Let $f$ be a function with infinite derivatives at $a=0$, and let $P_{n,f,0}=P_n$ be a Taylor polynomial of $f$ around $0$.
Let $m\in \mathbb N$. Prove that $Q(x)=P_{n}(x^{m})$ is a Taylor polynomial of $g(x)=f(x^{m})$ of degree $nm$ aroud $0$.
My proof:
Suppose $n\geq m$. Let $h(x)=x^{m}$, which is a Taylor polynomial of itself. Using rules of composition of Taylor polynomials we get:
$P_{nm,g,0}(x)=P_{nm,f\circ h,0}(x)=[P_{nm,f,0}\circ P_{m,h,0}]_{nm,0}=[P_{nm,f,0}(x^{m})]_{nm,0}=$ $=[\sum_{k=0}^{nm}\frac{f^{(k)}(0)}{k!}(x^{m})^{k}]_{nm,0}=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^{mk}=P_{n}(x^{m})=Q(x)$
What do I do in the case where $n<m$? I wouldn't be able to say that $h(x)$ is a Taylor polynomial of itself in that case.
Note that\begin{align}\lim_{x\to0}\frac{f(x^m)-P_n(x^m)}{x^{mn}}&=\lim_{x\to0}\frac{f(x^m)-P_n(x^m)}{(x^m)^n}\\&=\lim_{x\to0}\frac{f(x)-P_n(x)}{x^n}\\&=0.\end{align}So, since $\deg P_n(x^n)\leqslant mn$, $P_n(x^m)$ is the Taylor polynomial of order $mn$ of $f(x^m)$ near $0$.