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March 11

number of spanning trees of K_n-e

Hello. I want to use Cayley's formula to show that the number of spanning trees in the labeled graph K_n-e is (n-2)n^n-3. Here e is any edge in K_n. Can someone point me in the right direction please. Equivalently I want to figure out the number of spanning trees involving the edge e.--Shahab (talk) 06:23, 11 March 2009 (UTC)

You know there are $n^{n-2}$ spanning trees altogether. Choose one at random. What is the probability that it includes e? McKay (talk) 07:37, 11 March 2009 (UTC)

${\frac {n-1}{\frac {n(n-1)}{2}}}$. Now what?--Shahab (talk) 07:56, 11 March 2009 (UTC)OK I got it. Thanks for the tip. Cheers--Shahab (talk) 08:03, 11 March 2009 (UTC)

March 12

transcendental numbers and Khinchin's constant

In the biography of Gelfond, it states that before he proved his theorem few explicit transcendental numbers were known. Did not Liouville (constructively) prove the existence of infinitely many using certain decimal expansions?

In the description of Khinchine's constant, it reads "it has not been proven for any specific real number whose full continued fraction representation is not known". Should this not read "known"? For how can the determination be made for a specific number if the digits a0, a1, ... are unknown?

Thanks for any clarification —Preceding unsigned comment added by Aliotra (talk • contribs) 03:10, 12 March 2009 (UTC)

Yes, Liouville proved the existence of transcendental numbers (in fact of continuum-many of them). However, Liouville's proof only works for a class of numbers specifically invented for the purpose, so perhaps what is meant is that few naturally-occurring numbers were known to be transcendental. Algebraist 03:25, 12 March 2009 (UTC)

I take your point. The article at Mactutor about "accessible" real numbers appears not unrelated to this issue. "Naturally occurring" could be interpreted as "accessible", perhaps? I think the numbers Liouville used would be defined as accessible, though.

Looking through the history of the page on Khinchin's constant, I noted that someone else had suggested that the part which I questioned above, needed improvement (and I think you agreed). Is there an expert on number theory (perhaps the original author) to whom I can re-direct this question?142.27.68.72 (talk) 16:53, 16 March 2009 (UTC) 142.27.68.72 (talk) 16:51, 16 March 2009 (UTC)142.27.68.72 (talk) 16:55, 16 March 2009 (UTC)

Sorry about the unsigned post; I've created a user page. Aliotra (talk) 17:59, 14 March 2009 (UTC)

The best place to get expert advice on mathematical issues is at WT:WPM. My phrase 'naturally occurring' was not intended to be mathematically precise, and I'm not sure it can be made so. I would be interested to see an attempt to do so. If by 'accessible' you mean something like 'definable', though, then this doesn't do the job. For any sensible value of 'definable', every naturally-occurring real is definable, but Liouville's original number is definable, though certainly not naturally-occurring. Algebraist 17:01, 16 March 2009 (UTC)

Square root of a limit = limit of square root?

If g_n is a sequence of functions in L^2 and $\int _{0}^{1}|g_{n}|^{2}\to \int _{0}^{1}|g|^{2}$ is it also true that $\left(\int _{0}^{1}|g_{n}|^{2}\right)^{1/2}\to \left(\int _{0}^{1}|g|^{2}\right)^{1/2}$ I sure hope so because I need it to be true. I can also assume g_n and g are continuous if that matters. Thanks StatisticsMan (talk) 04:01, 12 March 2009 (UTC)

The square root of a limit of nonnegative real numbers equals the limit of the square roots, yes. Black Carrot (talk) 06:20, 12 March 2009 (UTC)

And the reason for that is that the square root function is continuous. — Emil J. 11:19, 12 March 2009 (UTC)

I wish that all assertions I make are true (in mathematics) but fortunately this is not the case (otherwise there wouldn't be much fun). --PST 09:25, 12 March 2009 (UTC)

Alright, thanks a lot. StatisticsMan (talk) 12:12, 12 March 2009 (UTC)

"Ecological" study etymology

In statistics, why is a study that deals with only aggregates called an ecological study when it doesn't necessarily relate to ecology? NeonMerlin 04:44, 12 March 2009 (UTC)

The word "ecology" here is being used similarly, but not indentically to the word "population" (see statistical population). In aggregate studies, statisticians work with a large number of aggregates of individuals: the word population would be inappropriate to describe each individual aggregate, because (1) the statistician is not sampling individuals from each aggregate, and (2) the statistician is sampling aggregates from the collection of all aggregates. In this context "population", being used technically (i.e., meaning "that which samples are taken from"), refers to the collection of all aggregates. So some word other than "population" is needed to describe each aggregate of individuals.

The word "ecology" is suggestive because individuals within a given aggregate tend to be highly correlated in the relevant dimensions, and form very messy, far-from-Gaussian distributions (or at least, the statisticians are not interested in the properties of the distributions within an aggregrate, so that messy distributions are expected and not problematic): for example, the example at ecological fallacy#Origin of concept could have a bimodal (at least) distribution of literacy rates within each state, for immigrants and non-immigrants. Although "ecology" is far from the perfect word to describe this, it seems to fit reasonably well. I'd guess that the usage of the word "ecology" arose because this concept arose first in biological studies (this is a guess, I really don't know), and was later generalized to other situations.

