Posted by **mathstudent** on Wednesday, January 16, 2008 at 7:39pm.

A trigonmetric polynomial of order n is t(x) =

c0 + c1 * cos x + c2 * cos 2x + ... + cn * cos nx

+ d1 * sin x + d2 * sin 2x + ... + dn * sin nx

The output vector space of such a function has the vector basis:

{ 1, cos x, cos 2x, ..., cos nx, sin x, sin 2x, ..., sin nx }

Use the Gram-Schmidt process to find an orthonormal basis using the inner-product:

<f,g> = definite integral over [0,2*pi] of: f(x) * g(x) dx

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The book gives the following answer for the orthonormal basis.

g0 = 1/sqrt(2*pi)

g1 = 1/sqrt(pi) * cos x

gn = 1/sqrt(pi) * cos nx

I've done the first two vectors using Gram-Schmidt and my answers don't match:

Original basis vector u0 = 1

Orthogonal basis vector v0 = 1

Orthonormal basis vector g0 = v0/|v0| = sqrt(2/pi)

Original basis vector u1 = cos x

Orthogonal basis vector v1 = u1 - <u1, g0> * g0

= cos x - sqrt(2/pi) * sqrt(2/pi) = cos x - 2/pi

Orthonormal basis vector g1 = v1/|v1| = (cos x - 2/pi) / sqrt((pi^2 - 8)/(4*pi))

= (cos x - 2/pi) * sqrt((4*pi)/(pi^2 - 8))

What am I doing wrong?

- math -
**Count Iblis**, Wednesday, January 16, 2008 at 7:49pm
Original basis vector u0 = 1

Orthogonal basis vector v0 = 1

Orthonormal basis vector g0 = v0/|v0|

|V0|^2 = 2 pi --->

g0 = 1/sqrt(2 pi)

Original basis vector u1 = cos x

Orthogonal basis vector v1 = u1 - <u1, g0> * g0 = cos x because

<cos(x), 1> = 0

- math -
**mathstudent**, Thursday, January 17, 2008 at 12:30pm
Thanks Count Iblis!

I was mistakenly integrating with pi/2 instead of 2*pi and every time I redid the problem, I just remade the same mistake without noticing it.

Your help pointed out the issue. Thanks so much!

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