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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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    $\begingroup$ See FAQ:how to ask $\endgroup$ Aug 27, 2013 at 15:45
  • $\begingroup$ Gedrox, why do you think this equation is true? $\endgroup$ Aug 27, 2013 at 15:51
  • $\begingroup$ At least for all $v$ and $n$ less than 50 it is true. $\endgroup$
    – Gedrox
    Aug 27, 2013 at 15:55
  • $\begingroup$ 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. $\endgroup$
    – Yemon Choi
    Aug 27, 2013 at 16:29
  • $\begingroup$ See my edit if it helps.. $\endgroup$
    – Gedrox
    Aug 27, 2013 at 16:48

3 Answers 3

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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 https://math.stackexchange.com/questions/104262/sum-of-reciprocals-of-derivative-of-polynomial-at-its-roots

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  • $\begingroup$ Thank you! Could you, please, guide me what theorems are used to convert problem from finite sum to contour integral? $\endgroup$
    – Gedrox
    Aug 28, 2013 at 7:27
  • $\begingroup$ 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 $R \to \infty$. $\endgroup$
    – Gedrox
    Aug 28, 2013 at 14:47
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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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    $\begingroup$ That's way too elegant. Surely there is something amiss? $\endgroup$ Aug 28, 2013 at 23:14
  • $\begingroup$ Bravo. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$ $\endgroup$ Aug 29, 2013 at 3:23
  • $\begingroup$ 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. $\endgroup$
    – Gedrox
    Aug 29, 2013 at 10:02
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(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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