Expanding Fourier Series to Different Periods

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In summary: I'll try to do the same with the first problem, and hopefully I'll receive the same success...Thanks!In summary, the function is periodic with period 3pi/2 and not 3pi. How come? sorry it is periodic with period 2pi.
  • #1
WannaBe22
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Homework Statement


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 continuous 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...


Homework Equations


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...

About 2- I'll be glad to receive some guidance in it...

Thanks a lot!
 

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  • #2
WannaBe22 said:
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...

the function is periodic with period 3pi/2 and not 3pi
 
  • #3
How come?
 
  • #4
sorry it is periodic with period 2pi
-pi/2 to pi/2
pi/2 to 3pi/2
completes a full revolution
so 2pi
 
  • #5
Welll...So guess I had a calculation mistake...
What about the second question?

Thanks a lot!
 
  • #6
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.
 
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  • #7
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
Try focusing on the interval [itex][0,\pi][/itex] and consider the symmetry of [itex]sin(nx)[/itex] about [itex]x=\pi/2[/itex].
 
  • #9
It's pretty obvious that in the first one we need to expand the function into an even one... But I really 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
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
Well, I'll try to answer:
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...
when n=3, the graph is anty-symmetric about this 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
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.
 
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  • #13
Thanks a lot man...Your guidance helped me a lot!
 

FAQ: Expanding Fourier Series to Different Periods

What is a Fourier series?

A Fourier series is a mathematical tool used to represent a periodic function as a sum of simpler trigonometric functions. It is named after French mathematician Joseph Fourier and is commonly used in signal processing, engineering, and physics.

How is a Fourier series calculated?

A Fourier series is calculated by finding the coefficients of the trigonometric functions in the series through integration. These coefficients can then be used to construct the series, which can approximate the original function with increasing accuracy as more terms are added.

What is the importance of Fourier series in science?

Fourier series are important in science because they allow us to analyze and understand complex periodic phenomena in terms of simpler trigonometric functions. This can be applied to a variety of fields such as signal processing, heat transfer, and quantum mechanics.

Can a non-periodic function be represented by a Fourier series?

No, a Fourier series can only represent periodic functions. However, there are extensions of Fourier series, such as the Fourier transform, which can be used to analyze non-periodic functions.

How does the number of terms in a Fourier series affect its accuracy?

The accuracy of a Fourier series increases as more terms are added, but it will never perfectly match the original function. The rate of convergence also depends on the smoothness of the function being represented. In general, the more discontinuities or sharp changes in the function, the more terms will be needed for a good approximation.

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