# Thread: A question on Fourier series

1. ## A question on Fourier series

I need to know an example of a function defined on the interval [0, pi] which cannot be expanded as a Fourier series. Thanks.

2. Originally Posted by curvature
I need to know an example of a function defined on the interval [0, pi] which cannot be expanded as a Fourier series. Thanks.
How about $\displaystyle 1/x^2$.

In particular how would you evaluate:

$\displaystyle \int_0^{\pi} \frac{\cos(2nx)}{x^2}~dx$

RonL

3. Thanks. But I was thinking of an example of function with infinitely many discontinuities or with infinitely many extrema, if any.

4. Originally Posted by curvature
Thanks. But I was thinking of an example of function with infinitely many discontinuities or with infinitely many extrema, if any.
$\displaystyle f(x) = \left\{ \begin{array}{c}0\mbox{ if }x\in \mathbb{Q} \\ 1\mbox{ if }x\not \in \mathbb{Q}\end{array} \right.$.

5. Originally Posted by ThePerfectHacker
$\displaystyle f(x) = \left\{ \begin{array}{c}0\mbox{ if }x\in \mathbb{Q} \\ 1\mbox{ if }x\not \in \mathbb{Q}\end{array} \right.$.
A Fourier series converges to the function a.e. so the zero Fourier series converges to this a.e. as $\displaystyle \mathbb{Q}$ is a null set. That is this function does have a Fourier series and it converges to the function a.e. as required.

(even the Fourier series of a function with a step discontinuity does not converge to the function at the step without a bit of jiggery pokery about what the value of the function is at the step)

RonL

6. Originally Posted by ThePerfectHacker
$\displaystyle f(x) = \left\{ \begin{array}{c}0\mbox{ if }x\in \mathbb{Q} \\ 1\mbox{ if }x\not \in \mathbb{Q}\end{array} \right.$.
Thanks. I know the function is called the Dirichlet function and I guess this bounded function cannot be expressed into a Fourier series, but not quite sure since no books I read tell the fact.

I wonder if examples like this motivated Dirichlet to propose the Dirichlet Conditon for Fourier series?