2
$\begingroup$

In the process of proving the Wrongskian identity for the Bessel function $J_\nu(x)J_{-\nu}'(x)-J_\nu'(x)J_{-\nu}(x)=-\frac{2\sin(\pi \nu)}{\pi x}$ (where the primes are differentiation by $x$), I came across the following sum: $$\sum_{n=0}^j\frac{j-2n+\nu}{n!(j-n)!\Gamma(n-\nu+1)\Gamma(j-n+\nu+1)}$$ where $\nu$ is not necessarily an integer. When $j=0$, this is equal to $\sin(\pi\nu)/\pi$ from the reflection relation for the gamma function. I am trying to show that for $j\geq 1$, this sum is equal to $0$. However, none of the typical methods for tackling this have succeeded, since $\nu$ is not an integer, and the terms do not cancel in an obvious manner. I have reduced it to the sum $$\frac{\sin(\pi \nu)}{\pi} \sum_{n=0}^j\frac{(-1)^n(j-2n+\nu)}{n!(j-n)!(\nu-n)_{j+1}}$$ where $(a)_n=a(a+1)...(a+n-1)$, which seems easier to work with, but I still do not get anywhere. Combining denominators gets $$\frac{\sin(\pi \nu)}{\pi\ j!(\nu-j)_{2j+1}} \sum_{n=0}^j (-1)^n(j-2n+\nu){{j}\choose{n}}(\nu-j)_{j-n}(\nu+j+1-n)_n $$ One can see that each power of $\nu$ cancels for specific $j$'s, but the general case is mess of Stirling numbers, and I feel like there is a better way of proving that this is $0$.

Edit: I have found my own answer, provided below, but other responses are still welcome.

$\endgroup$
0
$\begingroup$

I found a way to evaluate this sum using residues. In the following, let $b!$ denote $\Gamma(b+1)$ for non-integer $b$.

Rewrite the original sum as $$\sum_{n=0}^j \frac{2n-j-\nu}{n!(j-n)!(n-\nu)!(j-n+\nu)!} =\frac{1}{j!^2}\sum_{n=0}^j (2n-j-\nu)\binom{j}{n}\binom{j}{n-\nu}$$ Now we use $$\binom{a}{b}=\text{Res}_{x=0}\frac{(1+x)^a}{x^{b+1}},\ a\in\mathbb{Z}_{\geq 0}$$ Our sum then equals $$\text{Res}_{x=0}\frac{1}{j!^2}\sum_{n=0}^j (2n-j-\nu)\binom{j}{n}\frac{(1+x)^j}{x^{n-\nu+1}}=\text{Res}_{x=0}\frac{(1+x)^j}{j!^2 x^{1-\nu}}\sum_{n=0}^j (2n-(j+\nu))\binom{j}{n}\frac{1}{x^n}$$ Then we use $$\sum_{n=0}^j\binom{j}{n}x^n=(1+x)^j,\ \sum_{n=0}^j n\binom{j}{n}x^n=jx(1+x)^{j-1}$$ to see our sum is $$\text{Res}_{x=0}\frac{(1+x)^j}{j!^2 x^{1-\nu}}\left(2j\frac{1}{x}(1+1/x)^{j-1} -(j+\nu)(1+1/x)^j\right) =\text{Res}_{x=0}\frac{(1+x)^{2j-1}}{j!^2 x^{j+1-\nu}}(2j-(j+\nu)(1+x)) =\text{Res}_{x=0}\frac{(1+x)^{2j-1}}{j!^2 x^{j+1-\nu}}(j-\nu-(j+\nu)x)$$ $$=\frac{1}{j!^2}\left((j-\nu)\binom{2j-1}{j-\nu}-(j+\nu)\binom{2j-1}{j-\nu-1}\right)=\frac{1}{j!^2}\left((j-\nu)\binom{2j-1}{j-\nu}-(j-\nu)\binom{2j-1}{j-\nu}\right)=0$$ In evaluating the residue we assume $j>0$.

I am still interested in seeing if there are simpler or alternate techniques for showing this equality.

$\endgroup$
0
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{f}\pars{\nu} & \equiv \sum_{n = 0}^{j}\pars{2n - j - \nu}{j \choose n}{j \choose n - \nu} \\[1cm] \mrm{f}\pars{\nu} & = \sum_{n = 0}^{j}\bracks{2\pars{j - n} - j - \nu} {j \choose j - n}{j \choose \bracks{j - n} - \nu} = -\sum_{n = 0}^{j}\bracks{2n -j + \nu}{j \choose n}{j \choose n + \nu} \\[5mm] & = -\,\mrm{f}\pars{-\nu} \\[1cm] \mrm{f}\pars{\nu} & = \sum_{n = -\infty}^{\infty}\pars{2n - j - \nu}{j \choose n} {j \choose n - \nu} = \sum_{n = -\infty}^{\infty}\pars{2n - j + \nu}{j \choose n + \nu} {j \choose n} = \,\mrm{f}\pars{-\nu} \end{align}


$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{% \mrm{f}\pars{\nu} = -\,\mrm{f}\pars{-\nu}\quad\mbox{and}\quad \,\mrm{f}\pars{\nu} = \,\mrm{f}\pars{-\nu} \implies \bbox[15px,#eef,border:1px dotted navy]{\ds{\,\mrm{f}\pars{\nu} = 0}}}} $$

$\endgroup$
  • $\begingroup$ Interesting proof; however, it seems you assume that $\nu$ is an integer in showing that $f(\nu)$ is even, since you shift the index $n$ by $\nu$. (Correct me if I am wrong.) $\endgroup$ – Grant B. Mar 20 '17 at 6:21
  • $\begingroup$ @GrantB. Yes. You're right. It would be interesting to check the general result. I'll think about it. $\endgroup$ – Felix Marin Mar 20 '17 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.