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


1 Answer 1


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 $T_n(x^2)$.

  • $\begingroup$ Would you mind elaborating, why this proof is sufficient? Can you, with the last sum, just say that it is the taylor polynomial of $g(x)$? $\endgroup$
    – Jack4t3
    Mar 13, 2017 at 19:40
  • $\begingroup$ @Jack4t3 yes, it's a classical result. Assume $\sum_{k=0}^n a_k x^k + o(x^n)= \sum_{k=0}^n b_k x^k + o(x^n)$. For the sake of contradiction, suppose there is some $p$ such that $a_p\neq b_p$. Consider the smallest $p$ such that $a_p\neq b_p$. Then $a_px^p+o(x^p)=b_px^p+o(x^p)$, hence $(a_p-b_p)x^p = o(x^p)$ and $a_p-b_p=o(1)$. The left-hand side is a nonzero constant, whereas the right-hand side goes to $0$, a contradiction. $\endgroup$ Mar 13, 2017 at 19:47

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