# Math Help - Fourier series

1. ## Fourier series

How do you do an even expansion of a odd function?

2. Originally Posted by adam_leeds
How do you do an even expansion of a odd function?
Can you provide a bit of context for this? It doesn't make much sense on its own.

The only way in which I can think of this happening would be if you were given an odd function defined on half of a fundamental interval, and then you make it into an even function by defining it symmetrically on the other half of the interval.

For example, if you are given the odd function $f(x) = x^3$, defined on the interval $[0,\pi]$, then you can make it into an even function on the interval $[-\pi,\pi]$ by defining $f(-x) = f(x)$. That gives you the even function $|x^3|$, which will have an even Fourier expansion.

But if a function is odd on the whole fundamental interval then it can only have an odd Fourier expansion.

3. Originally Posted by Opalg
Can you provide a bit of context for this? It doesn't make much sense on its own.

The only way in which I can think of this happening would be if you were given an odd function defined on half of a fundamental interval, and then you make it into an even function by defining it symmetrically on the other half of the interval.

For example, if you are given the odd function $f(x) = x^3$, defined on the interval $[0,\pi]$, then you can make it into an even function on the interval $[-\pi,\pi]$ by defining $f(-x) = f(x)$. That gives you the even function $|x^3|$, which will have an even Fourier expansion.

But if a function is odd on the whole fundamental interval then it can only have an odd Fourier expansion.
thanks

i have to do this for sin(x) for 0 <= x <= pi

to find its fourier series over -pi <= x <= pi

so all i do is do the normal fourier series but between -pi and pi?

4. Originally Posted by adam_leeds
thanks

i have to do this for sin(x) for 0 <= x <= pi

to find its fourier series over -pi <= x <= pi

so all i do is do the normal fourier series but between -pi and pi?
If you want an even function that is equal to sin(x) on $[0,\pi]$ then you need the function |sin(x)| on $[\pi,\pi]$. To find its Fourier series, you need integrals of the form $\int_{-\pi}^\pi|\sin x|\cos nx\,dx$. But because you now have an even function, that will be equal to $2\!\!\int_0^\pi\sin x\cos nx\,dx$, and that is the way to calculate the Fourier coefficients.