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March 22

Statistics question: Estimating a probability distribution given sample trials

Is there a technique for estimating the probability distribution of a continous random variable given the results for many trials?

For instance, a continous random variable was rolled 10 times, and the results were 4.03, 1.99, 3.2, 3.119, 4.21, 0.87, 4.14, 2.02, 3.324, 4.39 - Find the probability distribution, or an approximation of the probability distribution.

The article on recursive Bayesian estimation originally looked promising, but it may not be relevant.--220.253.247.165 (talk) 09:50, 22 March 2010 (UTC)

There are many techniques for this - choosing one depends on the particular application. If you know absolutely nothing about the distribution, you can't do much better than the empirical cumulative distribution function. If you assume the density function is smooth, then a good way to estimate it is kernel density estimation. If you assume the distribution is from some parametric family, you can use the method of moments or preferably, the MLE. -- Meni Rosenfeld (talk) 11:17, 22 March 2010 (UTC)

Wouldn't it make sense to plot it first, then try to determine the type of distribution by inspection ? If I plot the ranges 0-1, 1-2, 2-3, 3-4, and 4-5, I get this:

  4^                 *
  3|             *
n 2|
  1| *   *   *
  0+------------------->
    0-1 1-2 2-3 3-4 4-5
            range

I also notice data clusters around 2, 3.2, and 4.2, although more data points are needed to confirm this. StuRat (talk) 12:58, 22 March 2010 (UTC)

Compute the cumulant generating function of the sample. Bo Jacoby (talk) 18:13, 22 March 2010 (UTC).

Wouldn't that just give the empirical distribution? -- Meni Rosenfeld (talk) 18:25, 22 March 2010 (UTC)

Yes, but the cumulants of the sample approximate the cumulants of the population. See also multiset. Bo Jacoby (talk) 18:44, 22 March 2010 (UTC).

Iterations of a multiplicative function

For those who are interested in such things, here is an oddment from recreational number theory that has been puzzling me (not homework, and not a competition problem either !).

If the prime factoriation of a positive integer n is

$n=\prod _{p_{i}{\text{ prime}}}p_{i}^{q_{i}}$

define the function f(n) as

$f(n)=\prod q_{i}^{p_{i}}{\text{ for }}n>1{\text{ ; }}f(1)=1$

Then f is a multiplicative function, although not completely multiplicative. We can iterate f; for example:

${\begin{aligned}f(24500)&=f((2^{2})(5^{3})(7^{2}))&=(2^{2})(3^{5})(2^{7})&=124461\\f(124461)&=f((2^{9})(3^{5}))&=(9^{2})(5^{3})&=10125\\f(10125)&=f((3^{4})(5^{3}))&=(4^{3})(3^{5})&=15552\\f(15552)&=f((2^{6})(3^{5}))&=(6^{2})(5^{3})&=4500\\f(4500)&=f((2^{2})(3^{2})(5^{3}))&=(2^{2})(2^{3})(3^{5})&=7776\\f(7776)&=f((2^{5})(3^{5}))&=(5^{2})(5^{3})&=3125\\f(3125)&=f(5^{5})&=5^{5}&=3125\\\end{aligned}}$

As the above sequence shows, f(n) may be less than, greater than or equal to n. However, the second iterate f(f(n)) always seems to be less than or equal to n.

Can you either prove that f(f(n)) ≤ n for all n or find a counterexample such that f(f(n)) > n. Gandalf61 (talk) 11:10, 22 March 2010 (UTC)

If p is an arbitrary prime number, then $f(p^{k})=k^{p}$. However, if we let $k=\prod _{i=1}^{m}q_{i}^{n_{i}}$ (q_i prime for all i), we can compute $f(f(p^{k}))=f(k^{p})=f(\prod _{i=1}^{m}q_{i}^{pn_{i}})=\prod _{i=1}^{m}(pn_{i})^{q_{i}}=p^{\sum _{i=1}^{n}q_{i}}\prod _{i=1}^{m}n_{i}^{q_{i}}=p^{\sum _{i=1}^{n}q_{i}}f(k)$. Now use the formula $f(f(p^{k}))=p^{\sum _{i=1}^{n}q_{i}}f(k)$ to test your conjecture. PST 15:03, 22 March 2010 (UTC)

Okay, I follow that - but I am not sure where it goes next. And what if n does not have the form p^k ? Gandalf61 (talk) 15:48, 22 March 2010 (UTC)

You want to show that $\prod _{i=1}^{m}q_{i}^{n_{i}}\geq \sum _{i=1}^{m}q_{i}(1+\log _{p}n_{i})$ for any p.

$2^{n-1}\geq 1+\log _{2}n$ for all integer n > 0, so then $q^{n}\geq q(1+\log _{2}n)$ for any prime q. The product of numbers ≥2 is at least their sum so $\prod _{i=1}^{m}q_{i}^{n_{i}}\geq \sum _{i=1}^{m}q_{i}^{n_{i}}\geq \sum _{i=1}^{m}q_{i}(1+\log _{2}n_{i})\geq \sum _{i=1}^{m}q_{i}(1+\log _{p}n_{i})$ which is what we want.

For terms that aren't of the form p^k, I don't think the argument should be too bad. The main idea is that for n = p^kq^k, with k prime f(f(n)) = (p + q)^k which is less, but there might be some difficult cases you have to deal with. Rckrone (talk) 22:36, 22 March 2010 (UTC)

Problem is that although f() is multiplicative, f(f()) is not - for example

$f(f(2^{5}\,3^{5}))\neq f(f(2^{5}))\,f(f(3^{5}))$

So a proof for prime powers does not easily extend to a proof for all n. Gandalf61 (talk) 14:47, 23 March 2010 (UTC)

What you can do though is show that for any a and b, f(f(a)f(b)) ≤ f(f(a))f(f(b)). Let $f(a)=\prod q_{i}^{n_{i}}$ and $f(b)=\prod q_{i}^{m_{i}}.$ The problematic terms are only the primes they have in common, so just consider those. $f(\prod q_{i}^{m_{i}+n_{i}})=\prod (m_{i}+n_{i})^{q_{i}}.$ Numbers such as f(a) and f(b) in the image of f must have all exponents m_i, n_i ≥2, so $\prod _{i}(m_{i}+n_{i})^{q_{i}}\leq \prod _{i}(m_{i}n_{i})^{q_{i}}=f(\prod _{i}q_{i}^{m_{i}})f(\prod _{i}q_{i}^{n_{i}}).$

With that result, for any number $n=\prod _{i=1}^{r}p_{i}^{k_{i}},$

$f(f(\prod _{i=1}^{r}p_{i}^{k_{i}}))=f(\prod _{i=1}^{r}f(p_{i}^{k_{i}}))\leq \prod _{i=1}^{r}f(f(p_{i}^{k_{i}}))\leq \prod _{i=1}^{r}p_{i}^{k_{i}}$ Rckrone (talk) 17:10, 23 March 2010 (UTC)

Inverse functions $e^{x}-e^{-x}$ and $\ln \left({\frac {x+{\sqrt {x+4}}}{2}}\right)$

I worked out that the second in my question's title was the inverse of the first by using the first and then substituting $a=e^{x}$, flipping my x-and y-values so I have $x={\frac {a^{2}-1}{a}}$, rearranging to get $a^{2}-xa-1=0$ and applying the quadratic equation and discarding the solution with the minus and finally taking the log to get back to y. But when I graphed these two functions on the online grapher at walterzorn.com, they only looked like mirror images over the line y=x for positive x-values. Are these two not inverses, meaning I did something wrong, or is there some restriction here I'm not remembering? I know you can't take the log of a negative value, but $x+{\sqrt {x+4}}$ doesn't go below zero until x equals approximately -1.56, where the asymptote of my second function is. 20.137.18.50 (talk) 13:23, 22 March 2010 (UTC)

Should be $\ln \left({\frac {x+{\sqrt {x^{2}+4}}}{2}}\right)$. Gandalf61 (talk) 13:36, 22 March 2010 (UTC)

Thanks, I see how I forgot to square my middle term. 20.137.18.50 (talk) 14:05, 22 March 2010 (UTC)

