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Proving a set of functions is orthogonal

  1. Oct 16, 2013 #1
    03izT8l.png

    Why is the math in the red box necessary? According to this definition, it isn't:

    kziTaTs.png
     
  2. jcsd
  3. Oct 16, 2013 #2

    tiny-tim

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    hi ainster31! :smile:
    sorry, i don't understand your question :redface:

    the red box proves that (φ0, φn) = 0 (for n ≠ 0)
     
  4. Oct 16, 2013 #3
    According to definition 12.1.3, a set of real-valued functions can be proven to be orthogonal if (φm, φn) = 0. So why is it necessary to prove (φ0, φn) = 0?
     
  5. Oct 16, 2013 #4

    dextercioby

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    m=0 is contained as a particular case for arbitrary m and n. It's no need to make the particular case. The proof goes directly by putting cos a = Re (e^ia).
     
  6. Oct 16, 2013 #5
    So you're saying it was unnecessary?

    That went over my head.
     
  7. Oct 16, 2013 #6

    tiny-tim

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    because φo is a member of the set :smile:
     
  8. Oct 16, 2013 #7

    dextercioby

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    That's exactly what I meant.
     
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