Series Expansion for [cos(x)-1]/x^2: Explained and Checked

[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!

 

[scurty]

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

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

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!

Ʃ[(-1)^(n)(x)^(2n-2)-(pi)^(2n)

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

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

 

[d.tran103]

Okay thanks!