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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?

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up vote 5 down vote accepted

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.

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Thanks for the link. How can I get the same result as Rogers-Ramanujan got. I try to learn the proof. – Mathlover Apr 10 '12 at 21:50
@Mathlover: I haven't really gone through the proofs myself, and I don't expect they will be easy. I suggest you follow the references. These will probably be covered in some number theory textbooks. – Aryabhata Apr 10 '12 at 21:55
@Mathlover: you'll want to look at the "Ramanujan's Lost Notebooks" series published by Springer-Verlag; I forget which, but one of the volumes goes into excruciating detail (through a modular forms detour) to derive a series from the CF... – J. M. Apr 14 '12 at 3:07

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