# Understanding Fourier Coefficients using PDE

1. Jan 24, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
In my PDE course we have a homework question stating the following:

Let ϑ(x) = x in the interval [-pi, pi ]. Find its Fourier Coefficients.

2. Relevant equations

From my notes on this type of question:

a_o = 2c_o = 1/pi * integral from -pi to pi [f(x) dx]

a_n = c_n + c_(-n) = 1/pi * integral from -pi to pi [f(x) cos(n*x) dx ]

b_n = i(c_n - c_(-n)) = 1/pi * integral from -pi to pi [f(x) sin(n*x) dx]

3. The attempt at a solution

Is it as simple as just a plug and chug based off my noes?

a_o's integration with f(x) = x just is x^2/(2*pi) from -pi to pi so we have
a_o = pi/2 - pi/2 = 0

a_n's integration is just equal to 0 as well.

b_n is just -2(-1)^n/n

So thus, the fourier coefficients here are b_n = [(-2)(-1)^n]/n
for n ≥ 1

Am I understanding the question properly?

2. Jan 24, 2016

### Dr. Courtney

Graph the sum of the first 10 terms and compare it to the original function.

3. Jan 24, 2016

### RJLiberator

Since f(x) = x is odd, we can safely say a_n is equal to 0 for n >= 0.

So, it seems I am right that b_n is the only coefficients.

The fourier expansion is thus
2sin(x)- sin(2x) + (2/3)sin(3x)/3 +...

So I guess my understanding on this problem seems to be getting better.
My question is thus, according to the question in the initial post, is a complete and safe way to answer this question by stating that the function is odd so a_n coefficients are 0, and so we observe
b_n = (2/n)(-1)^(n+1)

I just want to make sure I am answering this question with completeness.

4. Jan 24, 2016

### Dr. Courtney

I'd be happy with your answer if I was grading it. But I'd be happier with a graph showing the series agrees with the initial function.

My classes always have a heavy emphasis on assessment: showing your answer is right with a method independent of the method originally used to compute the answer.