor loadings are the eigenvectors of the covariance matrix of $\Delta R_i,i=1,\ldots,n$ and the variances of the factors are the squared eigenvalues.
Looking at the total variance explained by these it often turns out that $n$ rates can be described by the loadings to these 3 factors and the variances of these factors.
You can do this with spot rates and with forward rates. It would be interesting how a PCA of spot and forward rates together looks.
Note that such a reduction of dimensionality is an approximation and as always - take care doing it.
One of the first Google hits points to an article with more mathematical details:PRINCIPAL COMPONENT ANALYSIS by GRAEME WEST.
Problems of the interpretation are described here inPotential PCA interpretation problems for volatility smile dynamics by Reiswich and Tompkins.
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– Richard Feb 4 at 8:15
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http://mathoverflow.net/questions/82492?sort=newest
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## Nontrivial algorithm to check for polynomial symmetry?
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Hi.
As is known, a polynomial `$P \in K[x_1, \dots, x_n]$` is symmetric when permuting its variables always yields the same polynomial. This immediately yields an algorithm `$O(n!)$` to check for symmetry of a polynomial.
Are there known algorithms faster than `$O(n!)$` (perhaps using other bounds, like the degree) to decide if a polynomial of `$n$` variables is symmetric?
Thanks!
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The algorithm is $O(n!)$ if you can check whether two polynomials are identical in one step, but in general this is not trivial to do for large polynomials. – Qiaochu Yuan Dec 2 2011 at 18:00
@Qiaochu -- if you allow a one-sided error, then you can check if polynomials are equal by evaluating them at random points. This gives a lot more flexibility and speed than expanding the polynomials and checking them term-by-term. – David Harris Dec 2 2011 at 18:11
Surely you cannot estimate the running time of the algorithm independently of the size of the polynomial. Also the way it is represented is important. (As extreme cases, if it is given as a polynomial in the elementary symmetric polynomials, the check is trivial; if it is given as a black-box polynomial function, the check is impossible.) Very often the size of an expanded symmetric polynomial will dwarf the number $n!$. Please be more specific. – Marc van Leeuwen Dec 3 2011 at 14:01
## 2 Answers
One needs only check $n$ transpositions, since if each of the transpositions $(12),(23), \dots$ preserves a polynomial, then every permutation preserves that polynomial.
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9
Even less, $S_n$ is generated by two elements, $(12\cdots n)$ and $(12)$. – Ryan Budney Dec 2 2011 at 17:52
That's a huge speedup, thanks both of you :) – Federico Lebrón Dec 2 2011 at 18:01
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Will Sawin's answer necessitates $n-1$ passes over the polynomial $P$, checking for equality. Ryan's comment to Will's answer brings that down to $2$ passes, but with (somewhat) expensive operations to be done.
You can do it in a single pass with only cheap operations, assuming your polynomial $P$ is given in expanded form in the monomial basis. First, you know that, if it is symmetric, it can be rewritten as a polynomial in the symmetric polynomials. So, march through all the coefficients of $P$, figuring out the 'signature' of each monomial (i.e. set of degrees) you encounter; then make sure that the coefficient for each signature is constant, and that you encounter enough such monomials for each degree. This a linear pass on $P$, and storage $O(m)$ where $m$ is the number of different symmetric polynomials which actually occur in $P$.
Of course, if your polynomial is not presented in expanded form in the monomial basis, the above will not work.
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And if it IS presented in expanded form, it is going to be gigantic. The question is of most interest if the polynomial is sparse... – Igor Rivin Dec 2 2011 at 21:02
@Igor: my 'algorithm' will work especially well in the sparse case. Of course, if it is sparse in a non-monomial basis, or is represented in some other way (say via a straight line program), that would indeed be a problem. But we need the OP to clarify that. – Jacques Carette Dec 2 2011 at 21:40
@Jacques: yes, you are right. I was thinking of polynomials represented as black boxes or straight line programs, or?! – Igor Rivin Dec 3 2011 at 10:34
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http://mathhelpforum.com/calculus/156316-solving-delta-epsilon-proof-given-epsilon-print.html
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# Solving Delta Epsilon Proof (Given Epsilon)
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• September 15th 2010, 03:29 PM
penguinpwn
1 Attachment(s)
Solving Delta Epsilon Proof (Given Epsilon)
I read over the guide stickied at the top of the forum, but I did not understand the section on solving proofs when given a particular epsilon value. Could someone walk me through it with the attached problem?
• September 15th 2010, 03:42 PM
yeKciM
Quote:
Originally Posted by penguinpwn
I read over the guide stickied at the top of the forum, but I did not understand the section on solving proofs when given a particular epsilon value. Could someone walk me through it with the attached problem?
perhaps it's better to show you what is that $\epsilon$ region ...
you will (if not yet) have sequence which $a_n = (-1)^n$ (which don't converge) . why is that. well if you assume different , that for some $a \in \mathbb{R}$ there is $\lim_{n \to \infty } x_n = a$ now you realize that all members of that sequence are going to either one or negative one, that means that in any $\epsilon$ region around point "a" there should be both members of that sequence. that can't be true because if you chose any $\epsilon < \frac {1}{2}$ there is no chance that both that numbers be in some region $(a-\epsilon , a+ \epsilon)$ which length is less than one :D
All times are GMT -8. The time now is 05:17 PM.
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http://math.stackexchange.com/questions/138502/what-is-the-form-of-fx/138531
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# what is the form of $f(x)$
Given that area of OPB and OPA are same, could any one help me to find the the $f(x)$
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It depends whether you are supposed to use Fundamental Theorem of Calculus. Easiest to guess answer will be $kx^2$, calculate, get $k=4/3$. – André Nicolas Apr 29 '12 at 18:12
## 2 Answers
I will think of the dashed line as not being a boundary, more like an awkward hint, maybe to break up the integral, or maybe to integrate with respect to $y$. We integrate with respect to $y$. But breaking up is better, no fractional exponents.
It is easy to show that the area of $OPB$ is $(1/3)t^3$. Let us guess that the answer is $y=2k^2x^2$. Then $x=\frac{1}{k\sqrt{2}}y^{1/2}$.
The area of $OPA$ is $$\int_0^{2t^2} \left(\frac{1}{\sqrt{2}}y^{1/2}-\frac{1}{k\sqrt{2}}y^{1/2}\right)dy.$$ Calculate. We get $(1-1/k)(4/3)t^3$. This should be $(1/3)t^3$, so $1-1/k=1/4$, $k=4/3$, and therefore the equation is $y=(32/9)x^2$.
Now we can appeal to the geometrically obvious uniqueness. Or else we can let the inverse function of $f$ be $g$, set up the integral, and differentiate under the integral sign (Fundamental Theorem of Calculus). Same result.
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The area of OPB:
$$\int_0^t (2x^2 - x^2) \ dx = \int_0^t x^2 \ dx = \left[\frac{1}{3} x^3\right]_0^t = \frac{1}{3} t^3.$$
The area of OPA:
$$\int_0^s (f(x) - 2x^2) \ dx + \int_s^t (2t^2 - 2x^2) \ dx = \int_0^s f(x) \ dx + 2 (t - s) t^2 - \int_0^t 2x^2 \ dx \\ = \int_0^s f(x) \ dx + 2 t^3 - 2 t^2 s - \frac{1}{3} t^3 = \int_0^s f(x) \ dx + \frac{4}{3} t^3 - 2 t^2 s.$$
Since they should be equal, we get the condition
$$\int_0^s f(x) \ dx + t^3 - 2 t^2 s = 0 \quad \Longleftrightarrow \quad \int_0^s f(x) \ dx = t^2(2s - t).$$
Furthermore we have the conditions $f(0) = 0$, $f(s) = 2t^2$ and $f$ is increasing at least on $(0, s)$. Several functions will satisfy all these conditions.
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If we assume that $f$ is of the form $f(x) = ax^2$ for some $a$, then for fixed $t$ we need $s = \frac{3}{4}t$ to find a single solution $f(x) = \frac{32}{9} x^2$. For other pairs $(s,t)$ there is no solution. – TMM Apr 29 '12 at 18:30
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http://mathoverflow.net/questions/112258/express-weierstrass-g-2-and-g-3-in-terms-of-theta-functions-of-the-periods/112282
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## Express Weierstrass' g_2 and g_3 in terms of theta-functions of the periods
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If E is a complex elliptic curve defined as the quotient of C over a lattice generated by w_1 and w_2, then it can be also written in Weierstrass form y^2=4*x^3-g_2*x-g_3. The coefficients g_2 and g_3 can be computed as well-known Eisenstein sums, however, there is a better expression in terms of Jacobi theta-functions of w_1 and w_2 (and as a consequence, an expression for j-invariant via w_2/w_1). Unfortunately, I could not find that expression in the books on my bookshelf, like Koblitz, Milne, Silverman etc, and the only place on the Internet where I found such formulas, were two wiki-articles: http://en.wikipedia.org/wiki/Weierstrass's_elliptic_functions and http://en.wikipedia.org/wiki/Theta_function. These formulas, however, contradict each other, and neither of them, when run on computer, gives rational values for elliptic curves with complex multiplication and class number 1. Can anybody give a reliable link to correct formulas?
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## 3 Answers
I recommend Lester R. Ford's classic book "Automorphic Functions" for this. It gives very explicit formulae for all these quantities, in particular, the $J$-invariant, and it is written with great care in a very readable style.
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Thanks a lot. I'll try to find that book. If all else fails, I'll look nt up on Amazon. – potap Nov 13 at 13:39
I found a pdf file on the Internet. The book is VERY good, but it does not give any formulas for $j$ in terms of theta-functions of $\tau$. – potap Nov 13 at 14:35
Oh, sorry. I was working from memory because I don't have a copy of Ford here at home. I thought it was there. Did you try Herb Clemens' book A Scrapbook of Complex Curve Theory? Perhaps the formulae you want might be in there. If I didn't see it in Ford, then it's possible that I saw it in Clemens. – Robert Bryant Nov 13 at 15:32
@potap: I finally had a chance to look in Clemens' book, and it turns out that the formulae that you want (for the $j$-function in terms of theta functions) are in Chapter III. I'm sorry for mis-directing you to Ford. – Robert Bryant Nov 13 at 23:58
@Robert: Shame on me! I turned back on my chair and immediately found a paperback copy of that book in Russian translation. How come I did not think about it? (Imagine that I know Herb personally and I have even participated in a party at his home...) Thanks a lot! – potap Nov 14 at 4:45
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A good modern source in English is N. Akhiezer, Elements of the theory of elliptic functions. The formulas you are asking are in section 21, chapter IV.
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Thanks a lot. I managed to find a dejaview file of that book in Russian. This is exactly what I needed! The best. – potap Nov 13 at 14:53
The formulas at the Wikipedia article on the J-invariant seem to be correct. At least, the formula $g_2 = \frac{2\pi^4}{3}(\theta_{00}^8 + \theta_{01}^8 + \theta_{10}^8)$ yields $g_2 = 60G_4 = \frac{4\pi^4}{3}E_4$, as one would expect from the usual coordinate representation of the $E_8$ lattice.
I don't know what you mean when you say that the expression of $j$ in terms of Jacobi theta functions is "better". Each expression has its advantages.
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Thank you. But I think that $g_2$ cannot be expressed in terms of $\tau$ alone: it depends on both periods $\omega_1$ and $\omega_2$ and can change under homotheties! Do you mean that $\omega_1=1$ and $\omega_2=\tau$? By "better" I mean computational complexity, as I'm doing computer experiments. – potap Nov 13 at 10:06
The standard convention is $\omega_1 = 1$ and $\omega_2 = \tau$. You can adjust for arbitrary pairs of periods using the fact that weight $k$ forms are homogeneous of degree $-k$ under homothety. – S. Carnahan♦ Nov 13 at 13:43
Yes I know. However, you can take Weierstrass's $P$ and $P'$ for any lattice and write the relation between them which will depend both on $\omega_1$ and $\omega_2$ (the equivalence class of the curve will be defined by the ratio alone, of course). – potap Nov 13 at 14:13
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http://mathoverflow.net/questions/106522?sort=newest
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## Why can’t an explicit well-ordering of the reals be ruled out in ZF?
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The statement A = "There exists a well-ordering of the reals" is independent of ZF. My understanding is that the statement B = "There exists an explicit well-ordering of the reals" is also independent of ZF, yet this seems counterinutitive to me, because of the following line of reasoning: If B were true, then A would be provable in ZF, which is impossible since A is independent of ZF, so B must be false. This line of reasoning seems perfectly clear to me, and I see no reason why it cannot be carried out in ZF itself. But if the line of reasoning could be carried out in ZF, then that would mean that ZF implies that not B, contradicting the claim that B is independent of ZF.
So can anyone clarify how exactly ZF+B is consistent?
Any help would be greatly appreciated.
Thank You in Advance.
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1
See related question on definable well-orderings of the reals mathoverflow.net/questions/6593/… – Joel David Hamkins Sep 6 at 17:19
## 4 Answers
I suspect that the root of your confusion is an ambiguity in the word "explicit". Your statement B could mean, as Gerald Edgar suggests in a comment and as others seem to have assumed, "there exists a definable relation that well-orders the reals." This is consistent and does not imply the ZF-provability of statement A. But B could also mean "There is a definable relation that can be proved, in ZF, to well-order the reals", so that "explicit" involves both definability and provability. This version of B (when sufficiently precisely formulated) is not consistent with ZF. I think you had the latter version of B in mind when you used B to infer provability of A.
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I think your second formulation actually is consistent with ZF, since it is consistent with ZF that $\neg\text{Con}(\text{ZF})$, and in such a model, proofs from ZF are easy to come by. That is, if there are any models of ZF at all, then there are models of ZF that have a definable well-ordering of the reals such that, in that model, there is a proof from ZF that this definition does define a well-ordering of the reals. – Joel David Hamkins Sep 6 at 20:27
Joel, you're right. In fact, it seems that the second formulation of B is actually equivalent (in PA or appropriate weaker theories) to $\neg$Con(ZF). – Andreas Blass Sep 6 at 21:42
### 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.
From Gödel's proof of the consistency of AC: There is an explicit subset $A \subseteq \mathbb R$ and an explicit well-ordering defined on $A$. It is consistent with ZF (and even with ZFC) that $A = \mathbb R$. But of course it is not provable in ZFC that $A = \mathbb R$.
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Just to clarify, you are using "explicit" to mean "definable", right? – Trevor Wilson Sep 6 at 16:50
It sounds like you're taking the word "explicit" (which has no precise mathematical meaning) to mean "can be proved to exist by ZF". This is problematic because a theory (e.g., ZF) proves statements. It does not construct objects, or prove that a particular object exists, although it may prove statements asserting the existence of objects with certain properties.
For example, although every model of "ZF + $V=L$" has a definable wellordering of its reals, this does not help us construct wellorderings of the reals in models of "ZF + $V \ne L$". The most serious obstacle in this case is that the two models we are considering could have different sets of reals. So although ZF does prove that $L$ satisfies "there is a definable wellordering of the reals," this only gives us a definable wellordering of the reals of $L$. It is consistent that $L$ does not contain all the reals, and even that it contains only countably many reals.
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Here is the error in your reasoning "If B were true, then A would be provable in ZF, which is impossible since A is independent of ZF, so B must be false." The statement "B is independent of ZF" means that there are models of ZF where it is true (V=L) and models where it is false(almost everything else). You are right in saying that in a model where B is true, then A is true but that does not mean that A is provable from ZF. There are also models where A is true and B is false (i.e. if we assume choice that will give us a well ordering, but that does not mean that it is definable.)
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http://math.stackexchange.com/questions/37601/vector-calculus-integral-over-vector
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# vector calculus - Integral over vector
Our physics prof wrote the following equation:
$\int\frac{\vec{r}}{r^3}d\vec{r} = \int\frac{1}{r^2}dr$
This is logical as long as I argue that $\vec{r}$ and $d\vec{r}$ are parallel, which is why the dot product evaluates as $|\vec{r}||d\vec{r}| = r dr$ However then i tried to do it by hand:
$\vec{r}d\vec{r} = \left(\begin{array}{c}x\\y\\z\\ \end{array}\right)\left(\begin{array}{c}dx\\dy\\dz\\ \end{array}\right) = xdx + ydy + zdz$
but this is nowhere near
$rdr = \sqrt{x^2+y^2+z^2}\sqrt{dx^2 + dy^2 + dz^2}$
which is why I would like to ask you what i am doing wrong.
Thanks in advance
ftiaronsem
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Your $d\vec{r}$ is not necessarily radial (and thereby parallel to $\vec{r}$). – Fabian May 7 '11 at 12:09
hmm, why is that? I thought that this would always be true in clyndrical/spherical coordinates??? – ftiaronsem May 7 '11 at 13:13
but your $\displaystyle{d\vec{r} =\left(\begin{array}{c}dx\\dy\\dz\\ \end{array}\right) }$ is neither in cylindrical nor in spherical coordinates. – Fabian May 7 '11 at 13:25
ahh, yeah, ok convinced ^^. But even in that case I thought it would be true. Can you give a counter example? – ftiaronsem May 7 '11 at 14:58
The line integral in general depends on the whole line along which you integrate it (and not only on the endpoints). In general, $d\vec{r}$ points along the line and is not radially. – Fabian May 7 '11 at 15:18
show 1 more comment
## 1 Answer
Maybe the following helps (it seems to be sloppy notation of your prof):
The integral $$\int_{\mathbf{r}_a}^{\mathbf{r}_b}\frac{\vec{r} \cdot d\vec{r}}{r^3}$$ is a line-integral. The vector field $$\vec{E}(\vec{r})=\frac{\vec{r}}{r^3}$$ is the gradient of a scalar field $\phi(\vec{r}) = -r^{-1}$, i.e., $$\vec{E}(\vec{r}) = \vec{\nabla} \phi(\vec{r}).$$ Therefore, the line-integral is path independent and the result is given by $$\int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{E}(\vec{r}) \cdot d\vec{r} = \int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{\nabla} \phi(\vec{r}) \cdot d\vec{r} = \phi(\vec{r}_b) - \phi(\vec{r}_a) = r_a^{-1} - r_b^{-1},$$ which coincides with the result of your prof.
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thanks, for your answer, this is a nice way of solving the integral. I know nearly nothing about vector calculus so could you please explain why $\int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{\nabla} \phi(\vec{r}) \cdot d\vec{r} = \phi(\vec{r}_b) - \phi(\vec{r}_a)$? I tried to do it, which resulted at integrating $\int_{\mathbf{r}_a}^{\mathbf{r}_b}3d\phi$ after evaluating the dot product. – ftiaronsem May 7 '11 at 13:16
– Fabian May 7 '11 at 13:28
ahh, I see. Thanks a lot. Originally I planned on rewriting the integral and solving it afterwards. I understand your way of solving the problem now, but can you help spot the mistake in my consideration? Would gladly accept your answer then ;-) – ftiaronsem May 7 '11 at 15:50
– Fabian May 7 '11 at 15:56
Ok, thanks for this answer. I see now how it is supposed to be done. You said that there are some formulas that don't make a lot of sense in my post. Could you point out where? I would really appreciate to know where i am making a mistake in the above considerations. – ftiaronsem May 8 '11 at 9:01
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http://mathoverflow.net/revisions/39413/list
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## Return to Question
2 added 193 characters in body; edited title; added 1 characters in body
# Intuitive Proof of Cramer's Decomposition Theorem
Cramer's Theorem decomposition theorem states that if $X$ and $Y$ are independent real random variables and $X+Y$ has normal distribution, then both $X$ and $Y$ are normally distributed. I've seen a few proofs of this result which involve grunt work in the realm of complex analysis. I was wondering if there is an intuitive proof of this result. I would like to see a proof that exploits the unique nature of the normal distribution. For example, can one derive the result from the fact that on the real line, the normal distribution maximizes entropy for a given mean and standard deviation?
Also
Edit: The complex analysis proof I'm thinking of uses the fact that if $E[\exp(\alpha X^2)]<\infty$ for some $\alpha>0$ and the analytic continuation of the characteristic function of $X$ is nonzero, does Cramer's theorem extend to other distributions?then $X$ is normally distributed.
1
# Intuitive Proof of Cramer's Theorem
Cramer's Theorem states that if $X$ and $Y$ are independent real random variables and $X+Y$ has normal distribution, then both $X$ and $Y$ are normally distributed. I've seen a few proofs of this result which involve grunt work in the realm of complex analysis. I was wondering if there is an intuitive proof of this result. I would like to see a proof that exploits the unique nature of the normal distribution. For example, can one derive the result from the fact that on the real line, the normal distribution maximizes entropy for a given mean and standard deviation?
Also, does Cramer's theorem extend to other distributions?
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http://www.openwetware.org/index.php?title=User:Timothee_Flutre/Notebook/Postdoc/2012/01/02&oldid=607238
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# User:Timothee Flutre/Notebook/Postdoc/2012/01/02
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## Learn about the multivariate Normal and matrix calculus
(Caution, this is my own quick-and-dirty tutorial, see the references at the end for presentations by professional statisticians.)
• Motivation: when we measure items, we often have to measure several properties for each item. For instance, we extract cells from each individual in our study, and we measure the expression level of all genes. We hence have, for each individual, a vector of measurements (one per gene), which leads us to the world of multivariate statistics.
• Data: we have N observations, noted X1,X2,...,XN, each being of dimension P. This means that each Xn is a vector belonging to $\mathbb{R}^P$. Traditionally, we record all the data into an $N \times P$ matrix named X, with samples (observations) in rows and variables (dimensions) in columns. Also, it is usual to assume vectors are in column by default. Therefore, we can write $X = (X_1^T, ..., X_N^T)$.
• Descriptive statistics: the sample mean is the P-dimensional vector $\bar{X} = (\bar{X}_1, ..., \bar{X}_P)$ with $\bar{X}_p = X_{\bullet p} = \frac{1}{N} \sum_{n=1}^N X_{np}$. The sample covariance is the $P \times P$ matrix noted S2 with $S^2_{pq} = \frac{1}{N} \sum_{n=1}^N (X_{np} - \bar{X}_p)(X_{nq} - \bar{X}_q)$.
• Model: we suppose that the Xn are independent and identically distributed according to a multivariate Normal distribution NP(μ,Σ). μ is the P-dimensional mean vector, and Σ the $P \times P$ covariance matrix. If Σ is positive definite (which we will assume), the density function for a given observation is: $f(X_n|\mu,\Sigma) = (2 \pi)^{-P/2} |\Sigma|^{-1/2} exp(-\frac{1}{2} (X_n-\mu)^T \Sigma^{-1} (X_n-\mu))$, with | M | denoting the determinant of a matrix M and MT its transpose.
• Likelihood: as usual, we will start by writing down the likelihood of the data, the parameters being θ = (μ,Σ):
L(θ) = f(X | θ)
As the observations are independent:
$L(\theta) = \prod_{i=1}^N f(x_i | \theta)$
It is easier to work with the log-likelihood:
$l(\theta) = ln(L(\theta)) = \sum_{i=1}^N ln( f(x_i | \theta) )$
$l(\theta) = -\frac{NP}{2} ln(2\pi) - \frac{N}{2}ln(|\Sigma|) - \frac{1}{2} \sum_{i=1}^N (x_i-\mu)^T \Sigma^{-1} (x_i-\mu)$
• ML estimation: as usual, to find the maximum-likelihood estimates of the parameters, we need to derive the log-likelihood with respect to each parameter, and then take the values of the parameters at which the log-likelihood is zero. However, in the case of multivariate distributions, this requires knowing a bit of matrix calculus, which is not always straightforward...