Should be $\ln \left({\frac {x+{\sqrt {x^{2}-4}}}{2}}\right)$. (minus not plus in the square root) Staecker (talk) 13:46, 22 March 2010 (UTC)

Sorry- you're right. I screwed it up twice so it still looked right when I double-checked. Staecker (talk) 14:39, 22 March 2010 (UTC)

Since no one gave the WP link yet, I will: hyperbolic function#Inverse functions as logarithms. The inverse of $e^{x}-e^{-x}$ is then $\operatorname {arsinh} (x/2)=\log \left({\frac {x}{2}}+{\sqrt {\left({\frac {x}{2}}\right)^{2}+1}}\right)=\log \left({\frac {x+{\sqrt {x^{2}+4}}}{2}}\right)$.—Emil J. 14:54, 22 March 2010 (UTC)

You have some missing logs, of course. -- Meni Rosenfeld (talk) 17:13, 23 March 2010 (UTC)

Drat. Thanks for pointing it out.—Emil J. 11:37, 24 March 2010 (UTC)

If and only if symbol to use in thesis

I am wanting to use the symbol for iff in my theoretical framework but I am not sure which one would be the right one to use in this instance. Should it be ≡ ?--160.36.39.157 (talk) 14:27, 22 March 2010 (UTC)

If you must use a symbol, I've most often seen $\Leftrightarrow$. But English words are usually better in my opinion. Staecker (talk) 14:34, 22 March 2010 (UTC)

I agree, words are often better than symbols (I don't know about "usually better"). The abbreviation "iff" is very commonly used, as well. --Tango (talk) 15:35, 22 March 2010 (UTC)

See iff. -- SGBailey (talk) 16:28, 22 March 2010 (UTC)

I'd say the symbol is OK, but you need to define it first: "Throughout this thesis I will use 'iff' to mean 'if, and only if,' ". StuRat (talk) 17:08, 22 March 2010 (UTC)

I think "iff" is sufficiently widely used not to need to be defined. --Tango (talk) 17:11, 22 March 2010 (UTC)

I agree with that -- definitely do not define the word iff; that would be somewhere on the continuum from silly to offensive.

But I don't actually agree with using it. Iff is a blackboard abbreviation; it's not the right linguistic register for a thesis. That said, some good people do use it in print, but I still don't like it.

As for symbols versus words, it depends on whether you're writing prose or logical formulae. It's true that $\Leftrightarrow$ would be odd in prose, but it's just the right thing if the rest of the formula is in symbols. (Well, that or $\leftrightarrow$.)

If you find that the repetition of if and only if gets tedious, a couple of synonyms that you could throw in for elegant variation are just in case or exactly when, or you could reword to use necessary and sufficient. --Trovatore (talk) 17:20, 22 March 2010 (UTC)

You're both assuming that the audience has a math or science background. This could be a thesis in another area, with just a bit of math or science thrown in. StuRat (talk) 17:32, 22 March 2010 (UTC)

That strikes me as unlikely. --Trovatore (talk) 17:58, 22 March 2010 (UTC)

It is a thesis for M. S. in Agricultural Economics. I have my logic explained in words and I want to also show it in a equation form.--160.36.39.157 (talk) 18:58, 22 March 2010 (UTC)

Optimal location of a bridge

Does anybody suggest me an algorithm for finding an optimal location of a bridge joining two sets of populated places on either side of a river? —Preceding unsigned comment added by Amrahs (talk • contribs) 15:23, 22 March 2010 (UTC)

For a optimisation problem you need a cost function (some way of comparing different places) and a set of constraints (things that must be satisfied by the place). What algorithm is best will depend on what form those things take. --Tango (talk) 15:38, 22 March 2010 (UTC)

Obvious criteria to include are: minimising sum of (journey length over bridge) * (Relative journey frequency) and minimising construction cost of bridge (eg build it over the narrowest part of the river) -- 16:26, 22 March 2010 (UTC)

This would get so complex, with so many variable weighting factors, that human judgment might be better than an algorithm. Other factors might be the depth of the river in various locations, ground quality (bedrock or swamp ?) at the adjacent land, what would need to be demolished to build the bridge (new skyscraper or old slum ?), environmental impact, attractiveness of the bridge in various locations, effect on river traffic, etc. Something which they don't always consider, but should, is the ability to connect to existing highways. In many places you must exit the highway, drive on local roads to the bridge, cross the bridge, then drive on local roads again to get to the highway on the other side. And, if political reasons are included, like getting government money for your area that will otherwise go elsewhere, you can even justify a bridge to nowhere. StuRat (talk) 17:15, 22 March 2010 (UTC)

The quest for optimum is not worth while to pursue. Comparing two options, if one of them is clearly inferior, discard it. If the difference is not clear, pick either one. Using a lot of efford in making an unimportant decision is not rational behaviour. Bo Jacoby (talk) 08:55, 23 March 2010 (UTC).

Seems to me that such a decision can be very consequential to a lot of people. -- Meni Rosenfeld (talk) 10:25, 23 March 2010 (UTC)

I do agree. You could even build several bridges crossing the same river. Bo Jacoby (talk) 14:02, 23 March 2010 (UTC).

If the river was a straight line, and you were solely interested in the shortest distance between town A and town B, then the obvious place to put it would be where the line from A to B crossed the river. Too obvious? 92.29.120.231 (talk) 15:30, 23 March 2010 (UTC)

That neglects all the other factors mentioned so far. I'd say many of those other factors are far more important than the bridge being equidistant from both cities. For example, if you have swamps there ("precious wetlands", to environmentalists), and no roads, that may be a poor choice. StuRat (talk) 15:41, 23 March 2010 (UTC)

The bridge would only be equidistant if the towns or cities were the same perpendicular distance from the straight-line river. The OP did not give any criteria for assessing the merits of the bridge position. 92.29.120.231 (talk) 16:31, 23 March 2010 (UTC)

The way I understood the question, there are more than 2 towns. -- Meni Rosenfeld (talk) 17:12, 23 March 2010 (UTC)

If there were several cities on each side of the river, then the OP needs to define the road system interconnecting the cities. If roads cost nothing to build and you can have as many of them as you like, and there are no other considerations, then you just travel from one end of the river to another and at each point, sum the distances to each city. Choose the point that has the lowest sum of distances. 78.149.133.100 (talk) 21:52, 23 March 2010 (UTC)

Probably in reality you would connect the two road networks on either side of the river by joining the two closest points of them, since that would require building the least amount of new road. If this is for a game, then the roads could be constructed using Central place theory with some randomness added. 78.149.167.173 (talk) 21:39, 24 March 2010 (UTC)

March 23

How to parametrise distribution given a few CDF points

Hi. Let's say I have a random variable, X (that represents a certain loss level) and I have three points for the CDF. So I know that
F(0.5) = k₁
F(0.9) = k₂
F(0.95) = k₃
X has a normal distribution.
Where F is the CDF and the k's are known.

My questions are:

A) How do I find the parameters for the distribution?
B) What if X is lognormal (or some other distribution) - is it still possible to find the distribution?
Thank you! Mudupie (talk) 08:02, 23 March 2010 (UTC)

A) Use a table or a computer to find the standard scores corresponding to the CDF values. For example, if $k_{1}=0.6$ and $k_{2}=0.8$, you have $z_{1}=0.2533,\ z_{2}=0.8416$ so $0.5=\mu +0.2533\sigma ,\ 0.9=\mu +0.8416\sigma$. Solve for $\mu$ and $\sigma$.

