If $P_3$ is a vector space of third-degree polynomials.
It is known the basis for $P_3$ is ${( 1,x,x^2 , x^3})$
and $\langle p, q\rangle = \int_{0}^{1} p(x)q(x)\, dx.$ is a valid product on $P_3$
I am trying to use the Gram-Schmidt method to get a basis for $P_3$ which is orthonormal with respect to the
above inner product.
Even though I found partial solutions or similar problems the explanations are limited.
PS. I read the rules before posting my first question. Even though I found similar problems I didn't understand entirely the method and calculations.
Additional Sources
- the below exercise which has a partial solution, but I am not sure how to calculate the remaining values.
- this question which is similar but in $P_2$ Finding an orthonormal basis for the space $P_2$ with respect to a given inner product
I hope I did not violate any rule. It was my last hope to ask here since due to current conditions I can't ask my Teacher face to face.
Best Answer
Graham Schmidt.
Pick a vector, to make it a candidate for your first basis vector.
$w_0 = 1$
Normalize it. Since $\|w_0\| = 1$ we that step is already done.
$e_0 = w_0 = 1$
Your second basis vector.
$w_1 = x$
Subtract the projection of $e_1$ onto $x.$
$e_1^* = x - \langle e_1,x\rangle e_1$
$e_1^* = x - \int_0^1 x \ dx = x-\frac 12$
Normalize it...
$e_1 = \frac {e_1^*}{\|e_1^*\|}$
$\|e_1^*\|^2 = \langle e_1^*,e_1^*\rangle = \int_0^1 (x-\frac 12)^2 \ dx\\ \int_0^1 x^2 -x + \frac 14\ dx = \frac 13 - \frac 12 + \frac 14 = \frac 1{12}\\ e_1 = \sqrt {12} x - \sqrt 3$
$w_2 = x^2\\ e_2^* = w_2 - \langle e_0,w_2\rangle - \langle e_1,w_2\rangle$
Normalize it...
lather, rinse, repeat.