# Creating Fourier sine Series

1. Mar 3, 2016

### RJLiberator

1. The problem statement, all variables and given/known data

Find the Fourier series for the following function (0 ≤ x ≤ L):
y(x) = Ax(L-x)

2. Relevant equations

3. The attempt at a solution

1. We start with the sum from n to infinity of A_n*sin(n*pi*x/L) where An = B_n*Ax(l-x)

2. We have the integral from 0 to L of f(x)*sin(m*pi*x/L) dx

I really have no idea what to do, I am francticlly looking through notes and websites. I understand the Fourier sine series should be pretty easy to find, it's just plugging in values, but there are so many different equations/elements.

Let me try this solution:

f(x) = L/pi(sum from n = 1 to infinity of sin(n*pi*x/L)

Ah?

2. Mar 3, 2016

### Staff: Mentor

When you are asked the for the Fourier series of a function, your answer should be the coefficients appearing in the sum.

3. Mar 3, 2016

### RJLiberator

From my notes, would:

A_m = 2/L integral from 0 to L f(x)*sin(m*pi*x/L) dx be the answer then?

where f(x) = Ax(L-X)

4. Mar 3, 2016

### RJLiberator

Okay, let me say this:

sum from 1 to infinity of (b_n*sin(n*pi*x/L))

where b_n = 2/L integral from 0 to L (f(x)*sin(n*pi*x/L))

where f(x) = the function in question, namely Ax(L-x)

All together we have

The sum from n=1 to infinity of 2/L integral from 0 to L of (Ax(L-x))*sin^2(n*pi*x/L) dx

5. Mar 3, 2016

### Staff: Mentor

While formally correct, this doesn't answer the question. You have to find an expression for the b_n.

6. Mar 3, 2016

### RJLiberator

b_n = 2/L integral from 0 to L (f(x)*sin(n*pi*x/L))
where f(x) = the function in question, namely Ax(L-x)

so b_n = 2/L integral from 0 to L (Ax(L-x)*sin(n*pi*x/L))
Is that incorrect for b_n?

7. Mar 3, 2016

### Staff: Mentor

It's a start. Now you have to perform the integral.

8. Mar 3, 2016

### vela

Staff Emeritus
Now that you've been working with Fourier series for a while, it wouldn't hurt to go back and review the derivation of the various formulas (using one reference). If you understand the basics, all the variations/conventions will make more sense and won't seem so confusing.