# A problem on finding orthogonal basis and projection

Use the inner product <f,g> = integral f(x) g(x) dx from 0 to 1 for continuous functions on the inerval [0, 1]

a) Find an orthogonal basis for span = {x, x^2, x^3}

b) Project the function y = 3(x+x^2) onto this basis.
---------------------------------------------------------
I know the following:
Two vectors are orthogonal if their inner product = 0
A set of vectors is orthogonal if <v1,v2> = 0 where v1 and v2 are members of the set and v1 is not equal to v2
If S = {v1, v2, ..., vn} is a basis for inner product space and S is also an orthogonal set, then S is an orthogonal basis.

Regarding projection, I know that if W is a finite dimensional subspace of an inner product space V and W has an orthogonal basis S = {v1, v2, ..., vn} and that u is any vector in V then,
projection of u onto W = <u, v1> v1/||v1||^2 + <u, v2> v2/||v2||^2 + <u, v3> v3/||v3||^2 + ...<u, vn> vn/||vn||^2

I can calculate integrals, but I really do not know how to fit all these together for this problem. I am not sure how to start.

Sorry, but still I did not understand. Can you at least give major steps?

What are the column vectors of S ?

Is this correct for part a?

If r + s = n, then
<x^r, x^s> = ∫_0^1 x^n dx = x^(n+1)/(n+1) (0, 1)
= 1^(n+1)/(n+1) - 0^(n+1)/(n+1)
= 1/(n+1)

Let the orthogonal basis = {f1, f2, f3}

Put f1 = 1
f2 = t^2- (<t^2,t>)/(<t,t>)*t = t^2- (1/4)/(1/3)*t = t^2 - 3t/4
f3 = t^3- (<t^3,t>)/(<t,t>)*t - (<t^3,t^2>)/(<t^2,t^2>)*t^2
= t^3 - (1/5)/(1/3)*t - (1/6)/(1/5) * t^2 = t^3-3t/5-(5t^2)/6

The orthogonal basis = {1, t^2 - 3t/4, t^3-3t/5-(5t^2)/6}

--------------
How do I solve part b ?

spamiam
Let the orthogonal basis = {f1, f2, f3}

Put f1 = 1
Shouldn't that be $f_1 = t$ since you're orthogonalizing the basis $\{x, x^2, x^3\}$? You should set $f_1$ equal to the first vector of the original basis. Actually, it looks like you took $f_1 = t$ for the calculation of $f_2$...
f2 = t^2- (<t^2,t>)/(<t,t>)*t = t^2- (1/4)/(1/3)*t = t^2 - 3t/4
f3 = t^3- (<t^3,t>)/(<t,t>)*t - (<t^3,t^2>)/(<t^2,t^2>)*t^2
= t^3 - (1/5)/(1/3)*t - (1/6)/(1/5) * t^2 = t^3-3t/5-(5t^2)/6
Make sure you're following the formula I linked. You should be taking the projection along the orthogonal vectors you calculated previously, so you should have
$$f_3 = t^3 - \frac{\langle t^3, f_1 \rangle}{\|f_1\|^2} f_1 - \frac{\langle t^3, f_2 \rangle}{\|f_2\|^2} f_2$$

I think you calculated $f_2$ correctly, but not $f_3$.
How do I solve part b ?
You wrote the formula for this in your first post. Once you finish part (a), you can use it.