B) For lognormal specifically, you know that $\log X$ is normal so you solve problem A with $F(\log 0.5)=k_{1},\ F(\log 0.9)=k_{2}$. If you have some other parametric family, you just plug in the values for the CDF in terms of the parameters and solve. Whether the solution will have a nice closed-form expression, or you will have to solve it numerically, depends on the family. -- Meni Rosenfeld (talk) 10:34, 23 March 2010 (UTC)

nth iteration of a koch snowflake

For a koch snowflake in which iteration 0 (i.e. the triangle) has a side length of 1, is it true that iteration 3 has an area of ${\frac {94}{81{\sqrt {3}}}}$? Also, is the area of the nth iteration $A_{n}=A_{n-1}+{\frac {4^{n-2}{\sqrt {3}}}{3\times 9^{n-1}}}\quad n>1\,,A_{0}={\frac {\sqrt {3}}{4}}$, and if so, is there a non-recursive formula for the area of the nth iteration? --220.253.247.165 (talk) 08:02, 23 March 2010 (UTC)

Yes, it is a geometric series. Bo Jacoby (talk) 08:57, 23 March 2010 (UTC).

what's the best area for contributing something practical

what should i study, graph theory or what to contribute to actual projects later graf theory being microchips —Preceding unsigned comment added by 82.113.121.34 (talk) 18:23, 23 March 2010 (UTC)

Your question isn't clear. Can you rephrase it? -- Meni Rosenfeld (talk) 09:24, 24 March 2010 (UTC)

If you're interested in the design of integrated circuits specifically, it doesn't have too much to do with graph theory, it's more a matter of computer science and electrical engineering. Learning to program is probably the first step. Paul Stansifer 12:38, 24 March 2010 (UTC)

Distribution of a subset of a distribution

Suppose we have a variable x, drawn from some known probability distribution over [l,h], l<h, both real. Suppose then we're interested in y, which is the set of all x less than some constant m, l<m<h -- what is the distribution for y? I suspect that it's just the normalized distribution for x<m, but I'm not super good at statistics so I'd appreciate confirmation or correction as needed. 71.70.143.134 (talk) 18:29, 23 March 2010 (UTC)

It is. Algebraist 18:30, 23 March 2010 (UTC)

You're talking about the conditional probability distribution given the event that y < m. Michael Hardy (talk) 18:43, 23 March 2010 (UTC)

how much do words impede mathematics

a lot of times you see people struggling with concepts because they don't know what the subbranch of mathematics is called, or has been explored, they're just trying to do it ab ovo. Isn't this a case of words impeding mathematics, where if it were more systemized with elementary symbols, people could just see if this property has been explored before? there is no way to "guess" words, you can't just guess words like normal, perfect, complete, etc, when they have nothing to do with English usage. 82.113.121.38 (talk) 19:14, 23 March 2010 (UTC)

I fixed the title ("to" -> "do"). StuRat (talk) 19:28, 23 March 2010 (UTC)

You can't create a usable notation that covers everything. You have to invent new notations for each new branch of maths and those new notations are no easier to guess than words. --Tango (talk) 20:35, 23 March 2010 (UTC)

The general idea that words (or other symbolic expressions like musical or mathematical notation) shape thought (and vice versa) is known as the Sapir-Whorf Hypothesis. The degree to which this is really true is unclear; the current consensus seems to be roughly "maybe, a bit". In another vein, I just finished reading The Strangest Man, a biography of physicist Paul Dirac. It covered the different ways that different mathematical physicists thought about quantum theory - some used physical intuition, some thought about it algebraically, some with abstract timeline diagrams, and Dirac thought about it with a kind of projective geometry. So different mathematical minds used different approaches to understand the same shared concept; surely this will hold for ordinary students of maths too - some will benefit from diagrams, some from measurement, some from algebra and logic, some from thinking about the real-world meaning of the problem. There are some deeply-formal expressions of mathematics, such as Principia Mathematica, but they're deeply unsuitable for struggling learners. -- Finlay McWalter • Talk 20:54, 23 March 2010 (UTC)

I think it does happen, and has not much to do with words. There's a wide dependency graph between mathematical concepts, and if you're struggling with some problem in area X, even if you're relatively knowledgeable, it can take quite a bit of effort to find out that there's a body of theory in area Y that applies to your problem. Being around a big department with lots of different kinds of specialists to talk to helps. The Internet (including Wikipedia) also helps, since it makes surfing between topics much faster than was possible in the era of having to read a paper and then separately chase down each relevant citation from it in the library. 66.127.52.47 (talk) —Preceding undated comment added 01:02, 24 March 2010 (UTC).

But what the original poster seems to want is some way to standardize mathematical discourse, so that (to paraphrase maybe a bit reductively), you wouldn't have to think to see that, oh, this concept from general topology is kind of like that concept from algebra; it would just be obvious from the naming itself. That's not going to happen. To come up with such a standard you'd have to be able to predict what concepts mathematicians in all fields would think of and find useful. --Trovatore (talk) 01:17, 24 March 2010 (UTC)

Good mathematical notation does do that, to some extent—it's full of metaphors that suggest similarities between different things. For example, Leibniz's notation in calculus suggests that derivatives are like fractions, which suggests such properties as the chain rule and leads to the idea of differentials. The Cartesian product of sets is written with multiplicative notation such as × and ∏, suggesting similarities with the ordinary concept of multiplication. The set inclusion symbols ⊆ and ⊇ are analogous to the inequality symbols ≤ and ≥, which suggests that set inclusion induces a (partial) ordering. Similarly, the join and meet symbols ∨ and ∧ are similar to the union and intersection symbols ∪ and ∩, which suggests that join and meet can be thought of as generalized unions and intersections. Exclusive or is often denoted ⊕, because (thinking of true as 1 and false as 0) it can be thought of as addition in the finite field $\mathbb {F} _{2}$. Of course, all of this notation was invented by somebody who first saw the similarities between the concepts—it would be impossible (or at least highly unlikely) for someone to invent notation for a mathematical concept and only then see a similarity with another concept because it was suggested by the notation. —Bkell (talk) 10:36, 24 March 2010 (UTC)

A taxonomy of mathematical concepts, like giving chemicals names. Now there's an interesting idea. I suppose one could make a start with category theory. I think what I'd do for something like this is not actually give names to the concepts but tag them with types and attach them to linked concepts, and have a computer look for similar structures. A bit like matching up separate trees made by different people in one of those ancestry programs on the web that look for similar birthdays or names and which can be mapped partly to each other. Could be interesting, I'm not sure it would help much but occasionally things like that do pan out and are quite useful sometimes in ways you never expect. Dmcq (talk) 10:05, 24 March 2010 (UTC)

Questions about Birkhoff Normal form

I have an area preserving map on the plane with an elliptic fixed point. My questions are:

• where can I find the procedure to compute the coefficients of the Birkhoff normal form?
• are there sufficient conditions which implies that there is at least a coefficient different from 0?

--Pokipsy76 (talk) 21:27, 23 March 2010 (UTC)

March 24

How to display surds on the TI-84 Plus

Does anyone know? Is there a program which I can download which does it?--124.171.116.21 (talk) 02:32, 24 March 2010 (UTC)

the volume of a solid bounded by the lines x=0, y=0, z=0 and the plane x+y+z=9

Just checking webwork ... isn't the volume 729/2? I'm attacking this from several angles -- e.g. double integration just plain ole geometric sense (it should be half the cube 9x9x9). But the system says I'm wrong ... John Riemann Soong (talk) 09:01, 24 March 2010 (UTC)

If you had $x+y\leq 9,\ 0\leq z\leq 9$ you'd get a prism which is half the cube. Since you have $x+y+z\leq 9$ you get a pyramid which is one third of the prism, or only 1/6 of the cube. Integration also gives 121.5. -- Meni Rosenfeld (talk) 09:21, 24 March 2010 (UTC)

You've got a pyramid one of whose faces is half of the square. The volume of a pyramid is 1/3 × base × height. So it's 1/3 × (half the square) × (height) = (1/3) × (1/2) × (volume of the cube). Michael Hardy (talk) 14:25, 24 March 2010 (UTC)

Poincare Conjecture and Ricci flow

Perelman used Ricci flow in proposing a proof of Poincare's conjecture.

Eddington in about 1934 said he proved, or seemed to prove, (as Kumar seemed to earlier) that if Pythagoras' theorem is extended to four or more dimensions then the sign of the fourth or further term is negative. For example, time in Einstein's metric contributes with the opposite sign to the space terms, ds squared = (say) dx1 squared + dx2 squared + dx3 squared - dt squared. Eddington said he did this to say something about Dirac's spin. Eddington's proof (?) involved group theory, but I'm not sure of its elimination of alternatives, for I only saw a semi-popular account of it.

Does this modify or limit the applicability of the Ricci flow method to Perelman's proof of Poincare's conjecture? Is Poincare's conjecture itself affected by the different sign on the fourth dimension, in a metric?

