### Function Inner Product

Source: A linear algebra problem book

Problem:

Find the function f  ( member of span {1, sin x, cos x} ) that minimizes norm of ( sin 2x − f(x) ), where the norm comes from the inner product

< f , g > = integral over x from -pi to pi [ f(x)g(x) ]

Update (Oct 30, 2011):
Solution posted by Harsh Pareek (Graduate Student at UT Austin, CSE IITB 2011 Alumnus) in comments.

1. This is one of your easier ones, but it's so nice I couldn't help commenting.

First, you're asking us to find the point in P=span{1,sin x, cos x} closest to sin 2x.
Usually in vector spaces, the answer would be the projection of sin 2x on P. sin 2x is orthogonal to {1,sin x, cos x} so in this case we expect the element 0, and as we'll see the intuition goes through.

Let m = = norm(sin 2x)^2 + norm(f)^2 - 2

f(x) = A + B sin x + C cos x => = 0

So m is minimum when norm(f) is minimum, i.e. f=0.

In general, if you substitute sin 2x by g(x), our answer will be +sin x + cos x

2. @Harsh.. you probably tried to use latex.. Hence your comment is not well formed..

Let me try to decipher what you wrote:

First, you're asking us to find the point in P=span{1,sin x, cos x} closest to sin 2x.
Usually in vector spaces, the answer would be the projection of sin 2x on P. sin 2x is orthogonal to {1,sin x, cos x} so in this case we expect the element 0, and as we'll see the intuition goes through.

Let m = norm(sin2x - f(x))^2 = norm(sin 2x)^2 + norm(f)^2 - 2

f(x) = A + B sin x + C cos x

Hence, = 0 (Since, sin2x is orthogonal to all 1, sinx and cosx)

So m is minimum when norm(f) is minimum, i.e. f=0.

In general, if you substitute sin 2x by g(x), our answer will be the projection of g(x) onto P.

Is this what you meant? I think that is correct.