I'd welcome discussion from others. Eric. 131.215.158.184 (talk) 08:51, 12 March 2009 (UTC)

Calculation probability for draws

In the aftermath of last nights Champions League matches, I was wondering what the probability for a couple draws for the quarter-finals would be. So there are 4 English teams, 2 Spanish, 1 Portuguese, 1 German. So what's the probability of the four English teams draw each other. No English team draw another. Two of the English teams draw each other. Two Spanish teams draw each other. And the big one, the four English teams draw each other while the two Spanish teams draw each other. (this might perhaps be trivial math though I can't figure out the right formula for calculating) chandler · 09:37, 12 March 2009 (UTC)

If the english teams are ABCD, the spanish EF and the others GH, we can only get your last criteria if the draw is one of: AB CD EF GH, AC BD EF GH, AD BC EF GH. So that is 3 in 7*5*3 (A v One of 7; first alpha left v remaining one of 5; first alpha left v remaining one of 3; the last two play each other.), so 3 in 105 or 0.0286. -- SGBailey (talk) 10:24, 12 March 2009 (UTC)

The article Multiset#Polynomial_notation may be relevant. (?). Bo Jacoby (talk) 19:49, 12 March 2009 (UTC).

bound degree of a region in a planar graph

Hello. Suppose we are given a planar representation of a graph G, and for each of its regions we define the bound degree to be the number of edges enclosing it. Suppose also that it is given that the bound degree of each region is even. I want to show that the graph is bipartite. Obviously as a bipartite graph has no odd cycles, so what I want is to show that cycles involving edges encompassing more then 1 region are also even. How can I do that? Thanks--Shahab (talk) 12:39, 12 March 2009 (UTC)

This is a standard homework exercise, and I hope that is not your reason for asking. Add together the lengths of all the faces inside the cycle. McKay (talk) 21:34, 12 March 2009 (UTC)

No this isn't homework. I am a self learner. Anyway your hint was all I needed in this case. Thanks--Shahab (talk) 06:27, 13 March 2009 (UTC)

What is this probability distribution?

I've recently graphed the following probability distribution:

What is it? It seems to have a fat tail, but that doesn't concern me much; its the oddly mis-shapen nose at the top, and the log-linear sides that have my interest. linas (talk) 14:38, 12 March 2009 (UTC)

Compute the cumulants. Bo Jacoby (talk) 19:57, 12 March 2009 (UTC).

When I saw the title of this section and the graph, I assumed you meant the graph is that of the probability density. But you've labeled it "mutual information". Did you mean that the "mutual information" is itself a random variable? If log ƒ = a|x| then ƒ = e ^−a|x| and you've got a Laplace distribution. That gives you the log-linear sides, but not the funny shape at the top. Michael Hardy (talk) 20:00, 12 March 2009 (UTC)

Extract the constant term from a polynomial, with Maple9

Hi, excuse this naive question. I'm doing a computation with Maple9. As a result I have a huge trigonometric polynomial, and I want to extract the constant term. As a mathematician, I would just integrate over [-pi, pi], but this can not be the right answer. How can I just make it find the constant term? Thanks --131.114.72.215 (talk) 15:31, 12 March 2009 (UTC)

Integration from −π to π will kill all trigonometric terms except the constant term a which will contribute 2πa. So the constant term equals the integral divided by 2π. Is this answering your question? (I don't know Maple9). Bo Jacoby (talk) 20:04, 12 March 2009 (UTC).

Evaluate it at 0? 76.195.10.34 (talk) 21:55, 12 March 2009 (UTC)

That won't work. A trigonometric polynomial generally involves cosine terms that do not vanish when evaluated at 0. Michael Hardy (talk) 22:13, 12 March 2009 (UTC)

A (weird) solution, for the trigonometric polynomial T in x is:

subs(x=0, subs(cos=sin, T));

this transforms all "cos" in "sin" and then evaluates the result at 0, as suggested by 76. This seems much faster than integration (but I guess there is a better way). --pma (talk) 22:33, 12 March 2009 (UTC)

It's hard to give a useful answer without more information. It all depends on what form the expression is in. If you just want to pick out the terms in a sum F that have no sin or cos in them, remove(has,F,{sin,cos}) will do it. However, be careful because if there is only one term remove will operate on the operands of that term (this is a problem in Maple that is endlessly annoying). McKay (talk) 23:24, 12 March 2009 (UTC)

March 13

Products of Multiple Geometric Series

Does anybody know a way to simplify

:   n        m
: -----   ------ 
:  | |     \      
:  | |     /      (Ri)-j  
:  | |    ------
:  i=1      j=0                               ?