As a University of Sydney student, I had not much spare time to study or ask this.

122.152.132.156 (talk) 10:25, 24 March 2010 (UTC)

I should preface this by saying I don't know a whole lot about this topic. But the Poincare conjecture is purely a topological statement. It doesn't depend on any metric, even if a particular metric was used to prove it.

Also, the standard definition of a metric requires it be positive definite. Rckrone (talk) 17:47, 24 March 2010 (UTC)

Agree with above (I'm also not an expert here). Your question sounds similar to Marilyn vos Savant's infamous objection (later retracted) to Wiles' proof of Fermat's last theorem. She objected that the proof uses hyperbolic geometry which isn't allowed because the "real word" uses Euclidean geometry (which is only mostly true). The fact is that mathematicians use whatever abstract notions are appropriate for the setting, regardless of any considerations of the "real world", which is irrelevant in the context of an abstract proof. Staecker (talk) 22:41, 24 March 2010 (UTC)

The "metric" with -dt^2 is called the Minkowski metric which as Rckrone says is not a true metric in the usual mathematical sense. The geometric symmetry group of Minkowski space is the Poincaré group, if that's what you're thinking of. AFAIK it doesn't have anything to do with the Poincaré conjecture in topology. 66.127.52.47 (talk) 07:04, 25 March 2010 (UTC)

Beads on a necklace

If you had n beads, all different and unique, then would the number of unique orderings on the necklace, after allowing for rotations and reflection, be n!/2n ? 78.149.167.173 (talk) 22:09, 24 March 2010 (UTC)

No. If we plug in 2 beads, we get 2!/2(2) = 2/4 = 1/2. Obviously there isn't half a combination. Can you figure out what's wrong with your formula ? StuRat (talk) 00:49, 25 March 2010 (UTC)

The formula gives the correct answer for n=3 at least. 84.13.22.69 (talk) 14:09, 25 March 2010 (UTC)

You can cheat using the search function on wikipedia 9or google) and you'll find Necklace (combinatorics) Dmcq (talk) 01:24, 25 March 2010 (UTC)

The problem is fairly easy, and this article doesn't make it any easier at all. -- Meni Rosenfeld (talk) 09:58, 25 March 2010 (UTC)

You seem to be quite right, you'd need to know how to solve the problem above long before you ever turned to that article! Dmcq (talk) 14:30, 25 March 2010 (UTC)

The way I got the n!/2n was, if you had a line of n unique and different objects, then the number of different ways of ordering them would be n!. Join the ends together, then the number of different ways would be reduced by n as you could rotate the string of n beads into n positions, and by a further 2 because you could flip the necklace over. What is wrong with that reasoning? 84.13.22.69 (talk) 14:06, 25 March 2010 (UTC)

In fact that's quite reasonable but it still goes wrong for 0, 1 or 2 beads! :) For more than 2 you're quite right, you don't get exactly the same pattern flipping the necklace over. Dmcq (talk) 14:30, 25 March 2010 (UTC)

And the reason is that flipping the necklace over doesn't change the order, with 2 beads or less. StuRat (talk) 17:58, 25 March 2010 (UTC)

So is the formula correct provided that n>2 ? 84.13.22.69 (talk) 14:45, 25 March 2010 (UTC)

Indeed. -- Meni Rosenfeld (talk) 15:00, 25 March 2010 (UTC)

Would it be worth a mathematician noting that on the Bracelet (combinatorics) page, since itys not obvious to the casual reader that the complicatated formula given does reduce down to the above with the limitations given? 84.13.34.56 (talk) 13:14, 26 March 2010 (UTC)

I gave it a shot, using the material from this Q. Note that I'm not a mathematician. However, I think the original version having been written by a mathematicians was the problem, as this results in articles only readable by other mathematicians. StuRat (talk) 13:43, 26 March 2010 (UTC)

Unfortunately, you added it to the wrong article. Your example is about (what our articles refer to as) bracelets, not necklaces. Algebraist 13:46, 26 March 2010 (UTC)

Thanks, I moved it here: Bracelet_(combinatorics)#Example. Although, I don't care for the terms "necklace" and "bracelet", I much prefer "reversible necklace" and "nonreversible necklace", since those terms actually describe the difference. But, if the mathematicians have chosen the vague terms, we may be stuck with them. StuRat (talk) 13:56, 26 March 2010 (UTC)

So, if we allow rotation, but not flipping it over, what they call a "necklace", is the formula then n!/n (which is the same as (n-1)!) ? What are the restrictions on the range of n (obviously it can't be zero) ? StuRat (talk) 14:00, 26 March 2010 (UTC)

Except that the beads on a necklace are not assumed to have pairwise different colours, so this formula has nothing to do with the $N_{k}$ and $M_{k}$ counting functions in the article. The same problem is with bracelet counting by the way, so the example you added there is quite misleading, n!/2n has nothing to do with $B_{n}(n)$.—Emil J. 14:13, 26 March 2010 (UTC)

Please explain what "pairwise different colours" means, so we can bridge the gap. StuRat (talk) 14:20, 26 March 2010 (UTC)

Let n = 3, for example. The 3!/3 = 2 things you counted are 012 and 210 (as cyclic orders). The $N_{3}(3)=11$ necklaces are 000, 111, 222, 001, 002, 110, 112, 220, 221, 012, and 210. Is it clear?—Emil J. 14:26, 26 March 2010 (UTC)

It looks like my example assumes each bead is unique (which is stated in the first sentence), while, in general, this assumption is not made. That seems OK to me, as long as the assumption is stated clearly. I can clarify it a bit more, if you like. StuRat (talk) 14:35, 26 March 2010 (UTC)

I added an example to the necklace article, too: Necklace_(combinatorics)#Example. Is it OK ? Specifically, is calling it a "2D" necklace right, since it closes back on itself, or should it still be considered 1D ? StuRat (talk) 15:34, 27 March 2010 (UTC)

March 25

Circle

Hey all. I was surfing the web, looking for nothing in particular one night, and I came upon a short math quiz on some website. Most of the questions were pretty easy, but there's one that had me stuck. It goes something like, "I have three semicircles, with radii of 3, 2, and 1 unit respectively. The semicircle with radius 2 is externally tangent to the semicircle of radius 1, and both the semicircles of radii 1 and 2 are internally tangent to the one with radius 3. There is a circle which is externally tangent to the semicircles of radii 1 and 2, and internally tangent to that with radius 3. What is its radius?" At first I thought it had something to do with derivatives, but that just led me to a dead end. Thougths? —Preceding unsigned comment added by 99.13.219.136 (talk) 03:15, 25 March 2010 (UTC)

Just eyeballing it, it looks to be a bit smaller than the radii 1 circle, maybe (3^½)/2 or 0.866 ? StuRat (talk) 03:32, 25 March 2010 (UTC)

There is no diagram? Hope I'm getting this pictured correctly, the semi-circle (SC) of radius 3 contains SC 1, SC 2, and circle of unknown radius inside. Very interesting, I'll get back to you on that. --Kvasir (talk) 03:49, 25 March 2010 (UTC)

I'm getting hung up on why they're explicitly semicircles - the way I'm imagining things laid out (r1 & r2 side-by-side hills, r3 joining over them, the circle of unknown radius nestled in the valley between r1 & r2), the problem would be exactly the same if it was circles of radius 1, 2 & 3. The fact that they explicitly mention "semicircles" indicates that this may be a "gotcha" problem, where the "real" way to solve it is not the obvious one. -- 174.21.224.236 (talk) 05:08, 25 March 2010 (UTC)

I also drew a full R3, R1 and R2 circles thinking it wouldn't change the problem, except there would be 2 of the same unknown circles fitting on either sides of the R1 and R2 circles. I think one of the keys of solving this is to think outside the box and draw the full circles. I think it's still a geometry problem, not a system of equation type question. --Kvasir (talk) 06:05, 25 March 2010 (UTC)

The arrangement of semicircles you describe is called an arbelos. But I agree that in this case the fact that they're semicircles seems to be a bit of a distraction and it would be simpler to describe with just circles. Descartes' theorem is probably what you're looking for. —David Eppstein (talk) 07:16, 25 March 2010 (UTC)