• Matrix calculus: some useful formulas
d(f(u)) = f'(u)du, eg. useful here: d(ln( | Σ | )) = | Σ | − 1d( | Σ | )
d( | U | ) = | U | tr(U − 1dU)
d(U − 1) = − U − 1(dU)U − 1
• Technical details: from Magnus and Neudecker (third edition, Part Six, Chapter 15, Section 3, p.353). First, they re-write the log-likelihood, noting that (xi − μ)TΣ − 1(xi − μ) is a scalar, ie. a 1x1 matrix, and is therefore equal to its trace:
$\sum_{i=1}^N (x_i-\mu)^T \Sigma^{-1} (x_i-\mu) = \sum_{i=1}^N tr( (x_i-\mu)^T \Sigma^{-1} (x_i-\mu) )$
As the trace is invariant under cyclic permutations (tr(ABC) = tr(BCA) = tr(CAB)):
$\sum_{i=1}^N (x_i-\mu)^T \Sigma^{-1} (x_i-\mu) = \sum_{i=1}^N tr( \Sigma^{-1} (x_i-\mu) (x_i-\mu)^T )$
The trace is also a linear map (tr(A + B) = tr(A) + tr(B)):
$\sum_{i=1}^N (x_i-\mu)^T \Sigma^{-1} (x_i-\mu) = tr( \sum_{i=1}^N \Sigma^{-1} (x_i-\mu) (x_i-\mu)^T )$
And finally:
$\sum_{i=1}^N (x_i-\mu)^T \Sigma^{-1} (x_i-\mu) = tr( \Sigma^{-1} \sum_{i=1}^N (x_i-\mu) (x_i-\mu)^T )$
As a result:
$l(\theta) = -\frac{NP}{2} ln(2\pi) - \frac{N}{2}ln(|\Sigma|) - \frac{1}{2} tr(\Sigma^{-1} Z)$ with $Z=\sum_{i=1}^N(x_i-\mu)(x_i-\mu)^T$
We can now write the first differential of the log-likelihood:
$d l(\theta) = - \frac{N}{2} d(ln(|\Sigma|)) - \frac{1}{2} d(tr(\Sigma^{-1} Z))$
$d l(\theta) = - \frac{N}{2} |\Sigma|^{-1} d(|\Sigma|) - \frac{1}{2} tr(d(\Sigma^{-1}Z))$
$d l(\theta) = - \frac{N}{2} tr(\Sigma^{-1} d\Sigma) - \frac{1}{2} tr(d(\Sigma^{-1})Z) - \frac{1}{2} tr(\Sigma^{-1} dZ)$
$d l(\theta) = - \frac{N}{2} tr(\Sigma^{-1} d\Sigma) + \frac{1}{2} tr(\Sigma^{-1} (d\Sigma) \Sigma^{-1} Z) + \frac{1}{2} tr(\Sigma^{-1} (\sum_{i=1}^N (x_i - \mu) (d\mu)^T + \sum_{i=1}^N (d\mu) (x_i - \mu)^T))$
At this step in the book, I don't understand how we go from the line above to the line below:
$d l(\theta) = \frac{1}{2} tr(d\Sigma)\Sigma^{-1} (Z - N\Sigma) \Sigma^{-1} + (d\mu)^T \Sigma^{-1} \sum_{i=1}^N (x_i - \mu)$
$d l(\theta) = \frac{1}{2} tr(d\Sigma)\Sigma^{-1} (Z - N\Sigma) \Sigma^{-1} + N (d\mu)^T \Sigma^{-1} (\bar{x} - \mu)$
The first-order conditions (ie. when dl(θ) = 0) are:
$\hat{\Sigma}^{-1} (Z - N\hat{\Sigma}) \hat{\Sigma}^{-1} = 0$ and $\hat{\Sigma}^{-1} (\bar{x} - \hat{\mu}) = 0$
From which follow:
$\hat{\mu} = \bar{x} = \frac{1}{N} \sum_{i=1}^N x_i$
and:
$\hat{\Sigma} = \bar{S}_N = \frac{1}{N} \sum_{i=1}^N (x_i - \bar{x})(x_i - \bar{x})^T$
Note that $\bar{S}_N$ is a biased estimate of Σ. It is usually better to use the unbiased estimate $\bar{S}_{N-1}$.
• Sufficient statistics:
From Z defined above and $\bar{S}_{N-1}$ defined similarly as $\bar{S}_N$ but with N − 1 in the denominator, we can write the following:
$Z = \sum_{i=1}^N (x_i-\mu)(x_i-\mu)^T$
$Z = \sum_{i=1}^N (x_i - \bar{x} + \bar{x} - \mu)(x_i - \bar{x} + \bar{x} - \mu)^T$
$Z = \sum_{i=1}^N (x_i - \bar{x})(x_i - \bar{x})^T + \sum_{i=1}^N (\bar{x} - \mu)(\bar{x} - \mu)^T + \sum_{i=1}^N (x_i - \bar{x})(\bar{x} - \mu)^T + \sum_{i=1}^N (\bar{x} - \mu)(x_i - \bar{x})^T$
$Z = (N-1) \bar{S}_{N-1} + N (\bar{x} - \mu)(\bar{x} - \mu)^T$
Thus, by employing the same trick with the trace as above, we now have:
$L(\mu, \Sigma) = (2 \pi)^{-NP/2} |\Sigma|^{-N/2} exp \left( -\frac{N}{2}(\bar{x} - \mu)^T\Sigma^{-1}(\bar{x} - \mu) -\frac{N-1}{2}tr(\Sigma^{-1}\bar{S}_{N-1}) \right)$
The likelihood depends on the samples only through the pair $(\bar{x}, \bar{S}_{N-1})$. Thanks to the Factorization theorem, we can say that this pair of values is a sufficient statistic for (μ,Σ).
We can also transform a bit more the formula of the likelihood in order to find the distribution of this sufficient statistic:
$L(\mu, \Sigma) = (2 \pi)^{-(N-1)P/2} (2 \pi)^{-P/2} |\Sigma|^{-1/2} exp \left( -\frac{1}{2}(\bar{x} - \mu)^T(\frac{1}{N}\Sigma)^{-1}(\bar{x} - \mu) \right) \times |\Sigma|^{-(N-1)/2} exp \left(-\frac{N-1}{2}tr(\Sigma^{-1}\bar{S}_{N-1}) \right)$
$L(\mu, \Sigma) \propto N_P(\bar{x}; \mu, \Sigma/N) \times W_P(\bar{S}_{N-1}; \Sigma, N-1)$
The likelihood is only proportional because the first constant is not used in any of the two distributions and a few constants are missing (eg. the Gamma function appearing in the density of the Wishart distribution). This doesn't matter as we usually want to maximize the likelihood or compute a likelihood ratio.
• References:
• Magnus and Neudecker, Matrix differential calculus with applications in statistics and econometrics (2007)
• Wand, Vector differential calculus in statistics (The American Statistician, 2002)
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http://mathoverflow.net/questions/107904/the-odds-3-or-more-group-elements-commute/108392
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## The Odds 3 (or More) Group Elements Commute
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Some time ago I asked about the odds 2 group elements commute. I wonder about the odds that 3 group elements commute. Is there a "closed" formula for the sum
$$\frac{1}{|G|^3} \sum_{g,h,k} \delta([g,h]=1)\delta([h,k]=1)\delta([k,g]=1)$$
One approach, as mentioned in Kefeng Liu, might be to use the "Heat Kernel" for finite groups.
$$H(t,x,y) = \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \mathrm{dim}(\lambda) \chi_\lambda(xy^{-1}) e^{-t f(\lambda)}$$
If I'm not mistaken $f(\lambda)$ is the quadratic Casimir, but not sure. Really, for $t=0$ it reduces to the group theory identity:
$$\delta(xy^{-1})= \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \chi_\lambda(x) \overline{\chi_\lambda(y)} = \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \chi_\lambda(1) \chi_\lambda(xy^{-1})$$
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Could you fix the arXiv link to go to the abstract rather than directly to the PDF? Thank you! – Harry Altman Sep 29 at 3:25
## 2 Answers
There are $k(G)|G|$ commuting pairs $(y,z)$ of elements of $G.$ How many commuting triples $(x,y,z)$ of elements of $G$ are there? If we fix the first component $x$ of the triple, note that $x$ permutes (by conjugation) the commuting pairs of elements of $G,$ and the only such pairs it fixes are the commuting pairs which already have both components in $C_{G}(x).$ Hence $x$ fixes $k(C_{G}(x))|C_{G}(x)|$ commuting pairs, and the total number of commuting triples in $G \times G \times G$ is `$\sum_{x \in G} k(C_{G}(x))|C_{G}(x)|$`. If $G$ has $k$ conjugacy classes, with representatives `$\{ x_{i} : 1 \leq i \leq k \}$`, this may be rewritten as `$|G| \sum_{i=1}^{k} k(C_{G}(x_{i})).$` The probability you require, assuming a uniform distribution, is `$\frac{1}{|G|^{2}}\left( \sum_{i=1}^{k} k(C_{G}(x_{i})) \right)$`. The discussion (and Burnside's counting lemma) makes it clear that this is (number of orbits of $G$ by conjugation on commuting pairs of elements of $G$), divided by $|G|^{2}$.
Later edit: Since I noticed the "(or more)" in the question title, the pattern is now clear: let $c_{n}(G)$ denote the number of commuting (ordered) $n$-tuples of elements of $G$. Then we have `$c_{n+1}(G) = \sum_{x \in G} c_{n}(C_{G}(x))$`.
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For symmetric groups there is the beautiful identity $$\prod_{j=1}^\infty (1-u^j)^{-\sigma(j)} = \sum_{n=1}^\infty \frac{T(n)}{n!} u^n$$ where $T(n)$ is the number of triples of commuting elements in the symmetric group $S_n$ and $\sigma$ is the sum of divisors function. See arxiv.org/abs/1203.5079 by John Britnell for an elementary proof. – Mark Wildon Sep 23 at 18:10
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For the generalization to commuting $k$-tuples of elements in $S_n$ and related results, see Exercise 5.13 in Enumerative Combinatorics, vol. 2. – Richard Stanley Oct 2 at 0:38
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This question is studied in the article Isoclinism classes and commutativity degrees of finite groups by P. Lescot. The probability that $n+1$ elements of $G$ pairwise commute is called there the $n$-th commutativity degree $d_n(G)$. A kind of recursive formula for $d_n(G)$ in terms of $d_{n-1}$ of centralizers is proved (Lemma 4.1). Lescot also proves that if $G$ is not abelian then $d_n(G) \leq \frac{3 \cdot 2^n - 1}{2^{2n+1}}$ with equality if and only if $G$ is isoclinic to the quaternion group $Q_8$.
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http://mathhelpforum.com/trigonometry/147840-trig-identities.html
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# Thread:
1. ## Trig identities
14. a) Which equations are not identities? Justify your answers.
b) For those equations that are identities, state any restrictions on the variables.
i) $(1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$
Left Side
$= (1 - cos^2x)(1 - tan^2x)$
$= 1 - tan^2x - cos^2x + (-cos^2x)(-tan^2x)$
$= 1 - cos^2x - tan^2x + (cos^2x)(tan^2x)$
$= sin^2x - tan^2x + sin^2x$
$= (cos^2x)(tan^2x) - tan^2x + (cos^2x)(tan^2x)$
$= tan^2x(cos^2x - 1 + cos^2x)$
$= tan^2x(-sin^2x + cos^2x)$ Edit: Was an error made here?
$= tan^2x(-1)$
$= -tan^2x$
Right Side:
$= \frac{sin^2x - 2sin^4x}{1 - sin^2x}$
$= \frac{sin^2x - 2(sin^2x)(sin^2x)}{cos^2x}$
$= \frac{sin^2x(1 -2(1)(1))}{cos^2x}$
$= \frac{sin^2x(1 -2)}{cos^2x}$
$= \frac{sin^2x(-1)}{cos^2x}$
$= -tan^2x$
Left Side = Right Side
Therefore, this equation is an identity.
When I showed this solution to my teacher, I was told that had I used identities that we haven't learned yet. She said that on the exam, we were to only be able to use reciprocal identities and the Pythagorean identity $sin^2x + cos^2x = 1.$ She also told me that using these identities, I can make the $LS$ and $RS$ equal however with different values. Is there another way of proving this equation to be an identity?
Also, how do I know whether or not an equation will eventually end up as $LS = RS$? The only reason I know it with the few assigned questions is because there is an answer at the back of the book. Though on an exam I won't have access to these answers. Is it just trial and error? Because if it is, I can be going on for a pretty long time just trying to solve one question.
2. $(1-\cos^2{x})(1-\tan^2{x}) =<br />$
$\sin^2{x}\left(1 - \frac{\sin^2{x}}{\cos^2{x}}\right) =$
$\sin^2{x} - \frac{\sin^4{x}}{\cos^2{x}} =<br />$
$\frac{\sin^2{x}\cos^2{x}}{\cos^2{x}} - \frac{\sin^4{x}}{\cos^2{x}} =$
$\frac{\sin^2{x}\cos^2{x} - \sin^4{x}}{\cos^2{x}} =$
$\frac{\sin^2{x}(1 - \sin^2{x}) - \sin^4{x}}{1 - \sin^2{x}} =$
$\frac{\sin^2{x} - \sin^4{x} - \sin^4{x}}{1 - \sin^2{x}} =$
$\frac{\sin^2{x} - 2\sin^4{x}}{1 - \sin^2{x}}$
3. Originally Posted by RogueDemon
14. a) Which equations are not identities? Justify your answers.
b) For those equations that are identities, state any restrictions on the variables.
i) $(1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$
Left Side
$= (1 - cos^2x)(1 - tan^2x)$
$= 1 - tan^2x - cos^2x + (-cos^2x)(-tan^2x)$
$= 1 - cos^2x - tan^2x + (cos^2x)(tan^2x)$
$= sin^2x - tan^2x + sin^2x$
$= (cos^2x)(tan^2x) - tan^2x + (cos^2x)(tan^2x)$
$= tan^2x(cos^2x - 1 + cos^2x)$
$= tan^2x(-sin^2x + cos^2x)$ Edit: Was an error made here?
$= tan^2x(-1)$
$= -tan^2x$
Right Side:
$= \frac{sin^2x - 2sin^4x}{1 - sin^2x}$
$= \frac{sin^2x - 2(sin^2x)(sin^2x)}{cos^2x}$
$= \frac{sin^2x(1 -2(1)(1))}{cos^2x}$
$= \frac{sin^2x(1 -2)}{cos^2x}$
$= \frac{sin^2x(-1)}{cos^2x}$
$= -tan^2x$
Left Side = Right Side
Therefore, this equation is an identity.
When I showed this solution to my teacher, I was told that had I used identities that we haven't learned yet. She said that on the exam, we were to only be able to use reciprocal identities and the Pythagorean identity $sin^2x + cos^2x = 1.$ She also told me that using these identities, I can make the $LS$ and $RS$ equal however with different values. Is there another way of proving this equation to be an identity?
Also, how do I know whether or not an equation will eventually end up as $LS = RS$? The only reason I know it with the few assigned questions is because there is an answer at the back of the book. Though on an exam I won't have access to these answers. Is it just trial and error? Because if it is, I can be going on for a pretty long time just trying to solve one question.
When proving identities it is good practice to only change one side and manipulate it to get the other.
$<br /> (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}<br />$
I have put my steps as spoilers along with explanation.
From the LHS
Use FOIL
Spoiler:
$1-\tan^2(x) - \cos^2(x) + \sin^2(x)$
Multiply through by $\frac{\cos^2(x)}{\cos^2(x)}$ as this is the LCD of the expression.
Spoiler:
$\frac{\cos^2(x) - \sin^2(x) - \cos^4(x) + \sin^2(x) \cos^2(x)}{\cos^2(x)}$
Using the identity $\sin^2(x) + \cos^2(x) = 1$ rewrite any cos terms as sin.
Spoiler:
$\frac{(1-\sin^2(x)) - \sin^2(x) - (1-\sin^2(x))(1-\sin^2(x)) + \sin^2(x)(1-\sin^2(x))}{1- \sin^2(x)}$
Expand the brackets
Spoiler:
$\frac{1-2\sin^2(x) - 1 + 2\sin^2(x) - \sin^4(x) + \sin^2(x) - \sin^4(x)}{1-\sin^2(x)}$
Collect like terms and simplify
Spoiler:
$\frac{\sin^2(x) - 2\sin^4(x)}{1-\sin^2(x)}$
4. Hello, RogueDemon!
We should work on one side only
. . and try to make it equal the other side.
14. a) Which equations are not identities? Justify your answers.
. . .b) For those equations that are identities, state any restrictions on the variables.
$i)\;(1 - \cos^2\!x)(1 - \tan^2\!x) \:=\: \frac{\sin^2\!x - 2\sin^4\!x}{1 - \sin^2\!x}$
I started with the left side:
. . $(1-\cos^2\!x)(1-\tan^2\!x)$. . . . . . . . . .Replace $1-\cos^2\!x$ with $\sin^2\!x$
. . . . . $=\;\sin^2\!x(1 - \tan^2\!x)$. . . . . . . . Replace $\tan^2\!x$ with $\frac{\sin^2\!x}{\cos^2\!x}$
. . . . . $=\;\sin^2\!x\left(1-\frac{\sin^2\!x}{\cos^2\!x}\right)$. . . . . . . Subtract the fractions
. . . . . $=\;\sin^2\!x\left(\frac{\cos^2\!x - \sin^2\!x}{\cos^2\!x}\right)$. . . . . Replace $\cos^2\!x$ with $1-\sin^2\!x$
. . . . . $=\;\sin^2\!x\left(\frac{[1-\sin^2\!x]-\sin^2\!x}{\cos^2\!x}\right)$. . Combine like terms
. . . . . $=\;\sin^2\!x\left(\frac{1 - 2\sin^2\!x}{\cos^2\!x}\right)$. . . . . . .Multiply
. . . . . $=\;\frac{\sin^2\!x - 2\sin^4\!x}{\cos^2\!x}$. . . . . . . . . .Replace $\cos^2\!x$ with $1-\sin^2\!x$
. . . . . $=\;\frac{\sin^2\!x-2\sin^4\!x}{1-\sin^2\!x}$
5. There is an error in the OP. For the LHS, you cannot go from
$=\tan^{2}(x)(-\sin^{2}(x)+\cos^{2}(x))$
$=\tan^{2}(x)(-1)$,
because it is only when the $\sin^{2}(x)$ and the $\cos^{2}(x)$ have the same sign, that they sum to 1.
Regards.
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http://crypto.stackexchange.com/questions/5710/how-to-generate-a-random-integer-in-interval-1-2n-1-from-random-integer-in
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# How to generate a random integer in interval $[1, 2^n-1]$ from random integer in interval $[0, 2^n-1]$?
For a project I am working on, I have access to a CSPRNG that outputs a random integer in the interval $[0, 2^n-1]$ for any integer $n$ greater than 0. I cannot use the zero values, so I have my RNG code in a `while` loop that runs until the random number is `!==0`. For small $n$ it is very likely that the while loop will run more than once, even more than twice (for $n=8$, the probability is $1/65536$). While this is not the bottleneck in my program, I want to eliminate the `while` loop and replace it with a linear transformation on the the generated random integer to get the integers I need in $[1, 2^n-1]$.
Is there a transformation that can be performed on the interval $[0, 2^n-1]$ to yield integers in $[1, 2^n-1]$ while retaining the uniform distribution of the random numbers?
(Would this be better posted in math.se?)
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Are you sure that your PRNG really has an output range of $2^{n-1} + 1$ different values (e.g. that $2^{n-1}$ itself is a possible value)? This looks a bit unusual. – Paŭlo Ebermann♦ Dec 15 '12 at 19:02
@PaŭloEbermann you are correct, I entered the interval incorrectly. Fixed now, thank you. – ampersand Dec 15 '12 at 19:06
## 2 Answers
Assuming that $n > 1$ so as to avoid a trivial special case, here is no function $g(\cdot)$ (linear or otherwise) that will transform a discrete random variable $X$ uniformly distributed on $\{0,1, 2, \ldots, 2^{n}-1\}$ into a discrete random variable $Y$ uniformly distributed in $\{1, 2, \ldots, 2^{n}-1\}$. This is because each value of $X$ has probability $2^{-n}$ attached to it, and this gets mapped onto the $Y$ value $g(X)$. Thus, for any $m \in \{1, 2, \ldots, 2^{n}-1\}$, $P\{Y=m\}$ is necessarily an integer multiple of $2^{-n}$ including, possibly, a zero multiple. It follows that the best one could do is to get a random variable $Y$ that takes on all values in $\{1, 2, \ldots, 2^{n}-1\}$ except one with equal probability $2^{-n}$, and this exceptional value occurs with probability $2\times2^{-n}=2^{-(n-1)}$.
On the other hand, if $X$ were a continuous random variable uniformly distributed on the interval $[0,2^n-1]$, the answer would be easy: $Y = 1+\frac{2^n-2}{2^n-1}X$ is a continuous random variable uniformly distributed on the interval $[1,2^n-1]$.
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You must exclude $n=1$, because it is possible to generate a $Y$ taking the single value $1=2^1-1$ from pretty much anything. Also in "$n\in\{1,2,\ldots,2^{n}-1\}$" the left $n$ is not meant to be the same as the right $n$. My demonstration would be that in order for a suitable mapping to exist, $2^n-1$ (cardinality of destination) must divide $2^n$ (cardinality of source), and that's impossible for $n>1$. I upvoted nevertheless. – fgrieu Dec 17 '12 at 17:57
@fgrieu Thanks for the comment about $n > 1$, the bad choice of notation, and the upvote. I have edited my answer to incorporate your corrections. – Dilip Sarwate Dec 17 '12 at 19:16
No, there is no way to do it the way you want.
Just stick with the while loop. There is nothing wrong with using a while loop. As you seem to realize, the while loop will almost never execute more than one iteration, so it will perform extremely well. You don't say why you find the straightforward solution (the while loop) problematic, so I'm sticking with my answer: just use the while loop.
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http://physics.stackexchange.com/questions/31247/why-do-we-need-higgs-field-to-re-explain-mass-but-not-charge
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# Why do we need Higgs field to re-explain mass, but not charge?
We already had definition of mass based on gravitational interactions since before Higgs. It's similar to charge which is defined based on electromagnetic interactions of particles.
Why did Higgs need to introduce concept of universe-wide Higgs field to define mass, based on interactions with it? And nobody cared about the charge of an electron (for example), which is also basic attribute and constant?
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Why did you pick out the 19$^{\text{th}}$ century? – Nick Kidman Jul 4 '12 at 9:17
@Nick Edited the question to add pre-higgs era.. – Sachin Shekhar Jul 4 '12 at 9:22
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We do care about charge. Just that the Standard Model explains charge reasonably well already. On the other hand, without Higgs, we cannot explain mass (in the SM) – Manishearth♦ Jul 4 '12 at 10:25
@Manishearth In standard model, gravitation does exist with its hypothetical messenger particle "Gravitons". How can we not explain mass in terms of that? Post as answer with citations. – Sachin Shekhar Jul 4 '12 at 11:04
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Unfortunately, I don't have access to any research papers so I can't use citations too well--but I've nearly finished my answers, ask for citations on some of my claims if you feel like it after seeing it. The graviton and the Higgs explain two separate parts of mass. Mass is not really defined just by gravity, that is a schoolboy definition. Inertia is a factor as well--and it is inertia that the Higgs explains. – Manishearth♦ Jul 4 '12 at 11:08
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## 2 Answers
Actually, mass and charge are only superficially similar. Yes, they both appear in inverse square force laws, namely Newton's law of gravitation and Coulomb's law of electrostatic force, but both of those are approximations. Coulomb's law ignores quantum effects, which is a very slight approximation, but Newton's law ignores all of relativity, which makes a huge difference under certain circumstances. The true underlying theories, quantum electrodynamics and general relativity, are almost completely different.
Now, to address your questions directly (though admittedly, this would be a lot easier to explain with the math):
Why did Higgs need to introduce concept of universe-wide Higgs field to define mass based on interactions with it? And, no body cared about charge of electron (for example) which is also basic attribute and constant?
Think about this: in either Newtonian gravity or general relativity, mass is a property that you just assume an object has. Neither of those theories makes any attempt to explain where the mass of an object comes from; the mass is just something you plug into the equation to calculate a trajectory or a force.
The standard model is more ambitious than that, though: it wants to actually explain things, not just have them put into the theory by hand. It all starts with a principle called local gauge invariance. By going through the math, we find that the consequences of this principle correspond to many of the same properties we know particles to have. For example, one consequence of local gauge invariance is the fact that some particles have electric charge, and the existence of the electromagnetic force. Another consequence is that particles have "color charge" which leads to the existence of the strong force. It predicts the existence of antiparticles and the correct conservation laws that govern which elementary particle reactions can and cannot occur in nature.
But before the Higgs mechanism was discovered, the one thing the standard model did not predict was mass. In fact, all the particles it predicted to exist, which in almost all other respects matched known particles exactly, would be massless! Sure, we could tweak the standard model to force the particles to have mass, but there was no particular reason to do that (other than the fact that we know the particles have mass in real life). There was no simple principle that would require the theory to include mass the way local gauge invariance requires the theory to include electric charge, color charge, etc.
What Higgs and other scientists (Anderson, Brout, Englert, Guralnik, Hagen, Higgs, and Kibble) discovered is that the principle of spontaneous symmetry breaking does exactly that: it enables, and in fact requires, the particles of the standard model to have mass. The neat thing is that it only does this in combination with local gauge invariance, but that's kind of beside the point here. The important thing is that when you add spontaneous symmetry breaking in to the standard model, you get particles with mass, where before you had massless particles. In order to add spontaneous symmetry breaking, you need to add a field whose symmetry can be broken. That's where the Higgs field comes from.
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I picked charge because its popular, not because its inverse square force in classical physics. I could have picked color charge for example. – Sachin Shekhar Jul 5 '12 at 9:07
Why couldn't we pick gravitational field for spontaneous symmetry breaking? – Sachin Shekhar Jul 5 '12 at 9:09
Please, don't tell its not in Standard Model.. – Sachin Shekhar Jul 5 '12 at 9:09
Another thing: Newtonian gravity and General Relativity calculate mass based on gravitational interaction. Its just like we calculate charge based on electromagnetic interaction. If not, can't I just say that charge is a property a particle just has. – Sachin Shekhar Jul 5 '12 at 9:15
For one thing, the field whose symmetry is spontaneously broken corresponds to a massive, spin-1 particle. Gravity corresponds to a massless spin-2 particle - clearly not compatible. Besides, gravity is a nonrenormalizable interaction so it doesn't fit into the standard model. – David Zaslavsky♦ Jul 5 '12 at 10:52
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Once again, I am way out of my league in answering this. I may be wrong about many things here, comments appreciated
That was just a definition of mass. The Higgs explains where rest mass (but not gravity) comes from in a mathematically rigorous manner.
One of the attempts to explain how our universe works in a mathematically rigorous manner is the Standard Model. It tries to put all the fundamental forces (except gravity) under one umbrella, along with the particles in a comprehensive theory that explains the subatomic world in a consistent manner.
