$$f(x)=\dfrac{1}{1+\dfrac{x}{1+\dfrac{x^2}{1+\dfrac{x^3}{1+\dfrac{x^4}{\ddots}}}}}$$
How can a power series be found given the continued fraction $f(x)$?
I'm trying to find $f(x) =1+a_1x+\dfrac{a_2x^2}{2!}+\dfrac{a_3x^3}{3!}+\dfrac{a_4x^4}{4!}+\cdots$
I tried some ways to define $f(x)$ but I could not find the general patern. It goes to complex patern after $n=3$ in my approach. I think that I need another approach to problem.
$$\begin{align} &f_1(x)=\frac{1}{1+x}=1-x+x^2-x^3+\ldots\\ &f_2(x)=\frac{1}{1+\frac{x}{1+x^2}}=\frac{1+x^2}{1+x+x^2}\\ &f_3(x)=\frac{1}{1+\frac{x}{1+\frac{x^2}{1+x^3}}}=\frac{1+x^2+x^3}{1+x+x^2+x^3+x^4}\\ &\lim_{n\to \infty} {f_n(x)}=f(x) \end{align}$$
Could you please give me hand on how to find the patern of this function?
You are looking for the reciprocal of the classic Rogers-Ramanujan Continued fraction. A listing of some of the power series coefficients can be found here: OEIS. You might be able to find more references there.