I placed the circles in a Cartesian coordinate system so that C3 centre is (0,0) and C1 centre is (2,0), built a three equations system, solved it and got r and Cr centre... Err, is it OK to put the answer here...? --CiaPan (talk) 07:45, 25 March 2010 (UTC)

I expect Circles of Apollonius helps - but I haven't read it through yet.And Problem of Apollonius is definitely relevant. -- SGBailey (talk) 11:32, 25 March 2010 (UTC)

I sort of cheated with the help of Autocad, the radius of the circle is approximately 0.8571 unit. (AutoCad can draw a circle given 3 tangent points, so there must be some sort of logarithm.) I'm still trying to look at the geometric relationships there. Stay tuned. --Kvasir (talk) 14:43, 25 March 2010 (UTC)

That is accurate to 4 decimal places, but there are no logarithms involved, only rational numbers. Gandalf61 (talk) 15:19, 25 March 2010 (UTC)

oh yeah... I meant to say algorithm, hahah. --Kvasir (talk) 15:33, 25 March 2010 (UTC)

Hmm check out Bankoff circle. It's nice not needing to re-invent the wheel, so to speak. --Kvasir (talk) 15:42, 25 March 2010 (UTC)

That article is little more than a stub, and seems to be poorly written, in that the labels in the discussion don't appear to match those in the diagram (C₁, C₂, C₃, vs. C"₆ ?). I also can't follow what they mean by "r = AB/AC". From the diagram, "r = AB" would be correct. Finally, "R", I assume, is the radius of the red circle in the diagram, which is not the one we want. Would anyone be willing to fix this article and diagram ? StuRat (talk) 17:17, 25 March 2010 (UTC)

Yeah the Bankoff circle is not the circle referred to by the OP. The only thing that is helpful from the article is the diagram illustrating this particular problem here. --Kvasir (talk) 17:32, 25 March 2010 (UTC)

In case anyone has problem solving it, the exact value confirmed the previous estimation above using Descartes' Theorem suggested above. The answer is 6/7 ~ 0.8571. This was fun. --Kvasir (talk) 17:59, 25 March 2010 (UTC)

So my hand drawn sketch and eyeballing the answer were off a bit, but 0.866 is pretty good for that kind of estimate (just over 1% off). StuRat (talk) 18:34, 25 March 2010 (UTC)

Cute problem. I now think I can express the answer in what some might consider "closed form", although it's a bit hairy. Michael Hardy (talk) 21:41, 27 March 2010 (UTC)

A matrix M raised to the i-th power (i being imaginary), or the natural base raised to the M-th power (M being a matrix).

What is that? Eliko (talk) 10:36, 25 March 2010 (UTC)

The second one presumably refers to matrix exponential.—Emil J. 11:11, 25 March 2010 (UTC)

... and for the first one, if M has a matrix logarithm then the natural way to define Mⁱ would be Mⁱ = exp(i log(M)). Gandalf61 (talk) 11:15, 25 March 2010 (UTC)

...but that's not unique (even if it exists), is it?—Emil J. 11:21, 25 March 2010 (UTC)

No. Relatedly, e^M could be interpreted to mean exp(M log(e))=exp(M (1+2niπ)), which is also multivalued, with the standard matrix exponential as one of its values. Algebraist 11:42, 25 March 2010 (UTC)

For a sufficiently well-behaved matrix, you can use the principal logarithm. -- Meni Rosenfeld (talk) 11:46, 25 March 2010 (UTC)

So ? The i-th power of a real or complex number isn't unique either, for the same reason. Gandalf61 (talk) 12:16, 25 March 2010 (UTC)

Eliko, you say on your user page that you're a native speaker of English. Please get clear on this: it's a matrix, not a "matrice". The plural of "matrix" is "matrices". Michael Hardy (talk) 02:44, 27 March 2010 (UTC)

Yes, you are right. I've fixed it. Eliko (talk) 21:34, 27 March 2010 (UTC)

Dissections in Riemann Integration

Hi. I'm currently going through my notes on an Analysis course and am a bit confused on a Lemma regarding dissections. It says

"If D₁ and D₂ are any two dissections then $S(f,D_{1})\geq S(f,{D_{1}}\cup {D_{2}})\geq s(f,{D_{1}}\cup {D_{2}})\geq s(f,D_{2})$, where S(f, D) denotes the upper sum of f wrt D and s(f, D) the lower sum."

My main problem comes from the two extremes, ie that $S(f,D_{1})\geq s(f,D_{2})$. If we know nothing about these dissections then how can we make such a statement? Thanks. 92.11.43.155 (talk) 15:38, 25 March 2010 (UTC)

Any upper sum is greater than the integral, and any lower sum is less than the integral, so any upper sum is greater than any lower sum.

This of course is an intuitive explanation for cases when the integral exists - for a proof in the general case, that's exactly what the lemma gives. -- Meni Rosenfeld (talk) 16:48, 25 March 2010 (UTC)

Thank you Meni! 92.11.43.155 (talk) 17:42, 25 March 2010 (UTC)

Two dimensional representation of an arbitrary distance matrix

Given a set of items and a complete set of arbitrary pairwise "distances" between them (with the property that d(a,b) = d(b,a)), is there a technique for assigning 2D positions to them, such that the set of pairwise 2D euclidean distances "best" (e.g. in a least squares/expectation maximization sense) matches the set of starting distances? -- 174.21.224.236 (talk) 17:03, 25 March 2010 (UTC)

Would you want to keep all the distances, or toss out those that don't fit with the rest ? StuRat (talk) 18:30, 25 March 2010 (UTC)

Yes. From the distance matrix compute a kernel matrix, apply kernel PCA and take the first two components.

To find the kernel matrix, enumerate the items as $\{v_{1},\ldots ,v_{n}\}$. Let $S_{i}=\sum _{j}d(v_{i},v_{j})^{2}\;\!,$ $S={\frac {\sum _{i}S_{i}}{2n}}\;\!,$ $\|v_{i}\|^{2}={\frac {S_{i}-S}{n}}\;\!.$ Then $K_{i,j}=(\|v_{i}\|^{2}+\|v_{j}\|^{2}-d(v_{i},v_{j})^{2})/2$.

Note that your distance matrix should satisfy the triangle inequality - otherwise the results may be unpredictable.

This is closely related to Nonlinear dimensionality reduction, you may find some relevant information there. -- Meni Rosenfeld (talk) 19:32, 25 March 2010 (UTC)

Thanks! Nonlinear dimensionality reduction was what I was looking for. A few followup questions: In the proceedure you suggest, am I correct in stating that you transform the distance matrix with the kernel K, and then do PCA on the transformed matrix? If so, I'm a little confused on how to use the first two principle components to compute the (x, y) positions of the original points. For principle component vectors C₁ and C₂, is the coordinate for item v_i simply ( C_1,i, C_2,i ), or do I have to "backtransform" the components through the kernel? Additionally, is there a particular reason you selected kernel PCA from the list of techniques at nonlinear dimensionality reduction, and how did you select the kernel function? Finally, the matrix is only "distances" in that it's somewhat analogous to a separation between the two objects. I'm uncertain whether they will always satisfy the triangle inequality. Is there a (simple) way around that, or what do you mean by "unpredicable"? -- 174.31.194.126 (talk) 16:32, 26 March 2010 (UTC)

The orthogonal diagonalization of K will give you $K=UDU^{-1}$ where $U^{T}U=I$ and the diagonal entries of D are with decreasing magnitude. Letting $C_{1},\ C_{2}$ be the first two columns of U and $\lambda _{1},\ \lambda _{2}$ be the corresponding eigenvalues, the coordinates of point i are $\left((C_{1})_{i}{\sqrt {\lambda _{1}}},(C_{2})_{i}{\sqrt {\lambda _{2}}}\right)$.

The techniques of dimensionality reduction typically deal with data that is explicitly embedded in some high-dimensional Euclidean space, and they construct a distance matrix as part of the solution. If this is your "real" problem they may all be applicable, but you have given the distance matrix as part of the problem.