The theory, in the course of its evolution, has predicted many particles--including many of the quarks, the $W$ and $Z$ bosons, and of course the Higgs boson. All except the Higgs{*} have been experimentally confirmed.
All of these particles are necessary for the theory to work. The Higgs is a product of a neat mathematical trick (spontaneous symmetry breaking), that leads to the concept of "mass" blossoming into existence.
If the Higgs is not found, the whole theory does not work (or needs significant tweaking). IIRC, the SM initially predicted a massless $W$ particle and had inconsistencies--which were resolved by introducing spontaneous symmetry breaking. The original intention for introducing spontaneous symmetry breaking (and thus the Higgs), was to "fix" this issue in the electroweak interaction. From this point of view, the "Higgs makes particles massive" is more of a side-effect, an afterthought.
Mass is not exactly comparable to charge. Mass has two aspects--the gravitational aspect and the inertia aspect. EM forces are the "charge" analogues of gravity, but there is no such analogue for inertia.
Now, the Higgs is explaining the inertia aspect of mass, of which there is no electric counterpart. So there is no need to assume the existence of an electric analogue of the Higgs.
As for EM forces, they are already explained by the SM (the force is mediated by photons). The SM specifically leaves out gravity, but there is a hypothetical particle, the graviton, being researched--this may explain gravity as well.
As @David has mentioned, the above section isn't exactly correct. THe reason we don't need another particle for charge is that charge is already explained by the SM, mathematically (whereas, without Higgs, rest mass is something we just assume)
*As of today, this has probably changed--a particle that is similar to the Standard Model Higgs has been discovered by CERN
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I didn't care that a particle possesses inertia without interacting with a field. Higgs field comes in.. Thanks for the answer. – Sachin Shekhar Jul 4 '12 at 14:14
– Sachin Shekhar Jul 4 '12 at 15:03
@sachin Kostya's answer already explains the etymology, John's answer is about why it isn't a good name. I think you're good there :) – Manishearth♦ Jul 4 '12 at 17:16
Most of this isn't wrong, but (as I mentioned in chat) I would pick on the fact that the Higgs explains inertia. It doesn't, really, because inertia is related more to energy than to just rest mass. And the Higgs only has anything to do with rest mass, not energy. – David Zaslavsky♦ Jul 5 '12 at 8:28
@David Can you please link me to that chat? – Sachin Shekhar Jul 5 '12 at 9:17
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http://crypto.stackexchange.com/questions/3888/why-eg-gn-1-in-bilinear-pairings-holds?answertab=active
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why $e(g,g)^N=1$ in bilinear pairings holds?
I can't get the point of prime order bilinear pairings:$\mathbb{G}\times\mathbb{G}\rightarrow\mathbb{G}_T$,$g=$ generator of $\mathbb{G}$ , $N=p*q$, $p$ and $q$ primes and $e(g,g)^N=1$. why $e(g,g)^N=1$ holds? Why is it 1?
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1 Answer
If $N$ is the order of the group $\mathbb{G}_T$, then for any element $x \in \mathbb{G}_T$ we have that $x^N = 1$. This follows from the Lagrange theorem. Since $e(g,g) \in \mathbb{G}_T$, the same applies to it.
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So the reason is because: $e(g,g)^N=e(g^q,g^p)=e(1,1)=1$ Right? If i have e$(g^q,x) =e(1,x)$ this is equal to 1 also? – curious Sep 27 '12 at 13:14
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No, if the order of $g$ is the composite $N$, then $g^q$ and $g^p$ will be different than $1$. The reason that $e(g,g)^N = 1$ holds is that $e(g,g)$ is the generator of $\mathbb{G}_T$ and has order $N$. – Conrado PLG Sep 27 '12 at 16:22
$e(g^{a1},g^{a2})^N=1$ holds also? – curious Sep 27 '12 at 16:52
Yes, since $e(g^{a_1},g^{a_2})^N = e(g,g)^{N a_1 a_2} = (e(g,g)^N)^{a_1 a_2} = 1^{a_1 a_2} = 1$. – Conrado PLG Sep 27 '12 at 19:03
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http://mathoverflow.net/questions/67708/depth-of-intersection
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## depth of intersection
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi let $I$ be an ideal in $S=K[x_1,...,x_n]$. Can we compute depth $( I \cap K[x_1,...,x_r])$ with $r \leq n$ ?? Is there any relation between depth $I$ and depth $( I \cap K[x_1,...,x_r])$ ??
And what ca we tell about depth $(M \cap N)$ when M,N are S - modules ? Does it have any bounds?
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What do you mean by "depth"? That is usually associated to an ideal where the elements are taken from. Or do you mean the grade? For an ideal that is actually $\mathrm{depth}_I(S)$. Is that what you mean? – Sándor Kovács Jun 14 2011 at 12:48
en.wikipedia.org/wiki/Depth_%28ring_theory%29 this is the definition for depth – Andrei Jun 14 2011 at 14:01
and yes I mean depth$_S I$, at least that's the notation I know. – Andrei Jun 14 2011 at 14:06
I think that if we note $J=(x_{r+1},...,x_n)$ then $I \cap K[x_1,...,x_r] = \frac{I+J}{J}$ and then we get depth $(I \cap K[x_1,...,x_r])$ from depth lemma. In general for $M \cap N$ I don't see anything. – Andrei Jun 14 2011 at 20:30
A relation between $\text{depth} I$ and $\text{depth} I \cap K[x_1,\dots,x_r]$ can hardly exist. If $r=n/2$, say, then $I=(x_1,\dots,x_r)$ and $I' = (x_{r+1}, \dots, x_n)$ are two ideals, both containing a regular sequence of length $r$. In the first case the entire regular sequence survives, in the second case nothing of it survives. – Thomas Kahle Oct 12 2011 at 5:54
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http://physics.stackexchange.com/questions/8400/when-is-use-of-the-effective-mass-concept-appropriate
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# When is use of the 'effective mass' concept appropriate?
In textbooks the characteristic length scale of an exciton, or an electron bound to dopant atom, in silicon is calculated by analogy to the vacuum case.
Bohr radius in vacuum:
$$a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_0 e^2} = 0.53\ {\rm Å}$$
where $\varepsilon_0$ is the vacuum permittivity, $m_0$ is the electron mass, and $e$ is the electron charge.
Exciton bohr radius in material:
$$a = \frac{4 \pi \varepsilon_{\mathrm{r}} \varepsilon_0 \hbar^2}{m_e e^2} = a_0 \varepsilon_{\mathrm{r}} \frac{m_0}{m_e}$$
where $\varepsilon_{\mathrm{r}}$ is the relative permittivity, and $m_e$ is the effective mass of the electron in the material.
For silicon, these can be looked up in a properties table and are:
$$\varepsilon_{\mathrm{r}} = 11.7$$
$$m_e = 0.26 m_0$$
Which gives:
$$a = \frac{11.7}{0.26} 0.53\ {\rm Å} = 23.85\ {\rm Å}$$
However, experimental evidence suggests an exciton radius of $4.3\ \mathrm{nm}$ in silicon (according to Yoffe, AD. Advances in Physics, 42(2) pg 173, 1993). This seems quite a bit off to me.
Is the idea of effective mass not entirely appropriate for this analogy?
And more specifically, what about in heavy fermion systems where the effective mass of the electron can be even larger than a muon!? This would give an incredibly small scale for the exciton or defect bound radius.
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Interesting discovery with `\AA`. I've never actually used that macro myself; I guess the MathJaX developers thought it was obscure enough to leave out. `\overset{\circ}{\mathrm{A}}` is a decent substitute. – David Zaslavsky♦ Apr 11 '11 at 7:33
I have replaced this contrived sequence by the actual Unicode character, Å ... You may also get it as ... & A r i n g ; ... in HTML, without the dots and spaces between characters. – Luboš Motl Apr 11 '11 at 7:50
## 1 Answer
Dear John, note that 23.85 Å is equal to 2.385 nm, while the observed 4.3 nm is approximately two times larger.
There is a simple error in your calculation that exactly fixes the factor of two. Note that the actual calculation you should have done has a radius proportional to $1/m$ and the correct $m$ that you should substitute is the reduced mass of the two-body problem governing the relative position of the two particles.
http://en.wikipedia.org/wiki/Reduced_mass
The reduced mass is $m_1 m_2 / (m_1+m_2)$.
Now, the important point is that an exciton is not a bound state of the effective electron and a superheavy nucleus: instead, it is a bound state of an effective electron and an effective hole - a larger counterpart of the positronium (an electron-positron bound state).
http://en.wikipedia.org/wiki/Exciton
Assuming that both the electron and hole masses are equal, 0.26 $m_0$, the reduced (and still also effective) mass is 0.26/2 $m_0$ = 0.13 $m_0$, and the resulting $a$ is twice as big as your result, 4.77 nm - assuming that your arithmetics is right.
The deviation from 4.3 nm is not too large but I can only handwave if I were trying to pinpoint the most important source of the discrepancy. It could be a different effective mass of the hole; finite-size effects caused by the fact that the silicon atoms were not quite uniformly distributed inside the exciton, and so on.
Update
Oh, in fact, I noticed that your properties table does include a special figure of the effective hole's mass and it differs from the electron mass: 0.38 $m_0$. So the reduced mass is $$\frac{0.38\times 0.26}{0.38+0.26} m_0 = \frac{0.0988}{0.64} m_0 = 0.154 m_0$$ and the calculated radius is $$\frac{11.7}{0.154} \times 0.53\ Å = 40.3\ Å.$$ Well, this is 7 percent too small, much like the previous one was 7 percent too big. ;-)
Hydrogen atoms with composite heavy fermions
Concerning your second question, as you clearly realize, the calculated radius of the "atom" with such "heavy electrons" would be much smaller than the ordinary atom. This also proves that the assumptions of such a calculation fail: the heavy fermions (in condensed matter physics) are the result of the collective action of many atoms on the electron and its mass.
So the large mass of the heavy fermions is only appropriate for questions about physics at long distances - much longer than the ordinary atom. If you look at very short distances - a would-be tiny atom with the heavy fermion - you cannot use the long-distance or low-energy effective approximations of condensed matter physics. You have to return to the more fundamental, short-distance or high-energy description which sees electrons again.
At any rate, you will find out that there can be no supertiny atoms created out of the effective particles such as heavy fermions. The validity of all such phenomenological effective theories - such as those with heavy fermions - is limited to phenomena at distances longer than a certain specific cutoff and highly sub-atomic distances surely violate this condition, so one must use a more accurate theory than this effective theory, and in those more effective theories, most of the fancy emergent condensed matter objects disappear.
Non-relativistic effective theories
Just a disclaimer for particle physicists: in this condensed matter setup, we are talking about non-relativistic theories so the maximum allowed energy $E$ of quasiparticles doesn't have to be $pc$ where $p$ is the maximum allowed momentum in the effective theory. In other words, we can't assume $v/c=O(1)$. Quite on the contrary, the validity of such effective theories in condensed matter physics typically depends on the velocities' being much smaller than the speed of light, too.
So the mass of the heavy fermions is much greater than $m_0$ which would make $m_e c^2$ much greater than $m_0 c^2$; however, the latter is not a relevant formula for energy in non-relativistic theories. Instead, $p^2/2m_e$, which is (for heavy fermions) much smaller than the kinetic energy of electrons, is relevant. The maximum allowed $p$ of these quasiparticles is much larger than $\hbar/r_{\rm Bohr}$ - the de Broglie wavelength must be longer than the Bohr radius. That makes $p^2/2m_e$ really tiny relatively to the Hydrogen ionization energy.
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Oops, forgot to use reduced mass! So the exciton case would have a larger bohr radius ~ 4 nm, while the binding to a defect/dopant atom would give the smaller bohr radius ~ 2 nm. Am I understanding that alright? Now what about in the second part of the question regarding heavy fermion systems... it predicts such a small bohr radius can it really be correct!? – John Apr 11 '11 at 8:12
Yes, I think that the bound state to a dopant would have a radius near your 2 nm, one half of this exciton one. I will add a comment on heavy fermions. – Luboš Motl Apr 11 '11 at 8:19
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http://physics.stackexchange.com/questions/1402/is-it-possible-to-separate-the-poles-of-a-magnet/3560
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# Is it possible to separate the poles of a magnet?
It might seem common sense that when we split a magnet we get 2 magnets with their own N-S poles. But somehow, I find it hard to accept this fact.(Which I now know is stated by Gauss's Law)
I have had this doubt ever since reading about the quantum-field-theory and I know I might sound crazy but is it really impossible to separate the poles of a magnet?
What about black holes? They attract matter and everything else only inwards into them, right? So then, are they the only such case/exception?
Am I being naive? Is there some proof/explanation that can give closure to this question of an independently existing pole?
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## 8 Answers
Well, in order for this splitting to be possible, the magnet would have to be made of two magnetic monopoles (like charged particles, but with "magnetic charge" instead of electric charge) bound together. No known magnet is actually constructed this way; all real magnets that have been studied are made of either little current loops, or particles that have a spin magnetic moment (and these basically act like little current loops).
It's still an open question whether or not magnetic monopoles exist. Some theories predict that they should, but most have nothing to say about it either way. I am not aware of any theories that prohibit the existence of these monopoles. Quantum field theory in general falls in the second category; that is, there is nothing inherent in QFT that requires magnetic monopoles to exist or not exist.
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Of course the totalitarian "rule" would suggest either the existence of monopoles or some---as yet unknown---rule that prohibits them. Stay tuned. – dmckee♦ Nov 28 '10 at 23:45
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As I understand it, there are theories that hold that magnetic monopoles do exist, but in ridiculously small numbers-- a handful of them in the entire visible universe. Which would let you get the benefit of charge quantization without needing really large numbers of them running around. – Chad Orzel Nov 29 '10 at 0:38
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Also, there's a semi-famous experiment to look for magnetic monopoles that saw one signal of exactly the type they would've expected for a monopole within the first few hours of operation. And then never saw anything like it again. It was almost certainly a glitch, but it's kind of amusing to imagine that they really did see the one monopole in the Local Group right when they first turned their detector on... – Chad Orzel Nov 29 '10 at 0:40
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Monopoles are prohibited by Maxwell's equations, specifically $\nabla \cdot B = 0$. If you want monopoles you need to modify this particular equation. – Joe Fitzsimons Jan 22 '11 at 6:42
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@Joe: of course, but Maxwell's equations were designed the way they are to reflect the fact that magnetic monopoles are not observed in nature. I don't consider that a fundamental reason why they couldn't exist. – David Zaslavsky♦ Jan 22 '11 at 23:55
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This is basically how a magnet's atoms look like:
so, when you split it into two, you do not change anything but the length of the magnet. As you can see the North poles(Black sides or "K" as it's in a different language) face north and south poles face south in each part of the magnet.
IF there are such things as monopoles, then it is possible to define a magnetic charge which would allow us to separate a unit magnet into two monopoles. However there are no confirmations for magnetic monopoles so we'll have to accept magnets as a whole for now.
As for black holes, their attraction is gravitational, and not very different from the gravity earth applies on you.
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What you are talking about is the creation of magnetic monopoles (effectively magnetic charges). In classical electrodynamics, such objects are inconsistent with one of Maxwell's equations. Specifically the $\nabla \cdot B = 0$ equation specifically prohibits the existence of magnetic monopoles. In order for them to exist, you need to modify this equation so that it is proportional to the magnetic monopole density. While there are some unverified theories which could give rise to magnetic monopoles, all of our observations so far are consistent with the non-existence of monopoles.
Further, the way magnets that you may have encountered work is via the magnetic field induced by moving charge, not by the presence of monopoles. Usually this field is caused by the angular momentum of charged electrons.
Hope this clarifies matters.
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Seriously? A down vote with no comment? What exactly is incorrect with this answer? – Joe Fitzsimons Jan 23 '11 at 15:40
I suspect your problem is you may want to think about it rhetorically. Magnetic poles are really just a mental shortcut useful to provide a bit of intuition to something that is inherently just math. We don't have physical entities called mag poles, we have a magnetic field, and it works as if it were generated by currents (and maybe spin, which may or may not work like moving charge (a current)). So cut your magnet, and you have two similar, but shorter pieces, and a local concentration of field lines is usually called a "pole", and polarity refers to the signed value of the magnetic field normal to the surface.
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Cutting a magnet in two pieces is like cutting a vector: you can change its length but you cannot change its direction (you cannot make it directionless).
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Magnetic monopoles appear in some Grand Unified Theories and SUSY models of particle physics, but are not possible in the standard model, so we don't know if they really can exist.
However, even in stabdard electrodynamics you can create the illusion of magnetic monopoles by channelling the magnetic flux between two poles along a thin flux tube or "needle".
Recently some condensed matter physicists were able to use this possibility within a spin ice system to make it seem like magnetic monopoles were moving in the material. This effect was described on some blogs at the time so I'm just going to link to the detailed one at Backreaction rather than try to repeat it myself.
Nevertheless, a truly isolated magnetic monopole that could be thrown into a black hole leaving its twin behind is only possible if some GUT theories are correct. People have tried to observe them by watching the current in a coil for the effect of a monopole that passes through it, but without (confirmed) success.
In theory, if they exist they should have been created in abundance in the very early universe at the time of the GUT scale energies. The theory of cosmological inflation was originally proposed to explain why we don't see them. They would be there but in such small numbers that we can't find one. Our only hope would be accelerate and collide particles at the GUT scale of $10^{12}$ TeV but that is currently beyond our wildest dreams.
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Magnetic monopoles certainly exist. This does not require a GUT, they exist in any theory where the electromagnetic U(1) is compact (i.e. where charge is quantized). This follows only from the semiclassical behavior of black hole decay, and so does not require unknown physics.
The reason is essentially the one you state--- you can polarize a black hole in a strong magnetic field, and let it split by Hawking radiation into two oppositely magnetically charged black holes of opposite polarities. Magnetically charged black holes exist in classical General Relativity, as are arbitrary electric-magnetic charge ratio holes, and you can't forbid them, at least not for macroscopically sized black holes, without ruining the theory.
When you let the monopolar black holes decay, you find relatively light monopoles. The lightest monopoles will be lighter than its magnetic charge, so that two such monopoles will repel magnetically, not attract. Presumably, the monopole you find will be a (small multiple of) Dirac's magnetic monopole quantum.
To me, this is as certain as the existence of the Higgs. We haven't observed either one, but the theoretical argument is completely convincing.
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A few non-physics comments have been edited/deleted because of flags. Please keep it civil. – Qmechanic♦ Jan 4 at 20:04
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For the downvoters: the question asked specifically about whether black holes can split poles. Within GR, you can make a magnetically charged black hole solution, and there is no real problem because the nonzero divergence of B is hidden behind the horizon. In semiclassical gravity, you can therefore make magnetic black holes by making opposite magnetic charged BH pairs, and the result needs to decay to the lightest magnetic monopoles by Hawking radiation. This is such a clear requirement that it is incorrect in my opinion to dither on this--- modern physics predicts monopoles unambiguously. – Ron Maimon Jan 9 at 21:28
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Hi Ron, could you repost the physics you explained so nicely as an meta answer in your answer here? I wanted to reread it again and now I cant. – Dilaton Jan 14 at 0:07
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Nice answer,+1. In the same way as it is legitimate for more experimentally inclined people to dismiss the existance of magnetic monopoles, theoretical physicists are allowed to explaine why they are rather convinced of their existance from theoretical reasons. This is exactly what Ron nicely does in his answer, so there is nothing wrong writh it from a physics point of view. – Dilaton Jan 17 at 10:43
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To further explain my above comment, it is not necessary to have directly discovered a magnetic monopole (meaning a heavy elementary particle) as Sklivvz thinks, to be convinced that they have to exist (even though they may be to heavy to be accessible by the current technilic possibilities) from theoretical reasons, as Ron explains in his answer. – Dilaton Jan 17 at 22:09
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No, because found magnet is not composed of poles. Lacking magnetic charges, the creator had to cheat use currents of electric charges. These currents make two poles at once, and these are all magnets found in nature (and stores) today.
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This is not really precise. Actually, creator cheated in a little different way and he used intrinsic magnetic moment of elementary particles to produce magnetism. Some of the magnetism comes from orbital motion (and this can be interpreted as currents) but that is only relevant to diamagnetism and paramagnetism, not ferromagnetism. – Marek Nov 30 '10 at 21:10
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@marek Not only orbital motion means current; spin is, well, to some extent, a current, unless you have a point size. A spinning charged body has magnetic moment because that is electric current. Neutrons have magnetic moment and no electric charge, because it is a compound object which may have inner currents (even though it's strange to associate spin of elementary charged particles with current but this still the correct classical way to look at it) – Pavel Radzivilovsky Dec 1 '10 at 6:56
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that is precisely what I am saying. You are talking just classical perspective and I was correcting you to say that this is not quite correct. Arguably, most of the magnetism comes from the intrinsic magnetic moment of electrons (and this can be classically connected to their spin via gyromagnetic ratio, but that is not important). You can't ever explain that by currents. It is much better explained by solid state physics models and statistical models (like Ising model, Heisenberg model, etc.). Currents have nothing to do with origin of ferromagnetism. – Marek Dec 1 '10 at 8:46
## protected by Qmechanic♦Jan 4 at 12:47
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/156975/solving-p-vs-np-with-computer/157493
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# Solving P vs NP with computer
Is it possible to build a computer program that would (eventually) bring a solution to the P vs. NP question?
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## 3 Answers
Nobody knows. I suppose if there is a polynomial-time algorithm for 3-SAT (or some other NP-complete problem) then a computer could find it and prove P = NP. And if there is a proof that P isn't NP, well, I suppose a computer could find that, too. Why - are you looking for something to work on this summer?
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Not necessarily. It is also conceivable that there is an algorithm that always solves 3SAT in polynomial time, but that it is not provable that it does so. – Henning Makholm Jun 11 '12 at 13:50
I've studied different ways to construct algorithms for certain complexity classes ("syntactic" constructions). It somehow popped out of my head if we could use these constructions to approach the P vs. NP problem in the questions spirit. – rank Jun 11 '12 at 13:57
Hm.. @realazthat you should probably write an answer. – user2468 Sep 11 '12 at 0:22
The statement $\mathsf{P}\overset{?}{=}\mathsf{NP}$ (there is a polynomial time Turing machine correctly solving SAT on all inputs) is a first order statement with an alternation of quantifiers and first-order logic is not decidable. There is no algorithm that given such a sentence will tell you if the statement is true or false.
Regarding the provability in a certain theory (say ZFC) the theorems of an effective theory are computably enumerable, i.e. you can list all provable theorems in them and if the statement is provable you will eventually find it. However there is no computable upper bound on the space and time needed to find the proof and the proof can be huge (needing more symbols than all particles in the universe).
Using computers for deciding much simpler problems (propositional tautologies) is considered to be inefficient although there is a lot of theoretical and industrial interest in the topic (Google for "SAT solver"). The most advanced SAT-solvers fail to answer in reasonable time on simple tautologies like Pigeon Hole Principle (there is no injection from a set of size n+1 to a set of size n). They would take years to answer on the instance n=100.
Also note that given a first order formula $\varphi$ and a natural number $n$ in unary, deciding whether there is a size $n$ proof in first order logic for $\varphi$ is itself an $\mathsf{NP\text{-}complete}$ problem.
### Clarification regarding the relevance of the answer to the question:
Existence of a program to decide a particular instance is not the real question because obviously there always exists such a machine (though we may not know what it is or how to find it). Consider two programs, one always output "Yes", the other always output "No". So one of these two programs is capable of solving the problem correctly. Which one? We don't know!
So we there exists a program that answers a particular instance of a problem. But to write it we need to know the answer to the instance in the first place. That is not what people mean when they say can we write a program to solve a particular instance of a problem. What is mean is not existence. We want to write such a program without knowing the answer to the question. That is not helpful nor what is meant when one is asking for solving a problem using computer programs.
So naturally the problem is writing an algorithm that will find the answer to the problem (here provability) on this instance ($\mathsf{P}$ vs. $\mathsf{NP}$). But what we know about this particular instance that would make it different from any other instance of the problem? Not much. The algorithm needs to work on similar instances also. So the problem naturally generalized to the problem of finding an algorithm that solves this and similar instances.
If someone gives us a graph and asks us to check if it has a $\mathsf{NP\text{-}hard}$ property, we would answer by pointing out that the problem is $\mathsf{NP\text{-}hard}$ in general and is not believed to be solvable efficiently. (Though a generic instance hardness result would be more powerful than the general hardness result.)
Let's consider an example: if I give you an instance of a graph where there is not any clear property which would make it an easy instance it is not unusual to say the problem of checking if the graph has a Hamiltonian circuit is $\mathsf{NP\text{-}hard}$ in general. We don't know anything about the instance. However if the person tells us that the graph is supposed to have some nice properties, e.g. it is planner or it is the Peterson graph then we can say consider algorithms which work on instances that have such properties.
Now let's return to the question. AFAIK there is no evidence that this particular instance (of provability of a first-order arithmetic sentence in ZFC or PA) is easier than any other instance, so the statement about the general difficulty of the problem is relevant. If at some point someone gives us particular useful properties about the instance such that deciding the problem on such instances is easier then one might use computers after that point. In other words, the problem because a different one. But coming up with such properties and a particular algorithm efficiently solving instances with such properties are not done by computers.
(Think about the statement constructively not in the classical sense of existence, we want a computer program to decide these instances, not a non-constructive proof of existence of a program which no one knows what it is. As Bridges writes in his book on computability, in practice no one is going to pay any attention if you give them two algorithms and tell them one of the two algorithms correctly decides the problem they are interested in but no one knows which one.)
