# 2 Questions on Forier Series

1. Jan 14, 2010

### WannaBe22

1. The problem statement, all variables and given/known data
1. Find the fourier series of the function:
f(x)= 1 when -pi/2 < x < pi/2
-1 when pi/2 < x < 3pi/2

2. Let f be a functin that is defined in [0,pi/2]. f is continous at [0,pi/2] and has a piecewise-continous deriative.
In each case, determine how should we expand the definition of f to [-pi,pi] so that the fourier series will be as followed in the picture...

2. Relevant equations
3. The attempt at a solution
About 1- I can't figure out how I should expand this function...expanding it to a 3pi period gives me a result that doesn't make sense at all... Also when I've tried to consider the function as a function with 2pi period- The results I've received were unreasonable...

Thanks a lot!

#### Attached Files:

• ###### fourier.jpg
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2. Jan 14, 2010

### payumooli

the function is periodic with period 3pi/2 and not 3pi

3. Jan 14, 2010

### WannaBe22

How come?

4. Jan 14, 2010

### payumooli

sorry it is periodic with period 2pi
-pi/2 to pi/2
pi/2 to 3pi/2
completes a full revolution
so 2pi

5. Jan 14, 2010

### WannaBe22

Welll...So guess I had a calculation mistake...

Thanks a lot!

6. Jan 14, 2010

### HallsofIvy

If "as seen in the picture" refers to the attachment, not two crucial points:
"A" is a sum of even functions and so must be the Fourier series for an even function. "B" is a sum of odd functions and so must be the Fourier series for an odd function.

Last edited by a moderator: Jan 15, 2010
7. Jan 15, 2010

### WannaBe22

Yep, I know this...But I can't figure out how to make my expansion work with coefficients that are zero for every odd n .....
Hope you'll be able to explain this...

:(
Thanks

8. Jan 15, 2010

### vela

Staff Emeritus
Try focusing on the interval $[0,\pi]$ and consider the symmetry of $sin(nx)$ about $x=\pi/2$.

9. Jan 15, 2010

### WannaBe22

It's pretty obvious that in the first one we need to expand the function into an even one... But I realy can't figure out how to expand it so that the coefficients of the odd terms will be zero and only the coefficients at the even terms will remain...
I've tried "researching" several functions but I had no success in figuring out the regularity in the functions which their odd coefficients are zero...

I'll be delighted to receive further help on this question...The hint vela gave me didn't help at all...

Thanks a lot

10. Jan 15, 2010

### vela

Staff Emeritus
Maybe because I was talking about the second expansion. :) But the same idea works with the first problem. Try plotting cos(nx) for n=0,1,2,3 from x=0 to x=pi. Is there any symmetry to the graphs about the line x=pi/2, and if so, how does it depend on n?

11. Jan 15, 2010

### WannaBe22

when n=0, the graph is symmetric about the line x=pi/2,
when n=1, the graph of cos(x) is anty-symmetric about the line x=pi/2,
when n=2, the graph is symmetric about the line...

My assumtion is that when n is even- the graph cos(nx) is symmetric about the line
x=pi/2 and when n is odd- it's anty-symmetric...
Well, so we need to make the graph of f(x) in [pi/2 , pi ] symmetric about the line x=pi/2 and the odd coefficients will be zero? How can we prove it in a formal way?

Thanks a lot for your guidance!

12. Jan 15, 2010

### vela

Staff Emeritus
Look up how you prove that an odd function integrated over [-a,a] is zero. That's the strategy you want to take, I think. Use the same technique to show that the integral from 0 to pi is zero if f(x) has the right symmetry about x=pi/2, from which it should follow the integral from -pi to pi is zero. You'll also need to establish that cos(nx) has the required symmetry about x=pi/2, but that's probably just a trig identity.

Last edited: Jan 15, 2010
13. Jan 16, 2010

### WannaBe22

Thanks a lot man...Your guidance helped me a lot!