Thread: Fourier Series Coefficients and Symetry

1. Fourier Series Coefficients and Symetry

Given the series

and that f(x) = f(pi - x), what can one say about the coefficients an and bn? I know that due to orthogonality an=0 for odd functions and bn=0 for even functions, but I'm not sure what symmetry the function f(x) = f(pi-x) has.

2. Originally Posted by StaryNight
Given the series

and that f(x) = f(pi - x), what can one say about the coefficients an and bn? I know that due to orthogonality an=0 for odd functions and bn=0 for even functions, but I'm not sure what symmetry the function f(x) = f(pi-x) has.
Here are a few hints:
First the function is symmetric across the line $x=\frac{\pi}{2}$

Also since the inside of your trig function is of the form

$nx$ The generic form is $\displaystyle \frac{n \pi x}{L}$ This gives the function must be $2L$ peroidic.

Using this try to figure out if the function is odd or even!

3. Originally Posted by TheEmptySet
Here are a few hints:
First the function is symmetric across the line $x=\frac{\pi}{2}$

Also since the inside of your trig function is of the form

$nx$ The generic form is $\displaystyle \frac{n \pi x}{L}$ This gives the function must be $2L$ peroidic.

Using this try to figure out if the function is odd or even!
So the function f is even about the point x=pi/2. This means that a(n) will be 0 when cos(nx) is odd about x=pi/2 and b(n)=0 when sin(nx) is odd about x=pi/2.

$\newline&space;sin(nx)&space;=&space;-sin(n(\pi-x))&space;\newline&space;sin(nx)&space;=&space;sin(n(x-\pi))&space;\newline&space;sin(nx)&space;=&space;sin(nx-n\pi)&space;\newline&space;\therefore&space;n=2k$

A similar argument for cos gives a(2n+1) = b(2n) = 0. Is this the best way to tackle the problem?