See also my answer on CS.SE to Is it NP-hard to decide whether P=NP?
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Could you clarify what you mean with "There is no algorithm that given such a sentence will tell you..."? I understand that we can formulate the problem in FO (I guess using Kleene's T-predicate), but do you mean that whatever formulation we use, it will remain "undecidable"? How could one prove this? – rank Jun 13 '12 at 6:54
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– Tsuyoshi Ito Jun 13 '12 at 13:13
@TsuyoshiIto I think he was actually saying that the original question could be as hard (i.e. undecidable) as the general question. Unlike the Hamiltonian circuit problem on the Petersen graph, it is unknown whether $\mathsf{P}\overset{?}{=}\mathsf{NP}$ is decidable. – Quinn Culver Jun 13 '12 at 13:21
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@Quinn Culver: I am afraid that you are mixing up the decidability in the computability theory and the decidability (in a specific theory) in the formal logic theory. A single question (such as "is P=NP provable in ZFC?") cannot be undecidable in the computability sense. – Tsuyoshi Ito Jun 13 '12 at 13:43
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I downvoted for reasons already explained by Tsuyoshi Ito. The claim "First-order logic is not decidable. There is no algorithm that given such a sentence will tell you if the statement is true or false" is extremely misleading, because the second clause does not follow from the first. – MJD Jul 25 '12 at 2:22
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Just an additional note:
There is an existing algorithm that runs in polynomial time, iff (if and only if) $P=NP$. But with an absolutely huge constant. It basically works as follows:
````Iterate through every possible string:
Compile this string with your favorite compiler of your favorite language.
````
See if this program is a polynomial time algorithm to solve the problem. So, if P=NP, we will eventually hit the algorithm this way, after insanely large number of iterations. Note that the number of iterations is constant; not dependent on the size of the input, notwithstanding the times that we will get programs that just happen to run correctly. Also, note, that we don't know how long to let such a program run; therefore we bound it by a number of steps. See wikipedia/P_versus_NP_problem#Polynomial-time_algorithms for details. Given an NP-complete problem that is polynomial-time verifiable, we will get our solution to the problem. Though we might not know if the program that works for our particular problem works in general for all problems, ie. if it is the algorithm, in truth, this in itself can be considered the algorithm.
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How does one tell whether this program is a polynomial time algorithm to solve "the problem" (where I assume "the problem" is "the NP-complete problem of your choice")? For any particular instance of your NP-complete problem, you can verify that it gives a solution, but how can you be sure it always solve the problem. I just don't see how this answers the original question. – Gerry Myerson Sep 11 '12 at 4:02
Well, one can argue about figuring that out, but in truth you don't need to. Such an algorithm exists, since we are operating under the assumption $P=NP$, and you will eventually reach it. The time that this takes is not dependent on the size of the input; it is thus a "constant" amount of time (some constant hehe). Of course you might reach a program that just "happens" to work before reaching the correct-in-all-cases algorithm. – Realz Slaw Sep 11 '12 at 4:15
Just to be clear, this entire operation in itself would be the algorithm. It doesn't solve the orginal question in total; only if $P=NP$, we already solved the problem. Circular argument. I just thought it would be a good point, and was originally an edit in your answer. Jennifer Dylan suggested I make this a separate answer, so I did. – Realz Slaw Sep 11 '12 at 4:25
I changed the tone of the answer to "additional note", and also noted that it might not find the algorithm (even though in truth, this in itself can be considered the algorithm). – Realz Slaw Sep 11 '12 at 4:37
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http://mathoverflow.net/questions/13831?sort=newest
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## Limit for divergent sequences
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Sorry for the title, but I think it's funny. Can you write down a homomorphism (of additive groups)
$\mathbb{R}^\mathbb{N} \to \mathbb{R}$,
which is nontrivial and whose kernel contains the finite sequences? For example, on the subgroup of convergent sequences, we can take the limit. The question is not if such thing exists (according to the axiom of choice, $\mathbb{R}^\mathbb{N} / \mathbb{R}^{(\mathbb{N})}$ has a basis over $\mathbb{R}$, etc.). I want to write something down1 in order to play around with this "limit for divergent sequences", which might be helpful here. Possibly all of you immediately think that this is not possible, but for which reason? Perhaps it works somehow, but it's just complicated?
1in an informal sense. I'm not interested in a discussion about mathematical logic ;-).
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Are you demanding that the homomorphism agree with limits when they exist? In that case, the kernel contains not only finite sequences, but those with infinite support that converge to zero. – S. Carnahan♦ Feb 2 2010 at 15:51
no this is not imposed – Martin Brandenburg Feb 2 2010 at 15:59
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I think a logician can say this better than I can, but the precise meaning of "writing something down" seems to be at the center of your question. Therefore, it is strange to me that you don't want a discussion about mathematical logic here. – S. Carnahan♦ Feb 2 2010 at 16:07
the rules are simple: if you are a logician and can prove that in some formal sense this hom. cannot be written down, feel free indicate a proof. if you can write something down in an informal sense, please let me know. otherwise, nothing has to be said. – Martin Brandenburg Feb 2 2010 at 16:46
Wouldn't such a homomorphism allow you to construct a non-principal ultrafilter on N? – Qiaochu Yuan Feb 2 2010 at 18:23
show 2 more comments
## 3 Answers
Since $\mathbb{R}^\mathbb{N}$ with the product topology is a polish group and the set $F$ of finite sequences is dense in it, it follows that the only Baire measurable homomorphism $\mathbb{R}^\mathbb{N} \to \mathbb{R}$ that contains $F$ in its kernel is the trivial one. So "writing down" a nontrivial one will be pretty hard.
Now a bit of logic: It is consistent with $ZF$ that all subsets of (and hence all functions between) polish spaces are Baire measurable. So it is consistent with $ZF$ that the only homomorphism $\mathbb{R}^\mathbb{N} \to \mathbb{R}$ that contains $F$ in its kernel is the trivial one. This means that the use of (at least some of) the axiom of choice is unavoidable.
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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.
The construction I know which comes closest to answering your question is that of a Banach limit. This is a bounded linear functional on the Banach space $\ell^{\infty}$ of bounded sequences which extends the limit of a convergent sequence and has some other nice properties. Two problems:
1) You want a functional on the set of all sequences. For this you can take a Banach limit and extend it linearly, but not in any canonical way. This brings me to
2) The construction of a Banach limit and its extension as above use the Axiom of Choice in critical ways, which you seem not to want.
I must say though that your desire to "write something down" and an unwillingness to consider the implications of that phrase make your quest somewhat quixotic. It is a generally agreed upon principle that if a certain proposition can be shown to require the Axiom of Choice in the sense of not being provable from ZF set theory, then it is futile to try to "write something down" that gives a construction. So I think you should be interested in what set-theoretic algebraists have to say about homomorphisms of additive groups of vector spaces without assuming AC. What you want to do may (I'm not saying that it has) have been shown to be impossible. Wouldn't you want to know this?
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I strongly suspect that this is the best that can be done. Judging by the answers to my questions on duals, any such functional will restrict to an algebraic functional on l^\infty, but in some axiomatic systems, that's the continuous dual and (again, in some axiomatic systems), that's l^1. So as there are some axiomatic systems that say "It can't be done" (modulo the difference between l^0 and c_0), then in any axiomatic system you aren't going to be able to "write it down". – Andrew Stacey Feb 3 2010 at 17:25
I guess you need ultralimit :)
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a) ultrafilters are not concrete at all. b) R ist not compact. – Martin Brandenburg Feb 2 2010 at 16:04
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http://mathhelpforum.com/advanced-math-topics/203120-prove-four-vectors-coplanar.html
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# Thread:
1. ## Prove that four vectors are coplanar
I am given four vectors, A, B, C, and D, and need to show that they are coplanar.
I don't really know what to do. If it were three vectors, I know that I should scalar triple product them showing that the parallelepiped formed has zero volume, but when they add a fourth I'm not sure what to do.
2. ## Re: Prove that four vectors are coplanar
Would it work to show that A, B, and C are linearly dependent (and thus coplanar) and then do the same for B, C, and D?
3. ## Re: Prove that four vectors are coplanar
Are the vectors in $\mathbb{R}^3$? If so, you just need to show that the cross product of any two of those four vectors always results in a vector that is parallel (to any other choice of 2 vectors). That is, the set of all possible cross products should be vectors parallel to each other.
Your own idea above should work too, unless there's something counter-intuitive im not seeing.
4. ## Re: Prove that four vectors are coplanar
General approach that works for any number of vectors in any dimension of vector space:
Any set of vectors are coplanar if and only if the dimension of the subspace they span is less than or equal to 2.
For any list of vectors, the typical computational procedure of finding their linear span is by putting them in a matirx - each vector becomes a row - and then row reducing. (That's because the actions of row reducing a matrix are the same as the action of replacing a vecotr by some linear combination of it and other vectors in the linear span.) Once you've row reduced as far as possible, the non-zero rows remaining are necessarily linearly independent, and hence, when intepreted as vectors, are the basis of the linear span of the initial set of vectors. The number of basis vectors of the linear span is the dimension of the linear. If you're asking about coplanar, you're asking if the linear span has two or fewer basis vectors.
Thus, put your vectors into a matrix, and then row reduce as far as possible. If the number of non-zero rows is two or less, then your original vectors are coplanar, otherwise, they aren't coplanar.
Approach when your vectors are in $\mathbb{R}^{3}$:
Take the cross product of two of your vectors that produces a non-zero result (If that's impossible, then they're all coplanar - in fact, colinear). Call it N. Then all your vectors are coplanar if and only if all of them are in the plane perpendicular to the vector N (you already know the 1st two are in that plane). A vector is perpendicular to N if and only if its dot product with N is 0.
So in your case, with four vectors A, B, C, D in $\mathbb{R}^{3}$ they're all coplanar if and only if (assuming A and B are chosen so that AxB isn't 0):
<(A x B), C> = <(A x B), D> = 0. (There, <*,*> is the vector dot product, x is the vector cross product, and the N I refered to is AxB).
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http://stats.stackexchange.com/questions/43560/understanding-the-exponential-distribution
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Understanding the exponential distribution
I'm trying to wrap my head around the Exponential distribution and the meaning of its parameter. The parameter is the rate, right? So take, e.g., $$X\sim \exp(0.05)\,.$$ Now the probability of failure during the first time period is: $$P(X\le1)=1-P(X>1)=1-e^{-0.05}=0.04877\,.$$ Now I can do the math and get the correct result and so on, but cannot wrap my head around why the result should be slightly less than 5%, rather than exactly 5%. I don't get the intuition behind it.
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The rate $\lambda$ has nothing to do with the pdf value in $1$, the correspondence is simply a first order approximation $1-exp(-x) \approx x$... – Xi'an Nov 14 '12 at 11:06
– Xi'an Nov 14 '12 at 11:11
1 Answer
Suppose you have some mildly radioactive substance so that you expect to wait $20$ seconds between decays. That is a rate of $1/20$ per second. The average number of decays in one second is $1/20$.
One way to get an average count of $1/20$ would be if the count were $1$ with probability $1/20$ and $0$ with probability $19/20$. However, sometimes there are $2$ or more decays in that second. For the average count to be $1/20$, when it is sometimes $2$ or more, the probability that you get a count of $0$ must be greater than $19/20$. Therefore, the probability that you wait more than one second before the first decay is greater than $19/20$, and the probability that the first decay occurs within the first second is less than $1/20$.
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http://mathoverflow.net/questions/59563?sort=oldest
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## Configuration space of little disks inside a big disk
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The space of configurations of $k$ distinct points in the plane $$F(\mathbb{R}^2,k)=\lbrace(z_1,\ldots , z_k)\mid z_i\in \mathbb{R}^2, i\neq j\implies z_i\neq z_j\rbrace$$ is a well-studied object from several points of view. Paths in this space correspond to motions of a set of point particles moving around avoiding collisions, and its fundamental group is the pure braid group. It is not hard to prove that this space is homotopy equivalent to the configuration space of the unit disk $$F(D^2,k)=\lbrace(z_1,\ldots , z_k)\mid z_i\in D^2, i\neq j\implies z_i\neq z_j\rbrace$$
In real life however, particles, or any other objects which move around in some bounded domain without occupying the same space, have a positive radius, and so would be more realistically modelled by disks rather than points. This motivates the study of the spaces $$F(D^2,k;r)= \lbrace(z_1,\ldots , z_k)\mid z_i\in D^2, i\neq j\implies |z_i - z_j|>r\rbrace$$ where $r>0$. The homotopy type of this space is a function of $r$ and $k$. Fixing $k$ and varying $r$ gives a spectrum (is this the right word?) of homotopy types between $F(\mathbb{R}^2,k)$ and the empty space. It seems like an interesting (and difficult) problem to study the homotopy invariants as functions of $r$. For instance, for $k=3$ what is $\beta_1(r)$, the first Betti number of $F(D^2,k;r)$?
Question: Have these spaces and their homotopy invariants been studied before? If so, where?
Of course one can also ask the same question with disks of arbitrary dimensions.
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I wrote a detailed answer below including an example with hard disks in a square, but we also know some things about hard disks in a disk. Even though the boundary is smooth, things still get complicated pretty quickly. I have pictures of all the "critical configurations" we have found for disks in a disk, up to about seven disks, and I can email them on request. – Matthew Kahle Mar 25 2011 at 21:32
Persi Diaconis mentioned a similar problem viewed from a slightly different perspective in his paper "The Markov chain Monte-Carlo revolution" – Simon Lyons Mar 25 2011 at 21:54
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My answer was going to be "Ask Matt Kahle," but since he's already answered, I'll just link to what I wrote about this problem on my blog: in particular, its conceptual affinity with the "15 puzzle." quomodocumque.wordpress.com/2009/04/04/… – JSE Mar 26 2011 at 20:41
## 3 Answers
I have a number of results on hard disks in various types of regions, and preprints are in progress. The terminology "hard spheres" (or "hard disks" in dimension 2) comes from statistical mechanics, and I believe Fred Cohen is following my lead on this. (See for example the hard disks section of Persi Diaconis' survey article.)
--- With Gunnar Carlsson and Jackson Gorham, we did numerical experiments and computed the number of path components for 5 disks in a box, as the radius varies over all possible values. This is quite a complicated story already, as the number is not monotone or even unimodal in the radius. (This preprint is almost done, a rough copy is available on request.)
--- With Yuliy Baryshnikov and Peter Bubenik we develop a general Morse-theoretic framework and proved that in a square, if $r < 1/2n$ then the configuration space of $n$ disks of radius $r$ is homotopy equivalent to $n$ points in the plane. On the other hand this is tight: if $r> 1/2n$ then the natural inclusion map is not a homotopy equivalence. (Again this preprint is getting close to be being posted to the arXiv...) There is a much more general statement here.
--- I also have some results with Bob MacPherson about hard disks in a square and also in (the easier case of) an infinite strip. We have been talking to Fred Cohen about this a bit lately, who believes there may be connections to more classical configuration spaces.
I am slightly self-conscious about claiming results here, without first having posted the preprints to the arXiv, but I just wanted to state that there are a number of things known now, and I am working hard to get everything written up in a timely fashion. In the meantime I have some slides up from a talk I recently gave at UPenn.
Here is a concrete example, since that might be more satisfying.
It turns out for $3$ disks in a unit square:
For $0.25433 < r$, the configuration space is empty,
for $0.25000 < r < 0.25433$ it is homotopy equivalent to $24$ points,
for $0.20711 < r < 0.25000$ it is homotopy equivalent to $2$ circles,
for $0.16667 < r < 0.20711$ it is homotopy equivalent to a wedge of $13$ circles, and
for $r < 0.16667$ it is homotopy equivalent to the configuration space of $3$ points in the plane.
For $4$ disks in a square it looks like the topology changes $9$ or $10$ times, and for $5$ disks it looks like the topology might change $25$-$30$ times or more. The general idea is that certain types of "jammed" configurations act like critical points of a Morse function, and mark the only places where the topology can chance.
Update: two preprints have been posted to the arXiv.
Min-type Morse theory for configuration spaces of hard spheres (w/ Baryshnikov and Bubenik):
http://arxiv.org/abs/1108.3061
Computational topology for configuration spaces of hard disks (w/ Carlsson, Gorham, and Mason): http://arxiv.org/abs/1108.5719
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I think most of your inqualities are backwards? It's clear what you mean. – Theo Johnson-Freyd Mar 26 2011 at 5:18
Thankyou for your nice answer and for linking to the slides. Now I remember hearing Baryshnikov talk about this (my visual memory was sparked by your picture of "Kahle's corner"). So is it correct to say: (1) that these jammed configurations can't occur for disks in a disk or squares in a square; (2) that the models in (1) are equivalent, and harder than disks in a square? – Mark Grant Mar 26 2011 at 8:05
Nice answer. I think that squares in squares is easier than discs in a square, I'd very much like to know where discs in discs lies, is it harder than squares in squares (I'd imagine so). Is it only easier than discs in squares because it has the full orthogonal symmetry, rather than the dihedral symmetry? Finally, a fun application is to origami. Circle packings in squares give optimum folds of animals with a fixed number of 'legs'. See Lang's Origami Design Secrets. – James Griffin Mar 26 2011 at 12:52
@ Theo: Thanks, fixed. – Matthew Kahle Mar 26 2011 at 13:21
@ Mark, James: I think squares in a region is generally easier than disks in a region, at least if we assume that the sides stay parallel to the axes. From a lot of experimenting with small examples I believe that disks in a disk looks slightly easier than disks in a square at first, but that this is a little misleading --- it only takes a few more disks for the complications to begin. For example: with 3 disks in a disk, the topology only changes twice. But by the time you get up to 7 disks in a disk, it looks like the topology changes at least 18 times as the radius varies. – Matthew Kahle Mar 26 2011 at 13:26
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.
Yes, there has been a lot of work by Fred Cohen (University of Rochester, currently at IAS) on the subject. He has been giving talks about this (he calls it the hard sphere model), but I can't seem to find a relevant paper/preprint. Perhaps you can contact him directly. EDIT I have somehow managed to confuse Cohen's work and Matt Kahle's (if you know them both, you know how impressive a feat that is). Matt's answer is the true received wisdom.
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I would very much appreciate a good answer to this question, perhaps a follow up to Igor's answer as this is something that I have thought about before, but have not come across in the literature.
I quickly (perhaps too quickly) abandoned the discs model in favour of the little cubes model, or perhaps I should say the hard cubes model.
For hard 2-cubes and k=3 I think the homology groups are:
for r > 1/2: clearly 0
for 1/2 >= r > 1/3: H_0 = Z^6, H_1=Z^6 and H_i=0 for i>1
the three squares are effectively arranged in a circle which can be rotated, the order (and not just the cyclic order!) parametrises the 6 connected components.
for 1/3 >= r we get the usual configuration space: H_0 = Z, H_1 = Z^3, H_2 = Z^2 and H_i=0 for i>2.
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http://math.stackexchange.com/questions/9274/isomorphism-between-0-1-to-0-1
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# Isomorphism between [0,1] to (0,1)
I need to find an isomorphism between [0,1] to (0,1) can you help me with this please?
thanks. benny
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– Nuno Nov 26 '10 at 15:28
Also let me know if the tag elementary-set-theory here is inappropriate. – Nuno Nov 26 '10 at 15:36
## 6 Answers
The easiest way (if you can apply Cantor-Bernstein-Schroder theorem) is to find a bijection from $[0,1]$ onto a subset, say $[1/4,3/4]$; that is easy to do. Since the identity map from $(0,1)$ to $[0,1]$ is a bijection, it follows from the theorem that the two sets are isomorphic.
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– Jonas Meyer Nov 8 '10 at 0:40
The maps $(0,1) \to [0,1], x \mapsto x$ and $[0,1] \to (0,1), x \mapsto 1/4 + x/2$ are injective. The theorem of Cantor-Schröder-Bernstein implies that there is a bijection. It is actually possible to work out the proof of the theorem in these simple examples and get an explicit bijection, namely
$g : [0,1] \to (0,1), x \mapsto 1/2 \pm 1/2^{n+2}$ if $x = 1/2 \pm 1/2^{n+1}$ for some $n$, and $x \mapsto x$ else.
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You can identify a countably infinite collection of points starting with 0 and converging to some limit less than 1/2. Just shift each point one down the line. Now repeat from the top end. All points not involved go to themselves.
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Isn't this one of the tricks behind the Banach-Tarski decomposition? I seem to remember some discussion of this in a Mathematical Intelligencer article... – Steven Stadnicki Nov 7 '10 at 17:53
It is used. In Wagon's excellent book, The Banach-Tarski Paradox, one of the first things proved is equivalent to this, a bijection between the unit circle and the unit circle missing a point. He uses rotations that are multiples of 1 radian – Ross Millikan Nov 7 '10 at 18:00
What do you mean by "isomorphism"? Are you considering these sets with a given group or ring structure (if then, which one?), or as topological spaces (in which case you'd be looking for a "homeomorphism"), or just as sets?
I don't know of a canonical group structure on either set, but if you want your map between the two to be continuous you might have some luck by looking at the notion of compactness and how it behaves under continuous functions.
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sorry, I meant sets – benjamin Nov 7 '10 at 15:29
@user3224: an Isomorphism is a special Homomorphism, which is a structure-preserving map between two algebraic structures. so, what is the structure on those sets? or do you only want any bijective map? – comonad Nov 7 '10 at 15:42
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isomorphism in the category of sets = bijection, and this is meant here. sets are also algebraic structures (with empty signature). – Martin Brandenburg Nov 7 '10 at 16:28
If you just want a bijection there are many ways to do it. Here's a rather contrived one:
Let $\alpha$ be a limit ordinal of the same cardinality as $(0,1)$. By the Cantor-Bernstein-Schröder theorem there exists a bijective map $f: (0,1) \to \alpha$. Let $S$ denote the successor "function" (I know it's not a function - hence the quotes). Note that $\alpha$ does not have a maximal element, so it makes sense to define $\phi: [0,1] \to (0,1)$ by
$$\phi(x) = \begin{cases} (f^{-1} \circ S \circ S \circ f)(x) & \text{ for $x \in (0,1)$}, \\ f^{-1}(1) & \text{ for $x = 1$}, \\ f^{-1}(0) & \text{ for $x = 0$}. \end{cases}$$
Showing that $\phi$ is a bijection is an easy exercise. It's also completely useless and nonconstructive.
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"It's also completely useless and nonconstructive. " Indeed ;) – Martin Brandenburg Nov 7 '10 at 16:51
The bijection 1/x turns (0,1) into (1,$\infty$) which you can chop up into countably many (]'s. Also you can split [] into [)(] and these can be divide countably many times to bisect with the (]'s but it flips the ordering around.
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http://crypto.stackexchange.com/questions/2412/finding-the-lfsr-and-connection-polynomial-for-binary-sequence
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# Finding the LFSR and connection polynomial for binary sequence.
I have written a C implementation of the Berlekamp-Massey algorithm to work on finite fields of size any prime. It works on most input, except for the following binary GF(2) sequence:
$0110010101101$ producing LFSR $\langle{}7, 1 + x^3 + x^4 + x^6\rangle{}$
i.e. coefficients $c_1 = 0, c_2 = 0, c_3 = 1, c_4 = 1, c_5 = 0, c_6 = 1, c_7 = 0$
However, when using the recurrence relation
\begin{equation} s_j = (c_1s_{j-1} + c_2s_{j-2} + \cdots + c_Ls_{j-L}) \mbox{ for } j \geq L. \end{equation} to check the result, I get back:
0110010001111, which is obviously not right.
Using the Berlekamp-Massey Algorithm calculator they say the (I believe) characteristic polynomial should be $x^7 + x^4 + x^3 + x^1$. Which, according to my paper working, the reciprocal should indeed be $1 + x^3 + x^4 + x^6$.
What am I doing wrong? Where is my understanding lacking?
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## 2 Answers
This is probably just a difference in notation than any failure in understanding or implementation. Some people define recurrence relations with subscripts in reverse order than others; the original description given by Berlekamp in his 1968 book Algebraic Coding Theory began counting from $1$ instead of $0$ etc. Observe that $$x^7 + x^4 + x^3 + x^1 = x^7(1 + x^{-3} + x^{-4} + x^{-6})$$ in comparison to your $1 + x^3 + x^4 + x^6$ which you say is the correct reciprocal of what the web site's answer should be. So I would say that the web site seems to be following Berlekamp's original description and giving you an answer that is "off-by-one".
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http://math.stackexchange.com/questions/250154/why-the-sets-defined-in-the-cumulative-hierarchy-transitive?answertab=oldest
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# Why the Sets defined in the Cumulative Hierarchy transitive?
On Page 64, Set Theory, Jech(2006), define the following, by transfinite induction:
• $V_0=\emptyset$,
• $V_{\alpha+1}=P(V_{\alpha})$,
• $V_{\alpha}=\bigcup_{\beta<\alpha}V_\beta$, if $\alpha$ is a limit ordinal.
How can we prove $V_{\alpha}$ is transitive by induction.
I tried what if other set, say $\{\{\emptyset\}\}$, is disignated as $V_0$. It turns out the transitive property fails. So somehow I should incoorperate $V_0=\emptyset$ into the induction. But then I stared at it, I stared at it, I just don't know what should I do.
