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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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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.

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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.

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@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

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