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## Homework Statement

Find the value of A

_{n}and given that f(x) = 1 for 0 < x < L/2, find the sum of the infinite series.

## Homework Equations

## The Attempt at a Solution

The basis is chosen to be ##c_n = \sqrt{\frac{2}{L}}cos (\frac{n\pi }{L}x)## for cosine, and ##s_n = \sqrt{\frac{2}{L}}sin (\frac{n\pi}{L}x)## for sine.

[tex]|f\rangle = \sum_0^{\infty} \langle c_n|f\rangle |c_n\rangle + \sum_1^{\infty}\langle s_n|f\rangle|s_n\rangle[/tex]

Compare this with:

[tex]f = \sum_0^{\infty}c_n' cos(\frac{n\pi }{L}x) + \sum_1^{\infty}A_n sin(\frac{n\pi}{L}x)[/tex]

We are given that f

_{(x)}= 1 for 0 < x < L/2.

Therefore,

[tex]A_n = \sqrt{\frac{2}{L}}\langle s_n|f\rangle = \frac{2}{L}\int_0^{L/2}f_{(x)} sin(\frac{2\pi n}{L}x) dx[/tex]

[tex]A_n = \frac{1}{n\pi}\left[1 - cos(n\pi)\right][/tex]

[tex]A_n = \frac{1}{n\pi}\left[1 - (-1)^n\right][/tex]

For odd n, ##A_n = \frac{2}{(2n-1)\pi}##

Either the first series or the third series works for ##A_n##:

For 0 < x < L/2, f

_{(x)}= 1:

[tex]1 = \sum_{n=1}^{\infty} \frac{2}{(2n-1)\pi}sin\left(\frac{2\pi n}{L}x\right)[/tex]

Choosing x = L/4:

[tex]1 = \sum_{n=1}^{\infty} \frac{2}{(2n-1)\pi}sin\left(\frac{n\pi}{2}\right)[/tex]

[tex]1 = \sum_{n=1}^{\infty} \frac{2}{(2n-1)\pi} (-1)^{n+1}[/tex]

[tex]\frac{\pi}{2} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)} = 1 - \frac{1}{3} + \frac{1}{5} - ...[/tex]

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