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@Asafkaragila Sorry for incurring your inconvenience. I hope it is fixed now. – Metta World Peace Dec 3 '12 at 18:38
Much obliged! Thanks. – Asaf Karagila Dec 3 '12 at 18:41
## 2 Answers
First let us prove the following claim.
Claim I: If $x$ is transitive then $\cal P(x)$ is transitive.
Proof. Suppose that $z\in y\in\cal P(x)$, then $y\subseteq x$, and $z\in x$. However $x$ is transitive so $z\subseteq x$ as well. Therefore $z\in\cal P(x)$ as wanted. $\square$
Now we will prove this second claim as well.
Claim II: If $\mathcal X=\{x_i\mid i\in I\}$ is a $\subseteq$-chain of transitive sets then $\bigcup\cal X$ is transitive.
Proof. Let $x\in\bigcup\cal X$, then for some $i\in I$ we have that $x\in x_i$. By transitivity of $x_i$ we have that $x\subseteq x_i$ and therefore $x\subseteq\bigcup\cal X$, as wanted. $\square$
Now you can use Claim I for the successor steps and Claim II for limits, and of course fire off the induction with the proof that $\varnothing$ is transitive.
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Thank you for your answer with explicit use of two equvelents of transitivity. – Metta World Peace Dec 3 '12 at 19:35
If $V_{\alpha}$ is transitive, consider $x \in y \in P(V_{\alpha})$. Then $y$ is a subset of $V_\alpha$, so that $x$ is in a subset of $V_{\alpha}$ and hence is an element of $V_{\alpha}$, so must also be an element of $P(V_{\alpha})$. Thus $P(V_{\alpha}) = V_{\alpha+1}$ is transitive.
If $V_{\beta}$ are transitive for $\beta < \lambda$, then $\bigcup_{\beta < \lambda} V_{\beta} = V_{\lambda}$ is transitive because if $x \in y \in V_{\lambda}$ then $x \in y \in V_{\beta'}$ for some $\beta' < \lambda$, and since $V_{\beta'}$ was transitive, $x \in V_{\beta'}$, hence $x \in V_{\lambda}$.
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Thank you for your answer. – Metta World Peace Dec 3 '12 at 19:36
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http://correll.cs.colorado.edu/?p=1052
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Robots are systems that sense, actuate, and compute. So far, we have studied the basic physical principles of actuation, i.e., locomotion and manipulation. Understanding these principles is necessary to make and implement plans, e.g., using the A* or RRT for calculating shortest path for a mobile robot to travel on . Similarly, we need to understand the basic principles of robotic sensors that provide the data-basis for computation.
The goals of this lecture are
• provide an overview of sensors commonly used on robotic systems
• outline the physical principles that are responsible for uncertainty in sensor-based reasoning
# Robotic Sensors
The development of sensors is classically driven by industries other than robotics. These include submarines, automatically opening doors, safety devices for industry, servos for remote-controlled toys, and more recently the cell-phone, automobiles and gaming consoles. These industries are mostly responsible for making “exotic” sensors available at low cost by identifying mass-market applications, e.g., accelerometers and gyroscopes now being used in mass-market smart phones or the 3D depth sensor “Kinect” as part of its XBox gaming console. As we will see later on in this class, sensors are hard to classify by their application. In fact, most problems benefit from every possible source of information that they can obtain. For example, localization can be achieved by counting encoder increments, but also by measuring acceleration, or using vision. All of these approaches differ drastically in their precision and the kind of data that they provide, but none of them is able to completely solve the localization problem.
Exercise: Think about the kind of data that you could obtain from an encoder, an accelerometer, or a vision sensor on a non-holonomic robot. What are the fundamental differences?
Although an encoder is able to measure position, it is used in this function only on robotic arms. If the robot is non-holonomic, closed tours in its configuration space, i.e., robot motions that return the encoder values to their initial position, do not necessarily drive the robot back to its starting point. Encoders are therefore mainly useful to measure speed. An accelerometer instead, by definition, measures the derivative of speed. Vision, finally, allows to calculate the absolute position (or the integral of speed) if the environment is equipped with unique features. An additional fundamental difference between those three sensors is the amount and kind of data they provide. An accelerometer samples real-valued quantities that are digitized with some precision. An odometer instead delivers discrete values that correspond to encoder increments. Finally, a vision sensor delivers an array of digitized real-valued quantities (namely colors). Although the information content of this sensor exceeds that of the other sensors by far, cherry-picking the information that are really useful remains a hard, and largely unsolved, problem.
We will now study the sensors on robots that are available at CU Boulder.
## The iRobot Create
The iRobot Create is the research version of iRobots “Roomba” vacuum cleaner. The Create is equipped with the following sensors:
• Wheel encoders
• Infrared “cliff” sensors detecting stairs or other edges the robot could fall off
• An infrared “wall” sensor allowing the robot to keep its distance to a wall to its left
• A front-bumper to detect collisions
• A mechanical “pick-up” sensor detecting lift of the robot
• An infrared sensor that can detect and decode infrared messages sent by the charger.
The iRobot Create is a good example for making very efficient use of the little information the sensors above provide. This allows this product to be relatively cheap and perform reasonably well on the original application, floor-cleaning, it was made for. Particularly noteworthy is the homing mechanism that allows the robot to autonomously return to its charger: the charger emits three different infrared signals that have information modulated on them and can be recognized as the “red buoy”, the “green buoy” and a “force field”. The geometry of each field, tuned by careful design of the charging station, is made so that the robot is able to use the different fields for inferring its location relative to the charger. This process is explained in more detail on Page 18 of the Create reference manual.
In summary, the Create provides a series of digital on/off sensors, one analog sensor to detect the distance to the wall that is encoded with 12-bit resolution (values from 0 to 4096), and an infrared sensor that operates as a communication device by decoding information modulated on the infrared signal. How this modulation works, is not part of this class.
# The E-Puck Robot
The E-Puck robot is a miniature robot that was specifically developed for education. It is therefore equipped with a large number of sensors to provide the opportunity for hands-on experience. In particular, the E-Puck provides
• infrared distance sensors that double as light sensors
• three microphones aligned in a triangle
• a three-axis accelerometer
• a VGA camera that is down-sampled to 42 x 42 pixels
What an actual signal from these sensors looks like is described on the pages at each of the above links.
Exercise: Study the response from the E-Puck distance sensors and that of the accelerometer. Why do you think the response of the distance sensors is non-linear (compare this also with how you modeled the distance sensor in Webots in Exercise 1)? Why does the accelerometer show a constant offset on all three channels?
The distance sensors are made from relatively simple components, namely an infrared emitter and an infrared receiver. They work by emitting an infrared signal and then measuring the strength of the reflected signal. In the graph on the e-Puck website, large values correspond to large distance or little reflected light and small values correspond to short distances or a lot of reflected light. (This should be actually the other way round, but the designers of the E-Puck have wired up the sensor so that the response becomes more intuitive, i.e. low values for small distances and high values for large distances.) As light intensity decays at least quadratically with distance (and not linear), the response of the sensor is not linear, but more precise close to the obstacle. The accelerometer shows constant offset for the z-axis as it measures the constant acceleration of the earth’s gravitational field. As the robot is not perfectly horizontal due to a missing caster wheel, also values of the x and y axes are slightly offset. As for the distance sensor, values are reported between 0 and 4095. This is because the analog-digital converter on the robot has a resolution of 12-bit, which corresponds to chopping the analog voltage it measures into 4096 equal chunks. As the accelerometer is able to measure acceleration both along the positive and the negative axes, zero acceleration corresponds to a measurement of 2048.
## The PrairieDog Robot
The PrairieDog is a research platform developed at CU Boulder and MIT that extends on the iRobot Create platform. In addition to the sensors of the create, it is equipped with
• an USB web-cam
• a localization system that can decipher infra-red markers mounted at the ceiling for global localization
• a scanning laser distance sensor
• encoders in each of its 5-DOF arm joints
PrairieDog with a CrustCrawler manipulating arm
The scanning laser consists of a laser that scans the environment by being directed via a quickly rotating mirror and the distance at which the laser is reflected sampled at regular intervals. This process, and the resulting data, is nicely animated on the the wikipedia site that is linked from this text. How the distance is calculated is fundamentally different from an infrared distance sensor. Instead of measuring the strength (aka amplitude) of the reflected signal, laser range scanners measure the phase difference of the reflected wave. In order to do this, the emitted light is modulated with a wave-length that exceeds the maximum distance the scanner can measure. If you would use visible light and do this much slower, you would see a light that keeps getting brighter, then getting darker, briefly turns off and then starts getting brighter again. Thus, if you would plot the amplitude, i.e. its brightness, of the emitted signal vs. time you would see a wave that has zero-crossings when the light is dark. As light travels with the speed of light, this wave propagates through space with a constant distance (the wavelength) between its zero crossings. When it gets reflected, the same wave travels back (or at least parts of it that get scattered right back). For example, modern laser scanners emit signals with a frequency of 5 MHz (turning off 5 million times in one second). Together with the speed of light of approximately 300,000km/s, this leads to a wavelength of 60m and makes the laser scanner useful up to 30m.
When the laser is now at a distance that corresponds exactly to one half the wave-length, the reflected signal it measures will be dark at the exact same time its emitted wave goes through a zero-crossing. Going closer to the obstacle results in an offset that can be measured. As the emitter knows the shape of the wave it emitted, it can calculate the phase difference between emitted and received signal. Knowing the wave-length it can now calculate the distance. As this process is independent from ambient light (unless it has the exact same frequency as the laser being used), the estimates can be very precise. This is in contrast to a sensor that uses signal strength. As the signal strength decays at least quadratically, small errors, e.g. due to fluctuations in the power supply that drives the emitting light, noise in the analog-digital converter, or simply differences in the reflecting surface have drastic impact on the accuracy and precision (see below for a more formal definition of this term).
## The xBox Kinect
While scanning lasers have revolutionized robotics by providing a real-time view on the obstacles in the vicinity of the robot at fairly high accuracy, they could only provide a 2D slice of the world. For example a table would only be seen by its legs, not allowing the robot to judge whether it could pass underneath it. Although people have played with pivoting laser scanners, the vertical scanning speed is intrinsically limited by the horizontal scanning speed of the device. For example, a laser scanner that provides a 2D slice of the environment 10 times per second (or at 10Hz), would need 1s to provide 10 vertical slices. This is not fast enough for real-time navigation, and requires multiple lasers. For example, the Velodyne 3D laser scanner employs 64 (!) parallel scanning lasers and provides 2cm accuracy over 50m range.
An alternative solution to this problem are 3D depth sensors that work by projecting an infrared pattern into the environment (aka structured light), observe it via a camera, and quickly calculate the depth of the environment based on the observed deformation. A prominent example of this technology is the Kinect for Microsoft’s xBox, which made the technology affordable by targeting a mass-market application. Although neither the resolution nor range are comparable with state-of-the-art laser range scanners, the sensor can provide depth images of 640×480 points (resolution of a VGA camera) between 0.7 and 6m at 30Hz, which not only enables real-time navigation but also gesture recognition from people, opening up novel avenues in human robot interaction and the potential for affordable, highly capable robot platforms.
## Terminology
It is now time to introduce a more precise definition of terms such as “speed” and “resolution”, as well as additional taxonomy that is used in a robotic context. Roboticists differentiate between proprioceptive and exteroceptive sensors. Proprioceptive sensors measure quantities that are internal to the robot such as wheel-speed, current consumption, joint position or battery status. Exteroceptive sensors measure quantities from the environment, such as distance to a wall, the strength of ambient light or the pattern of a picture at the wall.
Roboticists also differentiate between active and passive sensors. Active sensors emit energy of some sort and measure the reaction of the environment. Passive sensors instead measure energy from the environment. For example, most distance sensors are active sensors (as they sense the reflection of a signal they emit), whereas an accelerometer, compass, or a push-button are passive sensors.
The difference between the upper and the lower limit of the quantity a sensor can measure its known as its range. This should not be confused with the dynamic range, which is the ratio between two quantities (usually used for sensors that sense light or sound). The minimal distance between two values a sensor can measure is known as its resolution. The resolution of a sensor is given by the device physics (e.g., a light detector can only count multiples of a quant), but usually limited by the analog-digital conversion process. The resolution of a sensor should not be confused with its accuracy or its precision (which are two different concepts). For example, whereas an infrared distance sensor might yield 4096 different values to encode distances from 0 to 10cm, which suggests a resolution of around 24 micrometers, its precision is far above that (in the order of millimeters) due to noise in the acquisition process.
Technically, a sensors accuracy is given by the difference between a sensors output $m$ and the true value $v$:
$accuracy=1-\frac{|m-v|}{v}$
A sensor’s precision instead is given by the ratio of range and statistical variance of the signal. Precision is therefore a measure of repeatability of a signal, whereas accuracy describes a systematic error that is introduced by the sensor physics.
The speed at which a sensor can provide measurements is known as its bandwidth. For example, if a sensor has a bandwidth of 10 Hz, it will provide a signal ten times a second. This is important to know, as querying the sensor more often is a waste of computational time and potentially misleading. In Webots you usually provide the bandwidth of each sensor when calling its enable function, which requires the update rate in ms.
## Additional Sensors
There are a few additional sensors that are important in current robotic systems, but are not used on the platforms discussed above.
Gyroscope: a gyroscope is an electro-mechanical device that can measure rotational orientation. It is complementary to the accelerometer that measures translational acceleration. Classically, a gyroscope consists of a rotating disc that could freely rotate in a system of pivots and gimbals. When moving the system, the inertial momentum keeps the original orientation of the disc, allowing to measure the orientation of the system relative to where the system was started. A variation of the gyroscope is the rate gyro, which measures rotational speed. The rate gyro can also be implemented using optics, which allows for extreme miniaturization. Rate gyros are for example used in Apple’s iPhone IV or in Nintendo’s latest generation Wii controller.
Compass: a compass measures absolute orientation with respect to the earth’s magnetic field. Unlike the mechanical compass that measures the magnetic field only in one dimension, last generation electronic sensors can measure the earth’s magnetic field in three axes. Although this information is unreliable indoors and gets disturbed by metal parts in the environment, a compass can be helpful to determine local orientation.
Inertial-Measurement Unit (IMU): together, accelerometers, gyroscopes, and magnetometers (compass) can provide an estimate of a robots acceleration, velocity, and orientation along all 6 dimensions, and drastically improve the accuracy of odometry. As the underlying sensors are not very precise, however, absolute values provided by the IMU quickly drift and require update by some sort of global positioning system.
Global Positioning System (GPS): GPS systems triangulate the receiver’s position with meter accuracy from estimating the position of satellites in the orbit that have known position. As such, there are mostly adapt for outdoor applications. There exist a series of equivalent systems for indoor robotic systems that operate either using active or passive beacons in the environment and can provide millimeter accuracy. As of now, there is no established standard in industry.
Ultra-sound distance sensor: An ultra-sound distance sensor operates by emitting an ultra-sound pulse and measures its reflection. Unlike a light-based sensor that measures the amplitude of the reflected signal, a sound-based sensor measures the time it took the sound to travel. This is possible, because sound travels at much lower speed (300m/s) than light (300,000km/s). The fact that the sensor actually has to wait for the signal to return leads to a trade-off between range and bandwidth. (Look these definitions up above before you read on.) In other words, allowing a longer range requires waiting longer, which in turn limits how often the sensor can provide a measurement. Although US distance sensors have become less and less common in robotics, they have an advantage over light-based sensors: instead of sending out a ray, the ultra-sound pulse results in a cone with an opening angle of 20 to 40 degrees. By this, US sensors are able to detect small obstacles without the requirement of directly hitting them with a ray. This property makes them the sensor of choice in automated parking helpers in cars.
Take-home lessons
• Most of a robot’s sensors either address the problem of robot localization or localizing and recognizing objects in its vicinity
• Each sensors has advantages and drawbacks that are quantified in its range, precision, accuracy, and bandwidth. Therefore, robust solutions to a problem can only be achieved by combining multiple sensors with differing operation principles.
• Solid-state sensors (i.e. without mechanical parts) can be miniaturized and cheaply manufactured in quantity. This has enabled a series of affordable IMUs and 3D depth sensors that will provide the data basis for localization and object recognition on mass-market robotic systems
### Share this:
• http://boulderhackerspace.com/ Daniel Zukowski
There was brief mention of Sebastian Thrun today in class. Along with Peter Norvig, he is providing his Intro to AI course at Stanford this semester for free online. The course starts in October and you can still register. The course has two levels of participation, one with just the videos and simple self-quiz exercises, and another level with homework assignments and exams. If you take the advanced track, you get a certificate and a grade at the end of the course. You can register for free at ai-class.com
There are also intro courses in Machine Learning and Databases being offered in the same manner.
• http://correll.cs.colorado.edu/?page_id=19 Nikolaus Correll
Kinect Hackers Are Changing the Future of Robotics
http://www.wired.com/magazine/2011/06/mf_kinect/all/1
• Nikolaus Correll
Rodney Brooks talks about “exponentials” in other industries that can enable breakthroughs in robotics:
http://fora.tv/2009/05/30/Rodney_Brooks_Remaking_Manufacturing_With_Robotics
"RT @todd_murphey: Thanks to @heatherknight, @ken_goldberg, and Pericle Salvini for the amazing #uncanny workshop at ICRA http://t.co/OzMwLc…" — correlllab
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http://mathoverflow.net/questions/2650/conjugacy-classes-in-finite-groups-that-remain-conjugacy-classes-when-restricted/2655
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Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In a forthcoming paper with Venkatesh and Westerland, we require the following funny definition. Let G be a finite group and c a conjugacy class in G. We say the pair (G,c) is nonsplitting if, for every subgroup H of G, the intersection of c with H is either a conjugacy class of H or is empty.
For example, G can be the dihedral group of order 2p and c the class of an involution.
The case where c is an involution is o special interest to us. One way to construct nonsplitting pairs is by taking G to be a semidirect product of N by (Z/2^k Z), where N has odd order, and c is the conjugacy class containing the involutions of G. Are these the only examples? In other words:
Question 1: Is there a nonsplitting pair (G,c) with c an involution but where the 2-Sylow subgroup of G is not cyclic?
Slightly less well-posed questions:
Question 2: Are there "interesting" examples of nonsplitting pairs with c not an involution? (The only example we have in mind is G = A_4, with c one of the classes of 3-cycles.)
Question 3: Does this notion have any connection with anything of pre-existing interest to people who study finite groups?
Update: Very good answers below already -- I should add that, for maximal "interestingness," the conjugacy class c should generate G. (This eliminates the examples where c is central in G, except in the case G = Z/2Z).
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13 Answers
I cheated and asked Richard Lyons this question (or at least, the reformulation of the problem, conjecturing that (G,c) is nonsplitting for an involution c with `<`c`>` generating G if and only if there exists an odd A such that G/A = Z/2). His response:
Good question! This is a famous (in my circles) theorem - the Glauberman Z^`*-`Theorem. (Z^*(G) is the preimage of the exponent 2 subgroup of the center of G/O(G), and O(G)=largest normal subgroup of G of odd order.)
Z^`*-`Theorem: If c is an involution of G then c\in Z^*(G) iff [c,g] has odd order for all g\in G iff for any Sylow 2-subgroup S of G containing c, c is the unique G-conjugate of itself in S.
The last property is absolutely fundamental for CFSG. The proof uses modular character theory for p=2. Attempts to do it with simpler tools have failed.
George Glauberman, Central Elements in Core-free Groups, Journal of Algebra 4, 1966, 403-420.
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Hott! Good thing you asked, this answer means that we weren't going to prove this on our own. – Noah Snyder Oct 29 2009 at 17:50
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I suppose that's exactly the kind of thing this site was designed to do. I'd cite this as the best example so far that MO works! – Sonia Balagopalan Oct 29 2009 at 17:51
If I've followed everything correctly here's what we've shown: If c consists of involutions and (G,c) is nonsplitting and G is generated by c then G is a semidirect product of an odd order normal subgroup N by Z/2. Furthermore any such semidirect product is automatically non-splitting, the only remaining question is whether there are natural conditions on N and the involution acting on N which explain whether Z/2 generates G. – Noah Snyder Oct 29 2009 at 18:31
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Cheers to FC for the important insight that a good approach to finite group theory problems is to ask a finite group theorist.... – JSE Oct 29 2009 at 20:48
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Another example, different in flavour from the others: Let G be the set of affine linear transformations x --> ax+b over a finite field, and c the conjugacy class of [ax] for a \neq 1.
Proof that this works:
The conjugacy class of ax is the maps x --> ax+i, for i in Z/p.
We need to show that, if H contains ax+i and ax+j then ax+i and ax+j are conjugate in H. Since H contains ax+i and ax+j, it contains their ratio, x+(j-i)/a. Therefore, H contains every map of the form x+k. Conjugating ax+i by x+(j-i)/(a-1) gives ax+j.
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Note that this example is included in FC's example (A = additive group of the field, Z/pZ = multiplicative group of the field; OK, this isn't prime order, but I think FC's argument still applies.) Also, when c is an involution here, a = -1, and one is back in the dihedral situation. – JSE Oct 27 2009 at 4:30
I was going to say that, but then I decided it actually was slightly different. I was using Sylow to force conjugation, somehow here what is being used here is the fact that the additive group has prime order... BTW, is your involution secretly complex conjugation? – Lavender Honey Oct 27 2009 at 4:39
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ALL INVOLUTIONS ARE SECRETLY COMPLEX CONJUGATION! (Well, sort of. In fact for us this involution ends up being identified with the involution on a hyperelliptic curve over a finite field; and indeed these particular double covers X -> P^1/F_q are ramified at a chosen point at infinity, so are of the type hyperellipticologists call "quadratic imaginary" -- so in that very loose sense, yeah, it's complex conjugation) – JSE Oct 27 2009 at 13:33
The classification of involutions on complex reductive Lie groups supports your claim very well, JSE. +1. – Allen Knutson May 2 2011 at 14:41
Suppose that A is a group of order coprime to p such that p | #Aut(A). Let G be the semidirect product which sits inside the sequence:
1 ---> A ---> G --(phi)--> Z/pZ --> 0;
Let c be (any) element of order p which maps to 1 in Z/pZ. If c is conjugate to c^j, then phi(c) = phi(c^j). Hence c is not conjugate to any power of itself.
Let H be a subgroup of G containing c (or a conjugate of c, the same argument applies). The element c generates a p-sylow P of H (and of G). It suffices to show that if gcg^-1 lies in H, then it is conjugate to c inside H. Note that gPg^-1 is a p-Sylow of H. Since all p-Sylows of H are conjugate, there exists an h such that gPg^-1 = hPh^-1, and thus h c^j h^-1 = gcg^-1. Yet we have seen that c^j is not conjugate to c inside G unless j = 1. Thus gcg^-1 = hch^-1 is conjugate to c inside H.
I just noticed that you wanted `<`c`>` to generate G. It's not immediately clear (to me) what condition on A one needs to impose to ensure this. Something like the automorphism has to be "sufficiently mixing". At the very worst, I guess, the group G' generated by `<`c`>` still has the property, by the same argument.
This works more generally if p || G and no element of order p is conjugate to a power of itself. (I think you know this already if p = 2.)
The case where the p-Sylow is not cyclic is probably trickier.
Examples: A = (Z/2Z)+(Z/2Z), p = 3. (This is A`_`4).
A = Quaternion Group, p = 3. (This is GL`_`2(F`_`3) = ~A`_`4, ~ = central extension).
A = M^37, M = monster group, p = 37.
FURTHER EDIT BECAUSE PREVIOUS EDIT WAS NONSENSE AND CLAIMED TO PROVE THAT P ADMITTED A NORMAL COMPLEMENT A.
If we can prove that a P-sylow admits a normal complement A, then (assuming that G is generated by `<`c`>`) we can assume that G/A, which is isomorphic to P, is generated by the image of `<`c`>`. Yet the only p-group generated by a single conjugacy class is cyclic. (Proof: P/Phi(P) is cyclic, where Phi(P) = Frattini subgroup).
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You ask for c to map to 1 in Z/p, but you never use this condition. Am I missing something? PS Nice answer! – David Speyer Oct 27 2009 at 11:38
1 is just a stand-in for "nonzero" here. – Noah Snyder Oct 27 2009 at 13:02
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Suppose (G,c) is non-splitting, that c has order p and the p||#G, then you must be in FCs setting. Taking H to be the centralizer of the subgroup generated by c, nonsplitting tells you that the centralizer is equal to the normalizer. By the Burnside normal complement theorem this tells you that the subgroup generated by c has a normal complement. – Noah Snyder Oct 27 2009 at 15:53
Just because c generates G doesn't necessarily mean that c \cap P generates P, I don't think (this is what I was confused about in earlier versions of my answer below). – Noah Snyder Oct 27 2009 at 21:02
Also I think you've misunderstood the statement of Frobenius's normal complement theorem. You need that any two elements which are conjugate in P are conjugate in G, not just that this works for one conjugacy class. See: groupprops.subwiki.org/wiki/… – Noah Snyder Oct 27 2009 at 21:05
show 1 more comment
Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.
In particular, if no powers of elements of c are in c (for example, if c consists of involutions) and any two distinct elements of c generate all of G then it follows that (G,c) is nonsplitting.
As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonsplitting for c an involution iff the product of any two elements of c has odd order.
