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Fourier Series

  1. Aug 3, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm calculating the coefficients for the Fourier series and I got to part where I can't simplify an any further but I know I have to.

    [tex]a_n = \frac{1}{2π}\Big[\frac{cos(n-1)π}{n-1}-\frac{cos(n+1)π}{n+1}-\frac{1}{n-1}+\frac{1}{n+1}\Big][/tex]


    2. Relevant equations


    3. The attempt at a solution
    I have found an example where the above line goes straight to:

    [tex]a_n=-\frac{1}{2π}(1+(-1)^n)\frac{2}{(n-1)(n+1)}[/tex]

    I don't know how you get from one line to the next. If anyone could explain this to me, it would be greatly appreciated.
     
  2. jcsd
  3. Aug 3, 2016 #2

    Orodruin

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    What is ##\cos(n\pi)##?
     
  4. Aug 3, 2016 #3
    -1 for n = odd and 1 for n = even.
     
  5. Aug 3, 2016 #4

    Orodruin

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    So how can you write this on a more concise form?
     
  6. Aug 3, 2016 #5
    [tex](-1)^n[/tex]
     
  7. Aug 3, 2016 #6

    Orodruin

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    So how can you apply this to your problem?
     
  8. Aug 3, 2016 #7
    I've tried subbing (-1)n in for both of the cos parts but it doesn't simply to what is given.
     
  9. Aug 3, 2016 #8

    Ray Vickson

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    Yes it does; start again.
     
  10. Aug 3, 2016 #9
    When I sub in and simplify I'm getting:

    [tex]a_n = \frac{1}{2π}\Big[\frac{(-1)^n + (-1)^n}{(n-1)(n+1)}\Big][/tex]
     
    Last edited: Aug 3, 2016
  11. Aug 3, 2016 #10

    Orodruin

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    Observe that you have ##\cos[(n-1)\pi]## and that you have some other terms too.
     
  12. Aug 3, 2016 #11
    [tex]a_n = \frac{1}{2π}\Big[\frac{(-1)^n}{n-1}-\frac{(-1)^n}{n+1}-\frac{1}{n-1}+\frac{1}{n+1}\Big][/tex]

    Is this correct?
     
  13. Aug 3, 2016 #12

    Orodruin

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    Why don't you try inserting a few values to check?
     
  14. Aug 3, 2016 #13
    What would I be checking for?
     
  15. Aug 3, 2016 #14

    Orodruin

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    That the values that come out of your formula agree with the original expression.
     
  16. Aug 3, 2016 #15
    If I put integers in for n will the result not be a single number?
     
  17. Aug 3, 2016 #16

    Orodruin

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    Yes, but it will be different for different n of course. I said "inserting a few values".
     
  18. Aug 3, 2016 #17
    [tex]n=0: a_n = 0[/tex]

    [tex]n=1: a_n = \frac{1}{2π} [/tex](has zero dividers though)

    [tex]n=2: a_n = \frac{1}{12π}[/tex]
     
  19. Aug 3, 2016 #18

    Orodruin

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    And if you use the same values in the original expression?
     
  20. Aug 3, 2016 #19
    But do I not want there to be n's in my result?
     
  21. Aug 4, 2016 #20

    Orodruin

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    Yes. But you want to check that you got the right result first. You asked:
    If there is any n for which your new expression does not equal the old, then it is not correct.
     
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