deriving an orthogonal base for polynomials of degree 3

Question

I am given the inner product space $\mathcal{P}_3$ with the ordered base 1, $X$, $X^2$. I have to apply the Gram-Schmidt process to derive an orthogonal base for $\mathcal{P}_3$. I got 1, $X$, $X^2$ - 2/3.

I used $\langle P(X),Q(X)\rangle$ = $\int\limits_{-1}^{1}$$P(t)Q(t)dt$ to derive this base and to check if these are orthogonal, but they aren't.

What could I have done wrong?

Thanks for your help..

Answer 1

Define $v_n(x) = x^n$, and

$$ {\rm proj}_u(v) = \frac{\langle u, v\rangle}{\langle u, u\rangle} u = \frac{\int_{-1}^{1}{\rm d} x~u(x)v(x)}{\int_{-1}^{1}{\rm d} x~u^2(x)} u(x) $$

applying the Gram-Schmidt process will lead to

\begin{align} & u_0(x) = v_0(x) = 1 \\ & u_1(x) = v_1(x) - {\rm proj}_{u_0}(v_1) = x - \underbrace{\frac{\int_{-1}^{1}{\rm d} x~x}{\int_{-1}^{1}{\rm d} x}}_{0} = x\\ & u_2(x) = v_2(x) - {\rm proj}_{u_0}(v_2) - {\rm proj}_{u_1}(v_2) = x^2 - \underbrace{\frac{\int_{-1}^{1}{\rm d} x~x^2}{\int_{-1}^{1}{\rm d} x}}_{1/3} - \underbrace{\frac{\int_{-1}^{1}{\rm d} x~x^3}{\int_{-1}^{1}{\rm d} x~x^2}}_{0} x =x^2-\color{red}{\frac{1}{3}} \end{align}

Answer 2

Since you didn't show us your computations, it's hard to tell. But note that both $1$ and $X^2$ are orthogonal to $X$. So, find an orthogonal basis of $\langle1,X\rangle$ and then add $X$ to it.