gram-schmidtinner-productslinear algebra
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}$$P(t)Q(t)dt$ to derive this base and to check if these are orthogonal, but they aren't.
What could I have done wrong?
Thanks for your help..
Here are nice notes for deriving some types of orthogonal polynomials. The polynomials you are trying to derive are knon as Hermite polynomials.
I can't tell from what you said what you implemented, this is an example called [Chebfun]. 1. It'd be better if you posted your code. Note this is some code for the Legendgre polynomials with pseudo-code.
x = chebfun('x',[-1 1]);
w = exp(pi*x);
N = 5;
P = OrthPoly(w,N);
function P = OrthPoly(w,N)
if isnumeric(w), w = chebfun(w,[-1 1]); end
d = w.ends; % the domain
x = chebfun('x',d); % linear chebfun
P = chebfun(1./sqrt(sum(w)),d); % the constant (normalised)
for k = 1:N;
xk = x.*P(:,k);
P(:,k+1) = xk;
for j = 1:k % Subtract out the components
C = sum(w.*xk.*P(:,j));
P(:,k+1) = P(:,k+1) - C*P(:,j);
end
P(:,k+1) = P(:,k+1)./sqrt(sum(w.*P(:,k+1).^2)); % normalise
end
end
It's actually the lanczos iteration..
Note this is from Trefethen..
Best Answer
Define $v_n(x) = x^n$, and
$$ {\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}