$\langle f,g\rangle = \int_a^b f(x)g(x) \,dx$, Show that $p_n$ is the monic polynomial of degree $n$ of smallest norm

linear algebranumerical methods

Let $p_0,p_1,p_2,\cdots$ be monic polynmials orthogonal with respect
to $\langle f,g\rangle = \int_a^b f(x)g(x) \,dx$, $p_i$ of degree $i$. Show that
$p_n$ is the monic polynomial of degree $n$ of smallest norm ($\lVert p\rVert =
\langle p,p\rangle^{1/2}$)

I'm studying numerical methods and we're seeing orthogonality of polynomials and projection of things onto subspaces. I need to solve the exercise above using these concepts. I don't know if I understand the exercise correctly. I think I must show that, given a degree $n$, then of all polynomials of this degree, the monic is the one with smallest norm. Am I rigth? I have no idea how to start.

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$p_i, i\leq n$ span the space of all polynomials of degree at most $n$. If $q$ is any monic polynomial of degree $n$ then $q-p_n$ is orthogonal to $p_n$ because degree of $q-p_n$ is at most $n-1$ and hence it belongs to the span of $p_i:i<n$. Hence $\|q\|^{2}=\|(q-p_n)+p_n\|^{2}=\|q-p_n\|^{2}+\|p_n\|^{2}$ which implies $\|q\|^{2} \geq \|p_n\|^{2}$.

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Algorithm for orthogonalizing polynomials with specific inner product

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..

Prove polynomial set $S=\{p_0(x), p_1(x), p_2(x), \cdots\}$ is linearly independent

a) To check for linear independence, suppose that

$$\sum_{i=0}^∞ a_ip_i(x)=0$$

where all but finitely many ai are zero. 1 We claim that all ai are zero. If this were not the case, we could pick the largest m such that am = 0. Now compare the coefficient of $x^m$ on both sides. On the left hand side it is am (since pm(x) = $x^m$ + . . .), while on the right hnd side it is 0. Hence am = 0, a contradiction.

b) The check that the pi’s span P(F): We use induction to prove that all polynomials of degree ≤ n are linear combinations of $p_0$(x), . . . , $p_n$(x). This is immediate for n = 0 since polynomials of degree 0 are constant, p(x) = $a_0$, while $p_0$(x) = 1, so p(x) = $a_0$$p_0$(x). Suppose teh claim is known up to degree n, and let p(x) be a polynomial of degree n + 1. Then p(x) = $a_{n+1}$$x_n$ + . . . + $a_{1}x$ + $a_0$. But $p_{n+1}$(x) = $x_{n+1}$ + . . . + $b_1$x + $b_0$.

This shows that q(x) = p(x) − $a_{n+1}$ $p_{n+1}(x)$ is a polynomial of degree n, hence by induction is a linear combination of p0(x), . . . , pn(x). It follows that p(x) = $a_{n+1}$ $p_{n+1}$(x) + q(x) is a linear combination of $p_0$(x), . . . , $p_{n+1}$(x)

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Algorithm for orthogonalizing polynomials with specific inner product

Prove polynomial set $S=\{p_0(x), p_1(x), p_2(x), \cdots\}$ is linearly independent

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