Represent $f(x)$ with $g(x)$ when the taylor expension has specific dependency

Question

If I have $f(x)$ and $g(x)$ like that:$$f(x)=\sum_{n=1}^{\infty} a_nx^n$$$$g(x)=\sum_{n=1}^{\infty} \frac {a_nx^n} {n}$$
How can I find u(x) such: $f(u(x))=g(x)$?
I also know that the series $(a_n)$ is convergent.

Answer 1

Assuming that $\,a_1\ne 0\,$ then we can find the series reversion $\, h(x) := f^{-1}(x)\,$ which satisfies $$ x = h(f(x)) = f(h(x)). \tag{1}$$ Then, defining $$u(x) := h(g(x)) \tag{2}$$ we find the expected $$ f(u(x)) = f(h(g(x)) = g(x). \tag{3}$$ The first 3 terms of the expansion of the power series are: $$ h(x) = \frac1{a_1}x - \frac{a_2}{a_1^3}x^2 +\frac{2a_2^2-a_1a_3}{a_1^5}x^3 + \dots \tag{4} $$ and $$ u(x) = x - \frac{a_2}{2a_1}x^2 +\frac{3a_2^2-2a_1a_3}{3a_1^2}x^3 + \dots \tag{5} $$ where I used a computer algebra system PARI/GP to find the power series with code similar to the following:

fx = a1*x + a2*x^2 + a3*x^3 + O(x^4);
gx = intformal(fx/x);
hx = serreverse(fx)
ux = subst(hx, x, gx)

Answer 2

Let's assume $f(x),g(x),u(x)$ to be formal power series.

Let $^{-1}$ denote the compositional inverse.

$$f(u(x))=g(x)$$

$$u(x)=f^{-1}(g(x))$$

$f^{-1}$ is the compositional inverse of $f$. $f^{-1}$ is a formal power series iff $a_1\neq 0$. You can find $f^{-1}$ then e.g. by Lagrange inversion theorem.

You can find the composition of $f^{-1}$ and $g$ e.g. by Faà di Bruno's formula.