Sorry about the ASCII approximations of symbols, but I don't speak LaTex... Lucas Brown 42 (talk) 19:21, 13 March 2009 (UTC)

Set r_i = 1/R_i. For finite m, I think you are stuck with just foiling it out along the lines of the multinomial theorem. For infinite m you get a formal equality:

$\prod _{i=1}^{n}\sum _{j=0}^{\infty }r_{i}^{j}=\sum _{j=0}^{\infty }e_{j}(r_{1},r_{2},\dots ,r_{n})$

where e_j is the elementary symmetric polynomial of degree j. JackSchmidt (talk) 19:39, 13 March 2009 (UTC)

That can't be right since for $e_{j}(r_{1},r_{2},\dots ,r_{n})$ we must have $j\leqslant n$. Dauto (talk) 18:42, 14 March 2009 (UTC)

Would you believe: For infinite m you get a formal equality:

$\prod _{i=1}^{n}\sum _{j=0}^{\infty }r_{i}^{j}=\sum _{j=0}^{\infty }h_{j}(r_{1},r_{2},\dots ,r_{n})$

where h_j is the complete homogeneous symmetric polynomial of degree j. Whichever one is the one with each variable appearing to each power. JackSchmidt (talk) 03:07, 15 March 2009 (UTC)

March 14

What!

What! No questions on this 3.14... day? Geesh, and on the Math desk of all things  ;-) -hydnjo (talk) 00:20, 15 March 2009 (UTC)

If you've got a question, ask it. You've done it before. All right-thinking mathematicians know that the twenty-second of July is the true pi day, anyway. Algebraist 00:42, 15 March 2009 (UTC)

Maybe you'll have more luck on e day. —Preceding unsigned comment added by 99.255.228.5 (talk) 01:13, 15 March 2009 (UTC)

Which is presumably either the 8th of March, the 11th of April, on the 19th of July. Algebraist 01:39, 15 March 2009 (UTC)

To be honest none of those convergents are good enough to deserve a dedicated day on the calendar. e will then suffer the fate of many other irrational numbers which are too rational to be celebrated. --Xedi^Talk 07:53, 15 March 2009 (UTC)

I'm a big fan of non-trivial-geometric-mean day... the next three being 2009 March 27th, 2010 April 25th, and 2010 May 20th. There's a few more in 2012, one in 2014, 2015, and 2018, and then none until the next century. (So long as the year is the geometric mean of the day and month....) Eric. 131.215.158.184 (talk) 07:46, 15 March 2009 (UTC)

Well now, that's better! -hydnjo (talk) 08:59, 15 March 2009 (UTC)

The question is (for me), where is this horror vacui coming from, talking about no-question days? On the contrary, I look forward the empty event, and will greet its content as the trivial object of the RD/M, something that we should miss... ;) --pma (talk) 09:29, 15 March 2009 (UTC)

That horror vacui would be coming from me. Empty days seem so unfull, much more so than even trivial barely math at all inquiries  ;-) hydnjo (talk) 10:54, 15 March 2009 (UTC)

March 15

Hilbert curve

It is clear (by a self-similarity argument) that the Hausdorff dimension of the Hilbert curve is 2. But what is its topological dimension ? Is it 1 because it is the limit set of a series of curves each of which has topological dimension 1 ? Or is it 2 because it is space-filling ? A reliable source would be useful - I can find not-necessarily-reliable sources for both answers ! Gandalf61 (talk) 09:19, 15 March 2009 (UTC)

As far as I know these concepts (Hausdorff dimensions and the various definitions of topological dimension) are attached to subsets rather that maps; in this case, the subset is just the image of the map, a square; so it's 2.--pma (talk) 09:39, 15 March 2009 (UTC)

That's what I thought. But this on-line textbook, talking about a minor variation on the Hilbert curve, says "Even though its Hausdorff-Besicovitch dimension is a whole number (D = 2) its topological dimension (DT = 1) is strictly less than this". Gandalf61 (talk) 12:08, 15 March 2009 (UTC)

I didn't find (after a very quick search) a definition of (top or HB) dimension for a curve in that on-line book; another possibility is that they mean the dimensions of the graph. In this case, the topological dimension of the graph of the Hilbert curve is 1, for it is a topological invariant, while the Hausdorff dimension is at least 2 (I guess 2 in fact), since in general for a Lipshitz map F we have $\scriptstyle \dim _{HB}(F(A))\leq \dim _{HB}(A)$; here the projection is 1 Lipschitz and maps the graph onto the square; so it's all right if the point was just exibiting a compact subset of an euclidean space with different topological and Hausdorff-Besicovitch dimensions. Nice book by the way; it may still contain some imprecisions. It says:

"This curve twists so much that it has infinite length. More remarkable is that it will ultimately visit every point in the unit square. Thus, there exists a continuous, one-to-one mapping from the points in the unit interval to the points in the unit plane."

Of course the Hilbert curve is definitely not 1-1 from the closed interval [0,1] to the unit square [0,1]x[0,1], and there is no such a curve, for it would be an homeomorphism.--pma (talk) 14:25, 15 March 2009 (UTC)

To be a homeomorphism it would need to have a continuous inverse, which it quite obviously doesn't. I think the map is bijective (the cardinalities are certainly the same, so it's not hard to believe). --Tango (talk) 14:33, 15 March 2009 (UTC)

I don't understand your comment, you are just repeating what I said. Sorry I did not read carefully what you wrote. I repeat: it is not bijective, otherwise it would be a homeomorphism, which is impossible as you are saying--pma (talk) 14:39, 15 March 2009 (UTC)

You are not reading what I said. A homeomorphism is not a "continuous bijection", it is a "continuous bijection with continuous inverse". The Hilbert curve is a continuous bijection without a continuous inverse. --Tango (talk) 14:48, 15 March 2009 (UTC)

