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Let $f(x) = \sum_{n \geq 0} a_n x^n = \frac{x-2x^3}{4x^4 - 5x^2 + 1}$

Now I need to identify a "concrete formula" for $a_n$. This should be done by using the following proposition:

For a sequence $a = (a_0,a_1,\cdots)$ with $a_i \in \mathbb{C}$ and a $d$-tuple $(\alpha_1,\cdots,\alpha_d) \in \mathbb{C}^d$ with $\alpha_d \neq 0$ applies:

  • $f_a(x) = \sum_{n\geq 0} a_n x = \frac{P(x)}{Q(x)}$ with $Q(x) = 1 + \alpha_1 t + \cdots + \alpha_d t^d$ and a polynomial $P(x)$ having degree $< d$.
  • $a_{n+d} + \alpha_1 a_{n+d-1} + \cdots + \alpha_d a_n = 0$ for $ n \geq 0$
  • For $n \geq 0$ applies: $a_n = \sum_{i=?}^k P_i(n) \sigma_i^n$ with $1 + \alpha_1 x + \cdots + \alpha_d x^d = \prod_{i=1}^k (1- \sigma_i x)^{d_i}$ so that $\sigma_i \neq \sigma_j, 1 \leq i < j \leq k$ and $P_i(t)$ is a polynomial having a degree < $d_i$.

Could you please help me to apply this on the initial definition of $f(x)$? How do I start and identify $P(x)$ and $Q(x)$? Is this a method that has its own name (so I could look it up somewhere)?

[Edit] I made an error during the precalculation for the term above, it's correct now. Is this already $P(x)$ and $Q(x)$ as $\deg(P(x)) < \deg (Q(x))$? What is the next step?

Thanks in advance!

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    @leonbloy yes, sorry2011-07-03

2 Answers 2

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Yes, $P(x)$ is $x-2x^3$, and $Q(x)$ is $4x^4-5x^2+1$ (although I note that at one place you have written $Q(x)$ as a polynomial in $t$ when it should be in $x$). The next step is to factor $Q(x)$ as a product of polynomials each of degree 1; that will give you the $\sigma_i$ in the last bullet point.

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    @muffel, the proposition you quoted doesn't tell you how to find $P_i$, but it does tell you that its degree is less than $d_i$. Now from the factorization you have done for $Q(x)$, each $d_i$ is 1, so each $P_i$ is constant. How do you find these 4 constants? One way is to find $a_0,a_1,a_2$, and $a_3$, then you have 4 equations in those 4 unknown constants.2011-07-04
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I see this: ${(1 - 2x)^2 + 2x\over-(2x -1)^3} = {(1 - 2x)^2 + 2x\over(1 - 2x)^3} = {1\over 1 - 2x} + {2x\over (1 - 2x)^3}. $ Finding a Taylor expansion of this centered at 0 can be done with using the standard tools.

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    You can just use the Geometric Series Theorem.2011-07-03