sequence genration

Question

I am having difficulty -normalizing sequence $1,x,x^3$ in $L^2([-1,0]\cup[1,2],dx)$ So I begin by choosing basis $1,x,x^2...$ and i need to use Gram-Schmidt? or how should i approach this? I have attempted to integrate x from -1 to 0 and add integral from 1 to 2, but I don't think I understand the problem

Answer 1

Let $f_n = x^n$ for $n=0,1,2,\ldots$. Define $u_1=f_1$ and $$u_{n+1} = f_{n+1} - \sum_{k=0}^n \frac{\langle f_{n+1},f_k\rangle}{\langle f_k,f_k\rangle}f_k,$$ where $$\langle f,g\rangle = \int_{[-1,0]\cup[1,2]}f(x)g(x)\ \mathsf dx. $$ Then $\{u_n\}$ is an orthogonal sequence ($\langle u_i,u_j\rangle=0$ for $i\ne j$) with the span of $\{u_0, \ldots, u_n\}$ equal to the span of $\{f_0,\ldots, f_n\}$. To normalize, set $$e_n = \frac{u_n}{\|f_n\|_2} $$ where $$\|f\|_2 = \int_{[-1,0]\cup[1,2]}f(x)^2\ \mathsf dx. $$ For example, $$u_2 = x - \frac{\int_{-1}^0 x\ \mathsf dx + \int_{1}^2 x\ \mathsf dx}{\int_{-1}^0 \mathsf dx + \int_{-1}^0 \mathsf dx}\cdot 1 = x-\frac12, $$ and $$\|u_2\|_2 = \int_{-1}^0 \left(x-\frac12\right)^2\ \mathsf dx + \int_{1}^2 \left(x-\frac12\right)^2\ \mathsf dx = \frac{13}6, $$ so $$e_2 = \frac6{13}\left(x-\frac12\right). $$