Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
I guess the term of the sum should be $a_n x^n$, right? –  leonbloy Jul 3 '11 at 12:42
@leonbloy yes, sorry –  muffel Jul 3 '11 at 21:01

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
That would be $4x^4 - 5x^2 + 1 = (-1+x)(1+x)(-1+2x)(1+2x)$. But what's next? What really unsettles me is the index $i$ of P in $a_n = \sum P_i(n) \sigma_i^n$. $a_n$ needs to be like $P_1(n) \cdot (-1+x) + P_2(n) \cdot (1+x) + P_3(n) \cdot (-1+2x) + P_4(n) \cdot (1+2x)$, but I do already have defined P(x), haven't I? –  muffel Jul 3 '11 at 21:19
@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. –  Gerry Myerson Jul 4 '11 at 0:06
thanks again, I now got it! –  muffel Jul 4 '11 at 8:49

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.

share|improve this answer
@nemathsadist thank you, but I haven't heart of Taylor expansions yet and need to work it out. Isn't there a way of just defining $P(x)$ and $Q(x)$? –  muffel Jul 3 '11 at 8:49
You can just use the Geometric Series Theorem. –  ncmathsadist Jul 3 '11 at 13:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.