# How do you find a power series representation for  f(x)=(x / (x^(2)-3x+2) ) and what is the radius of convergence?

Jul 30, 2016

$S \left(x\right) = {\sum}_{n = 0}^{\infty} \left(1 - \frac{1}{2} ^ n\right) {x}^{n}$.
This serie is convergent to $f \left(x\right)$ for $\left\mid x \right\mid < 1$

#### Explanation:

$f \left(x\right) = \frac{x}{{x}^{2} - 3 x + 2} = \frac{x}{\left(x - 1\right) \left(x - 2\right)} = \frac{A}{x - 1} + \frac{B}{x - 2}$

$\frac{A \left(x - 2\right) + B \left(x - 1\right)}{\left(x - 1\right) \left(x - 2\right)} = \frac{\left(A + B\right) x - \left(2 A + B\right)}{\left(x - 1\right) \left(x - 2\right)}$

Solving for $A , B$

{ (A+B=1), (2A+B=0) :}

we obtain $A = - 1 , B = 2$ So

$f \left(x\right) = - \frac{1}{x - 1} + \frac{2}{x - 2}$

Now considering that $\frac{{x}^{n + 1} - 1}{x - 1} = 1 + x + {x}^{2} + \cdots + {x}^{n}$

if $\left\mid x \right\mid < 1$ we know

${\lim}_{n \to \infty} \frac{{x}^{n + 1} - 1}{x - 1} = - \frac{1}{x - 1} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Using those results and supposing that $\left\mid x \right\mid < 1$ we propose

$S \left(x\right) = - \frac{1}{x - 1} + \frac{2}{x - 2} = {\sum}_{n = 0}^{\infty} {x}^{n} - {\sum}_{n = 0}^{\infty} {\left(\frac{x}{2}\right)}^{n}$

so $S \left(x\right) = {\sum}_{n = 0}^{\infty} \left(1 - \frac{1}{2} ^ n\right) {x}^{n}$.

This serie is convergent to $f \left(x\right)$ for $\left\mid x \right\mid < 1$

Supposing you need a series representation for $f \left(x\right)$ in the surroundings of $x = 3$ you can proceed as follows.

Making $y = x - 3$ then
$f \left(y\right) = - \frac{1}{y + 2} + \frac{2}{y + 1}$

so with $\left\mid y \right\mid < 1$

$f \left(y\right) = {\lim}_{n \to \infty} \left\{- \frac{1 - {\left(- \frac{y}{2}\right)}^{n}}{2 \left(1 - \left(- \frac{y}{2}\right)\right)} + 2 \frac{1 - {\left(- y\right)}^{n}}{1 - \left(- y\right)}\right\}$

or

$f \left(y\right) = - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- \frac{y}{2}\right)}^{n} + 2 {\sum}_{n = 0}^{n} {\left(- y\right)}^{n}$

Finally

$S \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \left(2 - \frac{1}{2} ^ \left\{n + 1\right\}\right) {\left(x - 3\right)}^{n}$

This series converges to $f \left(x\right)$ for $2 < x < 4$