# Power series and partial fractions

Let $$f(x) = \frac{x-1}{x^2-3x+2}$$ Factorise the denominator and use partial fractions to write $f$ as the sum of two functions whose power series you know. Write down the power series centered at $a=0$ and determine the radius of convergence.

It specifically asks for partial fractions. I tried partial fractions using $A, B$, etc. and I got what I would have got had I just factored and cancelled out the same terms. What is this exactly asking? How do I do this?

• Please type your questions using MathJax instead of posting pictures. Pictures are not searchable, and not accessible to persons using screen readers for example. Also, please show us what it was you got. As it is, it's hard to tell what you are asking, and your question is likely to be closed by the moderators, I should think. – saulspatz Apr 5 '18 at 1:43
• I got the same thing you got. I don't see how to write $f$ as the sum of two functions with partial fractions. Could be a typo perhaps. – Matthew Leingang Apr 5 '18 at 1:44

It's asking you to create a power series representation of the rational function. By using partial fractions, you can split into two:

$$A(x-2) + B(x-1) = x-1$$ $$x = 2 \implies B = 1$$ $$x = 1 \implies A = 0$$

You get:

$${A \over x-1} + {B\over x-2} = {x-1\over x^2-3x+2} \implies {1\over x-2}= {x-1\over x^2-3x+2}$$

And you're right; you could have factored, but the question specifically asks for partial fractions for who knows what reason. Now, recall the power series:

$${1\over 1-r} = 1+r+r^2+\cdots \quad |r|<1$$

Then apply:

$${1\over x-2} = {1\over -2\left(1-\frac x2\right)}=-\frac 12 {1\over 1-\frac x2}$$

Rewrite as the geometric power series where $r = \frac x2$. Also remember the restriction of $|r|<1$, which helps you determine the radius of convergence.