Power of a convergent series

Question

Let $ f(x)$ has the convergent power series representation of the form $$\sum_{n=0}^{\infty}a_nx^n$$. Then What is the power series representation of the function $\left[f(x)\right]^m$ for some integer $m$. Or, How to find $$\left(\sum_{n=0}^{\infty}a_nx^n\right)^m$$

Answer 1

A way to do it is this: the coefficient of $x^n$ in the expansion of the series is the sum of all the distinct products $a_{k_1}\cdot a_{k_2}\cdots a_{k_m}$ such that $k_1+k_2+\ldots+k_m=n$, that is

$$ \left(\sum_{n\geqslant 0}a_n x^n\right)^m=\sum_{n\geqslant 0}b_n x^n\tag1 $$

where

$$ b_n=\sum_{k_1+k_2+\ldots+k_m=n}\prod_{j=1}^m a_{k_j}\tag2 $$

The index of the sum on $(2)$ means that you get all the $m$-tuples of $\Bbb N_{\ge 0}^m$ such that the sum of it coordinates is equal to $n$.

Answer 2

The formula in the wikipedia link in the comment of @metamorphy is obtained from the differential equation for the power, if $g(x)=f(x)^m$, then $$ f(x)g'(x)=f(x)\cdot m\,f(x)^{m-1}f'(x)=m\,f'(x)g(x) $$ and comparing the coefficients of equal powers on both sides gives for the power $x^{n-1}$ $$ \sum_{k=0}^{n-1} a_k\cdot (n-k)b_{n-k}=m\sum_{k=1}^{n} ka_k\cdot b_{n-k} $$ so that $$ na_0b_n=\sum_{k=1}^{n} ((m+1)k-n)⋅ a_k\cdot b_{n-k} $$

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Similar calculations that are used in automatic/algorithmic differentiation in the propagation of truncated power series or Taylor series are

• How do I take high order derivatives for a Taylor expansion?
• $n$th derivative of $\sin(f(x))$