The kernel I've described is the matrix of inner products in the hi-dim Euclidean space in which the data is embedded (if there is one, and assuming wlog that the mean of all points is 0). So the output of the kernel PCA will be equivalent to the projection to a 2-dimensional subspace with the minimal total distance of the points from their projection.

If there is no triangle inequality, the points are not embedded in an Euclidean space and the theory behind PCA is void. The kernel matrix will have negative eigenvalues. It might still work and you may get good results, but I wouldn't count on it.

Multidimensional scaling is based on a different optimization criterion and it might be more suitable for arbitrary dissimilarity matrices (maybe the end result is the same for both, I'm not sure).

If the only distances that are meaningful are local, between nearby points, then Isomap and MVU are worth a look. Others might be good too but I have no familiarity with them (as opposed to these two with which I have little). -- Meni Rosenfeld (talk) 20:32, 27 March 2010 (UTC)

Normed field

I am looking at a problem in a book, which I know how to do, except that I wonder if the question is slightly off. The full question is:

Prove that in a normed field the following assertion holds: Let <a_n> be a Cauchy sequence, but not a null sequence. Prove there exist a number c > 0 and a positive integer N such that for all n > N either a_n > c or a_n < -c.

My question is, does the very last part make sense in an arbitrary normed field, a_n > c or a_n < -c? It does not say c is a real number. It just says it is a number, which could mean a field element. I mean, I think I can think of a counterexample, which would be the field of 2 elements with trivial norm and the sequence 1, 1, 1, 1, ... which is Cauchy but not null. However, there is no such c. I guess in such a field c > 0 may not even make sense? I don't know. What's going on here? Thanks! StatisticsMan (talk) 17:52, 25 March 2010 (UTC)

What, in the opinion of your book, is a normed field? Algebraist 17:59, 25 March 2010 (UTC)

(e/c) The statement as given only makes sense in an ordered field (interpreting c as a field element rather than a real number), and for it to hold you'd also need some compatibility conditions linking the order to the norm (which, I suspect, would imply that the field is Archimedean, and therefore a subfield of the reals, so you are back to square zero). Presumably the book wanted to say that the norm of a_n is at least c.—Emil J. 18:11, 25 March 2010 (UTC)

...where, with regard to Algebraist's question, I assumed that a normed field means a field equipped with an absolute value (algebra).—Emil J. 18:23, 25 March 2010 (UTC)

The book definition of a normed field is a field with a norm on it. The definition of field and norm are standard. Now that I've thought about it, I believe it's just that the author took a theorem about real numbers and didn't think to change it to work for a general normed field. So, I assume it should just be changed to "there exists a REAL number c > 0" and "such that for all n > N we have ||a_n|| > c". Thanks for your help. StatisticsMan (talk) 23:59, 25 March 2010 (UTC)

Periodic formula of an EKG shape?

Are there any formulas that would give me a repeating EKG type shape? --70.167.58.6 (talk) 19:07, 25 March 2010 (UTC)

A function of the form $f(t)=a_{0}+\sum _{k=1}^{n}(a_{k}\cos(k\omega t)+b_{k}\sin(k\omega t))$ should give a good fit. The parameters $\omega ,a_{k},b_{k}\;\!$ can be computed from data. See also Fourier series and Fourier transform. -- Meni Rosenfeld (talk) 19:40, 25 March 2010 (UTC)

In reality though, the heartbeat is never perfectly periodic; although it is often very close. We are all human and we all have quirks in our workings. A Fourier Series expansion is only any good for a short term model, under constant conditions. For example, the subject must not move, must not sleep, must not get excited, etc. •• Fly by Night (talk) 23:18, 25 March 2010 (UTC)

Reverse engineering a graph

This is a broader follow up to my question above. Are there any websites or apps that can reverse engineer a graph? For example, I draw my EKG shape in vector with Adobe Illustrator. Could I upload that image to a site that would spit out a trig formula that is similar to that curve? If I do it with a vector app, would there be some sort of PostScript data I could extract to make that curve as a trig function? --70.167.58.6 (talk) 19:12, 25 March 2010 (UTC)

Curve fitting has links, and says that it's a common feature for statistical programs. -- Finlay McWalter • Talk 19:16, 25 March 2010 (UTC)

Since the function is known to be roughly periodic, the generic techniques in that article are not the right tool for the job. You should attempt a Fourier transform. -- Meni Rosenfeld (talk) 19:44, 25 March 2010 (UTC)

Number of non-isomorphic graphs in n vertices?

Is there a known result to the following question; How many graphs are there in n vertices such that all of the graphs are non-isomorphic to each other graph? A math-wiki (talk) 23:49, 25 March 2010 (UTC)

OEIS: A000088. —David Eppstein (talk) 00:18, 26 March 2010 (UTC)

I figured as much, most of what that linked to was beyond my knowledge but em I correct in thinking the formula for the sequence they arrived at was $f(n)={\frac {2^{\frac {n(n-1)}{2}}}{n!}}$ A math-wiki (talk) 00:51, 26 March 2010 (UTC)

No. Stopping reading a formula at the first * sign is not a good route to understanding it. Algebraist 01:06, 26 March 2010 (UTC)

Then I must inquire as to what my result is missing. I arrived at that formula via a counting argument. I first asked how many possible edges (max) does a graph of n vertices have? This can by shown to be the sum of the first n-1 integers greater than 0 or ${\frac {n(n-1)}{2}}$ from the formula for sum of the first n integers. Then when thinking about the 'edge list' for any graph of n vertices any of those edges is either present or absent, and this there are 2 choices per edge in the 'list' and furthermore these choices are independent of each other. This gives $2^{\frac {n(n-1)}{2}}$ possible lists. This however counts the same graph numerous times as their are multiple ways to name some graphs (e.g. symmetric graphs) such that their 'edge list' remains unchanged, and furthermore there are different 'edge lists' that are, in fact, isomorphic to one another. The number of times each graph counted is precisely the number of ways its vertices can be named, since for any graph to be isomorphic to another, their must exist a bijection between their vertices that preserves their 'edge list.' Naming them from an ordered list (e.g. 1,2,3...) we can see their are n choices for the first name, n-1 for the second and so forth, so their are n! was to name each graph, so I concluded that $f(n)={\frac {2^{\frac {n(n-1)}{2}}}{n!}}$ gave the correct count for n vertices (not at most n, exactly n) It doesn't consider graphs with loops or with multiple edges between the same pair of vertices, but it does allow for non connected graphs and lone vertices with no edges connected with them. A math-wiki (talk) 02:00, 26 March 2010 (UTC)

If every graph were asymmetric (like the Frucht graph, having no nontrivial graph automorphisms), then ${\frac {2^{\frac {n(n-1)}{2}}}{n!}}$ would be the correct formula: there are $2^{\frac {n(n-1)}{2}}$ ways of writing down an adjacency matrix for a graph, and n! different adjacency matrices that result from the same graph. But when a graph has some symmetries, then there are fewer adjacency matrices that it can generate. Most graphs (in a technical sense, almost all of them) are asymmetric, so the formula is close to accurate, but it is not exact. The precise correction terms needed for situations like this, where you're trying to count things that may have symmetries, are given by the Pólya enumeration theorem.

There's also a very simple explanation for why your formula can't possibly be the correct answer: it's not an integer. The numerator has only factors of two in its prime factorization, while the denominator has other prime factors that can't be cancelled by anything in the numerator. As a non-integer it can't be the answer to a counting problem such as this one. —David Eppstein (talk) 02:39, 26 March 2010 (UTC)

March 26

unsolved problems

are these problems still unsolved?

1. Equichordal Points: Can a closed curve in the plane have more than one equichordal point?
2. Is pi+e irrational?
3. (Tiling the Unit Square)
   Will all the 1/k by 1/(k+1) rectangles, for k>0, fit together inside a 1 X 1 square?

