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.

One Answer

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

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