1. ## Orthogonal to orthonormal

I hava a orthogonal basis {1, t, $\displaystyle 3t^2-1$ , $\displaystyle 5t^3-3t$}

and this normalized is

{1,t, $\displaystyle 1/2(3t^2-1)$, $\displaystyle 1/2(5t^3-3t)$}

I dont understand how they have done this. Any help would be much appreciated. Thanks.

I hava a orthogonal basis {1, t, $\displaystyle 3t^2-1$ , $\displaystyle 5t^3-3t$}

and this normalized is

{1,t, $\displaystyle 1/2(3t^2-1)$, $\displaystyle 1/2(5t^3-3t)$}

I dont understand how they have done this. Any help would be much appreciated. Thanks.
You should tell us what interval $\displaystyle (a,b)$ these are supposed to be orthonormal over.

The usual norm for real functions over such an interval is:

$\displaystyle \|f\|_{_2}=\left[ \int_a^b f(x)^2\;dx \right]^{1/2}$

If you have not made any typos I could work out what interval is implied by the given normalisations, but then you could just tell us.

CB

3. In any case, once you have an orthogonal set of vectors, you can "normalize" them by dividing each vector by its length.

Apparently, your interval, [a, b], is such that "1" and "t" already have length 1 while "$\displaystyle 3t^2- 1$" and "$\displaystyle 5t^3- 3t$" both have length 1/2.

4. interval 1 and -1

interval 1 and -1
Then we already have a problem since:

$\displaystyle \left[\int_{-1}^1 1^2\;dt\right]^{1/2}=\sqrt{2}$

so normalising $\displaystyle f(t)=1$ gives $\displaystyle \widehat{f}(t)=\frac{f(t)}{\|f\|_2}=1/\|1\|_2=1/\sqrt{2}$

CB

I hava a orthogonal basis {1, t, $\displaystyle 3t^2-1$ , $\displaystyle 5t^3-3t$}
{1,t, $\displaystyle 1/2(3t^2-1)$, $\displaystyle 1/2(5t^3-3t)$}