Well, no, as you certainly know but are forgetting for a moment, a continuous bijective map between Hausdorff compact topological spaces is always a homeomorphism. The Hilbert curve is not 1-1, although the approximating curves are.--pma (talk) 14:56, 15 March 2009 (UTC)

You're right, I had forgotten that (assuming I ever knew it, which I probably did). After refreshing my memory of the subject, I can confirm you are entirely correct. I think my problem was caused by trying to use intuition to work with limits - never a good plan! The approximating curves are all injective, but the limiting curve is not (although it doesn't actually cross anywhere, just touches tangentially). --Tango (talk) 15:06, 15 March 2009 (UTC)

(rmv indent) Like pma, I don't think the Hilbert "curve" is even bijective. Be that as it may, we all seem to be agreed that the it is not a homeomorphism. So the argument that its image must have topological dimension 1 because the line has topological dimension 1 does not hold. However, I am not certain that I follow the distinction between the image of the Hilbert curve function and its graph. Is the graph the set of points {(t, f(t): 0 ≤ t ≤ 1}, where each f(t) is a point in R², whereas the image is the projection of that set onto the plane t=0 ? Gandalf61 (talk) 15:08, 15 March 2009 (UTC)

Exact; so the image is the square, and the graph is a compact in R³ homeomorphic to the interval [0,1] (again because the map $\scriptstyle [0,1]\ni t\mapsto (t,f(t))$ is continuous and bijective onto the graph, or even simplier, because the projection on the t component is a continuous inverse to it). That the Hilbert curve is not 1-1 can also be seen directly.--pma (talk) 15:38, 15 March 2009 (UTC)

How do I calculate probability of an event occurring consecutively over many repetitions?

I have an event with a 25% chance of occurring, and a 75% chance of NOT occurring. I'm trying to figure out the correct way to find out the odds/% chance of the event NOT occurring 65 times consecutively (or more, let's say 300 times). What is the name of what I'm trying to do, and more importantly, what is the correct formula for accomplishing this?

Sorry for such a simple question, but I was a music major, so math and I just aren't friendly. 74.163.76.31 (talk) 11:34, 15 March 2009 (UTC)

Don't be sorry; the relevant article is binomial distribution. Hope this helps. If you just need the result, the formula is:

P (the event not occurring x times consecutively) = (0.75)^x

--PST 13:32, 15 March 2009 (UTC)

Note that this formula gives the result as a decimal (where 0 means it will never happen and 1 means it will always happen). Multiply the result by 100 to make it into a percentage. StuRat (talk) 14:39, 15 March 2009 (UTC)

Thanks for the informative answer. One tiny problem, though. I've read the article (as well as every other one relevant to probability/statistics, it seems). While I understand the concept, I'll be honest, math and I not being friendly was an understatement. I have no earthly idea whatsoever how to apply the formula to find the answer. It looks simple enough to me, but I have no idea what to do with it.

If I understand you correctly (unlikely), I'm just looking for P = (0.75)⁶⁵ ? Is that right, and if so, how would one enter that on your average scientific calculator? Got a little more insight? :) 74.163.76.31 (talk) 22:14, 16 March 2009 (UTC)

I don't know what calculator you use, but I use Google. Algebraist 22:22, 16 March 2009 (UTC)

On your typical scientific calculator: .75x^y65×100, then hit the equals sign ("="), to get the answer as a percentage. One problem is that this will give a very small number, and it is likely to be in scientific or engineering notation. For example, I got 7.568e-7. This means you need to move the decimal point 7 places to the left. I get .000000768% when I do this. This works out to be so rare that it will only happen once in over 132 million trials of 65 events. For 300 events, the number rises to one so large I don't even know a name for it. StuRat (talk) 03:26, 17 March 2009 (UTC)

For 300 events, it's about one chance in thirty undecillion. Algebraist 13:13, 17 March 2009 (UTC)

Probability in Rummikub

For the benefit of those who don't know the game, I will restate it in more common terms. I have 2 decks of cards, each deck containing one joker, giving a total of 106 cards. I shuffle them together and keep on shuffling until they are randomized.
If I draw 14 cards, what is the probability that I will have exactly 1 pair (same suit and rank). Same for 2 pairs ... up to 7 pairs. Phil_burnstein (talk) 12:23, 15 March 2009 (UTC)

You have to count out the different possibilities. For example, to have ZERO pairs, you need one of each of the 13 ranks, plus one joker. How many ways can that happen? Well, there's 4 possible suits for each non-joker, plus 2 jokers. Then for one pair, look at how to get 13 cards with no pairs, plus a 14th card paired with one of the earlier 13, etc. Divide each of those by the ${104 \choose 14}$ possible deals. —Preceding unsigned comment added by 207.241.239.70 (talk) 02:12, 17 March 2009 (UTC)

Assuming that the deck consists of 53 pairs (52 pairs same suit, same rank plus 1 pair of jokers) I get:

Probability of 0 pairs ... 37.26%

Probability of 1 pair ..... 42.38%

Probability of 2 pairs ... 17.06%

Probability of 3 pairs ... 3.05%

Probability of 4 pairs ... 0.25%

Probability of 5 pairs ... approx 1 in 12,000

Probability of 6 pairs ... approx 1 in 10⁶

Probability of 7 pairs ... approx 1 in 7 × 10⁸

So you are 50 times more likely to guess all six numbers in the UK National Lottery than you are to get seven pairs in a Rummikub hand. Gandalf61 (talk) 09:13, 17 March 2009 (UTC)

