Proving a set of functions is orthogonal

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The discussion centers on the necessity of proving that (φ0, φn) = 0 for orthogonality of a set of functions, as defined in section 12.1.3. Participants debate whether this proof is required since m=0 is a specific case of arbitrary m and n. One contributor argues that the proof can be simplified by using the relationship cos a = Re(e^ia). Ultimately, the consensus leans towards the idea that proving (φ0, φn) = 0 is unnecessary, as φ0 is already included in the general case. The conversation highlights the nuances of mathematical proof and definitions in the context of orthogonality.
ainster31
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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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