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Unfortunately, the only groups generated by 2 elements of order 2 are Dihedral groups. – Lavender Honey Oct 27 2009 at 21:42
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Ah, good point. So "non-splitting" for involutions is just a condition on a bunch of dihedral subgroups. – Noah Snyder Oct 28 2009 at 0:41
I wonder if a Coxeter group perspective would be useful. Since G is generated by a bunch of transpositions it's a quotient of some (non-finite) Coxeter group. Is this Coxeter group also nonsplitting? – Noah Snyder Oct 28 2009 at 1:02
Re Noah's first comment here: oh, so when c is a class of involutions is non-splitting equivalent to "the product of any two members of c has odd order?" This suggests thinking about the subgroup of G generated by all products g_1g_2, with g_1, g_2 in c. I guess this is always either G or a normal subgroup of index 2? In the latter case one is surely done. – JSE Oct 28 2009 at 1:43
Sure, but what do you do with the former case? – Noah Snyder Oct 28 2009 at 4:06
show 4 more comments
Of course, if G is abelian, then the conjugacy classes of G are just the elements, and any pair (G, c) is nonsplitting. More generally, if x is in the center of G and c is the class of x, then (G, c) is nonsplitting. So the answer to Question 1 is yes. But I imagine that you're looking for more interesting examples to questions 1 and 2!
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Question 1 Here's a very simple example. Let $Q= < i,j,k >$ be the quaternion group. $-1$ is the unique involution, so is in its own conjugacy class. $(Q,-1)$ is nonsplitting, and $Q$ is its own Sylow 2-subgroup. In general, take any finite group with a unique involution. (These turn out to be cyclic, quaternion and 2 other kinds.)
I don't know what happens when you take a group with more than one involution.
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Suppose that elements of c have prime order p (for example, c consists of involutions). Let H be the subgroup of the Sylow p-group P generated by the intersection of c with P. Note that P normalizes H since H is generated by a conjugacy class in P.
Claim: H is a central cyclic subgroup of P
Proof: Let F be the Frattini subgroup of H (generated by commutators and pth powers). Since H/F is elementary abelian and generated by elements of a single conjugacy class in H (by nonsplitting), it follows that H/F is cyclic. But then by the Burnside basis theorem (a version of Nakayama's lemma for p-groups) H must also be cyclic. Since H is abelian, by nonsplitting it must be central in its normalizer (which includes P).
In fact by non-splitting H has to be central inside the normalizer of P. So we're in the situation where no two elements of c sit inside the same Sylow P. In particular the size of c divides the number of Sylow p-groups is the same as .
Studying groups where a centralizer of an involution contains the normalizer of a Sylow 2-group seems like the sort of thing the classification people might know something about. They were all about classifying groups where the centralizer of an involution has some property.
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If c is central but P is not normal, then (G,c) is splitting but there are fewer conjugates of c then P-Sylows. In other words, although H is central in N = normalizer of P, the centralizer of H could be larger. – Lavender Honey Oct 27 2009 at 22:31
Good point. Fixed! – Noah Snyder Oct 27 2009 at 22:40
(You should also say the the size of <c> divides the number of Sylow subgroups.) Off to an official dinner tonight, hope to finish the case P = Z/2+Z/2 during the speeches. – Lavender Honey Oct 27 2009 at 22:53
This is a new community wiki answer which people can edit instead of writing comments.
Noah wrote:
Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.
As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonplitting for c an involution iff the product of any two elements of c has odd order.
JSE suggests looking at the subgroup generated by all pairwise products of elements in c. This either generates the whole group or generates an index 2 subgroup (which is necessarily normal and has a complement generated by any element of c). FC noted that instead of looking at pairwise products you could instead look at pairwise commutants (since the commutant of two involutions is just the square of their product and in an odd order cyclic group the square of a generator generates).
Let's concentrate on the latter case.
when does a group A admit an involution i:A-->A such that i(a)a^-1 always has odd order, and {i(a)a^-1} generates for all A generate A.
[I think the only such groups have odd order. Also, Since A has an odd number of 2-Sylow subgroups, i preserves at least one Sylow P. Yet then i(a)a^-1 lies in P, and is thus trivial. Thus i fixes P. It follows that i preserves the normalizer N of P.--FC
I'd just run through the same argument myself before realizing this is just the fact that the centralizer of an element of c in the big group contains the 2-Sylow and its normalizer (as in the Frattini answer).--Noah]
Does this have anything to do with H^1(Z/2, A)? To flesh this out, the maps Z/2->A sending the nontrivial element to i(a)a^-1 are exactly the coboundaries. --Noah
Wait a sec, since every coboundary is a cocycle (or by a direct one-line computation) if y = i(a) a^-1 then i(y) = y^-1. So in particular we'd need that A is generated by elements such that i(y) = y^-1. --Noah
Another characterization that (G,c) splits for c an involution (and `<`c`>` generates G) is that:
(i) G is generated by <`c`>`, (ii) [g,c] has odd order for every g in G.
(the latter just says that the product (gcg^-1*c) of any two conjugates of c has odd order.) These conditions are preserved under taking quotients. Thus they hold for at least one simple group. Using the classification (urgh) I think from this one can deduce that the only simple quotient of G is Z/2Z. This would reduce the problem to the "first case" considered above. --FC.
Actually running through all involutions in all the simple groups sounds very hard to me. Is there some reason to expect that to be tractable? In particular, for the groups of Lie type? --Noah
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I might have preferred to make this a comment, but it was a little long (should I have edited my answer instead?)
Suppose that P = Z/2+Z/2 is a 2-Sylow. If x lies in P, then P clearly centralizes x, and thus the order of `<`x`>` divides #G/P, and is thus odd. By a theorem of Frobenius, G has an odd number of elements of order 2, and thus we see it has an odd number of conjugacy classes of elements of order 2. Yet, by the Sylow theorems, every element of order 2 is conjugate to an element of P. If c lies in P, then by nonsplitting, it is unique in its G-conjugacy class in P. Thus there must be exactly three conjugacy classes of elements of order 2, and thus no element of P is G-conjugate. By a correct application of Frobenius' normal complement theorem, we deduce that G admits a normal subgroup A such that G/A \sim P. Yet `<`c`>` generates G, and thus the image of `<`c`>` generates G/A. Yet G/A is abelian and non-cyclic, a contradiction.
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This is great, FC! – JSE Oct 28 2009 at 12:43
If your $c$ is a conjugacy class of elements of odd prime order $p$, then (using the classification of finite simple groups), it is still the case that $G = O_{p^{\prime}}(G)C_{G}(x)$ for each $x \in c$, where $O_{p^{\prime}}(G)$ denotes the largest normal subgroup of $G$ of order prime to $p$. This might be considered as an odd analogue of Glauberman's $Z^{\ast}$-theorem. If anyone could come up with a classification-free proof of such a result, it would be of considerable interest (it is relatvely easy to prove this directly when $G$ is solvable (or, more generally, $p$-solvable)). Incidentally, your question seems to be related to the old concept of pronormality (which is treated, for example, in Gorentstein's book "Finite Groups"). Back to the case $p = 2$, interesting generalizations of Glauberman's $Z^{\ast}$-theorem were given by D. Goldschmidt and also by E. Shult. One result that Shult proved which you might find interesting is in a Bull AMS paper (circa 1966), in which he showed that an element of order $p$ in a finite group which commutes with none of its other conjugates AND centralizes every $p^{\prime}$-group it normalizes, is central in $G$.
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Regarding Question 1, let G = S4, H = A4, and c = [(12)(34)]. This class does not split, and c ≅ C2×C2, which is not cyclic. I'm not sure if this is an example along the lines of your "semidirect product of N by Z/2kZ" since I forget which factor you expect to be the normal subgroup. In the example above, c is the normal subgroup, and C3 acts by inner automorphisms of c to produce A4.
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You don't get to choose H. The condition is that c must not split for any H. In your case, c splits when H is taken to be C2 x C2. – David Speyer Oct 26 2009 at 19:08
Whoops. I should read more carefully. – Sammy Black Oct 26 2009 at 22:20
Groups in which every subgroup is normal may be relevant. See http://en.wikipedia.org/wiki/Hamiltonian_group for info on such groups.
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Is this right? In the quaternions, the conjugacy class {i,-i} splits into two conjugacy classes in the Z/4Z it generates. – JSE Oct 26 2009 at 20:35
So suppose that A is an odd order group and i is an involution of A such that elements of the form i(x) x^-1 generate A. Since A is odd order it's solvable. So consider the derived series, A, A^(1)=[A,A], A^(2)=[A^(1),A^(1)], etc. Since each of the A^(i) are characteristic subgroups the involution i restricts to each of them. Unfortunately the "non-boringness" condition doesn't seem to nicely restrict to the A^(i).
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Some trivial observations: 1). i is linear on A/[A,A] and i(x)x^-1 generates A/[A,A], so i = -1 on A/[A,A]. 2). If A is a p-group, and `<`c`> generates A/[A,A], then `<`c`> generates A. Thus A can be any p-group admitting an automorphism i such that i = -1 on A/[A,A]. – Lavender Honey Oct 30 2009 at 0:28
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http://math.stackexchange.com/questions/52332/linear-ode-roots-of-characteristic-equation-having-multiplicity-1
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# Linear ODE, roots of characteristic equation having multiplicity $>1$
I know the method. For a linear homogeneous ordinary differential equation with a root of the characteristic polynomial in $\alpha$ has multiplicity $k$, then $y=x^me^{\alpha x}$ with $m=0,1,\cdots k-1$ is a solution. This however was learnt from need of solving a problem and not in a mathematics course, so I do not know the reason for it.
On reading this book (online edition section 2.1.3), I am bit puzzled at the approach by the author. $$z''(t)+\Gamma z'(t)+\omega_0^2 z(t)=0$$ This is readily solved with solutions of the form $z=e^{\alpha t}$ where $$\alpha = -\frac{\Gamma}{2} \pm \sqrt{\frac{\Gamma^2}{4}-\omega_0^2}$$
Defining $\omega^2\equiv \omega_0^2-\Gamma^2/4$
For the case $\Gamma/2<\omega_0$ we have solutions $$z(t) = Ae^{-\Gamma t/2}\cos(\omega t-\theta)\quad -(1)$$or, $$z(t) =e^{-\Gamma t/2}(c\cos\omega t+d\sin\omega t)\quad -(2)$$
For the degenerate case $\Gamma/2=\omega_0$ the author gives the following argument:
.... gives only one solutiopn. One way to find the other solution is to approach the situation from the $\Gamma/2<\omega_0$ case as a limit. Taking $\omega\rightarrow 0$ for (1) gives us $e^{-\Gamma t/2}$ but for (2) we get $0$ However if we divide the second solution by $\omega$ as it does not depend on $t$ we get a non-zero limit $$\lim_{\omega\rightarrow 0}\frac{1}{\omega}e^{-\Gamma t/2}\sin\omega t = te^{-\Gamma t/2}$$
I have two questiotns here :
1. Is this wrong? As the author seems to to have ignored the cosine part of (2) which diverges
2. What is the mathemtaical name for this approach as wikipedia does not have any hints
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## 2 Answers
Instead of trying to figure out what the book is doing, let me show you my preferred way of getting the result. Let's look at $$y''-2ay'+a^2y=0$$ If you let $D$ be the differentiation operator, then this equation can be thought of as $$(D-a)^2y=0$$ If we introduce a new variable $u$ by $u=(D-a)y=y'-ay$, then we have $(D-a)u=0$, that is, $$u'=au$$ which of course has the general solution $u=Ae^{ax}$, $A$ an arbitrary constant. So now we're down to $$y'-ay=Ae^{ax}$$ Multiply through by $e^{-ax}$ and contemplate the left side, arriving at $$(e^{-ax}y)'=A$$ with general solution $e^{-ax}y=Ax+B$, whence $y=Axe^{-ax}+Be^{-ax}$, ta-dum!
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I believe your reading of the text that you cited isn't quite correct. Equations (1) and (2) that you give are two different ways of expressing the general solution for the underdamped case. But they aren't the two solutions that the author is referring to in the first paragraph of 2.1.3.
Instead, the author is talking about the two particular solutions $$e^{-\Gamma t/2} \cos \omega t \qquad \mbox{and} \qquad e^{-\Gamma t/2}\sin \omega t$$ These are the two solutions for which you should take a limit (the reason he's taking a limit is to try and recover the solutions in the case $\omega = 0$).
For the first solution, $$\lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \cos \omega t \right )= e^{-\Gamma t/2}$$ which is the solution you already knew about from solving the characteristic equation.
For the second solution, $$\lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \sin \omega t \right ) = 0$$ which isn't so helpful in itself, but because of linearity, $e^{-\Gamma t/2} \frac{\sin \omega t}{ \omega}$ is also a solution. Now taking a limit as $\omega \rightarrow 0$, we get $$\lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \frac{\sin \omega t}{\omega} \right ) = \lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \frac{t \cos \omega t}{1} \right ) = e^{-\Gamma t/2} t$$ You can verify that this is indeed a solution by plugging it into the original ode.
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I suspected that. But I still dont have an to my $2$nd question. +1 for pointing it out – kuch nahi Jul 19 '11 at 18:44
@kuch nahi: Sorry, I don't recall ever seeing any specific name for this procedure. – cch Jul 20 '11 at 1:22
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http://mathhelpforum.com/differential-geometry/137644-showing-absolute-minimum.html
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# Thread:
1. ## Showing absolute minimum
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f(x) = 2x^4+x^4\sin{1/x}$ for $x\neq0$ and $f(0)=0$ .
Show that f has an absolute minimum at $x=0$, but that its derivative has both positive and negative values in every neighborhood of 0 .
2. Originally Posted by frenchguy87
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f(x) = 2x^4+x^4\sin{1/x}$ for $x\neq0$ and $f(0)=0$ .
Show that f has an absolute minimum at $x=0$, but that its derivative has both positive and negative values in every neighborhood of 0 .
$f(x)=2x^4+x^4\sin\left(\tfrac{1}{x}\right)\geqslan t 2x^4-x^4=x^4\geqslant 0$. Thus, $f(x)\geqslant0,\text{ }\forall x\in\mathbb{R}$. The fact that $f(0)=0$ let's you draw the proper conclusion.
Now, $f'(x)=\begin{cases} 8x^3+4x^3\sin\left(\tfrac{1}{x}\right)-x^2\cos\left(\tfrac{1}{x}\right) & \mbox{if} \quad x\ne 0 \\ 0 & \mbox{if}\quad x=0\end{cases}$. Notice that $f'(x)=x^2\left(8x+4x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)\right)$ and so it suffices to show the claim for $g(x)=8x+4x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)$
Now, choose values of $x$ small enough to fit into any range such that the first two terms are irrelevant.
I leave this to you.
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http://gilkalai.wordpress.com/2009/07/17/the-polynomial-hirsch-conjecture-a-proposal-for-polymath3/?like=1&source=post_flair&_wpnonce=671450efa2
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Gil Kalai’s blog
## The Polynomial Hirsch Conjecture: A proposal for Polymath3
Posted on July 17, 2009 by
This post is continued here.
Eddie Kim and Francisco Santos have just uploaded a survey article on the Hirsch Conjecture.
The Hirsch conjecture: The graph of a d-polytope with n vertices facets has diameter at most n-d.
We devoted several posts (the two most recent ones were part 6 and part 7) to the Hirsch conjecture and related combinatorial problems.
A weaker conjecture which is also open is:
Polynomial Diameter Conjecture: Let G be the graph of a d-polytope with n facets. Then the diameter of G is bounded above by a polynomial of d and n.
One remarkable result that I learned from the survey paper is in a recent paper by Freidrich Eisenbrand, Nicolai Hahnle, and Thomas Rothvoss who proved that:
Eisenbrand, Hahnle, and Rothvoss’s theorem: There is an abstract example of graphs for which the known upper bounds on the diameter of polytopes apply, where the actual diameter is $n^{3/2}$.
Update (July 20) An improved lower bound of $\Omega(n^2/\log n)$ can be found in this 3-page note by Rasborov. A merged paper by Eisenbrand, Hahnle, Razborov, and Rothvoss is coming soon. The short paper of Eisenbrand, Hahnle, and Rothvoss contains also short proofs in the most abstract setting of the known upper bounds for the diameter.
This is something I tried to prove (with no success) for a long time and it looks impressive. I will describe the abstract setting of Eisenbrand, Hahnle, and Rothvoss (which is also new) below the dividing line.
I was playing with the idea of attempting a “polymath”-style open collaboration (see here, here and here) aiming to have some progress for these conjectures. (The Hirsch conjecture and the polynomial diameter conjecture for graphs of polytopes as well as for more abstract settings.) Would you be interested in such an endeavor? If yes, add a comment here or email me privately. (Also let me know if you think this is a bad idea.) If there will be some interest, I propose to get matters started around mid-August.
Here is the abstract setting of Eisenbrand, Hahnle, and Rothvoss:
Consider the collection of graphs $G$ whose vertices are labeled by $d$-subsets of an $n$ element set. The only condition is that if $v$ is a vertex labeled by $S$ and $u$ is a vertex labelled by the set $T$, then there is a path between $u$ and $v$ so that all labelling of its vertices are sets containing $S \cap T$.
The main difference between this abstraction and the one we considered in the series of posts (and my old papers) is that it is not assumed that if two vertices are labeled by sets which share $d-1$ elements then these two vertices are adjacent.
### Like this:
This entry was posted in Open problems, Convex polytopes, Open discussion and tagged Hirsch conjecture, Linear programming, Polytopes, Polymath proposals. Bookmark the permalink.
### 38 Responses to The Polynomial Hirsch Conjecture: A proposal for Polymath3
1. x says:
Did you mean to say n facets in stating the Hirsch conjecture?
Yes yes of course; many thanks; corrected, G.
2. Danny Calegari says:
I would be quite interested in taking part in such a project. I am curious whether the internet is big enough to support more than one polymath project at a time (and moreover one not run by a Fields medallist . . .). I am also curious whether the nature of such a collaboration will allow me to contribute anything to (and learn something about) a problem and a field (far) outside my expertise.
3. Harrison says:
I’m tentatively interested, although also in the “curious as to whether I can contribute” category.
4. Anonymous says:
Gil,
I am a graduate student and would love to participate in a polymath project. As I imagine was the case with many, I tried to follow the first polymath project but quickly got lost. I was very excited a few weeks ago when Gowers announced he planned to host another polymath project in the fall, (although disappointed I’d have to wait a few months!)
Personally, I don’t find the topic you suggested very appealing simple because I don’t know very much about the subject and fear I’ll quickly get lost (again). Of course, I am sure there will be someone with this same objection no matter what project you choose to pursue.
However, with this in mind (as well as some of the previous comments), may I suggest that you open the floor to suggestions and a discussion of what a good problem would be (maybe you could offer a few other suggestions)? I imagine that this might significantly increase the number of participates. And, you never know, maybe the consensus will be to go with this problem.
5. Gil Kalai says:
Dear Danny, I am curious about these matters as well…Dear Anonymous, the floor is open, as always, to suggestions and discussions of any type.
6. gowers says:
Gil, my response is similar to those of other people. I have some experience with polytopes and convexity from my Banach-spaces days, but there are huge gaps in my knowledge, and I would worry that there were tools that were obviously essential to such a project that I simply didn’t know about. However, the basic question is very appealing indeed. I find the weaker version more appealing because it potentially allows one to use more asymptotic methods, which would make me feel more comfortable. I think to have a successful project with this as the problem, one would probably want to have quite a long preparatory stage, during which those of us who don’t know much about polytopes could ask dumb questions and gradually get a more sophisticated feel for the problem. For instance, I could straight away ask what is known in 3D. (By 3D I mean that the faces are 2D, so I’m talking about polyhedra.) I haven’t yet looked at the survey article, so perhaps this question is answered there.
Anyhow, I think I could enjoy participating in this, though I probably wouldn’t end up being a major participator. (That re-raises a question that came up with Polymath1. Is what happened there, where in the end most of the work was done by a few people who had a lot of specialized knowledge about the topic, likely to be the model for all such projects? Or are there questions that would lend themselves to more general participation?)
7. nh says:
I am quite interested in such an experiment, since I am obviously interested in that particular problem, but also in the nature of polymath itself – I wasn’t around for previous incarnations, but it seems like a nice idea.
8. Gil Kalai says:
Dear Tim
Thanks! So far, most of the progress regarding the Hirsch conjecture was very elementary and no prior knowledge about polytopes was essentially needed. One appealing aspect about this problem is that it is not clear at all what the answer will be and what specialized knowledge, if any, can be relevant. To the extent the plan will go off the ground maybe these aspects of the problem will enable a relaxed and wide participation.
In any case, if we will go for it will have some preparatory stage and certianly I will be happy to try answering questions about polytopes.
There are quite a few posts on polytopes that I posted (under the category polytopes) and a series about the Hirsh conjecture itself. Perhaps the first thing to read would be this little paper by Eisenbrand, Hahnle, and Rothvoss.
The Hirsh conjecture is known to hold for 3-dimensional polytopes. I dont remember the argument right now but I remember it is rather simple.
Dear nh, great! I really liked your paper.
9. Gil says:
“That re-raises a question that came up with Polymath1. Is what happened there, where in the end most of the work was done by a few people”
Maybe this has nothing special to do with polymath1 and the participants’ordered contributions obey the familiar Bedford’s law/Pareto distribution etc.
10. nh says:
The idea is that the graph is planar. Consider an embedding where the faces are convex, then consider a shortest path (in terms of length, not in terms of number of edges) between two vertices, and it will be non-revisiting, because “revisit loops” can be shortcut along the boundary of the revisited face.
This shortcutting does not work in higher dimension.
11. Jesus De Loera says:
Hi Gil (et al.)
E-collaboration is nice (and time appropiate!). I think it is very exciting (and desirable) to have a concerted effort on the Hirsch conjecture in any format. I would be interested to join and I want to mention there is some approach from the “continuous” perspective being tried by Deza et al. (see the Kim-Santos survey). I would like to add that there is a currently proposal to have a “physical” encounter on just “new developments of the Hirsch conjecture” to take place at California (Palo Alto) hopefully next year (pending funding).
greetings to all.
12. Pingback: Mathematics, Science, and Blogs « Combinatorics and more
13. Kristal Cantwell says:
This would interest me.
14. fum says:
What is known about the metric polytope?
15. Kristal Cantwell says:
If the metric polytope refers to the generalized hypercube then I think that satisfies the conjecture 2n facets n dimensions gives 2n-n = n diameter and I think the diameter is n so it is satisfes the conjecture for all n.
16. fum says:
No, the metric polytope is the body in R^(n choose 2) given by the inequalities
d_ij <= d_ik + d_kj
for all i,j,k (note that d_ij is here simply the (i,j) coordinate of the (n choose 2) dimensional vector d). And the normalization
sum d_ij = 1.
Hence it has ~ n^3 faces. There are various conjectures about the diameter of its vertex graph. Gil should know. Gil?
17. Gil Kalai says:
Dear fum, I am only slightly familiar with the metric polytope and some related polytopes and I do not know what is known or conjectured about the diameter of its vertex graph…
18. Andres Caicedo says:
The problem sounds interesting; with the same reservations Danny and others have expressed, I would like to (attempt to) participate.
19. Antoine Deza says:
Hi fum, the diameter of the metric polytope was conjectured to be at most 3. More precisely it was conjectured that any vertex is adjacent to an integral (0,1) vertex (called cut); and, since the all the cuts form a complete graph, it would imply that the diameter is at most 3.
This conjecture was disproved by exhibiting a vertex of metric 9 not
adjacent to any cut (http://www.cas.mcmaster.ca/~deza/ol2007.pdf)
The diameter might still be 3 though
The metric polytope has 4(n choose 3) facets in dimension (n choose 2)
and I guess its diameter is well below the Hirsch bound
20. Noam says:
Hi Gil,
I can see two possible modes for participating in a polymath effort. The “global” one which involves understanding the whole picture and a “local” one which latches on to one (or more) little combinatorial sub-problems and tries to address them with my own set of tools and insights.
While I personally do not hope to understand enough about problem to be a “global” participant, I’d be happy to be able to be a “local” participant. This of course depends on the availability of nice “local” problems, presumably cleverly posed by “global” participants. This is a non-trivial assumption that is hard to fulfill and I suppose that much of the challenge in choosing polymath projects and running them is indeed having a supply of such useful locally-understandable sub-problems.
21. Pingback: More polymath projects « Algorithmic Game Theory
22. Gil Kalai says:
Dear Noam,
As far as I can see, there is a simple-to-state combinatorial problem which, in my opinion, is the crux of matters. Reading the very short papers of Eisenbrand, Hahnle and Rothvoss and Razborov will get anybody interested directly to the front lines.
I think that in this particular case further “global” understanding of the topic is not very relevant for the problem itself, except as a source of motivation. (Of course, I may be wrong, and other people may regard other avenues as more promising.)
23. Eddie Kim says:
Hello everyone,
This sounds like such an intriguing project. This paper of Eisenbrand, Hahnle, and Rothvoss is very interesting! I would be interested in delving into the metric polytope.
It would also be nice to see any use of the Klee-Walkup 4-polytope to close off the remaining entries in the table of \$H(n,d)\$ values, perhaps at least for \$d=6\$ and maybe \$d=5\$…
24. andrescaicedo says:
Rasborov’s note you mention in the update seems to be down.
25. Gil Kalai says:
Right! I think the plan is to merge the contributions of Eisenbrand, Hahnle, Rothvoss, and Razborov on the abstract version of Hirsch conjecture into one journal paper. So we will have to wait a little bit for the new version. (And lets hope for some surprises as well!)