A more complete answer: Lets solve the problem of calculating the probability of getting $s\,$ pairs in $(2n)\,$ cards drawn out of $m\,$ pairs of cards (a total of $(2m)\,$ cards). Your question would be the case $m=53\,$, $n=7\,$, and $s=0,1,\dots 6,7\,$. Lets begin with the simpler case $s=0\,$. What's the number $n_{0}^{m,n}\,$ of different ways can we do that? That's simmilar to computing how many ways we can draw $2n\,$ cards out of a deck with $m\,$ different cards (since no repeats are allowed) which is given by ${m \choose 2n}$, but with the aditional caveat that each card could belong to either one of the two decks (of $m\,$ cards each). That gives us an extra factor of $2^{2n}\,$, and we get $n_{0}^{m,n}=2^{2n}{m \choose 2n}\,$, and the probability $p_{0}^{m,n}={\frac {n_{0}^{m,n}}{N}}\,$ where $N={2m \choose 2n}$ is the total number of possible ways we could have drawn the cards.

$p_{0}^{m,n}=2^{2n}{m \choose 2n}{2m \choose 2n}^{-1}={\frac {2^{2n}m!(2m-2n)!}{(m-2n)!(2m)!}}\,$.

Now for the general case, the number $n_{s}^{m,n}\,$ is given by the number of ways we can get $s\,$ pairs out of $m\,$ possible times the number $n_{0}^{m',n'}\,$ of drawing the remaining $2n'=2n-2s\,$ out of the remaining $2m'=2m-2s\,$ deck without any extra pairs.

$n_{s}^{m,n}={m \choose s}n_{0}^{m-s,n-s}={m \choose s}2^{2n-2s}{m-s \choose 2n-2s}$,

and the probability $p_{s}^{m,n}$ is given by

$p_{s}^{m,n}={\frac {n_{s}^{m,n}}{N}}={m \choose s}2^{2n-2s}{m-s \choose 2n-2s}{2m \choose 2n}^{-1}={\frac {2^{2n-2s}m!(2n)!(2m-2n)!}{s!(2n-2s)!(m+s-2n)!(2m)!}}$.

That formula can be used to reproduce Gandalf's table. Dauto (talk) 00:22, 18 March 2009 (UTC)

March 16

Self-Dual Tessellation

Hello, this question came up during my ongoing thesis. I have two material surfaces, with spikes on top of them. The spikes are regularly tessellated (like triangular, square or whatever). Both the surfaces have similar lattice of spikes (same structure), and the spikes have to fit into each other. If we don't take into account the profile of spikes, I want to ask how many geometrical combinations are possible, e.g. one is square tiling (spike centre on vertices of square for both the surfaces, and one surface will nicely fit into the other). Thanks. - DSachan (talk) 10:54, 16 March 2009 (UTC)

There are only three regular tessellations: with triangles, squares, and hexagons. I'm not sure what you mean by "we don't take into account the profile of spikes", so I'll ignore that part. Only the square and hexagonal tilings have the property you seek: the triangular-pyramidal spikes will not interlock nicely. One way to see this is that in the triangular tesselation, there are twice as many triangles as there are vertices (by density), that is, there are twice as many spikes as there are places to put the spikes.

If you're interested in other tilings besides the three regular tessellations the problem likely becomes trickier....

The term dual tiling refers to a different concept that what (I believe) you are asking. Wikipedia doesn't seem to have an article on dual tilings, but see duality (mathematics)#Geometric duality and dual graph. Eric. 131.215.158.184 (talk) 00:36, 17 March 2009 (UTC)

Converse to Heine-Cantor theorem

A continuous function on a compact metric space is uniformly continuous. is the converse- if all continuous function on metric space is uniformly continuous then the space itself is compact, true? 123.239.24.35 (talk) 18:03, 16 March 2009 (UTC)

True: if a metric space is not compact, it contains an infinite subset of points having distance from each other bounded away from zero, that you can use to build a continuous, not uniformly continuous function ---and unbounded too, showing that the converse of Weierstrass theorem is also true. You just have to play with the distance function of the space. PS: I added a header, that you may change, if you have a better one. --pma (talk) 18:21, 16 March 2009 (UTC)

Hint: Suppose the space is not compact; then one can find a countably infinite subset with no limit points. Let this subset be written as a sequence, x_n, for n a natural number. Define f on this subset by using the index of the terms in the sequence and you will get a function into the real numbers that is continuous but not uniformly continuous. You can construct f as pma said - play around with the index. One example of such an f would be f (x_n) = n; a function into the natural numbers. One you have found the function, you may conclude that f is continuous but not uniformly continuous. I would recommend that you think a bit harder. Sometimes with such proofs, having proved an easier result may help. For example, if you have proved that every metric (or normal pseudocompact space is countably compact, this proof would have been trivial. It is just a matter of experience. P.S. If you are interested in the proof that every pseudocompact space is countably compact (for metric, or slightly more weakly, normal spaces), use the Tietze extension theorem. This proof, i.e the proof of the converse of the Heine-Cantor theorem may help. P.S.S Pma, You may not necessarily be able to construct a sequence of points having distance from each other bounded away from 0 (assuming you mean you want to construct an unbounded sequence). For example, in the standard bounded metric, every subset is bounded. But you probably meant compact in which case you are of course right :). --PST 23:11, 17 March 2009 (UTC)

