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I am looking for a way to obtain the coefficient $c_k$ of $x^k$ in the expansion of $(a_{{0}}+a_{{1}}x+a_{{2}}{x}^{2}+a_{{3}}{x}^{3}+\cdots)^n$. I know it can be done by the multinomial theorem, but I am looking for an alternative expression.

It is easy to show by induction that the coefficient $c_k$ of $x^k$ is given by $c_k=\frac {\sum _{i=1}^{k} \left( in-k+i \right) a_{{i}}c_{{k-i}}}{ka_{{0}}}$.

However I wonder whether there exists a way to express $c_k$ in a closed form in the sense of not necessitating to calculate all of the preceding coefficients. I was hoping that someone here knew an answer to this.

Thank you very much in advance for any help.

R.J. Etienne
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    the Fourier series/discrete Fourier transform can compute the coefficients of $(\sum_{k=0}^K a_k x^k)^n$ without computing the coefficients of $(\sum_{k=0}^K a_k x^k)^m$ for $m =1 \ldots n-1$. if $n$ is a power of $2$ you can also do $\log n$ steps instead of $n$, and this generalizes also when $n$ is not a power of $2$ – reuns Apr 09 '16 at 15:54

2 Answers2

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Besides the representations via multinomial coefficients there are no other explicit expressions of the coefficients $[x^k]$ of $x^k$ of the $n$-the power of a general power series

\begin{align*} \left(\sum_{j=0}^{\infty}a_jx^j\right)^n \end{align*}

Your recurrence formula of $c_k$ is nice. It is also stated in the section Power series raised to powers of formal power series. Note, that in order to fully specify this recurrence, the initial condition $c_0=a^n$ is also stated as well as the restriction that the factors in the denominator have to be invertible.

Markus Scheuer
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Yes. For two powerseries, their product coefficient is readily defined in terms of their coefficients. Now iterate the calculation.

Pedro
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