How to find the inverse of a series?

Question

Imagine a series like $f(x) = \sum_{n=0}^{\infty } f_n(x)$. Is there any formula to find the $f^{-1}(x)$ as series expansion like $f^{-1}(x) = \sum_{n=0}^{\infty } g_n(x)$? Actually the the question is "How to find all of $g_n$ functions only using $f_n$ functions?"

For instance; $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$, and $\arcsin(x) = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} ... $

Of course we have a lot of information about $\sin(x)$. But what if the given function is a very custom function? For example how to find the inverse of this kind of function:

$$f(x) = \sum_{n=0}^{\infty } \frac{x^{11n+5}}{(3n+2)!}$$

In general, if $f(x) = \sum_{n=0}^{\infty } f_n(x)$, I am searching for an operator, $\text{Inv}[f_0, f_1, f_2, ... ] = [g_0, g_1, g_2, ...]$ such that $f^{-1}(x) = \sum_{n=0}^{\infty } g_n(x)$.

Answer 1

I will explain how this works with an example, namely $\sin x = x-x^3/6\pm\cdots$.

Write $y=\sin x$ so that

$$y = x-\frac{x^3}{6}+\frac{x^5}{120}\pm\cdots$$

We seek a formula $x=$ power series in $y$.

First, compute some powers of $y$.

\begin{align*} y^2&=x^2-\frac{x^4}{3}+\frac{2x^6}{45}\pm\cdots\\ y^3&=x^3-\frac{x^5}{2}\pm\cdots\\ &\qquad\qquad\vdots \end{align*} which I get by squaring/cubing/etc the series for $y$ above.

Now, by making a linear combination of these powers of $y$, we can eliminate the powers of $x$ on the RHS. First, the $x^3$-term is removed by taking $y+y^3/6$:

$$y+\frac{y^3}{6}=x-\frac{3x^5}{40}\pm\cdots$$

If we keep eliminating powers of $x$ on the RHS, we are left with $x$ alone, so

$$x=\sin^{-1} y=y+\frac{y^3}{6}\pm...$$

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Remark.

This method works for a power series $f(x)=\sum_{i>0} a_ix^i$ with $f(0)=0$ and $f'(0)=a_1\ne 0$.

Answer 2

$$f(x) = \sum_{n=0}^{\infty } \frac{x^{11n+5}}{(3n+2)!}$$ By the way $$f(x)=\frac{1}{3 x^{7/3}}\left(e^{x^{11/3}}-2 e^{-\frac{x^{11/3}}{2}} \sin \left(\frac{1}{6} \left(3 \sqrt{3} x^{11/3}+\pi \right)\right)\right)$$

Since we know all coefficients, it is sufficient to use the explicit formula for the $n^{\text{th}}$ term as given by Morse and Feshbach in year $1953$.

This could give for the inverse series $$\sum_{n=0}^\infty (-1)^{n+1} \,a_n\,t^{n+11}\qquad \text{with}\qquad t= \sqrt[5]{2x}$$ the very first coefficients being $$\left\{1,\frac{1}{300},\frac{367}{2520000},\frac{232339}{249480 00000},\frac{3431}{4892160000},\frac{31409801191}{5427359437 50000000}\right\}$$