Hey! I was completely wrong, sorry. For metric spaces, the converse of Weierstrass theorem does hold (i.e., if any continuous real-valued function is bounded then the space is compact) BUT the converse of the Heine-Cantor theorem does not. For a general non-compact metric space we can only build a continuous unbounded function. We do have a sequence $\scriptstyle (x_{n})_{n\in \mathbb {N} }$ with no converging subsequences. To semplify, let's say $\scriptstyle d(x_{n},x_{m})\geq 1$ for all $\scriptstyle n\neq m$. So we can consider the function $\scriptstyle f(x):=\sum _{n\in \mathbb {N} }n\left(1-nd(x,x_{n})\right)_{+}$; here $a_{+}$ denotes the positive part of the number $a$ as usual. It is a continuous function, because the sum is locally finite (precisely, for any x in X $\scriptstyle f_{|B(x,1/3)}$ is represented by the sum of finitely many terms). It is unbounded, because $f(x_{n})=n$. It also verifies $|f(x_{n})-f(x)|=n$ for all n and x such that $d(x_{n},x)=1/n$, which shows that f is not uniformly continuous, provided for any n there exists such a point x. But in general this is not true. Example: consider any infinite set X with the discrete distance: d(x,y)=0 or 1 according whether x=y or not. Then every function on X is uniformly continuous just because it vacuously satsisfies the $\epsilon ,\delta$ definition. However, for connected metric spaces the argument above can be completed, so the converse of Heine-Cantor actually holds for them. It is easy to characterize the metric spaces for which any continuous function is uniformly continuous; I don't see a nice way to state it though. --pma (talk) 08:24, 18 March 2009 (UTC)

Apologies for the misinterpretation! I didn't read your comment carefully enough; sorry about that. --PST 09:40, 18 March 2009 (UTC)

But in any case I was wrong, and I realized the mistake thanks to your post... besides, I'm glad that you fell in the same trap too ;) --pma (talk) 09:45, 18 March 2009 (UTC)

March 17

Elliptic and Parabolic equations

I am still confused what exactly are elliptic and parabolic differential equations, and why exactly are they called so. Could someone explain? deeptrivia (talk) 00:32, 17 March 2009 (UTC)

The operator of an elliptic pde, like the laplacian, has Fourier transform x^2 + ... + y^2 like a circle or ellipse. The operator of a parabolic pde, like the heat operator, has Fourier transform x^2 - y, like a parabola. The operator of a hyperbolic pde, like the wave operator, has Fourier transform x^2 - y^2 like a hyperbola. JackSchmidt (talk) 04:20, 17 March 2009 (UTC)

Two point boundary value problems

I want to get as much information as I can about nonlinear two-point third order boundary value problems (e.g., y ' ' ' + a*y' + b*cos(y) = 0, y(0) =y0, y(1) = y1, y'(1) =yp1). Some things I would like to know are: existence of solutions, uniqueness of solutions, analytical solutions (if possible), approximations, numerical solutions, range of the solution space, etc. Could anyone point me to some references on this? Thanks! deeptrivia (talk) 01:22, 17 March 2009 (UTC)

Surface name?

Does the undulating sheet like curve/surface formed by the equations $\sin(x)cos(y)\$ or $\sin(y)cos(x)\$, etc. have a paricular name? --Leif edling (talk) 07:00, 17 March 2009 (UTC)

You may talk about the equations $\scriptstyle z=\sin(x)\cos(y)\$ or $\scriptstyle z=\sin(y)\cos(x)\$. No, I do not know of particular names. Bo Jacoby (talk) 10:40, 17 March 2009 (UTC).

Oops! Well $\scriptstyle z=\sin(x)\cos(y)\$ was what I meant. --Leif edling (talk) 16:58, 17 March 2009 (UTC)

Chess problem

Consider a game of chess in which Black gives White queen odds: that is to say, the game is played exactly like a standard game of chess, except that Black starts without his queen. Given skillful play by both sides, White will win, of course, but for how many moves can Black delay checkmate? —Preceding unsigned comment added by 75.24.76.213 (talk) 08:39, 17 March 2009 (UTC)

I'd be amazed if this is known. In fact, I don't think anyone has even proved that white can force a win in this situation. Algebraist 13:08, 17 March 2009 (UTC)

On this page (warning: old and hard to load) dedicated to "difficult" computer problems, there was a note on the position where Black has only a Queen. While the page was about 12 years old, it pointed out that while computers could find mates in 12, it was suspected that there might be a mate in 11, but they couldn't prove or disprove it. While I suspect that today's computers could answer this question one way or the other, it shows how hard it is to "prove" a win.

