Power series raised to powers

Question

I wish to prove the formula stated in wikipedia or mentionned in this question: let us denote $ c_m^{(n)}$ the coefficients arising in the expansion of the $n$-th power $\displaystyle \left( \sum_{k=0}^\infty a_k X^k \right)^{\!n} =\, \sum_{m=0}^\infty c_m^{(n)} X^m$. The claim is that

$$\begin{align} c_0^{(n)} &= (a_0)^n,\\ c_m^{(n)} &= \frac{1}{m\, a_0} \sum_{k=1}^m (kn - m+k)\, a_{k}\, c_{m-k}^{(n)}, \quad \ m \geq 1. \end{align}\tag{1}\label{1} $$

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Let me first start with what seems natural but which leads to a different answer. I indeed added the $(n)$ index in $c_m^{(n)}$ because I would naively do a recurrence on $n$ by writing

$$ \begin{split} \sum_{m=0}^\infty c_m^{(n)} X^m & = \left( \sum_{k=0}^\infty a_k X^k \right)^{\!n-1} \times \left( \sum_{k=0}^\infty a_k X^k \right) = \sum_{m=0}^\infty c_m^{(n-1)} X^m \times \sum_{k=0}^\infty a_k X^k \\ &= \sum_{p=0}^\infty \left( \sum_{k=0}^p c_{p-k}^{(n-1)}\, a_k \right) X^p \end{split}$$ which gives $$ c_m^{(n)} = \sum_{k=0}^m c_{m-k}^{(n-1)}\, a_k \tag{2}\label{2}$$

For $\displaystyle n=1,\enspace c_m^{(1)} = a_m,\quad n=2,\enspace c_m^{(2)} = \sum_{k=0}^m a_{m-k}\, a_k\ .\quad$ For $ n=3$

$$ c_m^{(3)} = \sum_{j=0}^m c_{m-j}^{(2)}\, a_j = \sum_{j=0}^m \left( \sum_{k=0}^{m-j} a_{m-j-k}\,a_k \right) a_j = \sum_{j+k+l=m} a_l\, a_k\, a_j $$ It seems that the general solution is $\displaystyle c_m^{(n)} = \sum_{k_1+k_2+\cdots + k_n=m} a_{k_1}\, a_{k_2}\,\cdots\, a_{k_n} $, which is in fact just a notation, cf. also this question. In the first question above, they claim that it is possible to get a relation from the multinomial formula but that also is not clear to me: a first apparent (but in fact harmless) problem is that we do not raise a polynomial to the power $n$ but an infinite sum. Secondly, we want to regroup the answer by powers $X^m$ as opposed to a formula such as $\left(a_1\, X_1 + a_2 \, X_2 + \cdots + a_k\, X_k \right)^n$ for which we would look at the coefficient of a monomial of the form $X_{1}^{i_1}\, X_2^{i_2}\, \cdots X_n^{i_n}$

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Since (\ref{1}) is a relation between $c_m^{(n)}$ and $c_p^{(n)}$ for $p<m$ but the same $n$, we should look for an argument explaining this link. (\ref{2}) made a link between $c_m^{(n)}$ and $c_p^{(n-1)}$ for $p<m$.

Maybe, maybe, maybe we could insert formula (\ref{1}) within (\ref{2}) and prove heredity...

Answer 1

We consider \begin{align*} A(x)=\sum_{k=0}^\infty a_kx^k\qquad\qquad A^{n}(x)=\sum_{k=0}^\infty c_k^{(n)}x^k\qquad n\geq 1\tag{1} \end{align*}

Merging $c_m^{(n)}$ with the sum we show the following is valid for $n\geq 1$: \begin{align*} \color{blue}{c_0^{(n)} }&\color{blue}{= a_0^n}\\ \color{blue}{\sum_{k=0}^m(kn-m+k)a_kc_{m-k}^{(n)}}&\color{blue}{=0\qquad\qquad m\geq 1}\tag{2} \end{align*} It is convenient to use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ in a series. This way we can write for instance \begin{align*} [x^m]A(x)=[x^m]\sum_{k=0}^\infty a_kx^k=a_m\tag{3} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=0}^m}&\color{blue}{(kn-m+k)a_kc_{m-k}^{(n)}}\\ &=\sum_{k=0}^m\left((n+1)k-m\right)a_k[x^{m-k}]A^n(x)\tag{4.1}\\ &=[x^m]A^n(x)\sum_{k=0}^m\left((n+1)k-m\right)a_kx^k\tag{4.2}\\ &=(n+1)[x^m]A^n(x)\sum_{k=0}^\infty ka_kx^k -m[x^m]A^n(x)\sum_{k=0}^\infty a_kx^k\tag{4.3}\\\ &=(n+1)[x^m]A^n(x)x\frac{d}{dx}\left(A(x)\right)-m[x^m]A^n(x)A(x)\tag{4.4}\\ &=[x^{m-1}]\frac{d}{dx}\left(A^{n+1}(x)\right)-m[x^m]A^{n+1}(x)\tag{4.5}\\ &=[x^{m-1}]\frac{d}{dx}\left(\sum_{k=0}^{\infty}c_k^{(n+1)}x^k\right) -m[x^m]\sum_{k=0}^\infty c_k^{(n+1)}x^k\tag{4.6}\\ &=[x^{m-1}]\sum_{k=0}^\infty kc_k^{(n+1)}x^{k-1} -m c_m^{(n+1)}\tag{4.7}\\ &=mc_m^{(n+1)}-mc_m^{(n+1)}\\ &\,\,\color{blue}{=0} \end{align*} and the claim (2) follows.

Comment:

• In (4.1) we use the coefficient of operator (3) to work with $c_{m-k}^{(n)}$.

• In (4.2) we use the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

• In (4.3) we split the sum and we also set the upper limit of the sums to $\infty$ without changing anything, since the coefficient of operator skips terms with powers of $x$ greater $m$.

• In (4.4) we write the sum using the differential operator $\frac{d}{dx}$.

• In (4.5) we recall the rule $\frac{d}{dx}A^{n+1}(x)=(n+1)A^{n}(x)\frac{d}{dx}A(x)$.

• In (4.6) we write the series again using the notation from (1).

• In (4.7) and the line that follows we select the coefficient of $[x^m]$ resp. $[x^{m-1}]$.

Note: Here we did some kind of reverse engineering. In fact the starting point of (2) seems to be the useful functional identity (4.5) \begin{align*} \color{blue}{[x^{m-1}]\frac{d}{dx}\left(A^{n+1}(x)\right)=m[x^m]A^{n+1}(x)\qquad\qquad m\geq 1, n\geq 0} \end{align*}