—Preceding unsigned comment added by 208.79.15.130 (talk) 01:01, 26 March 2010 (UTC)

no(no) yes yes. I'm being minimalist today :) Dmcq (talk) 11:28, 26 March 2010 (UTC)

Dcmq, are you saying that the problem of whether pi plus e is irrational has been solved? Can you be specific? Who and when? Michael Hardy (talk) 02:42, 27 March 2010 (UTC)

I think he says yes, it is still unsolved. 66.127.52.47 (talk) 07:49, 27 March 2010 (UTC)

Ok,if pi+e is still unsolved problems,does this applied for all irrational numbers?and where can one look in wikipedia for this subject?respectfully. —Preceding unsigned comment added by 208.79.15.130 (talk) 19:47, 27 March 2010 (UTC)

See Irrational number. -- Meni Rosenfeld (talk) 21:05, 27 March 2010 (UTC)

Navier-Stokes equation

A University of Sydney student with not much spare time asks another question. If a metric is defined on solutions to the Navier-Stokes equation, then do 'distances' that diverge allow classification as turbulence? Then is fractal geometry or nonlinear dynamics the way to classify turbulence of Navier-Stokes solutions, a 'diverging' metric allowing turbulence to be inferred? Has this approach already been tried long ago? 122.152.132.156 (talk) 05:53, 26 March 2010 (UTC)

Wiener process properties

Hi, do you know how to calculate the expected value of the absolute value of the correlation between the time and the value of a Wiener process of some length t? I.e. for a sample function f what I mean would be

$\left\vert {\frac {Cov(time,value)}{Var(time)\cdot Var(value)}}\right\vert =\left\vert {\frac {{\frac {1}{t}}\cdot \int _{0}^{t}{ds\ s\cdot f(s)}-{\frac {t}{2}}\cdot {\frac {1}{t}}\cdot \int _{0}^{t}{ds\ f(s)})}{({\frac {t^{2}}{3}}-({\frac {t}{2}})^{2})\cdot ({\frac {1}{t}}\cdot \int _{0}^{t}{ds\ f(s)^{2}}-({\frac {1}{t}}\cdot \int _{0}^{t}{ds\ f(s)})^{2})}}\right\vert$

Thank you very much in advance. Hurugu (talk) 07:38, 26 March 2010 (UTC)

Two remarks that don't answer the question: (1) The correlation between X and Y is

${\frac {{\text{cov}}(X,Y)}{\sqrt {{\text{var}}(X){\text{var}}(Y)}}},$

with the radical in the denominator. That makes correlation dimensionless. (2) In order to speak of the correlation between time and something else, you'd need to think of time as a random variable, with a probability distribution. Without that, it's unclear at best what the question means. Michael Hardy (talk) 19:15, 26 March 2010 (UTC)

.... OK, from your formula is looks as if you're thinking of time as uniformly distributed between 0 and t. Michael Hardy (talk) 19:18, 26 March 2010 (UTC)

Your proposed way of computing the variance of the value of the Wiener process looks weird. Squaring the density function (if that's what you intend ƒ to be) is not done. And why integrate from 0 to the time. The value of a Wiener process is not constrained to lie in that interval. And do you mean the value at the time that's uniformly distributed in that interval? Or the value at time t? Or what? The question is unclear. Michael Hardy (talk) 19:21, 26 March 2010 (UTC)

Thank you for trying to answer, I am sorry for the confusing phrasing (and the missing square root in the denominator). $f$ is not meant to be interpreted as a density function. Maybe I can explain the question better in terms of a discrete problem: Take a sample function of the Wiener process, and then pick the values at the times ${\frac {1}{2}}{\frac {t}{n}}$, $(1+{\frac {1}{2}}){\frac {t}{n}}$, $(2+{\frac {1}{2}}){\frac {t}{n}}$, ..., $(n-{\frac {1}{2}}){\frac {t}{n}}$ for some natural number $n$. Together with the corresponding times they can be thought of as points in a plane, like the patterns in the top right image in the article Correlation. I am looking for the expected value of the absolute value of this correlation, for the limit $n\rightarrow \infty$. Hurugu (talk) 22:19, 26 March 2010 (UTC)

OK, I think the question is clear now. In effect the time is uniformly distributed between 0 and t and you mean the value of the Wiener process at that random time. Maybe more later.... Michael Hardy (talk) 00:17, 27 March 2010 (UTC)

One of the complications is that you said "absolute value".

If (capital) T is the time that is a random variable uniformly distributed between 0 and (lower-case) t, then cov(T, B_T) = 0. But for any particular sample path, the correlation is not 0; rather it is a random variable whose expected value is 0. But now we want the expected value of its absolute value. To be continued..... Michael Hardy (talk) 01:43, 27 March 2010 (UTC)

....and now I see that by ƒ you say you mean the sample function. Now that that's clear, the correlation you wrote above makes sense. Michael Hardy (talk) 02:36, 27 March 2010 (UTC)

This now seems like a harder problem than I initially thought it was. Possibly it can only be done numerically. More later maybe.... Michael Hardy (talk) 18:54, 27 March 2010 (UTC)

Opposite Category of Sets

Let S be the category of sets. Consider the category with sets as objects and morphisms A -> B as preimages of functions f from B to A, is this the same as the opposite of S? Thanks:) 66.202.66.78 (talk) 10:26, 26 March 2010 (UTC)

What do you mean by preimage of f?—Emil J. 11:16, 26 March 2010 (UTC)

OK, here's a guess: If ƒ:B → A, what if we say

P(B) is the set of all subsets of B,

P(A) is the set of all subsets of A,

P(ƒ) is the function from P(A) to P(B) defined by

$P(f)(C)=\{b\in B:f(b)\in C\}{\text{ for }}C\in P(A).\,$

Does that give us an opposite category? Michael Hardy (talk) 02:27, 27 March 2010 (UTC)

PS: I don't know any standard definition of the concept of "preimage" of a function. Michael Hardy (talk) 02:30, 27 March 2010 (UTC)

Michael Hardy, I think your talking about the contravariant power set functor P from SET to SET. This is not the same as the opposite of SET, which has exactly the same objects and arrows as SET, with the domain of a function f:A->B equal to B instead of A. Composition is reversed, so fg (in the opposite category) is equal to gf (in SET). However P is covariant as a functor from the opposite of SET to SET. I don't know what the OP meant by preimage of a function. Money is tight (talk) 06:40, 27 March 2010 (UTC)

Do you generally take it to be part of the definition of opposite category that it has exactly the same objects? I had thought the category of Boolean algebras and Boolean homomorphisms is the opposite of the category of Stone spaces and continuous functions. Michael Hardy (talk) 18:56, 27 March 2010 (UTC)

The opposite of a category is just the same category with arrows reversed. Stone duality states that the category of Boolean algebras is equivalent to the opposite of the category of Stone spaces (and vice versa), not equal. Algebraist 19:04, 27 March 2010 (UTC)

Finite differences in exponential powers

Hi. I took positive integers starting from 1 to the power of y, in the order of 1^y - 2^y, 3^y - 2^y and so on. I did this up to an exponent of 6, found the differences between the powers (first differences), then the differences between those (second differences), and so on, always subtracting the previous number from the next number. For example, the first differences for x₂² - x₁²... starting from 1² - 0 were 1, 3, 5, 7, 9, 11, and so on. Here are some of the things I've found by doing this:

x²

• First differences are odd numbers.
• Second differences are all 2.

x³

• First differences are prime numbers, or multiples of prime numbers.
• Second differences begin at 6 and are all multiples of 6.
• Third differences are all 6.

x⁴

• First differences have a final digit of 1, 5, or 9.
• The remaining inverse pyramid of differences appear to be almost random; some examples include 14, 40, 306, 820, 154, 38, 112, 148, -74, 24, 4, -106, 324, -576, -1116, etc.
• Fourth differences end with 4, 8 or 6.

x⁵

• All first differences have a final digit of 1.
• All second, third and fourth differences have a final digit of 0.
• All fourth differences are multiples of 120, and start at 240.
• All fifth differences are 120.

x⁶

• All first differences are odd.
• All second differences have a final digit of 2, and a second-final digit of 6, 0 or 1.
• All remaining differences end with 0.
• All sixth differences are >700 and <800.

