Fourier Series: Can even functions be changed to odd?

[thelema418]

When creating a Fourier series for a function f(x), I consider whether the function is odd or even first. Yet, often these functions are in the positive region [0, L].

Since f(x) is only defined in this region, can I change the function to get a desired parity? By example, my concern originated with the function f(x) = x \sin x. This is an even function, but I could modify the function as f(x) = |x| \sin x to make it odd while retaining the desired information in [0, L].

Can this be done? And are there problems with doing this?

Thanks.

 

[micromass]

What do you mean with "retaining the desired information"? What exactly is it that you want to do?

 

[thelema418]

What do you mean with "retaining the desired information"? What exactly is it that you want to do?

Meaning that x \sin x on [0,\pi] is the same as |x| \sin x on [0,\pi].

The application is usually for solving heat equations. So the region is usually x = 0 to x = L in the problems I see. Again, another example: When I work out all the details of a solution by eigenfunction expansion based on homogeneous Dirichlet boundary conditions, I end up with a Fourier sine series equal to some function f(x), where f(x) is the initial condition u(x,0). But to have a Fourier sine series, f(x) must be an odd function, right?

 

[micromass]

Meaning that x \sin x on [0,\pi] is the same as |x| \sin x on [0,\pi].

The application is usually for solving heat equations. So the region is usually x = 0 to x = L in the problems I see. Again, another example: When I work out all the details of a solution by eigenfunction expansion based on homogeneous Dirichlet boundary conditions, I end up with a Fourier sine series equal to some function f(x), where f(x) is the initial condition u(x,0). But to have a Fourier sine series, f(x) must be an odd function, right?

OK. You're right then. The right way to find a sine series converging to $f$ is to extend it to an odd function on entire $[-L,L]$. So you'll have to work with $|x|\sin(x)$ on $[-\pi,\pi]$, like you suggested.