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Find a power series around $a=0$ for the function $f:\mathbb{R}\backslash\{2,3\} \to \mathbb{R}$ with $f(x) = \frac{1}{x^2-5x+6}$.

It is

$f(x) = \frac{1}{x^2-5x+6} = \frac{1}{2-x}\frac{1}{3-x} = \frac{1}{1-(x-1)}\frac{1}{1-(x-2)}$

Now I could use the geometric series but I need to take care of the range of $x$ then, right?

$= \sum_{n=0}^{\infty} (x-1)^n \cdot \sum_{n=0}^{\infty} (x-2)^n$

But anyway, I don't know how to proceed from there.. Any hints

1 Answers 1

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Hint: Before attempting to expand, express $\frac{1}{x^2-5x+6}$ as $\frac{A}{2-x}+\frac{B}{3-x}$, using the partial fractions idea. (The idea has many uses, it's not just for integration!)

After the partial fractions process, you will want to use a method other than the one you proposed. Perhaps (with different numbers) something like $\frac{1}{7-3x}=\frac{1}{7}\frac{1}{1-\frac{3x}{7}}.$

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    @menfredk: The general first term is $\frac{1}{2}\left(\frac{x}{2}\right)^n$. This has radius of convergence $2$. The other guy is similar. It has radius of convergence $3$. The sum (difference) has radius of convergence the **smaller** of $2$ and $3$. So the whole thing has radius of convergence $2$. Where you have $x$ it sould be $x^n$.2012-06-05