Normalizing an Orthogonal Basis

[beetle2]

Homework Statement

I have used the gram schmidt process to find an orthogonal basis for {1,t,t^2}
which is
<br /> (1,x,x^2 - \frac{2}{3})<br />

How to i normalize these

Homework Equations

e_1=\frac{u_1}{|u_1|}

The Attempt at a Solution

e_1=\frac{1}{\sqrt{\int_{-1}^{1}f(1)g(1)}}=\frac{1}{\sqrt{2}}


e_2=\frac{x}{\sqrt{\int_{-1}^{1}f(x)g(x)}}=\frac{x}{\sqrt{\frac{2}{3}}}

e_3=\frac{x}{\sqrt{\int_{-1}^{1}f(x^2-3)g(x^2-3)}}=\frac{x^2}{\sqrt{\frac{-62}{45}}}

 

[foxjwill]

Just divide each by its norm.

 

[Mark44]

There is a problem with your third function. It is orthogonal to x, but not to 1. Each function has to be orthogonal to each other function in your set.
\int_{-1}^1 1 (x^2 - 2/3) dx~=~\left[\frac{x^3}{3} - (2/3)x\right]_{-1}^1~=~ -2/3

Also, is the function x^2 - 2/3 or x^2 - 3? You are using both in your calculations.

 

[Mark44]

One other thing. The norm of a function is the square root of the inner product of it and itself.

e_2=\frac{x}{\sqrt{\int_{-1}^{1}f(x)g(x)}}=\frac{x}{\sqrt{\frac{2}{3}}}

You shouldn't have f(x) and g(x) in there for <x, x>, since f(x) = g(x) = x. It should be like this:
e_2=\frac{x}{\sqrt{\int_{-1}^{1}x*xdx}}= ...

 

[beetle2]

thanks your right my u3 should be x^2-\frac{1}{3}
which makes
e_3=\frac{x^{2}-\frac{1}{3}}{\sqrt{\int_{-1}^{1}x^{2}-\frac{1}{3}*x^{2}-\frac{1}{3}dx}}= ...

= \frac{x^2-\frac{1}{3}}{\sqrt{frac{8}{45}}}

hows that look

The innner product of x^2-3 and \frac{1}{\sqrt{2}} = 0 so they're orthogonal

 

[Mark44]

Much better.

For the quantity in the radical, you left off the \ before frac in your LaTeX code. That's why it looks like it does. Should look like this:
\frac{x^2-\frac{1}{3}}{\sqrt{\frac{8}{45}}}

 

[beetle2]

Thanks for your help guys much appreciated