# Homework Help: Fourier Series

1. Oct 18, 2009

### BustedBreaks

Let $$f(x):= x$$ when 0<x<1. Find the Fourier series for f if:
a) f is 1 periodic
b) f is even and 2 periodic
c) f is odd and 2 periodic.

I am very lost and behind. I'm reading through my notes and book and hopefully will be able to to this soon, but can anyone give me a hint or just explain how the differences between a, b, and c?

I get a, that is f(x) = x from 0 to 1, and that graph repeats for every interval of one because it is 1-periodic.

I don't get how f can be even and 2 periodic and also be = x on 0 to 1

or odd and 2 periodic an be = to x on 0 to 1...

2. Oct 18, 2009

### LCKurtz

You are given f(x) = x on (0,1). If f is to have period 2 you need its definition on an interval of length 2. So what might you want it to be on (-1,0)? Do you see what you might want it to be on (-1,0) if you wanted the function to be odd on (-1,1)? Or if you wanted it to be even on (-1,1)? Do you know what the graph must look like on (-1,1) if it has to agree with x on (0,1) and be odd? even?

These are different functions on (-1,1) that happen to agree on (0,1). And if you write their Fourier Series, one of them will have only cosine terms and one only sine terms.

3. Oct 18, 2009

### BustedBreaks

I am having trouble understanding this.

If it is even and 2-periodic are you saying that it would be basically the graph of |x| on (-1,1)?

for odd just the graph of x on (-1,1)?

What confuses me if this is right is that it says in the question that the interval is (0,1)

Also can you confirm that for a) it would be just the function x from (0,1) repeated for ever interval of 1 before and after that?

Last edited: Oct 18, 2009
4. Oct 18, 2009

### LCKurtz

Yes, that is exactly what I am saying. If you plot your 1p periodic function and these two 2p periodic function for a couple of periods, you will see that they are three different periodic functions that all happen to agree on (0,1). They will have different Fourier series, but they will all sum to x on (0,1).

5. Oct 18, 2009

### LCKurtz

And I should have added, since you have to actually find the series, that in the last two cases you have special "half range" formulas for the an and bn. They are each half the work because half the coefficients are zero compared to the 1p function, which is neither even nor odd.