So, my question is, what is the significance of this? Can it be applied to other areas of mathematics, and do we have an article excliciptly on this phenomenon? Thanks. ~AH1^(TCU) 11:47, 26 March 2010 (UTC)

here's an idea...work out $(n+1)^{2}-n^{2}$. Use that to work out why the diffs are odd, and the second differences are 2. Then try the same thing for other powers. If you start out with n^k, and take differences, what is the highest power of n that appears? What does this mean if you take differences k times? Tinfoilcat (talk) 12:12, 26 March 2010 (UTC)

... and check your working. 4th differences of x⁴ sequence all have the same value; and, in general, nth differences of xⁿ sequence will be constant. Gandalf61 (talk) 13:04, 26 March 2010 (UTC)

Could this problem have any applications for dimensions above 3, or even manifolds? ~AH1^(TCU) 00:13, 27 March 2010 (UTC)

"Multiples of prime numbers"?? What number is not a "multiple of a prime number"? Michael Hardy (talk) 01:44, 27 March 2010 (UTC)

Try Finite difference as a starting point. These patterns are well known but are rarely covered in standard math curricula, at least at an elementary level. This leads to them being rediscovered frequently. The nth differences of xⁿ is n!.--RDBury (talk) 03:03, 27 March 2010 (UTC)

I need your knowledge in all mathematical realms you are familiar with.

Let's assume that the concept of "even fraction" is defined as an irreducible fraction whose denominator is even. Unfortunately, no unique definiendum is received from this definition, even not from any of its sub-definitions referring to a given even denominator. Let's assume we would like to receive a unique definiendum. Fortunately, we know that there is a 'natural' surjection - from the class of pairs of an even denominator with an odd nominator - on the class of even fractions which are received by the original definition; Thanks to this surjection, we can receive a unique definiendum of "even fraction", by replacing the previous "indefinite" definition by a "definite" definition, which is received by deviding the original definition into sub-definitions, each of which defines the (unique) "even fraction" as the (unique) irreducible fraction whose given denominator is even and whose given nominator is odd, in such a way that this devision of the original definition into sub-definitions - succeeds to preserve the original class of "even fractions" received by the original "indefinite" definition.

Ignoring the very issue of "even fractions" (and fractions at all), do you know of any (well-known) similar process, in any mathematical realm you are familiar with? i.e., a process in which the classical definition of Y (whatever Y is) does not yield a unique definiendum; however, thanks to the existence of a 'natural' surjection (whatever this 'naturality' means) - from the class of X's - on the class of Y's received from the original definition, we can receive a unique Y, by replacing the previous "indefinite" definition by a "definite" definition, which is received by deviding the original definition into sub-definitions, each of which refers to a given X (by which Y can be defined uniquely), in such a way that this devision of the original definition into sub-definitions - succeeds to preserve the original class of Y's received by the original "indefinite" definition.

HOOTmag (talk) 12:19, 26 March 2010 (UTC)

Category theory is the closest, but overall it sounds to me more like All your base are belong to us :) Dmcq (talk) 18:07, 26 March 2010 (UTC)

More details? Does category theory deal with the uniqueness of definiendum? HOOTmag (talk) 18:42, 27 March 2010 (UTC)

I'd have guessed "even fraction" would mean one where the numerator is even, since then even numbers would be even fractions. Michael Hardy (talk) 19:11, 26 March 2010 (UTC)

The examples given here for "odd fractions" are f/3, f/5, f/7, etc. HOOTmag (talk) 18:42, 27 March 2010 (UTC)

I'm having trouble understanding what you're getting at, but maybe you're just talking about a "coding" or "construction" (as in constructive logic)? 66.127.52.47 (talk) 23:17, 26 March 2010 (UTC)

Any connection between my request and "coding" or "construction"? HOOTmag (talk) 18:42, 27 March 2010 (UTC)

In your example, you code the even fraction as a pair of integers, (even, odd). 66.127.52.47 (talk) 18:49, 27 March 2010 (UTC)

I code nothing. Look again at my first paragraph, where I explain how I explicitly define the even fraction (without coding anything), so that I receive a unique definiendum. HOOTmag (talk) 19:55, 27 March 2010 (UTC)

Hm. We could also say that you gave an axiomatic definition of an even fraction, then described a structure (the set of (even,odd) integer pairs) that interprets the definition. Does that help? There is a topic called constructive type theory that might also reach towards what you might be getting at. Unfortunately, our article about it is very technical. 66.127.52.47 (talk) 02:26, 28 March 2010 (UTC)

Curve Sketching

Hello. For some function f, f(a) = r, where r is a constant term; f'(x) is undifferentiable at x = a; $\lim _{x\rightarrow a^{-}}f'(x)=-\infty$; $\lim _{x\rightarrow a^{+}}f'(x)=\infty$. While sketching a graph with the information above, is there a plausible function where $\lim _{x\rightarrow a^{-}}f(x)=-\infty$ and $\lim _{x\rightarrow a^{-}}f(x)=-\infty$ while f(a) = r? If so, name the function. Thanks in advance. --Mayfare (talk) 20:21, 26 March 2010 (UTC)

There are definitely functions that satisfy the conditions you mentioned: f'(x) is undifferentiable at x = a; $\lim _{x\rightarrow a^{1-}}f'(x)=-\infty$; $\lim _{x\rightarrow a^{1+}}f'(x)=\infty$. There is the classic $f(x)=1/(x-a)$ for example.

You can then define the function where it is defined at f(a)=r (a single point) such that you have a piecewise function. --Kvasir (talk) 22:01, 26 March 2010 (UTC)

Can there be a cusp at x = a with the information in the first sentence? --Mayfare (talk) 12:50, 27 March 2010 (UTC)

Yeah, that's possible too, and if you want the function to be continuous then it would have to be that way. For example let $f(x)={\sqrt {R^{2}-(x-a+R)^{2}}}+r$ for a-R ≤ x ≤ a and $f(x)={\sqrt {R^{2}-(x-a-R)^{2}}}+r$ for a ≤ x ≤ a+R, for some radius R > 0, which is two quarter circles up against each other. Rckrone (talk) 17:45, 27 March 2010 (UTC)

March 27

Lattice of quotient groups

Instead of using normal subgroups, we'll think of all partitions of a group corresponding to congruence relations. If I partially order them by Q1 is smaller than Q2 iff Q1 is finer than Q2, does the set form a complete lattice? Subgroups do... Money is tight (talk) 07:06, 27 March 2010 (UTC)

A congruence relation on a group is the same thing as a normal subgroup, isn't it? Algebraist 10:48, 27 March 2010 (UTC)

A partition is uniquely determined by the normal subgroup of elements congruent to the identify, so I think we can just look at those. A partition is finer than another one if the corresponding normal subgroups are contained in each other (one-line proof omitted). The converse is obviously true (the normal subgroup is one of the parts). Therefore, the set of partitions of a group will be a complete lattice iff the set of normal subgroups of a group is one, which it is. So the answer to your question is: yes. --Tango (talk) 20:29, 27 March 2010 (UTC)

maruschka doll and half-life - fractals?

Are these things fractals? What if you find two dolls within each maruschka doll, would it be a fractal? And the half-life of radioactive materials, if I plot it, is it a fractal?--Quest09 (talk) 18:32, 27 March 2010 (UTC)

I'd say the nesting dolls are fractal, especially if each is identical to the others, but they aren't always like that: [1]. I'm not quite following the half-life argument. Are you talking about the decay of radioactive elements ? There has to be some pattern which repeats at different scales, and I'm not sure if a radioactive decay curve is complex enough to be called a "pattern". After all, a straight line could be considered fractal if you wanted to include it. StuRat (talk) 18:43, 27 March 2010 (UTC)

I think a fractal is a set whose fractal dimension is not an integer. 66.127.52.47 (talk) 01:53, 28 March 2010 (UTC)

March 28

How much should my game cost?

Say I developed a games for the PC which I wishes to sell by mailorder.

Here are my values

FixedCost = $100000 (I send 100 000 dollars in a year to develop the game)

MarginalCost = $3 (it cost me $3 to stamp out a DVD-ROM and mail it )

DemandCurve or SalesVolume(price) =1000000*(1/(1+price)) (The amount of game units sold for each price in dollars)

When I tried to calculate the sales price for my game (to maximize my profits). My calculations kept on crashing saying I should sell my game at $999 999 dollars per game which is a ridiculous amount for sell a computer game to teenagers. What is wrong with my calculations?

122.107.207.98 (talk) 03:10, 28 March 2010 (UTC)