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Given $f=\sum_{n}c_{n}x^{n}$, if we know the first $n$ terms of this sum, what can we say about the terms for $1/f$? I feel like we should be able to recover some of them. I believe it has to do with letting $1/f = \sum_{n}b_{n}x^{n}$ and then solving for the constants $b$ in $f \cdot (1/f)=1$. So if we have $$f=1+3x+2x^{2}+9x^{3}+...$$ how can we use just these terms to find the coefficients for terms in $1/f$? I have been doing, $$(1+3x+2x^{2}+9x^{3}+...)(b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+...)=1$$ However, I'm not sure how to finish it up. Is there a simple algorithm I could use to recover those $b$'s? Am I missing something obvious? Thank you!

Alex M.
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James Thent
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    I think I got it. If we assume $b_{0}=1$, we can say the other non-constant terms must add up to 0 after we multiply through, and by simply multiplying through we can get relations to solve those terms so that each coefficient in the new sum = 0 in terms of our known terms and the b's. – James Thent Oct 28 '15 at 16:50
  • I was writing an answer to explain precisely that. $+1$ for finding yourself the answer. – Julián Aguirre Oct 28 '15 at 16:52
  • Thank you! I just had a moment where I couldn't make the connection that the non-constant terms have to somehow be 0 after we multiply through. – James Thent Oct 28 '15 at 16:53

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You are not missing anything, your approach is correct. Perform the multiplications, groups the terms by the powers of $x$ and note that the coefficient of $x^0$ must be $1$, and the coefficients of $x^k$ for $k \ge 1$ must be $0$.

Explicitly, $b_0 + (3b_0 + b_1) x + (b_2 + 3b_1 + 2b_0) x^2 + \dots = 1$, whence $b_0 = 1$, so that $b_1 = -3$, so that $b_2 = 7$ etc. If you know the coefficients of $f$ up to order $n$, then you will also be able to find out the coefficients of $\frac 1 f$ up to the same order $n$.

Alex M.
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    More generally we can show that if the constant term $a_0$ of $f$ is nonzero, the truncated system of equations for series coefficients of $1/f$ up to degree $n$ is "triangular" and nonsingular. While this establishes a formal power series expansion for $1/f$, further argument is needed to establish its radius of convergence. Hint: Where will the nearest singularity of $1/f$ be given the series for $f$ converges in some disk about the origin? – hardmath Oct 28 '15 at 17:12
  • @hardmath: Agreed. Note, though, that the question claims that only the first $n$ terms of the development of $f$ are known, which rules out matters regarding power series, formal or not. – Alex M. Oct 28 '15 at 17:18
  • My gloss on your nice answer was esp. wordy, even for me, because I was making two points. First, we always get the reciprocal power series coefficients to order $n$, and nothing else, even in "special cases". Second, and more extraneously, even though nothing further is given of power series coefficients for $f$, we might know something about $f$ itself that helps to establish convergence of the power series for $1/f$. – hardmath Oct 28 '15 at 17:29