# Composition of two power series. We can compose two power series and get one power series. But why?

Mathematics Asked by tchappy ha on December 8, 2020

I am reading "Lectures on Complex Function Theory" by Takaaki Nomura.

In this book, the author wrote like this:

Let $$f(z) = sum_{n=1}^infty a_n z^n$$.
Let $$g(w) = sum_{n=0}^infty b_n w^n$$.
Because $$f(0) = 0$$, when $$|z|$$ is sufficiently small, $$f(z)$$ is inside the circle of convergence of $$g(w)$$.
So, $$g(f(z))$$ is defined on a set.
$$g(f(z)) = b_0 + b_1 (a_1 z + a_2 z^2 + cdots) + b_2 (a_1 z + a_2 z^2 + cdots)^2 + b_3 (a_1 z + a_2 z^2 + cdots)^3 \= b_0 + b_1 a_1 z + (b_1 a_2 + b_2 a_1^2) z^2 + (b_1 a_3 + 2 b_2 a_1 a_2 + b_3 a_1^3) z^3 + cdots.$$

The author didn’t explain why the second equality holds.
Is the second equality obvious?
If not obvious, please give me a proof.

All “inner” series $$(a_1 z + a_2 z^2 + cdots)^k$$ are expanded, and then the coefficients for each $$z^n$$ are collected. This can be justified strictly because only finitely many terms contribute to each $$z^n$$ term.

To keep is simple, I'll demonstrate it for the first three terms. Start with $$f(z) = a_1 z + a_2 z^2 + O(z^3)$$ for $$z to 0$$.

Then $$g(w) = b_0 + b_1 w + b_2 w^2 + O(w^3)$$ for $$w to 0$$. Substituting $$w = f(z)$$ gives $$g(f(z)) = b_0 + b_1 f(z) + b_2 f(z)^2 + O(f(z)^3) \ = b_0 + b_1 f(z) + b_2 f(z)^2 + O(z^3)$$ because $$f(z) = O(z)$$ for $$z to 0$$. It follows that $$g(f(z)) = b_0 + b_1(a_1 z + a_2 z^2 + O(z^3))z + b_2 (a_1 z+ a_2 z^2 + O(z^3))^2 + O(z^3) \ = b_0 + b_1 a_1 z + (b_1 a_2 + b_2 a_1^2)z^2 + O(z^3)$$ for $$z to 0$$. All these rearrangements are valid because they operate only on only finitely many terms.

On the other hand, $$g(f(z))$$ is holomorphic in a neighborhood of the origin and can therefore be developed into a power series. The above calculation shows that the first three terms of the power series are $$b_0 + b_1 a_1 z + (b_1 a_2 + b_2 a_1^2)z^2 , .$$

Answered by Martin R on December 8, 2020