I am trying to simplify an expression and find a closed form for $$\sum_{m=0}^l \binom{s-m}{s-l} \binom{s-1+m}{s-1}x^m$$

How could I get rid of this summation?

  • $\begingroup$ Can you use the method I showed in mathoverflow.net/questions/257349/equality-with-binomials $\endgroup$ – T. Amdeberhan Dec 20 '16 at 13:26
  • $\begingroup$ And what is $p$? $\endgroup$ – T. Amdeberhan Dec 20 '16 at 13:28
  • $\begingroup$ There we knew what to prove, but here we don't know the constant C yet. I know Wilf-Zeilberger method can be used to derive some expressions, however I am not familiar with the procedure. $\endgroup$ – LuHell Dec 20 '16 at 13:30
  • $\begingroup$ p is unknown as well $\endgroup$ – LuHell Dec 20 '16 at 13:30
  • $\begingroup$ I'm not understanding this question. Is this supposed to be an identity that holds true for any $x$? Because that seems clearly false. $\endgroup$ – Todd Trimble Dec 20 '16 at 13:44

You may argue as GH from MO from your other post.

  • the coefficient of $y^m$ in $(1-xy)^{-s}$ equals $\binom{s+m-1}{s-1}x^m$;
  • the coefficient of $y^{\ell-m}$ in $(1-y)^{\ell-s-1}$ equals $\binom{s-m}{s-\ell}$.

Therefore, the sum on your LHS equals to

  • the coefficient of $y^{\ell}$ in $(1-xy)^{-s}(1-y)^{\ell-s-1}$.

Unfortunately, this has no closed form. How can we be sure? To this end, denote your sum by $$f(\ell):=\sum_{m=0}^\ell\binom{s+m-1}{s-1}\binom{s-m}{s-\ell}x^m.$$ As I explained the WZ-method in the other post, the procedure generates a recurrence. However, this time it is a three-term relation $$(\ell+2)f(\ell+2)+(-sx-\ell x-s+2\ell-x+2)f(\ell+1)+(x-1)(s-\ell)f(\ell)=0$$ which reveals that $f(s)$ can not have a closed form.

If you're not interested in the sum, then formulate this as a contour integral. Let $\gamma$ be a closed path (oriented positive) around $z=0$, and apply Cauchy's Integral Formula: $$f(\ell)=\frac1{2\pi i}\int_{\gamma}\frac{dz} {z^{\ell+1}(1-xz)^s(1-z)^{s+1-\ell}}.$$ On a positive note, we can derive a generating function for the sequence $f(\ell)$: $$\sum_{\ell=0}^{\infty}f(\ell)y^{\ell}=\left(\frac{(1+y)^2}{1+y-xy}\right)^s.$$ To see this, start by interchanging summations to proceed as follows: \begin{align} \sum_{\ell\geq0}f(\ell)y^{\ell} &=\sum_{m\geq0}\binom{s+m-1}mx^m\sum_{\ell=m}^s\binom{s-m}{\ell-m}y^{\ell} \\ &=\sum_{m\geq0}\binom{s+m-1}mx^my^m(1+y)^{s-m} \\ &=(1+y)^s\sum_{m=0}^{\infty}\binom{s+m-1}m\left(\frac{yx}{1+y}\right)^m \\ &=(1+y)^s\left(1-\frac{yx}{1+y}\right)^{-s} \\ &=\left(\frac{(1+y)^2}{1+y-xy}\right)^s. \end{align}

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  • $\begingroup$ Your f(l) is the case where x=1, then the expression reduces to $f(l)=\binom{2s}{2s-l}$. So there is a closed form there. $\endgroup$ – LuHell Dec 20 '16 at 14:07
  • $\begingroup$ Make sense. However, I fail to see why there could not be a closed expression. $\endgroup$ – LuHell Dec 20 '16 at 14:09
  • $\begingroup$ If there was a closed form, then the minimal recurrence for $f(\ell)$ would have been of first order, not second order. $\endgroup$ – T. Amdeberhan Dec 20 '16 at 14:14
  • $\begingroup$ In the very last expression, outer $^2$ must be $^s$, right? $\endgroup$ – მამუკა ჯიბლაძე Dec 20 '16 at 17:18

Mathematica says: $$\binom{s}{s-l} \, _2F_1(-l,s;-s;x).$$

(without hypergeometrics for special values of $x,$ like $x=1:$

$$\frac{(-1)^l (l-2 s-1)!}{l! (-2 s-1)!}$$

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  • $\begingroup$ The case $x=1$, has been dealt with at mathoverflow.net/questions/257349/equality-with-binomials $\endgroup$ – T. Amdeberhan Dec 20 '16 at 16:38
  • $\begingroup$ Yes. The point is to get rid of the hypergeometrics for any x's. It seems hopeless. $\endgroup$ – LuHell Dec 21 '16 at 16:05
  • 2
    $\begingroup$ As I mentioned in my answer, we can't expect to remove the hypergeometric sum because there in no "closed form" answer. $\endgroup$ – T. Amdeberhan Dec 21 '16 at 17:40

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