# Thread: quick fourier series question

1. ## quick fourier series question

(2/(n^2))(cos(nPi)-cos(-nPi))

can be written as (2/(n^2))*2 for Even n
and (2/(n^2))*(-2) for n ODD

so this means it can be written (4/(n^2))(-1)^n

How? I cant get my head around why. if i sub in an even n and odd n say 2 and 1 i do not get 2 or -2! help

2. Originally Posted by sheep99
(2/(n^2))(cos(nPi)-cos(-nPi))

can be written as (2/(n^2))*2 for Even n
and (2/(n^2))*(-2) for n ODD

so this means it can be written (4/(n^2))(-1)^n

How? I cant get my head around why. if i sub in an even n and odd n say 2 and 1 i do not get 2 or -2! help
Are you sure that it's $\displaystyle \cos(n\pi)-\cos(-n\pi)$ and not $\displaystyle \cos(n\pi)+\cos(-n\pi)$?

Because $\displaystyle \cos(n\pi)-\cos(-n\pi) = 0$ for all $\displaystyle n$ because $\displaystyle \cos(n\pi)=\cos(-n\pi)$.

3. its how its written in my mark scheme, your totally right when i use + works out beautifully, the lecturer has just wasted about 10mins of my life. thanks!