Timeline for Square of Binomial Coefficient
Current License: CC BY-SA 2.5
23 events
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Jan 25, 2011 at 9:49 | comment | added | sigma_z_1980 | @Didier: yeah it's really helpful. Cheer for this. | |
Jan 25, 2011 at 7:18 | comment | added | Did | @sigma Re the derivation of closed expressions for such sums, a "must read" is the book generatingfunctionology by Herb Wilf, freely available on his webpage math.upenn.edu/~wilf. | |
Jan 25, 2011 at 7:14 | comment | added | Did | @sigma Re your first question: this is what I had in mind directing you to the Wolfram page. By the residue theorem, to extract the $x^n$ coefficient of a polynomial $R(x)$ one must integrate $R(z)\mathrm{d}z/(2\pi\mathrm{i}z^{n+1})$ on any positively oriented loop around the origin in the complex plane. If you do that for $R(x)=(1+x)^n(1+p^2x)^n$, an easy change of variable in the integral yields exactly the defining expression of Legendre polynomials. Et voilà ! | |
Jan 24, 2011 at 6:13 | comment | added | sigma_z_1980 | @Didier: thanks this did help a lot. I'm just interested, is this the same as Legendre polynomial in a way as Igor derived in Mathematica? Can you recommend some literature on derivation on closed expressions using binomial coefficients? | |
Jan 24, 2011 at 0:55 | comment | added | Did | (cont'd) But this means that the coefficient of $x^k$ in $P(x)$ must be ${n \choose k}p^{2k}$ instead of ${n \choose k}$. You know how to get this modification, do everything as before except that you replace $x$ by $p^2x$ in $P(x)$ and you are done. | |
Jan 24, 2011 at 0:55 | comment | added | Did | (cont'd) The term $x^n$ in the product $P(x)Q(x)$ is obtained by multiplying a term $x^k$ in $P(x)$ by a term $x^{n-k}$ in $Q(x)$ and then by performing the sum over $k$ of these contributions. Hence the sum of ${n \choose k}^2$ is the coefficient of $x^n$ in $P(x)Q(x)$ (which is simply ${2n\choose n}$, as you know). Now, this was just a warm up because you want these factors $p^{2k}$ in your sum. (to be cont'd) | |
Jan 24, 2011 at 0:55 | comment | added | Did | @sigma: Sorry I was not more explicit. Here is a way to get at the result. Gerry is right, one works in the space $\mathbb{R}[x]$ of the polynomials wih one indeterminate $x$. Let us forget the coefficient $p$ for a moment and note that ${n \choose k}^2={n \choose k}{n \choose n-k}$, that ${n \choose k}$ is the coefficient of $x^k$ in $P(x)=(1+x)^n$ and that ${n \choose n-k}$ is the coefficient of $x^{n-k}$ in $Q(x)=(1+x)^n$ (yes I know, $P(x)=Q(x)$ but just wait and you will see). (to be cont'd) | |
Jan 23, 2011 at 23:35 | comment | added | sigma_z_1980 | @Gerry: I am, I didn't quite get though how Didier arrived at this solution, so it got me confused. | |
Jan 23, 2011 at 22:34 | comment | added | Gerry Myerson | @sigma, I reckon $x$ is an indeterminate, a placeholder, something to hang a coefficient on. Are you not familiar with the idea of generating functions? | |
Jan 23, 2011 at 21:33 | vote | accept | sigma_z_1980 | ||
Jan 23, 2011 at 21:29 | comment | added | sigma_z_1980 | @Didier, I'd be glad if you explained a little bit more how you arrived at this solution. Thanks. What is x btw? | |
Jan 23, 2011 at 16:05 | comment | added | Did | And (surprise, surprise...) Legendre polynomials may be defined as contour integrals of a given function with respect to $\mathrm{d}z/z^n$ in the complex plane, see mathworld.wolfram.com/LegendrePolynomial.html. | |
Jan 23, 2011 at 15:36 | comment | added | Did | Using the notation $n=\ell/2$, your first sum is the coefficient of $x^n$ in the polynomial $(1+x)^n(1+p^2x)^n$. As such, it can be evaluated as the integral of a complex valued rational function on the unit circle of the complex plane. Not that his helps much... As regards the $_2F_1$ expression provided by Mathematica, I believe this function is the exact expansion along increasing powers of $p$ which is written in the original question. :-) | |
Jan 23, 2011 at 11:26 | comment | added | Gerry Myerson | @sigma, $F_1$ is a hypergeometric function, look it up. | |
Jan 23, 2011 at 7:18 | comment | added | sigma_z_1980 | @Igor: OK thanks. But why is it 'not always enlightning', and what is the $F_{1}$ function in the solution? thanks | |
Jan 23, 2011 at 4:30 | comment | added | Igor Rivin | @sigma-z-1980: The general way to evaluate summations like this is "Gosper's algorithm" or its generalization due to Wilf and Zeilberger ("the WZ method"). I think the best place to learn about this is Petkovsek/Zeilberger's book titled "A=B". As you see in my answer, it is easy to wind up with the answer as some sort of a hypergeometric function, which is not always enlightening. You might also look at Kovacic's paper on solving 2nd order ODEs (in the mid-eighties sometime). | |
Jan 23, 2011 at 4:10 | comment | added | sigma_z_1980 | Igor, is there any wat to see how this formula was derived? I have no expreience with Mathematica | |
Jan 23, 2011 at 4:08 | comment | added | sigma_z_1980 | OK, let's say $p=\frac{1}{l}$ or $1-\frac{1}{l}$. Does this help. I mean I hope to find such formula as a closed form for arbitrary p. | |
Jan 23, 2011 at 2:39 | comment | added | Gerry Myerson | @Thierry, if the binomial coefficient weren't squared, there would be the closed-form expression $(1+p^2)^{\ell/2}$, so maybe there's one here, too. | |
Jan 23, 2011 at 1:53 | answer | added | Igor Rivin | timeline score: 5 | |
Jan 23, 2011 at 1:48 | comment | added | Thierry Zell | I'm confused: why do you think there is hope to find a closed-form formula for your expression without knowing the exact dependence between p and l? | |
Jan 23, 2011 at 1:11 | history | edited | Yemon Choi | CC BY-SA 2.5 |
cleaned up LaTeX
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Jan 23, 2011 at 0:55 | history | asked | sigma_z_1980 | CC BY-SA 2.5 |