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Orthogonality relations of functions e^(2 pi i n x)

  1. Oct 26, 2006 #1
    I know that the functions [tex]e^{2 \pi inx}[/tex] for [tex]n \in \mathbb{Z}[/tex] are a base in the space of functions whith period 1. How do I derive the orthogonality relations for these functions?
     
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  3. Oct 26, 2006 #2

    HallsofIvy

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    By using the definition of "orthogonality"! For functions in L2 (square integrable functions- [itex]\int f^2(x)dx[/itex] exists), the functions over which the Fourier series are defined, the inner product is [itex]\int f(x)g(x)dx[/itex]. To show that [itex]e^{2\pi inx}[/itex] for n any integer are orthogonal show that
    [itex]\int e^{2\pi inx}e^{2\pi imx} dx[/itex] is 0 as long as [itex]m \ne n[/itex].
     
  4. Oct 30, 2006 #3
    What limits should I use for integration?
     
  5. Oct 30, 2006 #4
    I don't get it. The integral is never zero with m different from n. I guess I should use some weight function?
     
  6. Oct 30, 2006 #5

    OlderDan

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    Looks to me like -1/2 to +1/2 will do nicely.
    Can you show an example?
     
  7. Oct 31, 2006 #6
    [tex]\int_{- \frac{1}{2}}^{\frac{1}{2}}f_n^*(x)f_m(x)dx = \int_{- \frac{1}{2}}^{\frac{1}{2}}e^{-2 \pi inx}e^{2 \pi imx}dx = \int_{- \frac{1}{2}}^{\frac{1}{2}}e^{2 \pi ix(m-n)}dx = [/tex]

    [tex]= \left[ \frac{e^{2 \pi ix(m-n)}}{2 \pi i(m-n)} \right]_{- \frac{1}{2}}^{\frac{1}{2}} = \frac{e^{\pi i (m-n)} - e^{- \pi i (m-n)}}{2 \pi i (m-n)} = \frac{1}{\pi (m-n)} \sin{\pi (m-n)} = 0[/tex]

    Because m and n are integers. But this is only a solution for [-1/2,1/2] ? What if the interval is [0,1]? The I get

    [tex]I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}[/tex]

    and this is only zero for m = n...

    So this shows that the functions are orthogonal only on the interval [-1/2,1/2] and this is my solution. Am I right?
     
    Last edited: Oct 31, 2006
  8. Oct 31, 2006 #7

    OlderDan

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    Factor an [tex]e^{\pi i (m-n)}[/tex] out of your I. Does that help?
     
  9. Oct 31, 2006 #8
    Then I is zero if [tex]e^{\pi i (m-n)} = 0[/tex] or if [tex]-\pi i(m-n) = 2[/tex].
    The first one gives [tex]m-n = \frac{2k+1}{2}[/tex] and the second one gives [tex]m - n = \frac{2i}{\pi}[/tex]?
     
  10. Oct 31, 2006 #9

    OlderDan

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    OK. This is it.

    [tex]I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}[/tex]

    [tex] I = e^{\pi i(m - n)} \left[ {\frac{{e^{\pi i(m - n)} - e^{ - \pi i(m - n)} }}{{2\pi i(m - n)}}} \right] = e^{\pi i(m - n)} \left[ {\frac{{\sin \pi (m - n)}}{{\pi (m - n)}}} \right] [/tex]

    Compare this to your result for the [-1/2, 1/2] limits. In the form you had it, you cannot say I is zero when m = n; it is indeterminate. In this form, you can see the sine term is zero, but (sinx)/x is 1 for x = 0. When m = n the exponential is 1 and (sinx)/x is 1.
     
    Last edited: Oct 31, 2006
  11. Oct 31, 2006 #10
    So how do I do this on a general interval [a,b] where b - a = 1? How should I interpret the problem "Derive the orthogonality relations for these functions"?
     
  12. Oct 31, 2006 #11

    OlderDan

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    I think the factoring step I did will work for any interval. Give it a try.
     
  13. Oct 31, 2006 #12
    So, if I use a general interval [a,b] with b-a=1, I get

    [tex]I = e^{\pi i (m-n)(2a+1)} \frac{\sin{\pi (m-n)}}{\pi (m-n)} = \left\{\begin{array}{cc}0,&\mbox{ if }m \ne n\\1, & \mbox{ if } m = n\end{array}\right[/tex]

    and everyone is happy.
     
  14. Oct 31, 2006 #13

    OlderDan

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    I can't speak for everyone, but it works for me.
     
  15. Oct 31, 2006 #14
    Thanks for your help. =)
     
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