Proving a set of functions is orthogonal

  • Thread starter ainster31
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Main Question or Discussion Point

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

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Answers and Replies

  • #2
tiny-tim
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hi ainster31! :smile:
Why is the math in the red box necessary? According to this definition, it isn't:

kziTaTs.png
sorry, i don't understand your question :redface:

the red box proves that (φ0, φn) = 0 (for n ≠ 0)
 
  • #3
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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)
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?
 
  • #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).
 
  • #5
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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.
So you're saying it was unnecessary?

The proof goes directly by putting cos a = Re (e^ia).
That went over my head.
 
  • #6
tiny-tim
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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?
because φo is a member of the set :smile:
 
  • #7
dextercioby
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So you're saying it was unnecessary?[...]
That's exactly what I meant.
 

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