Can somebody check my work on this Fourier Series problem?

  • #1
169
1

Homework Statement


pix.jpg


Homework Equations



img1.gif

img3.gif



The Attempt at a Solution



Since P=2L, L=1 ?

a_o = 1/2 [ ∫(from -1 to 0) -dx + ∫(from 0 to 1) dx ] = 1/2 [ (0-1) + (1-0) ] = 1/2(0) = 0

a_n = - ∫ (from -1 to 0) cosnπx dx + ∫ (from 0 to 1) cosnπx dx = 0

b_n = - ∫ (from -1 to 0) sinnπx dx + (from 0 to 1) sinnπx dx = 2/nπ - 2/nπ*(cosnπ) = 2nπ / (1-cosnπ) = { 4/nπ n is odd ; 0 is even



The problem is my teacher has this as his answer:
pic2.jpg


Am i doing something incorrectly?
 
Last edited:

Answers and Replies

  • #2
Dr. Courtney
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Why not check by graphing the original function and the Fourier series?
 
  • #3
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Why not check by graphing the original function and the Fourier series?
I understand you can check through graphing, but I just wanted some verification that my math is correct?
 
  • #4
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Wait L is suppose to equal 2 correct? Not 1?
 
  • #5
Ray Vickson
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Wait L is suppose to equal 2 correct? Not 1?
If you mean that one period of the function goes from ##-L## to ##+L## then yes, of course ##L = 2##. That was implied in the question.
 
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  • #6
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If you mean that one period of the function goes from ##-L## to ##+L## then yes, of course ##L = 2##. That was implied in the question.
I figured...
For my b_n I got 2/(n*pi) [1 - cos(n*pi / 2 ).

Is this correct?
 
  • #7
vela
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I figured...
For my b_n I got 2/(n*pi) [1 - cos(n*pi / 2 ).

Is this correct?
Why don't you do as suggested and graph the resulting series? That'll tell you immediately if you got the right series. If you still want someone to check your work, you need to show it. Just posting the answer is next to useless.
 

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