Proving a set of functions is orthogonal

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Discussion Overview

The discussion revolves around the necessity of proving that a specific function, φ0, is orthogonal to other functions, φn, within a set of real-valued functions. Participants are examining the implications of a definition related to orthogonality and the relevance of particular cases in mathematical proofs.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question the necessity of proving (φ0, φn) = 0 for n ≠ 0, suggesting that it may not be required by the definition of orthogonality.
  • Others argue that since φ0 is a member of the set, it is necessary to establish its orthogonality with other functions.
  • One participant proposes that the proof can be simplified by using the relationship cos a = Re (e^ia), implying that a direct approach may suffice.
  • There is confusion expressed by participants regarding the clarity of the argument and the implications of the proof's structure.

Areas of Agreement / Disagreement

Participants exhibit disagreement on whether proving (φ0, φn) = 0 is necessary, with some asserting it is required and others suggesting it is not. The discussion remains unresolved regarding the necessity of this proof.

Contextual Notes

Participants reference a specific definition (12.1.3) related to orthogonality, but there is no consensus on how it applies to the case of φ0. The discussion highlights potential ambiguities in the proof process and the interpretation of mathematical definitions.

ainster31
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03izT8l.png


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

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hi ainster31! :smile:
ainster31 said:
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)
 
tiny-tim said:
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?
 
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).
 
dextercioby said:
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?

dextercioby said:
The proof goes directly by putting cos a = Re (e^ia).

That went over my head.
 
ainster31 said:
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:
 
ainster31 said:
So you're saying it was unnecessary?[...]

That's exactly what I meant.
 

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