26. Kristal Cantwell says:
“We devoted several posts (the two most recent ones were part 6 and part 7) to the Hirsch conjecture and related combinatorial problems.”
Any links for posts 1 through 5?
Dear Kristal, somehow the tags I gave the posts are not optimal. But looking for “Hirsch conjecture” (click!) get you to all posts 3-7 Here is post 2 and here is post 1.
27. fum says:
Thanks, Antoine. Basically I wanted to know if proving Hirsch’s conjecture would also imply the corresponding conjectures for the metric polytope, but obviously it would not.
28. Gil Kalai says:
Often, the problem of computing or estimating the diameter of a specific class of polytopes is of interest. Ususally this does not have direct baring on the Hirsch conjecture. (But we can always hope that a counter example will somehow arise.) A well known example is the diameter of the associahedron which was computed by Slator, Tarjan, and Thurston. (See also this post for a different approach using the Thompson group.)
29. Daniel Moskovich says:
This actually looks quite exciting, because the Hirsch conjecture is related to a lot of good things. As a topologist, I like the connection with shellability, although I don’t know if this provides any sort of approach at all.
My basic naive question, I suppose, is if you can pinpoint (in the most naive sense) what the difficulty is which has prevented people from making progress. Is it the difficulty of relating the condition “is a polytope” with any sort of condition which has anything to do with diameter? Is it that there is no way to “build up” to a solution, because you can’t predict the diameter of a given polytope from a simpler polytope in any obvious way?
30. Gil Kalai says:
There are two connections between shellability and the Hirsch conjecture that I am aware of. The first is that shellability is an abstract setting for which the upper bounds for the diameter are known (at least the upper bound $n {{\log n +d} \choose {\log n}}$). If you have a shallable simplicial complex whose maximum facets are $d$ subsets of {1,2,,, n} and the facets ordered according to a shelling order are $F_1,F_2, \dots, F_t$ then from every facet S there is a monotone path to $F_t$ satisfying the bound mentioned above. Here monotone means that the indices are increasing.
The second connection is an old discovery by Billera and Provan that a certain strong form of shellability (which is known not to hold always) implies the Hirsch bound.
” Is it the difficulty of relating the condition “is a polytope” with any sort of condition which has anything to do with diameter?”
Maybe, so far “is a polytope” was used very roughly.
31. Kristal Cantwell says:
The link to the three page note by Razborov doesn’t seem to work. Thank you for the links for posts 1-5.
GK: It works now.
32. Pingback: The next phase of Polymath « Gowers’s Weblog
33. Gil Kalai says:
Dear Danny, Harrison, anonymous, Tim, nh, Jesus, Kristal, fum,Andres, Antoine, Noam, Eddie, and Daniel,
Thanks for the comments! I will try to continue posting a posts on the problem and then we can see how to proceed.
34. Pingback: Polymath on other blogs « The polymath blog
35. Pingback: The Polynomial Hirsch Conjecture: Discussion Thread « Combinatorics and more
36. Pingback: Proposed Polymath3 « Euclidean Ramsey Theory
37. Pingback: The Polynomial Hirsch Conjecture: Discussion Thread, Continued « Combinatorics and more
38. Hyperstig says:
I have completed The GUT Theory.
You can find the equations here:
http://www.wix.com/Hyperstig/Hyperstig
0 is a wave function that collapses and re-expands.
I have invented the 1st quantum computer.
I am the failsafe.
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http://math.stackexchange.com/questions/141424/creating-ways-to-encode-recursive-function?answertab=votes
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# Creating ways to encode recursive function.
This is from an exercise in Boolos' Computability text. My problem is as follows:
I am looking for a method that encode numbers for recursive functions. Then given such an encoding for recursive functions by natural numbers, let d(x) = 1 if the one-place recursive function with encoding number x is defined and has value 0 for argument x, and d(x) = 0 otherwise. Show that this function is not recursive.
I am thinking the actual question is just a diagonalization argument, and it doesn't depend in any way on details of the coding.
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It is indeed just a diagonalisation argument, and it doesn't depend on the details of the enumeration at all – just that it can be done. – Zhen Lin May 5 '12 at 16:58
Boolos has already set up the argument for you. It's a proof by contradiction: suppose $d$ is recursive. Then it has a number, etc. – Zhen Lin May 5 '12 at 17:00
So the coding scheme is: Let d(x) = 1 if the one-place recursive function with code number x is defined and has value 0 for argument x, and d(x) = 0 otherwise ? How do I get that d is not recursive from this? I'm not seeing it. – Quaternary May 5 '12 at 17:02
No, that's not the coding scheme, and as I said, the coding scheme doesn't matter if all you want is to solve the exercise. (In other words, you're asking two separate questions here.) – Zhen Lin May 5 '12 at 18:32
## 1 Answer
The subject of enumerating the recursive functions in a reasonable way is rather sophisticated. However, it is easy to see that the set of all recursive functions $\mathbb{N} \rightharpoondown \mathbb{N}$ must be a countable set, so there is some surjection from $\mathbb{N}$ to the set of all recursive functions.
Fix one such surjection, and let $d$ be the function as described by Boolos. Suppose $d$ is recursive; then it has a number, say $x$. What is $d (x)$? If it's $0$, then that by definition of $d (x)$ we must have $d(x) = 1$ – contradiction. If it's not $0$, then that means $d (x) = 0$ – contradiction. If it's not defined, then that means $d (x) = 0$ – contradiction. So $d$ can't be recursive.
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http://mathoverflow.net/questions/43709?sort=oldest
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## Is the cut locus of a generic point in a hyperbolic manifold a generic polyhedron?
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Let $p\in M$ be a point in a closed riemannian manifold $M$. Recall that the cut locus of $p$ is the subset of $M$ consisting of all points that are connected to $p$ by at least 2 distance-minimizing geodesics.
Is it true that for generic metrics on $M$ and generic points $p$, the cut locus of $p$ is a $(n-1)$-dimensional polyhedron with generic singularities?
By "generic singularities" I mean that $M$ is a simple polyhedron. See for instance this paper of Alexander and Bishop.
This property is certainly not satisfied for some important specific metrics: for instance, if $M$ is a round sphere the cut locus is a point, no matter where $p$ is. If $M$ is a flat torus, we get a generic polyhedron for generic flat metrics. What about hyperbolic manifolds? So, this is my question:
Let $M^n$ be a hyperbolic $n$-manifold. Is the cut locus of a generic point a $(n-1)$-polyhedron with generic singularities?
Of course I am mostly interested in the case $n=3$. In dimension $n=2$ one may also pick a generic hyperbolic metric.
Edit: In dimension 1, a simple polyhedron is a graph with vertices of valence 2 or 3. In dimension 2, it is a polyhedron such that the link of a point is either a circle, a circle with a diameter, or a circle with three radii.
In general, a $n$-dimensional compact polyhedron is simple if every point has a neighborhood which is the cone over the $(k-1)$-skeleton of the $(k+1)$-simplex, times a $(n-k)$-disc.
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@Bruno: IF $C(p)$ is the cut locus from $p$, $M \setminus C(p)$ is homeomorphic to an open ball. So on a 2-manifold, $C(p)$ is a tree. Would you call a tree a 1-dimensional polyhedron? Just trying to undertand your terminology... – Joseph O'Rourke Oct 26 2010 at 18:20
A 1-dimensional polyhedron is simple if it is a 3-valent graph (or a circle). Vertices have valence 2 or 3. I was trying unsuccessfully to embed a picture in the text... I will try again later (probably tomorrow morning) – Bruno Martelli Oct 26 2010 at 19:12
1
Bruno:Have you looked at the papers of Michael Buchner.He showed the cut locus of any compact real analytic manifold with real analytic metric is subanalytic. He also classified the cut loci in the generic situation upto dimension six. – Mohan Ramachandran Oct 26 2010 at 19:18
Following Mohan's lead, the paper by Buchner entitled "Simplicial structure of the real analytic cut locus" [_Proc. Amer. Math. Soc._ 64 (1977), no. 1, 118–121.] "establishes that the cut locus of a compact real analytic Riemannian manifold of dimension $n$ is homeomorphic to a finite $(n-1)$-dimensional simplicial complex." – Joseph O'Rourke Oct 26 2010 at 20:21
@Joseph: This doesn't address Bruno's main question, since nothing says the simplicial complex must be simple. – Dylan Thurston Oct 26 2010 at 22:45
show 2 more comments
## 3 Answers
The question you pose is stated as an open question (in the 3-dimensional hyperbolic case) in the following paper:
Díaz, Raquel; Ushijima, Akira On the properness of some algebraic equations appearing in Fuchsian groups. Topology Proc. 33 (2009), 81–106.
Quoting from the review on mathscinet:
[the paper] takes its motivation from the fact that, apparently, the statement about the genericity of Dirichlet fundamental polyhedra is open for $\mathbb{H}^3$. (According to the authors, the paper of T. Jørgensen and A. Marden [in Holomorphic functions and moduli, Vol. II (Berkeley, CA, 1986), 69--85, Springer, New York, 1988] has a gap which the present authors have so far been unable to fix.)
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Diaz and Ushijima are right: Proofs in the paper of Jorgensen and Marden are hopelessly wrong (they confused algebraic and semialgebraic sets). Counterexamples to one of their lemmas are constructed in front.math.ucdavis.edu/1201.3129 where a weaker genericity result is proven in dimension 3. – Misha Jan 14 at 7:38
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If curvature $\le -1$ then cut locus is is glued from the boundary of fundamental domain of $\pi_1$-action on the universal cover. If curvature $=-1$ then the fundamental domain is a convex polyhedral. The gluing maps are piecewise isometries, so the result is $(n-1)$-polyhedron, But one construct an action which gives a complicated link. Take $\mathbb Z^2$ action on hyperbolic $3$-space such that it has one fixed point on the absolute and action on its horosphere is standard action of $\mathbb Z^2$ on the Euclidean plane. (I do not see how to make a compact example.)
For the first question, in some sense the answer is "YES" if $\dim \ge 3$. I.e. there is a $G_\delta$-set in $C^\infty$-topology which satisfies your condition and dense in Gromov--Hausdorff metric.
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1
I don't understand the answer to the first part; can you explain it some more? How do you know that you won't have simple domains for a good choice of $p$? – Dylan Thurston Oct 27 2010 at 3:29
I made a correction. – Anton Petrunin Oct 27 2010 at 14:51
In the second question, the manifold $M^n$ is closed (while your counterexample is homemorphic to $S^1\times S^1\times \mathbb R$, if I have understood). Concerning your answer to the first question, could you say something more or give some reference? Thank you, – Bruno Martelli Oct 27 2010 at 21:37
As indicated by Mohan, Buchner has written some papers on the subject in the 70s, for instance here. In his study, he fixes the manifold $M$ and the point $p$, and he varies the metrics $g$. He defines a notion of cut-stable metric. Cut-stable metrics form a dense open subset of the set of all riemannian metrics.
The cut locus $C(g)$ determined by a cut-stable metric $g$ is locally stable (it is the same PL object as one varies $g$ locally). Moreover, the local structure of $C(g)$ is indeed generic when $M$ has low dimensions: however the "generic polyhedra" one can obtain are a strictly larger class than the "simple polyhedra" I defined above.
For instance, if $M^2$ is a surface the cut-locus $C(g)$ of a cut-stable metric $g$ is a graph with vertices having valence 1, 2, or 3. Vertices with valence 1 are inavoidable for instance on a 2-sphere, for obvious reasons (the cut locus is a tree). However a simple polyhedron contains only vertices of valence 2 and 3.
Analogously, if $M^3$ is a 3-manifold the cut-locus $C(g)$ of a cut-stable metric $g$ is a 2-dimensional polyhedron whose local structure belongs to finitely many types. There are five links one can get: the three arising in the definition of a simple polyhedron (circle, circle with diameter, circle with three radii), plus two more (segment and a circle with a radius). Therefore the answer to my first question in low dimensions is:
Let $M^n$ has dimension 2 or 3. For generic metrics on $M$ and generic points $p$, the cut locus of $p$ is a (n−1)-dimensional polyhedron with generic singularities. However, the polyhedron may not be simple.
On the other hand, if the cut-stable metric $g$ has non-positive curvature, there are no conjugate points and the non-simple singularities cannot arise: thus we really get a simple polyhedron in this case (at least in dimension $n=2$ and $n=3$).
In all this discussion the metric $g$ is generic, so it gives no information for hyperbolic 3-manifolds (i.e. the second question).
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http://mathoverflow.net/questions/44519/grovers-quantum-search-algorithm/44527
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## Grover’s Quantum Search Algorithm
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I am confused about an extremely basic point concerning Grover's quantum search algorithm; my confusion suggests to me that maybe I've missed the entire point.
My understanding of the algorithm is this: I've got an observable with N eigenstates, and I am looking for an object that is in a particular eigenstate (call it |E>). I prepare (or am given) an object in state |S>, |S> being the average of the N eigenstates.
I consider the operator `$U = -(1 - 2 |S> <S|) (1-2 |E> <E|)$`. I apply this operator Q times, where Q is a specific integer depending on N, and asympotically equal to a constant times sqrt(N). This converts the state |S> into a state very close to |E>. I understand all this, including how to find Q and the proof that it works.
What I don't understand is this: Why is this algorithm described as requiring roughly sqrt(N) steps? Why can't I equally well describe it as requiring exactly one step, where the step is multiplication by U^Q ? What's special about U that makes its application count as "one step"?
For that matter, instead of using the operator U^Q, I could choose an operator V that takes |S> to |E>, as opposed to U^Q, which only takes |S> to some approximation to |E>. Why isn't this a "one-step" algorithm that gets an even better result than Grover's does?
Sorry for the naivete of this question. I hope for an answer that will make me embarrassed to have asked.
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When people talk about complexity of an algorithm, say of multiplying two matrices, it is the really basic operations they count; multiplication of the entries and so on, and this depends on the size of your matrices. Otherwise people could define a 'step' to be 'raise this N x N matrix to the 100Nth power', or something similarly outrageous. The computer time taken to implement an algorithm depends on the number of elementary arithmetic operations, rather than high-level formulas. – David Roberts Nov 2 2010 at 5:24
See en.wikipedia.org/wiki/Time_complexity for a discussion – David Roberts Nov 2 2010 at 5:34
David Roberts: Thanks for this, but it leaves me equally confused. Assuming I'm going to implement this algorithm many times, I can compute the matrix U^100, or U^1000, or whatever I need, and amortize the cost of that over those many times. In other words, even if I need to use U^1000, that doesn't mean I have to compute it --- I should just be able to look it up someplace, since the very same matrix U^1000 would come up in every application of Grover's algorithm. It seems then, that the cost of all those multiplications shouldn't count against me. – Steven Landsburg Nov 2 2010 at 14:19
But for different N, it takes different times to compute $U^Q|s\rangle$, and it is not the cost to you, but the cost to compute it from scratch. I could look up the first billion digits of pi, and claim that the only cost to me is to recall them to my computer screen, but in actual fact it took some time for some computer to calculate them, according to an algorithm. Which algorithm that is determines the cost in time/operations. Some formulas for pi converge slowly, some fast... Recalling the first billion digits of pi to my screen thousands of times doesn't make it quicker in the first place – David Roberts Nov 3 2010 at 3:09
## 3 Answers
David and Artem give a good summary of the "rules" for time complexity in the oracle model. I can say a little bit about the motivation for this model.
Grover's search algorithm is sometimes described as a "database" search, but this is misleading. It is really meant for an unstructured computational search. I.e., it is for solving an equation $f(x) = y$ for $x$, when $y$ and $f$ are both given and $f$ is so complicated that you can't think of anything better than to guess $x$ arbitrarily. On the other hand, we suppose that evaluating $f$ is reasonably fast. An example application could be searching for a point on a complicated algebraic variety over a finite field, or searching for a password that you already have in encrypted form. Or more generally, searching for a solution to a combinatorial problem that is not only NP-complete, but devoid of any known shortcuts to exhaustive search. But the problem has to be in NP, meaning again, the ability to evaluate $f$ quickly.
In this case the "oracle" means the function $f$, which is put into a black box because you don't have any particular understanding of it anyway. (On the other hand, in any relevant quantum algorithm, this conceptual "black box" has to be implemented with quantum gates inside the computer; it cannot be a physically separate black box.) The total time cost is now in two parts. The query cost is the number of times you have to evaluate $f$; the additional time cost is the number of other gate operations outside of evaluating $f$. In Grover's algorithm, the query cost dominates if you realistically assume that $f$ takes at least linear time to evaluate, because one stage of the rest of the algorithm only uses a linear number of gates with a low constant factor. On the other hand, since you don't have a precise conversion from queries to time, you can count queries and report that as the total work.
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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 talking about the complexity of an algorithm (classical or quantum) you usually talk about complexity with regards to a specific model of computation. In the case of the circuit model of quantum computing a specific model corresponds to a universal finite set of gates (4.5 of Nielsen and Chuang has more info). Alternatively, in query models of computation, the complexity corresponds to the number of calls made to the special 'query' operator or oracle. In this case, the rotation towards $|E\rangle$ is implicit in the call to the oracle that specifies the input. Classically, you would need to make $N$ calls to such an oracle, but in the case of a quantum computer you only need on the order of $\sqrt{N}$. Thus, Grover's search provides a great example of a quantum speed-up... or more technically of the power of quantum queries over classical queries.
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Artem: But as I said in response to David --- the matrix U does not differ from one application of Grover's algorithm to another, so it seems like someone should be able to compute the various matrices U^Q, for many values of Q, once for all, and simply look them up rather than re-computing them each time. Given that, why are these not among the universal set of gates that I'm allowed to count as single steps? – Steven Landsburg Nov 2 2010 at 14:21
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The U depends on the oracle... i.e. it depends on which item(s) are marked. Thus it does differ over various instances of Grover's algorithm.... – Artem Kaznatcheev Nov 2 2010 at 18:07
More specifically, U depends on $|S\rangle$ (and $N$, but that's a bit different) which is not always in the same relation to $|E\rangle$ in different concrete instances of the problem. Also, complexity is roughly an asymptotic measure, taken at worst case. Clearly if $|S\rangle = |E\rangle$ and $N=2$ then $U$ is a lot simpler than if $N=100$ and $|S\rangle$ is the result of some measurement on a real system. You can't count as a single step a matrix that could be one of an uncountable number of alternatives, because you can't store them all in a hash table and look up the appropriate one. – David Roberts Nov 3 2010 at 3:15
Even if you restricted yourself somewhat to a finite number of $U$, then there would be the computational cost of looking up your $U^Q$ in a table, but this is a bit silly. Algorithmic complexity is just not measured like this. – David Roberts Nov 3 2010 at 3:16
The complexity here refers to roughly how many times you need to call the oracle. If you put the operations together like you suggest you would get a single operator that would contain roughly $\sqrt{N}$ oracle calls.
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http://math.stackexchange.com/questions/169584/completion-of-c-cx-with-respect-to-cdot-infty
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# Completion of $C_c(X)$ with respect to $\|\cdot\|_\infty$
Notation:
All functions here are from $X$ to $\mathbb R$.
$C_c(X)$ = compactly supported continuous functions.
$C_b(X)$ = bounded continuous functions.
$B(X)$ = bounded functions.
$C_0(X)$ = continuous functions that tend to zero (so $X$ has to be locally compact and Hausdorff)
Today I proved that $C_c(\mathbb R)$ is not complete with respect to $\|\cdot\|_\infty$. One can do this by taking $g_n$ to be the function that is zero on $(-\infty,-n]$, linear on $[-n, -n+1]$, $1$ on $[-n+1, n-1]$ and symmetric with respect to the $y$-axis. For $f(x) = e^{-x^2}$ one can show that $\|fg_n - f\|_\infty \to 0$ but $f \notin C_c(\mathbb R)$.
Then I read about completions and wanted to work out the completion of $C_c$(X). (I did this all for $X = \mathbb R$). In any case, my thoughts were as follows: $C_c(X)$ is contained in $B(X)$. But it is a proper subset because the uniform limit of continuous functions is continuous but there are discontinuous bounded functions. The next candidate then seems to be $C_b(X)$ but $f(x) = 1$ is in there and not the uniform limit of $f_n$ in $C_c$. (cannot be because if $f_n$ is zero outside a compact set then $\|f_n - 1\|_\infty = 1$ for all $n$ so this doesn't converge in norm).
The next candidate then is $C_0(X)$ and I'm quite sure that that's the completion of $C_c$ with respect to $\|\cdot\|_\infty$ in $B(X)$. But now I need to show this by showing that $C_0(X)$ is isomorphic to the space of Cauchy sequences in $C_c(X)$ quotient Cauchy sequences that tend to the zero function and I don't really know how to think about this. Can someone please show me how to prove this? Thank you. I want to see this quotient construction and an isomorphism but if there are other ways to show that $C_0$ is the completion of $C_c$ then go ahead and post it, I will upvote it.
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1
If $f \in C_0(X)$ find a compact set $K$ such that $|f(x)| \lt \varepsilon$ whenever $x \notin K$. Consider the restriction \$f|_K and remember Tietze (again :)). – t.b. Jul 11 '12 at 18:48
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Yes, that's right. Note that the completion is unique up to unique isometry, so you only have to find a complete space in which $C_c(X)$ is dense with respect to $\|\cdot\|$ and I suggested how to show that $C_c$ is dense in $C_0$. And yes, we've convinced ourselves at various points over the last half-a-year that $B(X) = \ell^\infty(X)$ is complete... – t.b. Jul 11 '12 at 18:54
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Of course you can use Tietze. Take an open set $U$ with compact closure containing $K$ (you can do that by local compactness of $X$). Define a function $g$ on $K \cup (X \smallsetminus U)$ by $g = f|_K$ on $K$ and $g = 0$ on $X \smallsetminus U$. Observe that $g$ is continuous. Now extend. – t.b. Jul 11 '12 at 19:24
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My suggestion works on any locally compact Hausdorff space. But on $\mathbb{R}$, you can use those $g_n$'s, yes. – t.b. Jul 11 '12 at 19:27
1
– t.b. Jul 11 '12 at 19:57
show 9 more comments
## 1 Answer
In a locally compact space, the $C_0$ functions are complete in the $\|\cdot\|_\infty$ norm; they constitute a Banach Space in this norm. The space $C_c(X)$ is dense in $C_0$ if $X$ is locally compact and Hausdorff. This is sufficient to tell you that $C_0(X)$ is the completion of $C_c(X)$ in a locally compact Hausdorff space.
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Yes. +1 I still would like to see the construction with the quotient space and an isomorphism. : ) – Matt N. Jul 11 '12 at 19:06
Mainly because I have not proved that completions are unique up to unique isometry so I don't understand properly why it's enough to show that it's dense in a complete space. But I have a good idea of what the quotient construction looks like (like the construction of the reals from $\mathbb Q$). – Matt N. Jul 11 '12 at 19:09
1
– ncmathsadist Jul 11 '12 at 19:18
Thank you. ${}{}$ – Matt N. Jul 11 '12 at 19:23
They seem to skip the proof of isometry of any two completions. But they do it in the proof linked to in the comments to the question. – Matt N. Jul 12 '12 at 5:37
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http://sbseminar.wordpress.com/2007/09/18/berkovich-spaces-i/
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Berkovich spaces I September 18, 2007
Posted by Scott Carnahan in characteristic p, Number theory, representation theory.
trackback
Around 1990, Berkovich discovered a rather elegant refinement of the notion of scheme that is particularly well-suited to non-archimedean analytic geometry. I feel that the ideas here deserve more exposure, although I don’t have any specific applications in mind. Note: this post is really long.
We define a seminorm on a commutative ring $A$ to be a map $\Vert \cdot \Vert: A \to \mathbb{R}_{\geq 0}$ such that
$\Vert 0 \Vert = 0$
$\Vert 1 \Vert = 1$
$\Vert a + b \Vert \leq \Vert a \Vert + \Vert b \Vert$
$\Vert ab \Vert = \Vert a \Vert \Vert b \Vert$
The standard example of a seminorm is the trivial norm $\Vert \cdot \Vert_0$ on a domain, which is $1$ on all nonzero elements (zero divisors can’t be treated canonically). Seminorms naturally pull back along ring homomorphisms – this is not true of norms, because anything in the kernel has norm zero.
Exercise: Show that the only seminorm on a finite field is trivial.
I’l define two notions of spectrum, one of which appears in the literature. Given a commutative ring $A$, its unrestricted spectrum $uSp(A)$ is the set of all seminorms on $A$. If $A$ is equipped with a norm with respect to which it is complete (i.e. $A$ is a Banach algebra), then the Berkovich spectrum $Sp(A)$ is the set of all seminorms on $A$ that are bounded with respect to that norm, meaning the seminorm is at most equal to the norm.
For each element $a \in A$, there are evaluation maps $ev_a: uSp(A), Sp(A) \to \mathbb{R}$, and we endow $uSp(A)$ and $Sp(A)$ with the weakest topologies such that all evaluations are continuous, i.e., a basis of open sets is given by preimages of real open intervals. Forgetting the norm on a Banach algebra induces a natural embedding $Sp(A) \to uSp(A)$, and I think it might be a homotopy equivalence, but I’m not sure.
