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I'm looking for the power series for $f(x)=\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$

My approach: the given function is a combination of two problems. first i made some transformations, so the function looks easier.


Now i have two polynomials. i thought the Problem might be even easier, if thinking about the function as:

$$\frac{3x^{2}-4x+9}{x^3+x^2-5x+3)}= (3x^{2}-4x+9)\cdot \frac{1}{(x^3+x^2-5x+3)}$$

Assuming the power series of $3x^{2}-4x+9$ is just $3x^{2}-4x+9$ itself. I hoped, i could find the power series by multiplying the series of the both easier functions.. yeah, i am stuck.

$\sum\limits_{n=0}^{\infty}a_{n}\cdot x^{n}=(3x^{2}-4x+9)\cdot \ ...?... =$ Solution

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Have you thought about the fact that (3x^2-4x+9) = (3x^2-4x+1+8) = (3x-1)(x-1)+8 ? – rbm May 11 '13 at 12:27
you mean i should form the Expression this way: $\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=\frac{(3x-1)(x-1)+8}{x^3+x^2-5x+3}=\frac{2}{(x‌​-1)^2}+\frac{3}{x+1}$ ? – Toralf Westström May 11 '13 at 12:41
up vote 2 down vote accepted

You can use the partial fraction decomposition: $$ \frac{3x^{2}-4x+9}{(x-1)^2(x+3)}= \frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{1+\frac{1}{3}x} $$ and sum up the series you get, which are known.

If you do the computation, you find $A=0$, $B=2$ and $C=1$, so

$$ \frac{3x^{2}-4x+9}{(x-1)^2(x+3)}= \frac{2}{(1-x)^2}+\frac{1}{1+\frac{1}{3}x} $$

The development of $(1-x)^{-2}$ can be deduced from the fact that $$ \frac{1}{1-x}=\sum_{n\ge0}x^n $$ so, by deriving, we get $$ \frac{1}{(1-x)^2}=\sum_{n\ge1}nx^{n-1}=\sum_{n\ge0}(n+1)x^n $$

The power series for the other term is again easy: $$ \frac{1}{1+\frac{1}{3}x}=\sum_{n\ge0}\frac{(-1)^n}{3^n}x^n $$ so your power series development is

$$ \frac{3x^{2}-4x+9}{(x-1)^2(x+3)}= \sum_{n\ge0}\left(2n+2+\frac{(-1)^n}{3^n}\right)x^n $$

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$\frac{2}{(x‌​-1)^2}+\frac{3}{x+1}$ ? – Toralf Westström May 11 '13 at 12:57
@ToralfWestström No; you have $x+3$ in the denominator of the original fraction. – egreg May 11 '13 at 13:01
but where did you get the $1+\frac{1}{3}x$ from? – Toralf Westström May 11 '13 at 13:18
@ToralfWestström Just observe that $\frac{C}{1+x/3}=\frac{3C}{3+x}$; the form with $1+x/3$ makes it easier to write down the power series. – egreg May 11 '13 at 13:19
i still don't get it. how can i make partial fraction from the given function to this form? – Toralf Westström May 11 '13 at 13:36

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