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This question leads to an application of the inclusion‒exclusion principle leading to this sum: $$ \sum_{k=0}^n (-1)^k \binom n k (n-k)^x = (-1)^n \sum_{k=0}^n (-1)^k \binom n k k^x $$ $$ \text{e.g. } \quad 1\cdot 6^{10} - 6\cdot 5^{10} +15\cdot 4^{10} - 6\cdot3^{10} + 1\cdot2^{10} $$ Is there either a closed form or some sort of limiting form as $n\to\infty\text{?}$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{x} = \pars{-1}^{n}\sum_{k=0}^{n}\pars{-1}^{k}{n \choose k}k^{x}:\ {\Large ?}}$.

\begin{align} &\bbox[10px,#ffd]{\ds{% \pars{-1}^{n}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}k^{x}}} = \pars{-1}^{n}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k} \bracks{x!\oint_{\verts{z} = 1} {\expo{kz} \over z^{x + 1}}\,{\dd z \over 2\pi\ic}} \\[5mm] = &\ \pars{-1}^{n}\, x!\oint_{\verts{z} = 1}{1 \over z^{x + 1}} \sum_{k = 0}^{n}{n \choose k}\pars{-\expo{z}}^{k}\,{\dd z \over 2\pi\ic} = \pars{-1}^{n}\, x!\oint_{\verts{z} = 1}{\pars{1 - \expo{z}}^{n} \over z^{x + 1}} \,{\dd z \over 2\pi\ic} \\[5mm] = &\ x!\oint_{\verts{z} = 1}{\pars{\expo{z} - 1}^{n} \over z^{x + 1}} \,{\dd z \over 2\pi\ic} \end{align}

With the identity $\ds{\pars{\expo{z} - 1}^{n} = n!\sum_{\ell = 0}^{\infty}{\ell \brace n}{z^{\ell} \over \ell!}}$ where $\ds{{\ell \brace n}}$ is a Stirling Number of the Second Kind:

\begin{align} &\bbox[10px,#ffd]{\ds{% \pars{-1}^{n}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}k^{x}}} = x!\, n!\sum_{\ell = 0}^{\infty}{\ell \brace n}{1 \over \ell!}\ \overbrace{\oint_{\verts{z} = 1}{1 \over z^{x - \ell + 1}} \,{\dd z \over 2\pi\ic}}^{\ds{\delta_{\ell x}}} = \bbx{n!\, {x \brace n}} \end{align}

Note that $\ds{{x \brace n}_{\ x\ <\ n} = 0}$

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  • $\begingroup$ Interesting read. (+1). $\endgroup$ – Marko Riedel Mar 12 '18 at 20:39
  • $\begingroup$ @MarkoRiedel Thanks. Stirling ${\brace}$ are quite useful... $\endgroup$ – Felix Marin Mar 13 '18 at 0:19

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