Solving Basis Functions Homework w/ Constants A_n & B_n

[Fizz_Geek]

Homework Statement

Given x in the interval [0, [tex]\pi[/tex]], let [tex]\phi[/tex][tex]_{0}[/tex](x) = 1, and [tex]\Phi[/tex][tex]_{n}[/tex] (x) = sin ((2n-1)x).

Show that there are constants:
{A[tex]_{n}[/tex]}[tex]^{n=0}_{\infty}[/tex] and {B[tex]_{n}[/tex]}[tex]^{n=0}_{\infty}[/tex]

such that:

[tex]\sum[/tex][tex]^{n=0}_{\infty}[/tex]A[tex]_{n}[/tex][tex]\phi[/tex][tex]_{n}[/tex]=[tex]\sum[/tex][tex]^{n=0}_{\infty}[/tex]B[tex]_{n}[/tex][tex]\phi[/tex][tex]_{n}[/tex]

But A[tex]_{n}[/tex] [tex]\neq[/tex] B[tex]_{n}[/tex] [tex]\forall[/tex]n


All the n's should be subscripts. None are powers.

Relevant equations

I really don't know where to start. Any push in the right direction would be greatly appreciated.

 

[JonF]

note [tex]\Phi[/tex]_n won't give you cyclic results unless your x is a rational multiple of pi, making this a bit tricky.

Luckily -1<=[tex]\Phi[/tex]_n <=1.

So I would define try something like this.

Define A_n = a_n , where a_n *[tex]\Phi[/tex]_n = (-1)^n/n^2
This would mean the A_n series converges.

Then i would define B_n so it reorders the terms of your A_n series. So it would converge and be equal.

 

[Fizz_Geek]

Thanks for the reply!

I thought to do something like that, and I can understand that both series would converge, but how would they be equal?

 

[JonF]

They would only be equal as you took the limit to infinity, which is all your proof requires. Since the A_n series is absolutely convergent all of it's rearrangements will be equal. Note, you need to make A_n absolutely convergent, or this won't be true.