# An identity involving sum of probably binomial coefficients

How could I prove that $$\sum _{m=v}^n \left(\left(\prod _{k=v}^{m-1} \frac{k^2}{m^2-k^2}\right)\left(\prod _{k=m+1}^n \frac{k^2}{k^2-m^2}\right)(-1)^{m-v}\right)=1$$ or, simplified, $$\sum _{m=v}^n \prod _{k=v, k \neq m}^{n} \frac{k^2}{k^2-m^2}=1$$

for any positive integers $v$ and $n$, $v \leq n$? I feel this could be somehow related to binomial coefficient identities.

Why I want it to be true?

I got this problem while generating formula for eigenvalues of matrix of special type. I noticed that this $$\sum _{m=u}^n \frac{2(-1)^{m-1}(n!)^2}{m^2(n-m)!(n+m)!} \frac{m (m+u-1)!}{u (2u-1)! (m-u)!}(-4)^{u-1}$$ can be simplified to this $$\frac{2((u-1)!)^24^{u-1}}{(2u)!}$$

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See FAQ:how to ask – Anthony Quas Aug 27 '13 at 15:45
Gedrox, why do you think this equation is true? – Tom Leinster Aug 27 '13 at 15:51
At least for all $v$ and $n$ less than 50 it is true. – Gedrox Aug 27 '13 at 15:55
I think what @TomLeinster means is: what led you to come up with this formula? As your question stands, it gives no indication of the ideas that led you to this formula, nor any hint as to why you want the formula to be true. – Yemon Choi Aug 27 '13 at 16:29
See my edit if it helps.. – Gedrox Aug 27 '13 at 16:48

Consider the contour integral of $$\frac{1}{z} \prod_{k=v}^{n} \frac{k^2}{k^2-z^2}$$ over a circle of large radius centered at $0$. Since the integrand is small as $|z|\to \infty$ the answer must go to zero as the radius goes to infinity. But inside the circle there are poles at $z=0$ and $z= \pm k$ for $k$ from $v$ to $n$. Computing the residues here gives your identity.

Edit in response to OP's comment: The proof above uses complex analysis and the residue theorem; consult any introductory book in that subject. Alternatively, note that if $P(x)$ is a polynomial of degree $n$ with distinct roots $r_1$, $\ldots$, $r_n$ then $$Q(x)=\sum_{j=1}^{n} \frac{1}{P^{\prime}(r_j)} \frac{P(x)}{(x-r_j)}$$ is a polynomial of degree $n-1$ which also equals $1$ for all the $n$-points $x=r_j$. Therefore $Q(x)$ is identically $1$. Your identity follows by taking $P(x) = \prod_{k=v}^{n} (k^2-x^2)$, and evaluating $Q(x)=1$ at $x=0$. The general identity $Q(x)=1$ mentioned above is classical, and was discussed on MSE: see http://math.stackexchange.com/questions/104262/sum-of-reciprocals-of-derivative-of-polynomial-at-its-roots

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Thank you! Could you, please, guide me what theorems are used to convert problem from finite sum to contour integral? – Gedrox Aug 28 '13 at 7:27
Ok, I've got it! Residue theorem is used on the function you mentioned. And yes -- it's easy to prove that contour integral tends to zero when $R \to \infty$. – Gedrox Aug 28 '13 at 14:47

Consider the degree $n - v$ polynomial that interpolates the points $(x_i, y_i) = ((v + i)^2, 1)$, with $i = 0, \ldots , n - v$. This polynomial is $y = 1$, so the Lagrange interpolation formula gives $$\sum_{i = 0}^{n - v} \prod_{j \neq i} \frac{x - x_j}{x_i - x_j} = 1 .$$ Setting $x = 0$ gives the identity in the simplified second form.

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That's way too elegant. Surely there is something amiss? – The Masked Avenger Aug 28 '13 at 23:14
Bravo. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$ – Steven Landsburg Aug 29 '13 at 3:23
Seems legit. Very nice! Still I would add explanation that polynomial of order $n-v$ which has value $1$ in $n-v+1$ different points in fact is constant function of 1. – Gedrox Aug 29 '13 at 10:02

(Posted as an answer because it's difficult to make this readable in a comment):

The $m$th term (except for the sign) simplifies to $$2\binom{n}{n-m}\binom{m-1}{v-1}\binom{m+v-1}{v-1}\over\binom{n+m}{m}$$

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