**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.

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

ReplyDeleteFirst, 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

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

ReplyDeleteLet 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.