Proof of Fourier Series Symmetry/Antisymmetry

[PhysicsMark]

Homework Statement

Suppose, in turn, that the periodic function is symmetric or antisymmetric about the point x=a. Show that the Fourier series contains, respectively, only $$cos(k_{n}(x-a))$$ (including the $$a_0$$) or $$sin(k_{n}(x-a))$$ terms.

Homework Equations

The Fourier expansion for the periodic function, f(x):
$$f(x)=a_0 + \sum_{n=1}^{\infty}(a_{n}cos(k_{n}x) + b_{n}sin(k_{n}x))$$

$$a_0=\frac{1}{\lambda}\int_{x_0}^{x_0+\lambda}f(x) dx$$


$$a_n=\frac{2}{\lambda}\int_{x_0}^{x_0+\lambda}f(x)cos(k_{n}x) dx, n\neq0$$

$$b_n=\frac{2}{\lambda}\int_{x_0}^{x_0+\lambda}f(x)sin(k_{n}x) dx, (b_0=0)$$

$$k= \frac{2n\pi}{\lambda}\qquad \lambda=period$$

Hint: An integral centered on a point where the integrand is antisymmetric will vanish.

The Attempt at a Solution

I believe I understand what they are saying, but I do not know how to show/prove it. I know that cosine is symmetric about a point, while sine is not. Initially I was thinking that since sine is antisymmetric, for f(x) to be symmetric, the sine terms cannot exist in the expansion (which is exactly what the problem states). I was thinking the same for the opposite case regarding cosines.

The hint leads me to believe I am supposed to set up an integral. However, I do not understand how.

If I am to set up an integral, the best I can figure is that the lower limit should be $$a-\lambda$$ and the upper limit should be $$a+\lambda$$

I think this because for the integral to be centered at point "a", we would need to go 1 full period in both directions.

 

[PhysicsMark]

Something just dawned on me. Does anyone think the following is correct?:

If f(x) is symmetric, then b_n is 0 since sine is antisymmetric. Therefore, the sin term in the Fourier series drops out.

If f(x) is antisymmetric, then a_n is 0 since f(x) is antisymmetric and the cosine term in the Fourier series drops out.

 

[vela]

Yup, that's exactly what the hint was getting at.

 

[PhysicsMark]

Thanks! I can't believe I burned almost 3 hours on that, only to have the answer as soon as I posted it.