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Looking at an earlier post Finding the power series of a rational function, I am trying to get a closed formula for the n'th coefficient in the Taylor series of the rational function (1-x)/(1-2x-x^3). Is it possible to use any of the tricks in that post to not only obtain specific coefficients, but an expression for the n'th coefficient ?If T(x) is the Taylor polynomial I am looking at the equality (1-x) = (1-2x-x^3)*T(x) and differentiating, but I am not able to see a pattern giving me an explicit formula for the coefficients.

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Thank you. That ought to make it doable for me. – user5944 Jan 17 '11 at 14:58
You can add comments to your own questions and (I think) answers to your own questions by clicking the "add comment" link. – Qiaochu Yuan Jan 17 '11 at 14:58
@user5944 @derman you seem to have one registered and one unregistered account. Please let one of the moderators know if you would like them merged. – Willie Wong Jan 17 '11 at 15:10

1 Answer 1

Denoting the coefficients of $T(x)$ as $t_i$ (so $T(x) = t_0 + t_1 x + ...$), consider the coefficient of $x^a$ in $(1 - 2x - x^3) T(x)$. It shouldn't be hard to show that it's $t_a - 2t_{a-1} - t_{a-3}$.

So we have $t_0 - 2t_{-1} - t_{-3} = 1$, $t_1 - 2t_0 - t_{-2} = -1$, and $a > 1 \Rightarrow t_a - 2t_{a-1} - t_{a-3} = 0$. You can get $t_0$, $t_1$. Do you know how to solve the discrete recurrence $t_{a+3} = 2t_{a+2} + t_{a}$?

(If not, here's a hint: suppose you had a number $\alpha$ such that $t_a = \alpha^a$ satisfied the recurrence. What constraint on $\alpha$ can you prove? Now consider a linear combination of the possible such solutions, $t_a = b_0 \alpha_0^a + ... + b_n \alpha_n^a$, and solve $n$ simultaneous equations from your base cases $t_0$, $t_1$, and use $0 = t_{-1} = t_{-2} = ...$ if necessary).

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