Question

Solve using Fourier series.

Answer 1

With
`2L=\pi`, the Fourier series expansion of
`f(x)` is


`\displaystyle f(x)\sim\frac{a_0}2+\sum_(n\ge1)a_n\cos\frac{n\pi x}L+\sum_(n\ge1)b_n\sin\frac{n\pi x}L`

`\displaystyle f(x)\sim\frac{a_0}2+\sum_(n\ge1)a_n\cos2nx+\sum_(n\ge1)b_n\sin2nx`

where the coefficients are obtained by computing


`\displaystyle a_0=\frac1L\int_0^(2L)f(x)\,\mathrm dx`

`\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx`


`\displaystyle a_n=\frac1L\int_0^(2L)f(x)\cos\frac{n\pi x}L\,\mathrm dx`

`\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx`


`\displaystyle b_n=\frac1L\int_0^(2L)f(x)\sin\frac{n\pi x}L\,\mathrm dx`

`\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx`

You should end up with


`a_0=0`

`a_n=0`
(both due to the fact that
`f(x)` is odd)

`b_n=\frac1{3n}\left(2-\cos\frac{2n\pi}3-\cos\frac{4n\pi}3\right)`

Now the problem is that this expansion does not match the given one. As a matter of fact, since
`f(x)` is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of
`f(x)`, which is to say we're actually considering the function


`\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3<|x|\le\frac{2\pi}3\\-\frac\pi3&\text{for }\frac{2\pi}3<|x|\le\pi\end{cases}`

and enforcing a period of
`2L=2\pi`. Now, you should find that


`\varphi(x)\sim\frac2{\sqrt3}\left(\cos x-\frac{\cos5x}5+\frac{\cos7x}7-(\cos11x)/(11)+\cdots\right)`

The value of the sum can then be verified by choosing
`x=0`, which gives


`\varphi(0)=\frac\pi3=\frac2{\sqrt3}\left(1-\frac15+\frac17-\frac1{11}+\cdots\right)`

`\implies\frac\pi{2\sqrt3}=1-\frac15+\frac17-\frac1{11}+\cdots`

as required.