Wikipedia:Reference desk/Archives/Mathematics/2010 January 19

= January 19 =

Policy recommendation for comparing citation totals or h-index for any mathematician
On Talk:Steve Shnider, a discussion on testing his notability has included calculating his h-index. Using the Web of Knowledge has been challenged and MathSciNet/Mathematics Reviews suggested. Is there a policy for how we compare citation totals or h-index for mathematicians and where the data should come from?

PS. If anyone happens to be familiar with Shnider's work, a comment on notability would be helpful on the talk page.—Ash (talk) 09:44, 19 January 2010 (UTC)
 * /dev/random is an excelent source of such data. Freely available, too. — Emil J. 11:43, 19 January 2010 (UTC)
 * Talk:Steve Shnider has been raised; perhaps your critical viewpoint (I assume your comment relates to the idea of using an h-index) would be helpful in reaching a consensus...—Ash (talk) 11:47, 19 January 2010 (UTC)

Variation on equals sign
I came across a variation on the well-known defining rule $$Z=Z^2+c$$, which rather than the simple equals sign had one with short extra bits - the upper line bent backwards at its RH end to head northwest, as it were, while the lower line bent backwards at its LH end to head southeast. I can't find an image to display, hence the description. Does this symbol have a standard meaning, say an assignment statement applied repeatedly?→86.132.164.11 (talk) 11:08, 19 January 2010 (UTC)


 * This ≅ ? It's Unicode, so &#x2245; and &#x2245; should display it too.-- JohnBlackburne wordsdeeds 11:28, 19 January 2010 (UTC)


 * (e/c) Do you mean $$\rightleftharpoons$$? I don't think it has a standard meaning in mathematics. I've seen somebody use it for definitions, and even for equivalence, but neither seems to be widespread usage. A Google search suggests that it means something in chemistry, but that's probably not what you want. $$Z\rightleftharpoons Z^2+c$$ looks like the transformation rule whose iteration defines the Mandelbrot set, so it may mean just that: a transformation rule (though I would rather write it either as $$Z\mapsto Z^2+c$$ or $$Z_{n+1}=Z_n^2+c$$ or something like that). You should check the context, it may define the symbol somewhere. — Emil J. 11:36, 19 January 2010 (UTC)


 * It was exactly this $$Z\rightleftharpoons Z^2+c$$, thanks. It appeared in a TV programme so the meaning was just up to the viewer. I assumed that it had a particular significance, but it seems that the producer just appropriated it to mean iteration of the transformation rule.→86.132.164.11 (talk) 12:24, 19 January 2010 (UTC)
 * Films and TV shows routinely use nonsensical formulas whenever they want to impress the audience with the characters' awesome mathiness. The people who created the formulas for the show like to use lots of symbols which look exotic, even when they have no meaning. So somebody probably said "hey- let's use that crazy hooked equal sign- it'll look cool". Staecker (talk) 13:54, 19 January 2010 (UTC)
 * And now that I read it more carefully it sounds like maybe this was a legitimate documentary-style math program? About fractals? In that case forget what I said. The hooks are probably implying iteration, like you said. But like the others said, this isn't a standard meaning. Staecker (talk) 13:56, 19 January 2010 (UTC)
 * Even if it was indeed an informational program, I suspect that they still favored formulae by virtue of their perceived mathiness. -- Meni Rosenfeld (talk) 15:48, 19 January 2010 (UTC)
 * (OP) The programme was called The Secret Life of Chaos on UK BBC and though fairly undemanding was pitched at a respectable level. The narrator was Jim Al-Khalili, Professor of Physics at the University of Surrey. As said above, the notation was obviously used to mean iteration, but nobody here seems to recognise it as a standard usage.→86.132.164.11 (talk) 16:24, 19 January 2010 (UTC)
 * Yes, it seems quite common to use this sign in TV programs to indicate iteration. I saw that program, but it wasn't the first time I saw that symbol used in a TV program. I agree with EmilJ, one standard notation to use in place would be $$z_{n+1}=z_n^2+c$$ --XediTalk 10:51, 20 January 2010 (UTC)

P.G.F. question
I am stuck on part of a question about probability-generating functions. In part (i), you had to derive the P.G.F. of $$X\sim \mbox{Poisson} \left (\lambda \right)$$, and I got $$G_X(t)=e^{\lambda \left ( t-1 \right )}$$, which is right, I believe. Then you need to derive the distribution for $$X+Y$$, where $$Y\sim \mbox{Poisson} \left (\mu \right)$$, via $$G_{X+Y}\left(t\right)$$; I got $$X+Y\sim \mbox{Poisson} \left (\lambda+\mu \right)$$, so that $$P\left(X+Y=n\right)=\frac{e^{-\left(\lambda+\mu\right)}\cdot\left(\lambda+\mu\right)^n}{n!}$$. Now, part (iv) with which I'm having problems: "Use the distributions of $X$, $Y$ and $X+Y$ to find the conditional probability that $X=x$ given that $X+Y=n$, where $n$ is a non-negative integer. Deduce that the conditional distribution of $X$ given that $X+Y=n$ is binomial with parameters $n$ and $\frac{\lambda}{\lambda+\mu}$."