I don't doubt that Queen odds is lost with best play, but it will be some time before it's proven, let alone "Mate in ?" determined. (BTW, the main site of that Queen problem has quite a few interesting positions and commentary.) Baccyak4H (Yak!) 15:38, 17 March 2009 (UTC)

So if it's so hard to solve, it would perhaps make an interesting game. I have seen board games with unequal armies, but never one in which the win condition for the weaker side was simply to survive for a given number of moves. —Preceding unsigned comment added by 75.37.236.230 (talk) 20:04, 17 March 2009 (UTC)

White has a winning position... But finding a winning line is almost as hard as solving chess... For the record, after a minute or two my engine can find nothing but a very comfortable position for white (+11.28, depth of 20 half-moves). Pallida  Mors 15:35, 17 March 2009 (UTC)

It depends on how well each side plays. I don't know whether you can define "equal ability". --PST 23:29, 17 March 2009 (UTC)

We aren't talking about 'equal ability', we're talking about perfect play, which is easy to define. Algebraist 23:38, 17 March 2009 (UTC)

Expected value of length of a chord

If I were to randomly construct a chord in the unit circle, what is the probability that the length of the chord is greater than ${\sqrt {3}}$? What is the expected value of the length of the chord? AMorris (talk)●(contribs) 09:23, 17 March 2009 (UTC)

The answer depends on the random method of construction of the chord. You may for example (A) pick a random point inside the unit circle as the center of the chord, or you may (B) pick a random point on the circumference of the circle as one end point of the chord and pick a random arc between 0 and 360 degrees for defining the other. In case A the chord is greater than $\scriptstyle {\sqrt {3}}$ when the midpoint is inside a circle of radius 1/2, and the probability is 1/4. In case B the the chord is greater than $\scriptstyle {\sqrt {3}}$ when the arc is between 120 and 240 degrees, and the probability is 1/3. So your question is not well posed and the answer is not well defined. Bo Jacoby (talk) 10:28, 17 March 2009 (UTC).

(ec) It depends what you mean by "randomly" - see Bertrand's_paradox. AndrewWTaylor (talk) 10:30, 17 March 2009 (UTC)

Or in other words, it depends on the probability distribution of the random chord. Michael Hardy (talk) 20:22, 18 March 2009 (UTC)

Trapezoid solving an angle

I don't know how to solve for angle MNC. I know that it is an isosceles trapezoid and I know that angle B is 88 degrees. That means that angle A is 88 degrees, and angles C and D are 92 degrees. But I don't know how to solve for angle MNC, could you guys help me out? BTW- M is the midpoint of AD, and N is the midpoint of BC. I just drew the horizontal line in there. Thanks for the help! --71.98.25.121 (talk) 23:31, 17 March 2009 (UTC)

By what you are saying MNCD is also an isosceles trapezoid and the MNC=ABC=88 degrees. Dauto (talk) 00:28, 18 March 2009 (UTC)

Please remember that we do not do homework here, although it is customary to provide guidance or a hint if the OP has shown an effort to solve the problem themselves (as this poster has done). A few helpful comments or directed questions would have been better. ("Give a man to fish...") -- Tcncv (talk) 01:09, 18 March 2009 (UTC)

Yes, thank you for your post. I am aware that you do not do homework here. My assignment included over 40 problems; I only asked about one part of a problem I had already tried but didn't understand. Thank you for your explanation Dauto. --71.98.25.121 (talk) 01:37, 18 March 2009 (UTC)

My apologies, I did not mean to be condescending, and I think you misunderstand me – my note was not directed at your question, but at the way it was answered. Questions such as yours are welcome, especially since you had shown that you had already made a good faith effort at working it yourself (which I acknowledged above). My point was that a helpful hint that allows you to find the solution is preferable to giving the answer outright. At least that is my understanding of our policy here. And yes, Dauto is a regular contributor of much useful information here, such as the detailed analysis in the Probability in Rummikub topic above. -- Tcncv (talk) 02:39, 18 March 2009 (UTC)

Since AM and BN have equal lengths and MAB and NBA have equal angles, the distances of the two points M and N below the line AB must be equal; hence the line MN is parallel to AB. Two lines parallel to each other must both meet a line crossing both of them at the same angle. That should answer your question. Michael Hardy (talk) 20:20, 18 March 2009 (UTC)

Thank you for your responses. I'm guessing Michael Hardy's response is the "correct" one. And I'm sorry, Tcncv, for being so rude. I was having a bad day. Not much of an excuse, but I shouldn't have taken it out on you. --71.117.36.70 (talk) 21:02, 18 March 2009 (UTC)

No Problem. I suspect I am not alone in that I enjoy seeing questions such as yours come along. They add variety and give us amateurs a chance to exercise our minds while making ourselves useful. Thank you. -- Tcncv (talk) 00:23, 19 March 2009 (UTC)

March 18

F-statistic application

In comparing of two models(regressions), using F-statistic, why they should be nested in each other? —Preceding unsigned comment added by 217.219.166.149 (talk) 10:42, 18 March 2009 (UTC)

If the models aren't nested, then there is no way to build a hypothesis test to carry out with F-values or generic ANOVAs, as far as I know. Hence, your F-statistics serve merely as fitness measures, but nothing more than that (you may dischard one model, but not compare one with another if both seem to adjust well).
On the other hand, if models are nested, possibilities of variance analysis for direct comparisons are straightforward, as with the case of significance of additional variables, etc.
There are some ways of comparing non-nested models: for instance, analysis of information measures, such as AIC, etc. I'd like to mention the article Model selection, as it can be of your interest. Pallida  Mors 16:33, 18 March 2009 (UTC)

