Proving a set of functions is orthogonal

[ainster31]

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

 

[tiny-tim]

hi ainster31!

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

sorry, i don't understand your question …

the red box proves that (φ₀, φ_n) = 0 (for n ≠ 0)

 

[ainster31]

hi ainster31!


sorry, i don't understand your question …

the red box proves that (φ₀, φ_n) = 0 (for n ≠ 0)

According to definition 12.1.3, a set of real-valued functions can be proven to be orthogonal if (φ_m, φ_n) = 0. So why is it necessary to prove (φ₀, φ_n) = 0?

 

[dextercioby]

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).

 

[ainster31]

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.

 

[tiny-tim]

According to definition 12.1.3, a set of real-valued functions can be proven to be orthogonal if (φ_m, φ_n) = 0. So why is it necessary to prove (φ₀, φ_n) = 0?

because φ_o is a member of the set

 

[dextercioby]

So you're saying it was unnecessary?[...]

That's exactly what I meant.