I have a function $f: \mathbb R \rightarrow \mathbb R $ and its taylor polynomial $T_n$ of degree $n$, at $x_0 = 0$. I want to show that $$x \mapsto T_n(x^2)$$ is the taylor polynomial of $g(x) := f(x^2)$ with degree $2n$. I wrote out the definition of the taylor series at $x_0 = 0$ with $x= x^2$ and I get :
$$x \mapsto \sum_{i=0}^n \frac{f^{(i)}(0)}{i!}x^{2i} + \frac{f^{(n+1)}((\theta x)^2)}{(n+1)!}x^{2(n+1)}$$
What do I have to do from here? Just modify the sum so that I end up with $2n$ as upper bound? Any help is greatly appreciated
With Peano form of the remainder:
$$f(x)=\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k + o(x^n)$$
Replacing $x$ with $x^2$, $$f(x^2) =\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^{2k} + o(x^{2n})=\sum_{\substack{k=0\\ k \text{ even }}}^{2n} \frac{f^{(k/2)}(0)}{(k/2)!}x^{k} + o(x^{2n})$$
Therefore, $\displaystyle \sum_{\substack{k=0\\ k \text{ even }}}^{2n} \frac{f^{(k/2)}(0)}{(k/2)!}x^{k}$ is the Taylor polynomial with degree $2n$ for $f(x^2)$.
Note that it's exactly the same $T_n(x^2)$.