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Let's assume that we have two functions $f,g: \mathbb{R} \rightarrow \mathbb{R}$ of class $C^\infty$ such that $f(x)=g(x^p)$ for $x \in \mathbb{R}$, where $p \in \mathbb{N}$ is fixed.

I wish to calculate $g^{(n})(0)$ for $n \in \mathbb{N}$.

In particular case when $f(x)=\sum_{n=0}^\infty c_n x^n$, $f(x)=\sum_{n=0}^\infty d_n x^n$ for $x \in \mathbb{R}$, from $f(x)=g(x^p)$ for $x \in \mathbb{R}$, we receive that $c_n=0$ when $n$ is not of the form $n=kp$, where $k\in \mathbb{N}\cup\{0\}$. Hence $$f(x)=\sum_{n=0}^\infty c_{pn}x^{pn}=f(x)=g(x^p)=\sum_{n=0}^\infty d_n x^{pn} \textrm{ for } x\in \mathbb{R},$$

$$\frac{g^{(n)}(0)}{n!}=d_n=c_{pn}=\frac{f^{(pn)}(0)}{(pn)!} \textrm{ for } n \in \mathbb{N} \cup \{0\}.$$

It suggests that $$g^{(n)}(0)=\frac{n!}{(np)!} f^{(np)}(0) \textrm{ for } n\in \mathbb{N}\cup \{0\}.$$

I don't know how to prove it in general case, when $f,g$ are not analytic.

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up vote 1 down vote accepted

By the chain rule, $f'(x) = px^{p-1} g'(x^p)$, then by product and chain rule $f''(x) = p(p-1)x^{p-2} g'(x^p) + p^2x^{2(p-1)} g''(x^p)$. You may already see the pattern that leads after more steps to $f^{(p)}(x) = p!\cdot x^0 \cdot g'(x^p)+x^2\cdot (\ldots)$, hence $f^{(p)}(0) = p!g'(0)$, that is the first instance of your claim.

As a matter of fact, the power series argument is valid even without knowing wheter the power series have positive radius of convergence (or, if they converge, converge against the given functions). You may simply think of using an "array" to store the sequence of derivatives at $0$ that is conveniently and suggestively written as $\sum {a_n\over n!} x^n$ instead of just $(a_0, a_1, a_2, \ldots)$. The array belonging to a derivative is obtained exactly the same way as you would formally differentiate the power series; the same holds for other operations.

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