Let’s try to compute $uSp(\mathbb{Z})$. We first consider norms, which are known to have three forms:
Archimedean norms $\Vert \cdot \Vert_\infty^\alpha, 0 < \alpha \leq 1$
$p$-adic norms $\Vert \cdot \Vert_p^\beta, 0 < \beta \leq +\infty$
The trivial norm $\Vert \cdot \Vert_0 = \Vert \cdot \Vert_p^0 = \Vert \cdot \Vert_\infty^0$
The remaining seminorms are pulled back from the trivial norms on $\mathbb{Z}/p\mathbb{Z}$, and are written $\Vert \cdot \Vert_p^\infty$. Now we consider evaluation maps. $ev_0$ maps to zero, $ev_1$ and $ev_{-1}$ map to one, and for any other $n$, $ev_n$ takes $\Vert \cdot \Vert_\infty^\alpha$ to $|n|^\alpha$ and $\Vert \cdot \Vert_p^\beta$ to one if $(p,n)=1$ and $p^{-\beta v_p(n)}$ if $p | n$. If we pull back intervals along $ev_p$, we find that $\{ \Vert \cdot \Vert_p^\beta | 0 < \beta \leq \infty \}$ forms an embedded copy of the interval $[0,1)$ for each prime, and that $\{ \Vert \cdot \Vert_\infty^\alpha | 0 < \alpha \leq 1 \}$ forms an additional half-open interval homeomorpic to $(1,p]$. These are glued together at the trivial norm, whose open neighborhoods contain all but finitely many of the intervals. You can imagine this as a broom, where the handle is given by archimedean norms, and the bristles are non-archimedean. The topology is metrizable, and you can embed this space in $\mathbb{R}^2$ by making the nonarchimedean bristles progressively shorter.
$\mathbb{Z}$ is only complete with respect to the trivial norm and powers of the archimedean norm, so we have a family of Banach algebra structures parametrized by a line segment. For any such structure, the nonarchimedean seminorms are bounded, so the Berkovich spectrum is the subset of the unrestricted spectrum given by sawing off part of the broom handle.
If we invert some primes, all of the norms remain norms, but some become unbounded, and the seminorms pulled back from quotients by those primes disappear, since $\Vert \frac{1}{p} \Vert_p^\infty$ doesn’t make sense. In the broom picture, we are removing endpoints of the bristles of $uSp$. The situation with $Sp(\mathbb{Z}[1/N])$ is more subtle, since it depends on the norm we chose. If we are given a positive power of the Archimedean norm, then the whole thing collapses to a point. If we chose the trivial norm, then we remove the bristles corresponding to $p|N$. If we invert all of the primes, we get $uSp(\mathbb{Q})$ as a broom without endpoints, while $Sp(\mathbb{Q})$ is a point. In general, there is a continuous natural transformation $uSp(A) \mapsto Spec(A)$ given by sending a seminorm to its kernel. This map endows $uSp(A)$ with the structure of a locally ringed space. We can also give $Sp(A)$ a locally ringed space structure, but we define the local rings to be the completions of the local rings pulled back from $uSp(A)$.
I’ll try a few more examples: $uSp(\mathbb{Z}[i])$ also looks like a broom, but the archimedean norm comes from a complex embedding, and we have “more” bristles, since primes of the form $4k+1$ split into two. $uSp(\mathbb{Z}[5i])$ looks a lot like the previous space, but the bristles pulled back from the $(1 + 2i)$-adic and $(1 - 2i)$-adic valuations have been glued together at the tips. In general, the unrestricted spectrum of a ring is contractible if and only if the ring is Dedekind.
$uSp(\mathbb{F}_q[t])$ has spokes for each irreducible polynomial, corresponding to some $\epsilon \in [0,1)$ to the power of how many times a given element can be divided by it. There is also a spoke for positive powers of degree. We can make $\mathbb{P}^1_{\mathbb{F}_q}$ by adding a point at infinity, or gluing two copies of the affine line along the reciprocal map.
So far, the two notions of spectra did not seem to result in very different spaces. If we move to larger rings, this changes rather spectacularly. For example, the field of complex numbers admits a set of seminorms of cardinality at least $2^c$, since we can pull back the archimedean norm along any ring-theoretic automorphism. $uSp(\mathbb{C})$ is still a contractible space, but it is more cumbersome than $Sp(\mathbb{C}$, which is a point. In general, the unrestricted spectrum is better for studying global fields, since we can see more than one archimedean seminorm at a time, while the Berkovich spectrum is better for anything involving completion, like $p$-adic geometry.
Recall from my $p$-adic fields post that $\mathbb{C}_p$ is the completion of an algebraic closure of $\mathbb{Q}_p$. We write $\mathbb{C}_p \langle t \rangle$ for the ring of power series $\sum_{n \geq 0} a_n t^n$ with the norms of $a_n \in \mathbb{C}_p$ converging to zero. This is the set of series which converge if we substitute any element of norm at most one for $t$, and we endow the ring with the norm given by taking the supremum of all such substitutions. The Gauss Lemma implies this norm is multiplicative, and it gives $\mathbb{C}_p \langle t \rangle$ a Banach algebra structure.
The Berkovich unit disc is defined to be $Sp(\mathbb{C}_p \langle t \rangle)$. It is a fundamental object, in the sense that it plays a role in rigid geometry analogous to that of $\mathbb{A}^1$ – varieties are defined by solutions to equations on products of discs.
Let’s try to find some points on the disc, i.e., seminorms on $\mathbb{C}_p \langle t \rangle$. As noted above, the sup norm on the points of norm at most one gives a distinguished point, called the Gauss point. One can also take any element $a \in \mathbb{C}_p$ of norm at most one, and a real number $r$ between zero and one (not necessarily in the value group), and set $\Vert f \Vert_{B(a,r)} = sup \{ f(x) | \, |x-a| < r \}$. More generally, one can take any nested sequence of balls of decreasing radius, and take the limit seminorm. Berkovich’s classification theorem asserts that all seminorms come from some nested sequence, and two such norms are equivalent if and only if the sequences are cofinal (meaning any term in one sequence has a successor in the other sequence). We can categorize the seminorms into four types:
Type I points correspond to nested sequences for which the radius goes to zero. Since $\mathbb{C}_p$ is complete, these are just evaluation at an element. These are called classical points.
Type II points correspond to nested sequences whose intersection is a ball whose radius lies in the value group, i.e., the seminorm is just sup over the ball. These are called rational points.
Type III points correspond to nested sequences whose intersection is a ball whose radius is not in the value group, but it is still given by sup over a ball. These points exist because absolute values are rational powers of $|p|$.
Type IV points correspond to nested sequences whose intersection is empty. These points exist because $\mathbb{C}_p$ is not spherically complete.
How do we view this as a topological space? If one ball contains another, we connect the corresponding sup seminorms with a line segment, which is made of the seminorms of balls of intermediate radius. We get a sort of tree structure expressing the containment partial ordering on balls in the disc. Unlike an ordinary tree, this one has infinitely many branches coming out of a dense subset of any line segment (geometric group theorists call this an $\mathbb{R}$-tree). In particular, the type II points are incident to “edges” that are in natural bijection with points on the projective line over the residue field $\overline{\mathbb{F}_p}$. The connected open neighborhoods of any point $x$ are given by choosing finitely many points and excluding the parts of the tree for which a path to $x$ passes through those points, so they tend to be quite large – large enough that the unit disc
is compact.
If we glue two copies of the unit disc along the unit circle, we get another tree, but the Gauss point no longer plays a distinguished role. This space is called $\mathbb{P}^1_{Berk}$, and it is an equivariantly compactified union of Bruhat-Tits buildings for $PGL_2$ over finite extensions of $\mathbb{Q}_p$. If you switch to power series fields, restriction of scalars gives a natural bijection between type II points with integer valuation and points on the affine Grassmannian. You can get many natural representations of groups over local fields by choosing sheaves on these ind-buildings with good equivariance properties. Apparently the Harris-Taylor proof of local Langlands correspondence for $GL_n$ used étale cohomology on these spaces, although I must confess that I never got that far into their book.
More reading:
Matt Baker gave some lectures about harmonic functions on $\mathbb{P}^1_{Berk}$, and you can find notes on the Arizona Winter School page. Also on that page you can find a Grothendieck-style (well-written, thorough, no examples) treatment of Berkovich spaces by Conrad starting on page 33. I didn’t say anything about how to glue affinoids (spaces of the form $Sp(A)$) to form general rigid analytic spaces because it is rather subtle, but you can find it here.
Look up Berkovich on MathSciNet (you may need an institutional subscription).
There are books on Bruhat-Tits buildings by Ronan and Brown. You can also try the original papers by Bruhat and Tits, but they are long and French. Schneider and Stuhler proved a neat derived equivalence between smooth reps and sheaves on buildings. Bezrukavnikov’s thesis is similar, but I found it difficult to read.
Harris-Taylor has a lot of information, but you have to enjoy pain.
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http://mathoverflow.net/revisions/8649/list
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## Return to Answer
2 added 766 characters in body; added 91 characters in body; deleted 5 characters in body
For many questions, the easiest way to see the nuts and bolts of a projective variety $V \subseteq P^n$ is to look at its cone $CV \subseteq A^{n+1}$. After all, the graded ring whose Proj is $V$ is the same as the ungraded ring whose Spec is $CV$. Obviously, there is almost always a singularity at the origin; but if you ignore that point, the other singular points all correspond between $V$ and $CV$. You can also think of the grading as geometrically represented by multiplication by $k^*$, if you are working over an algebraically closed field $k$. (Because the homogeneous polynomials are then eigenvectors of that group action.) You can think of $V$ as obtained from $CV$ by and then dividing by scalar multiplication.
The atlas-of-charts analysis of a projective variety is certainly important, but to some extent it is meant as an introduction to intrinsic algebraic geometry rather than as the best computational tool.
Your second question is reviewed in Wikipedia. As Wikipedia explains, Hironaka's big theorem was that it is possible to resolve all singularities of a variety by iterated blowups along subvarieties. I do not know a lot about this theory, but if so many capable mathematicians went to so much trouble to find a method, then surely there is no simple method.
On the other hand for curves, there is a stunning method that I learned about (or maybe relearned) just recently. Again according to Wikipedia, taking the integral closure of the coordinate ring of an affine curve, or the graded coordinate ring of a projective curve, solves everything. The claim is that it always removes the singularities of codimension 1, which are the only kind that a curve has.
1
For many questions, the easiest way to see the nuts and bolts of a projective variety $V \subseteq P^n$ is to look at its cone $CV \subseteq A^{n+1}$. After all, the graded ring whose Proj is $V$ is the same as the ungraded ring whose Spec is $CV$. Obviously, there is almost always a singularity at the origin; but if you ignore that point, the other singular points all correspond between $V$ and $CV$. You can also think of the grading as geometrically represented by multiplication by $k^*$, if you are working over an algebraically closed field $k$. (Because the homogeneous polynomials are then eigenvectors of that group action.) You can think of $V$ as obtained from $CV$ by and then dividing by scalar multiplication.
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http://cs.stackexchange.com/questions/6771/lr1-items-look-ahead
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LR(1) - Items, Look Ahead
I am having diffuculties understanding the principle of lookahead in LR(1) - items. How do I compute the lookahead sets ?
Say for an example that I have the following grammar:
S -> AB A -> aAb | b B -> d
Then the first state will look like this:
````S -> .AB , {look ahead}
A -> .aAb, {look ahead}
A -> .b, {look ahead}
````
I now what look aheads are, but I don't know how to compute them. I have googled for answers but there isn't any webpage that explains this in a simple manner.
Thanks in advance
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1 Answer
There are two ways that lookaheads 'come into being'. The first is that the start production $S' \rightarrow S$ has lookahead \$\$$in the initial state of the $LR(1)$ automaton. Hence, $S' \rightarrow \bullet S, \$$ is an item in the initial state.
Pendantry note: we therefore accept in the state containing the item $S' \rightarrow S \bullet, \$$on lookahead $\$$ and we pad any input with$\.
The rest of the lookaheads are computed from lookaheads which we have already computed. If we have in some state an item $A \rightarrow \alpha \bullet X \beta, l$ (so $l$ is the lookahead and $X$ is a nonterminal) and another production $B \rightarrow \gamma$, then for every $a \in \mathsf{FIRST}(X \beta l)$ we add in the completion step for that state the item $B \rightarrow \bullet \gamma, a$.
In other words, the lookahead is some terminal that can appear as the first terminal in a string derived from $X \beta l$. As $l$ is a terminal and appears at the end of $X \beta l$, this means that $X \beta l$ does not derive the empty string (so we don't need $\mathsf{FOLLOW}$). Also note that we only ever derive items and therefore new lookaheads from items that already have lookaheads, so there is no problem there.
The precise definition of $\mathsf{FIRST}$ is: $\mathsf{FIRST}(a \alpha) = \{a\}$ if $a$ is a terminal, $\mathsf{FIRST}(A \alpha) = \mathsf{FIRST}(A)$ if $A$ is a nonterminal and $A$ does not derive the empty string, and $\mathsf{FIRST}(A \alpha) = \mathsf{FIRST}(A) \bigcup \mathsf{FIRST}(\alpha)$ if it does. $\mathsf{FIRST}(A)$ is the well-known relation denoting the first terminals in the strings derivable from $A$ (which is also used in $LL(1)$).
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http://mathoverflow.net/questions/121066/inequality-with-eulers-totient-function/121100
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## Inequality with Euler’s totient function
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Numerical analysis of the first several hundred n suggests the following inequality:
$\varphi(3^n-2) \ge 2\cdot3^{n-1}$
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Up to a small additive factor, your inequality says that if $p_1,\dots,p_k$ are distinct primes dividing a number of the form $3^n-2$, then $(1-p_1^{-1})\cdots(1-p_k^{-1})\ge2/3$. I suspect this is an instance of mathoverflow.net/questions/120514 . – Emil Jeřábek Feb 7 at 14:09
I can't see the connection to mathoverflow.net/questions/120514? – Werner Aumayr Feb 7 at 14:18
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The connection is only vague, just that some things are just naturally expected to be somewhat rare, a couple of hundreds is not much to test. In particular, you need an n such that 3^n - 2 is divisible by primes such that Emil J.'s product is less than 2/3. The only reason this is an issue at all is that you cannot have 2 and 3 for obvious reasons, neiter 11 nor 13, and also not 5 and 7 at the same time. So things get a bit large. Calculations got slightly too complex for me for just doing them so, but with Emil J. idea and some calculation with congruences one should find a counter ex. – quid Feb 7 at 15:07
It seems not to hold for n=1. Also, if you find a counterexample, both Emil's k and your n will be substantially larger than a thousand. Gerhard "Ask Me About System Design" Paseman, 2013.02.07 – Gerhard Paseman Feb 7 at 15:59
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@Gerhard Paseman: now while I might have been a bit too optimistic regarding how easy or not it is to find a counterexample but that one would need substantially more than a thousand different primes (the k) seems unlikely to me. Did you really mean this? – quid Feb 7 at 16:09
show 7 more comments
## 1 Answer
Take $n=382315009082231724951830011$. Then $3^n-2$ is divisible by the primes $5, 19, 23, 47, 71, 97, 149, 167, 173, 263, 359, 383, 389, 461, 479, 503, 557$. Furthermore, $\varphi(3^n-2)<2/3\cdot(3^n-2)<2\cdot 3^{n-1}$.
The following Sage code verifies this examples. I believe that this is close to a minimal counterexample.
```sage: n=382315009082231724951830011
sage: l=[5, 19, 23, 47, 71, 97, 149, 167, 173, 263, 359, 383, 389, 461, 479, 503, 557]
sage: set(3.powermod(n,p) for p in l)
set([2])
sage: prod(1-1/p for p in l).n()
0.666250824539016
```
As there are some speculations about how to find such an example, here is the (not really clever) Sage code which greedily collects the congruences for $n$ which do not contradict each other:
```p,n,Q = 5,3,4
lp = [p]
s = 1-1/p
while True:
p=p.next_prime()
e=IntegerModRing(p)(3).multiplicative_order()
l=[z for z in [1..e-1] if 3.powermod(z,p) == 2]
if len(l) == 0:
continue
b=l[0]
if (b-n) % e.gcd(Q) != 0:
continue
QQ=Q.lcm(e)
n=CRT_list([n,b],[Q,e]) % QQ
Q=QQ
s*=(1-1/p)
lp.append(p)
print p,s.n()
if 2/3>s:
print n,lp
break
```
-
Perfect! Thank you so much! – Werner Aumayr Feb 7 at 18:08
4
@Werner: If you like the answer, you might consider accepting it. – Peter Mueller Feb 7 at 19:07
I would like the answer more if I had a good lower bound on k, the number of distinct prime factors of the counterexample. Can you at least trial divide by candidates up to 3^20 or even perhaps 3^50 to get an idea? Gerhard "Would Be Ever So Grateful" Paseman, 2013.02.07 – Gerhard Paseman Feb 7 at 20:51
Thanks for this answer! After I initially saw the question I got quite curious what would be the outcome. Did you also try what happens if you start with 7 insted of 5? I would be quite curious how it compares. ps for Gerhard Paseman: regarding the k again, it occurs to me we were also talking about somewhat different things (which is my fault) but to me the k here is seventeen, if I counted right in any case the number of primes one had to take into account. But sure the 'rest' seems huge so there might be quite a few additional factors 'hiding'. – quid Feb 7 at 21:38
2
@Gerhard: I believe there is no way to factor $3^n-2$, this number has more that $10^{26}$ decimal digits! @quid: Indeed, if one starts with $7$, things look quite differently. Up to $p<70000$, this greedy approach yields $425$ compatible primes, starting with $7, 17, 23, 47, 71, 167,\ldots$, yet the product of the $1-1/p$ for these primes is still bigger than $0.692$. – Peter Mueller Feb 7 at 22:40
show 2 more comments
|
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|
http://physics.stackexchange.com/questions/56589/calculating-the-mass-equivalency-of-a-song?answertab=oldest
|
# Calculating the mass equivalency of a song?
I've recently become fascinated with the idea of sound energy having a theoretical equivalent mass. I've read over this thread:
Do light and sound waves have mass
I understand this part: $m_{eq}=E/c^2$ and $E=A\rho \xi^2\omega^2$
Where I am getting tripped up is the way to measure $E$.
Essentially, I want to measure $E$ and eventually $m$ of a song (for it's entire duration). The only equipment I have is an iPhone app that measures intensity (dB) (apparently it's fairly accurate for levels below 100 dB). I plan to play the "music" at about 80dB over speakers in a room that is 50'x30'x15' and I will assume the temperature is at room temperature. If your wondering, it's for a conceptual audio art piece.
-
– dmckee♦ Mar 11 at 21:49
– Qmechanic♦ Mar 11 at 22:13
Thanks dmckee. Qmechanic-an interesting indeed! Thx! – user21863 Mar 12 at 4:32
## 2 Answers
A song has no mass equivalence. The sound waves from playing a song do---to the extent that they carry energy, and you can relate an equivalent mass1 to energy.
To cut to the chase, it would probably be easiest to look up how much power (energy per unit time) your speakers produce, then multiply that by the duration of the song.
If you measure the 'intensity' of the song with your iphone, that tells you the power-density at the location of the phone---while the speakers are actually generating sound-waves in all directions. Through an elaborate series of calculations and approximations, you could convert this to an energy. For completeness:
The number of decibels actually measures the pressure of the sound waves, which you can convert to an amplitude. You can estimate the frequency ($\omega$) as, something like, the higher frequency end of the human audible range (possibly adjusted for the type of song). The total power is then computed by finding the surface-area of a sphere around the source ($A$), at the distance you are measuring.
Again, using the power of the speakers will be more accurate.
-
FWIW, people ask about measuring dB SPL on the iPhone. You would need to calibrate your iPhone first. Use a sound level meter instead. 50 bucks from radio shack. – Bjorn Roche Mar 12 at 3:32
You could also set an upper bound if you know the phone's battery capacity (usually written on the battery) and how long you get out of a full charge playing the song on repeat. You can estimate the energy required to play the song by
$$\text{Energy per song} = \text{Battery capacity} \times \frac{\text{Song length}}{\text{Battery life}}$$
This is an upper limit because because, obviously, not all of the battery power goes into playing the song. In fact most will go into other things like wifi and heat. If you want a more accurate estimate you can turn off as many other things as possible. If waiting for the phone to die isn't doable then you could run down say 30% of the battery and correct for that, but discharge curves aren't necessarily linear.
-
|
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|
http://mathoverflow.net/questions/54758/does-the-following-series-converge/54892
|
## Does the following series converge?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Does the following series converge? $\sum_{n=1}^{\infty} \vert \sin n \vert ^{n}$
-
1
Probably not. If instead of taking sin n, you take at the n'th stage sin x_n where x_n is uniformly distributed in [0, 2\pi] (the density of n 'mod' 2*\pi should behave the same), you get that |x_n - \pi/2| < 1/n infinitely often. Each such occurrence contributes O(1) to the sum. Also, not sure MO is the right venue for these type of questions. – Or Zuk Feb 8 2011 at 13:30
7
It's even worse than that. $\sin (\pi/2 \pm \epsilon) \geq 1 - \epsilon^2/2$. So, just the weaker bound $|n - (2k+1)\pi/2| < 1/\sqrt{n}$ gives $(\sin n)^n > e^{-1/2}$. I would guess that it is not too bad to show that $n$ is infinitely often this close to an odd multiple of $\pi/2$, but I don't see the details right now. – David Speyer Feb 8 2011 at 13:55
9
"Chebyshev's Theorem" (Khinchin's continued fraction book) says that for arbitrary irrational $\alpha$ and real $\beta$ the inequality `$|\alpha x-y-\beta|<3/x$` has infinitely many integer solutions. From this follows easily that $n$ is infinitely often as close to an odd multiple of $\pi/2$ as David's argument requires. – SJR Feb 8 2011 at 14:41
4
this looks too much like homework to me. – Igor Rivin Feb 8 2011 at 15:20
1
It reminds me an other innocent looking series: $$\sum\frac{(-1)^n}{\ln n+\cos n}.$$ It converges, but to prove it you need to know a bit about rational approximations of $\pi$. – Anton Petrunin Feb 8 2011 at 17:16
show 2 more comments
## 2 Answers
The question has basically been answered in the comments by David Speyer and SJR. It is a theorem of Chebyshev that that for any irrational $\alpha$ and any real $\beta$, the inequality $$|\alpha n - k - \beta| < 3/n$$ has infinitely many solutions. In particular, take $\alpha = 1/(2\pi)$ and $\beta = \frac12$. Then one gets that $n$ is so close to an odd multiple of $\pi$ that $|\sin n|^n$ converges to 1 for these values. Even if you took $|\sin n|^{n^2}$, these values would be bounded away from 0. Certainly if the terms of a series do not converge to 0, then the series does not converge.
-
### 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.
Since $\pi$ is transcendental (so also $\frac{\pi}{2}$ and $\frac{3\pi}{2}$), $\forall n \in \mathbb{N} , |\sin{n}|<1$. In another hand, $\sum_{n=2}^\infty|\sin{n}|^n <\sum_{n=2}^\infty|\sin{n}|^2$ which converges (because $\sum_{n=1}^\infty a^n$ converges if $|a| < 1$.
So, $\sum_{n=2}^\infty|\sin{n}|^n$ converges.
-
This doesn't work because the comparison to a geometric series is not apt; since $|\sin(n)|$ could be arbitrarily close to 1 infinitely often (so that $|\sin(n)|^n$ could also be arbitrarily close). – Stanley Yao Xiao Feb 8 2011 at 18:02
2
The reply is not correct because $\vert \sin n \vert^{2}$ has a fixed exponent and a basis dependent on $n$, the latter being infinitely many time close to 1, therefore the comparison with the geometric series has no meaning. Surely $\sum \vert \sin n \vert^{2}$ diverges, because $\vert \sin n \vert$ is infinitely many times close to 1, therefore $\vert \sin n \vert^{2}$ is infinitely many times bigger than $\frac{1}{2}$, so that the series diverges. – Fabio Feb 8 2011 at 18:10
I'm ashamed of this answer. That's what happens when I don't sleep enough. – Lamine Feb 14 2011 at 12:16
5
It may be wrong, but -9 votes? Really? – Yemon Choi Feb 14 2011 at 12:25
1
Scratching out the wrong answer is a classy move. – userN Feb 14 2011 at 16:45
|
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|
http://math.stackexchange.com/questions/103470/how-can-i-find-the-number-of-the-shortest-paths-between-two-points-on-a-2d-latti
|
# How can I find the number of the shortest paths between two points on a 2D lattice grid?
How do you find the number of the shortest distances between two points on a grid where you can only move one unit up, down, left, or right? Is there a formula for this?
Eg. The shortest path between $(0,0)$ and $(1,1)$ can be $(0,0)\to (0,1)\to (1,1)$ or $(0,0)\to(1,0)\to(1,1)$, so there are two shortest paths.
-
1
– Jalaj Jan 29 '12 at 12:25
## 1 Answer
Any shortest path from $(0,0)$ to $(m,n)$ includes $m$ steps in the x axis and $n$ steps in the y axis; what governs the shape of such a path is the order in which we choose to go to the right and up. Since we have $m+n$ total steps to take, we just need to choose which $m$ will be to the right. This is counted by the binomial coefficient $\binom{m+n}{m} = \binom{m+n}{n}$.
-
You need to assume that $m, n$ are non-negative here, but up to reflection this loses no generality. – Qiaochu Yuan Jan 29 '12 at 2:58
|
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|
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