Power series raised to an exponent...where does Wikipedia get this formula?

Question

On Wikipedia, they claim that $$ \left(\sum_{k=0}^\infty a_k x^k\right)^N $$ is another power series with $c_0 = a_0^N$ and $$c_m = \frac{1}{ma_0} \sum_{k=1}^m (kN-m+k) a_k c_{m-k}.$$ I tried proving this by induction but it's an absolute nightmare. Can someone provide some kind of motivation for where this power series expansion comes from?

Answer 1

That formula follows from the form of the derivative: $$ c'(x)=N·a(x)^{N-1}·a'(x)\implies a(x)·c'(x)=N·c(x)·a'(x) $$ and then looking at the coefficients of $x^{m-1}$ $$ \sum_{j=1}^m a_{m-j}·(j)c_{j}=N·\sum_{j=0}^{m-1}c_j(m-j)a_{m-j} \implies ma_0·c_m=\sum_{j=0}^{m-1} (Nm-Nj-j)a_{m-j}c_j $$ or set $k=m-j$ to obtain the given formula.

Answer 2

If you want a non-recursive solution you may be interested in this. If we let $f(x) = \sum_{k=0}^{\infty}a_k x^k$, then we have that \begin{align*} f(x)^n &= \left( \sum_{k=0}^{\infty}a_k x^k \right)^n \\ &= \sum_{k=0}^{\infty} \left(\sum_{\substack{0 \leq r_1,\ldots,r_n \leq k \\ r_1 + \cdots + r_n = k}}a_{r_1} \cdots a_{r_n}\right) x^k \\ \end{align*} Now for any specific choice of $r_1,\ldots,r_n$ satisfying the condition $$0 \leq r_1,\ldots, r_n \leq k,\quad r_1 + \cdots + r_n = k$$ there are, say, $m \leq n$ many distinct elements in the set $S = \{r_1,\cdots,r_n\}$, call these distinct elements $$s_1,\ldots,s_m$$ and define $$N(s_i) = \# \text{ of times $s_i$ appears in $(r_1,\ldots,r_n)$}$$ then we see that the number of times $$a_{r_1}\cdots a_{r_n}$$ is counted is equal to $$C(r_1,\ldots,r_n) := \frac{n!}{N(s_1)! \cdots N(s_m)!} \quad \text{(The multinomial coefficient)}.$$ Thus we have $$f(x)^n = \sum_{k = 0}^{\infty} \left( \sum_{\substack{0 \leq r_1 \leq\ldots \leq r_n \leq k \\ r_1 + \cdots + r_n = k}}C(r_1,\ldots,r_n)a_{r_1} \cdots a_{r_n} \right) x^k$$ Notice that now the inner sum is over a smaller set and hence (in theory) makes the calculation easier.