# Series expression

1. Apr 25, 2012

### d.tran103

Hey, I'm going over series expansions and was wondering if someone could check my work and tell me if my work is correct. If not, could you explain it to me? I couldn't find any example like this problem in my book so I'm posting it online. Here it is,

The closed form series expansion for cos(x) is Ʃ[(-1)^n(x)^2n]/(2n)!. Use this series to find a series expression for [cos(x)-1]/x^2.

Okay here's what I did:

[cos(x)-cos(pi)]/(x)^(2)

(x)^(-2)*Ʃ[(-1)^(n)(x)^(2n)-(pi)^(2n)]/(2n)!

Ʃ[(-1)^(n)(x)^(2n-2)-(pi)^(2n)

Is that the way I'm supposed to do it? Thanks!

2. Apr 25, 2012

### scurty

This step is wrong, it should be $x^{-2} \cdot \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}+(-1)^n \pi^{2n}}{(2n)!}$

Remember, $\cos{\pi} = -1$, and you also forgot that extra factor of the $(-1)^n$. Also, you didn't multiply your $x^{-2}$ term in the sum properly in your attempt, make sure you do so this time!

Alternatively, you could break up your expression to find the series expansion of $\displaystyle \frac{\cos{x}}{x^2} - \frac{1}{x^2}$.

3. Apr 26, 2012

Okay thanks!