Question

How would you find this infinite series-like thing?

Answer 1

\[y=\sum_{n=1}^{\infty}\frac{4}{n^2}cosnx\]

Answer 2

This is part of the confusion that came out when I (naively maybe) tried to Fourier transform x^2.

Answer 3

hmm...fourier series

Answer 4

More completely, I got:
\[x^2=\frac{4 \pi^2}{3}+\sum_{n=1}^{\infty}\frac{4}{n^2}cosnx+\sum_{n=1}^{\infty}\frac{-4 \pi}{n}sinnx\]

Answer 5

So if you just saw one of those series, you could only use the Fourier series to ascertain its value?
Also, is there any meaning of divergence or convergence in this scenario?

Answer 6

yes i think just fourier series

Answer 7

but x^2 is an even function how u get sin terms in its fourier series

Answer 8

http://upload.wikimedia.org/wikipedia/en/math/f/5/3/f53a89111da355fff38eb41c1f41fb6d.png

Answer 9

oh i think u change period from \( -\pi \le x \le \pi \) to \( 0 \le x \le 2\pi \) ha?

Answer 10

That's what I did

Answer 11

so thats ok :)

Answer 12

What would it mean if this series were to diverge?

Answer 13

these series converge quite quickly

Answer 14

I realise that, but my question was concerning these types of infinite series (which I haven't encountered before). What I meant was what does it mean that\[\sum_{n=0}^{\infty}n^2cosx\] diverges (if it does- if it doesn't assume that I inserted one which actually diverged)

Answer 15

use squeeze theorem for series to show it's bounded.

Answer 16

\[\sum_1^\infty {-4 \over n^2 } \leq \sum_{1}^\infty {4 \cos nx \over n^2} \leq \sum_1^\infty {4 \over n^2 } \\
\sum_1^\infty{1 \over n^2} = {\pi^2 \over 6}\]