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I know it's a simple question, but I keep getting different general formulas for the coefficients when I am trying to use the multinomial theorem for the following: $$ \left(\sum_{k=0}^{M}\frac{(-x^2)^k}{4^kn^{k/2}k!(1+k)!}\right)^n $$ Please help me to calculate it.

Thank you.

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Unforntunately it's not $(1-k)!$ (W|A), but where it come from? Bessel functions? At least Wolfram says that... –  draks ... Jun 25 '12 at 21:43
It comes from the calculation of integral with Bessel J_1 function. I;ve changed the sum to the finite, maybe it will be easier... –  Alex K. Jun 25 '12 at 21:47
Wolfram can give you some examples. Can one of your formulas reproduce that? –  draks ... Jun 25 '12 at 21:53
Thank you. Well, I've got formulas for n=2 and n=3 right. But I need to get a general formula for the coefficients and I am getting it wrong... –  Alex K. Jun 25 '12 at 22:15
Why not posting your solutions? –  draks ... Jun 26 '12 at 16:00
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1 Answer

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{% \bracks{\,\sum_{k = 0}^{M}{(-x^{2})^{k} \over 4^{k}n^{k/2}k!(1 + k)!}}^{n}:\ {\large ?}.\quad}$ Let $\ds{\quad z \equiv {-x^{2} \over 4n^{1/2}}\quad}$ and $\ds{\quad a_{k} \equiv {1 \over k!\pars{1 + k}!}}$.

$$ \mbox{Then}\,,\quad \bracks{\,\sum_{k = 0}^{M}{(-x^{2})^{k} \over 4^{k}n^{k/2}k!(1 + k)!}}^{n} = \pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}\tag{1} $$

\begin{align} &\pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}= \sum_{k_{1} = 0}^{M}a_{k_{1}}z^{k_{1}}\sum_{k_{2} = 0}^{M}a_{k_{2}}z^{k_{2}} \cdots\sum_{k_{n} = 0}^{M}a_{k_{n}}z^{k_{n}}\sum_{\ell = 0}^{nM} \delta_{\ell,k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{n}} = \sum_{\ell = 0}^{nM}A_{\ell}z^{\ell} \end{align} where

$$ \left.A_{\ell} \equiv \sum_{k_{1}, k_{2},\ldots,k_{n} = 0}^{M} a_{k_{1}}a_{k_{2}}\ldots a_{k_{n}}\right\vert_{\sum\limits_{i\ =\ 1}^{n}k_{i}\ =\ \ell} $$
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