How to prove that the Maclaurin expansion of $f(x^n)$ is equal to the Maclaurin expansion of $f(t)$ with substitution $t=x^n$?

Question

It seems to be intuitively true, however I still cannot find a satisfying proof to convince myself. At the same time, is there any more general statement about this fact?

Answer 1

An outline and a reference:

There is a general formula for the $k$th derivative of $f\circ g$ known as Faà di Bruno's formula. Use this to write down a formula for the $k$th derivative of $f(x^n)$ in terms of derivatives of $f$.

For a Maclaurin series, you must evaluate these derivatives at $x=0$. If you have used Faà di Bruno carefully, you will be able to show that evaluating at $x=0$ results in $0$ except when $k$ is a multiple of $n$. And you will also be able to show that when $k$ is a multiple of $n$, the derivative works out to exactly what it needs to be to match the result of simply substituting $t=x^n$ into the Maclaurin series for $f(t)$.

Answer 2

Hint:

The derivatives of $f(x^n)$ are

$$\frac{df(x^n)}{dx}=nx^{n-1}f'(x^n),\\ \frac{d^2f(x^n)}{dx^2}=n^2x^{2n-2}f''(x^n)+n(n-1)x^{n-2}f'(x^n),\\ \frac{d^3f(x^n)}{dx^3}=n^3x^{3n-3}f'''(x^n)+2n^2(n-1)x^{2n-3}f''(x^n)+n(n-1)(n-2)x^{n-3}f'(x^n),\\ \cdots$$

Hence, when evaluating at $x=0$, the first nonzero derivative is

$$\frac{d^{(n)}f(x^n)}{dx}(0)=n!f'(0).$$

Continuing this way you should find the next nonzero to be

$$\frac{d^{(2n)}f(x^{n})}{dx}(0)=\frac{(2n)!}{n!}f''(0).$$