I've tried working forwards, then working backwards, but my answers don't quite meet in the middle!

Working forwards

$$P \left(X=x|X+Y=n \right)$$

$$=\frac{P\left[\left(X=x\right)\cap\left(X+Y=n\right)\right]}{P\left(X+Y=n\right)}$$

Next, the only interpretation I can find of $$=P\left[\left(X=x\right)\cap\left(X+Y=n\right)\right]$$ is to say that if $$X=x$$ and $$X+Y=n$$ then $$Y=n-x$$. Thus I get:

$$\frac{P\left(Y=n-x\right)}{P\left(X+Y=n\right)}$$

$$=\frac{\frac{e^{-\mu}\cdot\mu^{n-x}}{\left(n-x\right)!}}{\frac{e^{-\left(\lambda+\mu\right)}\cdot\left(\lambda+\mu\right)^n}{n!}}$$

$$=\frac{e^{-\mu}\cdot\mu^{n-x}}{\left(n-x\right)!}\cdot\frac{n!}{e^{-\left(\lambda+\mu\right)}\cdot\left(\lambda+\mu\right)^n}$$ $$=\frac{e^{-\mu}\mu^{n-x}n!}{\left(n-x\right)!e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^n}$$

$${\color{blue}=\frac{e^{\lambda}\mu^{n-x}n!}{\left(n-x\right)!\left(\lambda+\mu\right)^n}}$$, and that's as far as I'm able to simplify it. Next,

Working backwards

If $$X\sim B\left(n,\frac{\lambda}{\lambda+\mu}\right)$$ then $$P\left(X=x\right)={_nC_x}\left(\frac{1}{1}-\frac{\lambda}{\lambda+\mu}\right)^{n-x}\left(\frac{\lambda}{\lambda+\mu}\right)^x$$

$$=\frac{n!}{x!\left(n-x\right)!}\left(\frac{\lambda+\mu-\lambda}{\lambda+\mu}\right)^{n-x}\left(\frac{\lambda}{\lambda+\mu}\right)^x$$

$$=\frac{n!}{x!\left(n-x\right)!}\left(\frac{\mu}{\lambda+\mu}\right)^{n-x}\left(\frac{\lambda}{\lambda+\mu}\right)^x$$

$$=\frac{n!}{x!\left(n-x \right)!}\cdot\frac{\mu^{n-x}}{\left(\lambda+\mu\right)^{n-x}}\cdot\frac{\lambda^x}{\left(\lambda+\mu\right)^x}$$

$$=\frac{n!\mu^{n-x}\lambda^x}{x!\left(n-x\right)!\left(\lambda+\mu\right)^{n-x}\left(\lambda+\mu\right)^x}$$ $${\color{blue}=\frac{n!\mu^{n-x}\lambda^x}{x!\left(n-x\right)!\left(\lambda+\mu\right)^n}}$$, and here end my powers of simplification.

As you can see, some of the terms are the same in the two expressions I manage to arrive at, but I am out by a few factors. Please advise me on where I have gone wrong. Thanks in advance.  It Is Me Here   t / c 17:27, 19 January 2010 (UTC)


 * In your working forward step, you make a very crucial mistake. In the numerator of your conditional probability, note that $$P((X=x)\cap(X+Y=n))=P(X=x)P(Y=n-x)$$ by the independence of $$X$$ and $$Y$$. Everything should work out fine starting from there. Nm420 (talk) 18:51, 19 January 2010 (UTC)


 * Thanks very much!  It Is Me Here   t / c 21:39, 19 January 2010 (UTC)

See also Cumulant. Bo Jacoby (talk) 22:53, 19 January 2010 (UTC).

3 rotations about the vertices of a triangle: identity map
Hi all,

I've just started a geometry course this term and I'm trying out a few questions but I'm struggling with the following:

Let R(P,$$\theta$$) denote the clockwise rotation of $$\mathbb{R}^2$$ through an angle $$\theta$$ about a point P. If A, B, C are the vertices, labeled clockwise, of a triangle in $$\mathbb{R}^2$$, prove that R(A,$$\theta$$)R(B,$$\phi$$)R(C,$$\psi$$) is the identity if and only if $$\theta=2\alpha$$, $$\phi=2\beta$$ and $$\psi=2\gamma$$, where $$\alpha,\,\beta$$ and $$\gamma$$ are the angles at the vertices A, B and C of the triangle.

Now I think I can show the 'if', just by showing that the rotations reflect A in BC then back again, and similarly for B and C, so the rotation fixes 3 points so must be the identity (I hope!) - however, I'm completely stuck on the 'iff'. I can't see any nice way to approach this at all, and I'd rather not churn through mounds of vectors if anyone can suggest an elegant proof I might utilise? I can't seem to get my head around it!

Many thanks, 82.6.96.22 (talk) 18:02, 19 January 2010 (UTC)


 * The circle round B radius AB and around C radius CA intersect at only two points. Which incidentally shows there's another set of angles satisfying the conditions - set them all to 0. Dmcq (talk) 18:16, 19 January 2010 (UTC)