AIME

Does anybody know where to find solutions for the 2009 AIME? Lucas Brown 42 (talk) 18:25, 18 March 2009 (UTC)

http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml should have them at some point. JackSchmidt (talk) 02:59, 19 March 2009 (UTC)

Prime number problem

You don't need to give me the whole answer, but I need to know how to get started. Find all the prime numbers p with the property that 8p^4-123 is also a prime. Thanks! 76.248.244.232 (talk) 22:43, 18 March 2009 (UTC)

A clever lucky use of modular arithmetic gives the answer immediately. (Ask if you want a more detailed hint, or if you are not familiar with modular arithmetic.) Eric. 131.215.158.184 (talk) 00:25, 19 March 2009 (UTC)

Not quite immediately. You still have to check 4877 by hand. Algebraist 00:37, 19 March 2009 (UTC)

True... I went the lazy way and just used "factor" on my command line, though. Eric. 131.215.158.184 (talk) 10:55, 19 March 2009 (UTC)

Sorry, I have no clue what to do with that. Guess-and-check shows that 2 works, and 5 gives the prime 4877, but I also need to explain my answer and why they're the only ones.76.248.244.232 (talk) 01:38, 19 March 2009 (UTC)

As Eric said, the slick solution uses a little modular arithmetic. If you don't know anything about that, it might help to think about the what the last digit of 8p^4-123 could be (this is the same thing as doing modular arithmetic modulo 10, but some people find it easier to think about). Algebraist 02:19, 19 March 2009 (UTC)

The hint I was holding back is Fermat's Little Theorem. You don't need Fermat's Little Theorem to solve the problem (and I found the approach through trial and error), but being familiar with it does suggest a particular line of reasoning: knowing Fermat's Little Theorem, what can we say about p^4? Eric. 131.215.158.184 (talk) 10:55, 19 March 2009 (UTC)

March 19

Functional Convergence

In a recent thread, if I understand correctly, pma says that $\textstyle \sum _{1\leq k\leq n}\log \cos(k^{2}{\sqrt {n^{5}}}s)$ converges pointwise to $\textstyle -s^{2}/10$ as n approaces infinity. How would you prove that? Black Carrot (talk) 07:53, 19 March 2009 (UTC)

Won't the limit depend on what branch of log you choose for negative arguments? Algebraist 10:23, 19 March 2009 (UTC)

My apologies: I made a misprint there (now corrected): the change of variables was $t=n^{-5/2}s$, with a minus in the exponent (this is consistent with the line below, that had it). So the term ${\sqrt {n^{5}}}$ is at the denominator, and the argument of log goes to 1 (actually, in that computation it was always positive). Do you see how to do it now?--pma (talk) 12:40, 19 March 2009 (UTC)

Here it is:

• Write the second order Taylor expansion for $\textstyle \log \cos(x)$ at 0, with remainder in Peano form: so, for all $x\in [0,\pi /2)$

$\log \cos(x)=-{\frac {x^{2}}{2}}+o(x^{2})$, as $x\to 0$.

• For any s we only have to consider the integers $n>4s^{2}/\pi ^{2}$. Replace $x={\frac {k^{2}}{\sqrt {n^{5}}}}s={\frac {s}{\sqrt {n}}}\left({\frac {k}{n}}\right)^{2}$ in the expansion above, getting

$\log \cos \left({\frac {k^{2}}{\sqrt {n^{5}}}}s\right)=-{\frac {s^{2}}{2n}}\left({\frac {k}{n}}\right)^{4}+o({\frac {1}{n}})$, as $n\to \infty$, and uniformly for all $1\leq k\leq n$.

• Summing over all $1\leq k\leq n$

$\textstyle \sum _{1\leq k\leq n}\log \cos({\frac {k^{2}}{\sqrt {n^{5}}}}s)=-{\frac {s^{2}}{2n}}\sum _{1\leq k\leq n}\left({\frac {k}{n}}\right)^{4}+o(1)$, as $n\to \infty$.

• Then you may observe that $\scriptstyle {\frac {1}{n}}\sum _{1\leq k\leq n}\left({\frac {k}{n}}\right)^{4}$ is the Riemann sum for the integral of $x^{4}$ on [0,1] (or use the formula for $\scriptstyle \sum _{1\leq k\leq n}k^{4}$) and conclude that the whole thing is $\textstyle -{\frac {s^{2}}{10}}+o(1)$.

Warning: I have re-edited this answer, to make it more simple and clear (hopefully) --pma (talk) 13:40, 19 March 2009 (UTC)

Differential Equation

How should one go about solving this equation.

$y{\frac {dy}{dx}}=xy{\frac {d^{2}y}{dx^{2}}}+x{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}$

92.9.236.44 (talk) 20:30, 19 March 2009 (UTC)

The right hand side is $x{\frac {d}{dx}}{\bigg (}y{\frac {dy}{dx}}{\bigg )}$. Does that help? —JAO • T • C 20:48, 19 March 2009 (UTC)

Ah yes. It seems to yeild a solution of the form $y={\sqrt {Ax^{2}+B}}$ does that seem correct? —Preceding unsigned comment added by 92.9.236.44 (talk) 21:01, 19 March 2009 (UTC)