Orthogonality of sin functions?

MarkH747 · #1 $Old$ April 30th, 2009, 09:21 AM

Hey, I'm looking through a solution to a problem and I'm not sure how this line went to the next.

Apologies for the writing. (Not my solution :P).

The reason it gave is (orthogonality of sin functions) which I don't understand.

Dn is a constant btw.

Thanks a lot.

Mark.

Chris L T521 · #2 $Old$ April 30th, 2009, 09:37 AM

Quote:

Originally Posted by MarkH747 $View Post$

Hey, I'm looking through a solution to a problem and I'm not sure how this line went to the next.

Apologies for the writing. (Not my solution :P).

The reason it gave is (orthogonality of sin functions) which I don't understand.

Dn is a constant btw.

Thanks a lot.

Mark.

It should be noted that $\sum_{n=1}^{\infty}D_n \sin\!\left(n\pi x\right)$ is a Fourier Sine Series, where the coefficient is $D_n=\frac{2}{L}\int_0^L f\!\left(x\right)\sin\!\left(n\pi x\right)\,dx$ . From what I see, $0<L<1$ and $f\!\left(x\right)=x(2-x)$ , so it turns out that $D_n=2\int_0^1x\left(2-x\right)\sin\!\left(n\pi x\right)\,dx$

Does that make sense?

HallsofIvy · #3 $Old$ April 30th, 2009, 09:46 AM

The "orthogonality of sine functions" refers to the fact that $\int_0^{2\pi} sin(nx)sin(mx)dx= 0$ as long as $m\ne n$ . You can prove that by using the trig identity sin(nx)sin(mx)= (1/2)[cos((n+m)x+ cos((n-m)x)].

The term "orthogonality" comes from thinking of integrable functions, periodic on $[0, 2\pi]$ as a vector space having {sin x, cos x, sin 2x, cos 2x, ...} as "basis" and inner product $<f, g>= \int_0^{2\pi} f(x)g(x)dx$ .

MarkH747 · #4 $Old$ April 30th, 2009, 11:48 AM

Thanks a lot. Makes a bit more sense now.