Fourier series problem: f (x) = x if −π/ 2 < x < π/ 2, 0 if π /2 < x < 3π /2

[promitheus]

Fourier series problem: f (x) = x if −π/ 2 < x < π/ 2, 0 if π /2 < x < 3π /2

Hi,

Just learning about the Fourier series and I'm stuck on an assignment problem.

Here is the question.

Use properties of odd and even functions in finding the Fourier series for
f (x) =
x if −π/ 2 < x < π/ 2
0 if π /2 < x < 3π /2

f (x + 2π ) = f (x).

I've attached my attempt. Could someone please give me feedback on if I've solved it correctly?

I am particularly having trouble with the Fourier series in the form of Sum of bn. sin nx as I'm getting values of b1=2/pi , b2= -1/2 , b3= -2/(9pi), b4= -1/4

I'm not sure if this is making sense, so thanks for your patience and help.

 

[j-astron]

Hi prometheus,

You are missing a variable in your handwritten solution. The symbols n and x are not the same thing. The letter 'x' is the dependent variable of the function, which is continuous. The letter 'n' is an integer that tells you which of the harmonics you are dealing with. The harmonics are the sinusoids with frequencies that are integer multiples of each other, which sum up to give you the original function.

The equation for the Fourier series coefficient (in the odd function case) is supposed to be given by:

\(\displaystyle b_n = \displaystyle \dfrac{2}{P}\int_{x_0}^{x_0+P} f(x) \sin\left(\dfrac{2\pi n x}{P}\right)\,dx \)

In this case, the period \(\displaystyle P\) is \(\displaystyle 2\pi\). As you did, we'll take our arbitrary starting point for the integration to be \(\displaystyle x_0 = -\pi\) so that the integral ranges over \(\displaystyle [-\pi, \pi]\). That gives us an integral

\(\displaystyle b_n = \displaystyle \dfrac{2}{2\pi}\int_{-\pi}^{\pi} f(x) \sin\left(\dfrac{2\pi n x}{2\pi}\right)\,dx = \dfrac{1}{\pi}\int_{-\pi/2}^{\pi/2} x \sin\left(n x\right)\,dx \)

where the last step follows from the fact that the function f(x) is zero outside the specified range. I don't think you need any fanciness with odd or even functions to evaluate this last integral, right? You can probably just do it by parts?