Complete series expansion of $\frac{1}{1+f(x)}$

Question

I have to find, in a closed form, the n-th coefficients of the series expansion for small $x$ of $$ \frac{1}{1+f(x)} $$ knowing that $f(x)=\sum_{n=1}^{\infty} a_nx^{n}$. I know that, by Taylor's theorem, the answer is $D^{n}[\frac{1}{1+f(x)}]$ where D is the derivative operator, but the derivative becomes very difficult. Also, I tried to use the geometric series writing: $$ \frac{1}{1+f(x)} = \sum_{n=0}^{\infty}(-1)^{n} f(x)^{n} $$ but I couldn't find an easy way to use the multinomial theorem.

Thanks

Answer 1

Hint:

Write

$$y(1+f)=1.$$

By differentiation,

$$y'(1+f)+yf'=0$$ $$y''(1+f)+2y'f'+yf''=0$$ $$y'''(1+f)+3y''f'+3y'f''+f'''=0$$ $$\cdots$$

and taken at $x=0$,

$$y_0=1$$ $$y'_0+y_0a_1=0$$ $$y''_0+2y'_0a_1+2y_0a_2=0$$ $$y'''_0+3y''_0a_1+3\cdot2y'_0a_2+3!y_0a_3=0$$ $$\cdots$$ or $$y_0=1$$ $$y'_0+a_1=0$$ $$y''_0-2a_1^2+2a_2=0$$ $$y'''_0+3\cdot2(a_1^2-a_2)a_1-3\cdot2a_1a_2+3!a_3=0$$ $$\cdots$$

Hard to see regularity emerge, maybe with more terms... A recursive formula is possible, from

$$\sum_{k=0}^n \binom nky_0^{(n-k)}k!a_k=0$$ or

$$y_0^{(n)}=-n!\sum_{k=1}^n \frac{y_0^{(n-k)}a_k}{(n-k)!}.$$

Answer 2

I tried to use Faa di Bruno's formula, following the idea of @RobertIstrael. We have: $$ f(x)=\sum_{n=1}^{\infty} a_nx^{n} $$ with $a_n=\frac{D^{n}[f(x)]}{n!}$ for $x=0$. We have also: $$ g(f(x)) = \frac{1}{1+f(x)} = \sum_{n=0}^{\infty} w_nx^{n} $$ and we know that $w_n=\frac{D^{n}[g(f(x))]}{n!}$ calculated in $x=0$. Then using the Faa di Bruno's formula with the same notation of wiki's page, but with the rule of $f(x)$ and $g(x)$ exchanged we have: $$ D^{n}[f(x)] = \sum_{k=1}^{n} \frac{(-1)^{k-1}(k-1)!}{(1+f(x))^{k}}B_{n,k}(f'(x),f''(x),...,f^{n-k+1}(x)) $$ When we calculate this expression for $x=0$ we have $$ w_n=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k-1}(k-1)!B_{n,k}(a_1,2a_2,...,(n-k+1)!a_{n-k+1}) $$ It is not very simple and beautiful, but I do not know if it can do better.