$F(x)$ is a formal power series. Find $G(x)$ if $F(G(x)) = x.$

Question

$F(x) = \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$. Find $G(x) = \sum_{n=0}^{\infty}g_nx^{n}$ if $F(G(x)) = x$. Specifically find a formula for $g_n$

I am learning about formal power series, so please don't use $\sin x = F(x)$ which I think is true here, hence we can easily get $G(x)$ using this way.

I tried $F(G(x)) = \sum_{n=0}^{\infty}(-1)^{n}\frac{G(x)^{2n+1}}{(2n+1)!}$ and somehow getting the coefficient of x as 1 and all other cofficients of $x^j$ as $0$. I am unable to do so. If I need to provide the tedious computation which is not leading me anywhere, I am happy to do so.

Answer 1

Hint: Differentiate to get $F'(G(x))G'(x) = 1$. Then differentiate again to get

$F''(G(x))G'(x)+F'(G(x))G''(x) = 0$

but note $F''(G(x)) = -F(G(x))$ (this secretly is true because $F$ is sine, but of course can be verified by differentiating the power series), so

$F'(G(x))G''(x) = -xG'(x)$ and multiplying both sides by $G'(x)$ gives

$G''(x) = -x(G'(x))^2$

which we can use to solve for the coefficients of $G(x)$.

Answer 2

Your $G(x)=\sum_{n\geq 1} g_n x^n$ should start from $n=1$ and not 0. Then e.g. $$ x = (g_1 x+g_2 x^2 + g_3 x^3+ \cdots) - (g_1 x + g_2 x^2+ \cdots)^3/3! + (g_1 x +\cdots)^5/5!+ \cdots$$ comes with a natural grading and allows you to read off $g_1=1$, $g_2=0$, $g_3=1/6$,...

Answer 3

The method for computing the coefficients of the inverse function is known as Lagrange inversion. In this particular case, we are inverting the sine function, hence we are interested in the power series around $x=0$ of $G(x)=\arcsin(x)$, and we may exploit the fact that the Taylor series of $$ G'(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2} $$ can be computed from the extended binomial theorem: $$ G'(x) = \sum_{n\geq 0}\binom{-1/2}{n}(-1)^n x^{2n}=\sum_{n\geq 0}\binom{2n}{n}\frac{x^{2n}}{4^n} $$ and by termwise integration: $$ G(x) = \sum_{n\geq 0}\binom{2n}{n}\frac{x^{2n+1}}{(2n+1)4^n}.$$ The radius of convergence of such a power series is $1$, by Weierstrass M-test or the fact that the closest singularities to the origin of $G'(x)=\frac{1}{\sqrt{1-x^2}}$ lie at $x=\pm 1$.

Answer 4

In this answer, it is shown that for a formal power series $$ x+\sum_{n=2}^\infty a_nx^n $$ the inverse series is $$ x+\sum_{n=2}^\infty b_nx^n $$ where the $b_n$ can be computed inductively using the composition formula $$ 0=a_n+b_n+\left[x^n\right]\sum_{k=2}^{n-1}b_k\left(x+a_2x^2+a_3x^3+\dots+a_{n-1}x^{n-1}\right)^k $$ This may be arduous, but it works.