gram-schmidtlegendre polynomialsreal-analysis
If we start with the polynomials $1, x, …, x^n$ on $[-1,1]$ and apply the Gram-Schmidt process with $\lt f\cdot g\gt =\int\limits_{-1}^{1} f(x)g(x)dx$. How can I prove that the new polynomials ${L_0,…,L_n}$ are Legendre polynomials:
$$L_n(x)=\dfrac{1}{2^nn!}\dfrac{d^n}{dx^n}[(x^2-1)^n]$$
with the Gram-Scmidt process?
This an answer to the version of the question given in Mike's comment. Note first that for any polynomials $p$ and $q$ we have $(xp,q)=(p,xq)$. Second, note that $P_n$ is orthogonal to all polynomials of degree less than $n$.
So if $m \le n-2$, then $$ (xP_n,P_m) = (P_n,xP_m) = 0. $$
But we know that $xP_n$ must be a linear combination of $P_0,\ldots,P_n,P_{n+1}$ (because this holds for any polynomial of degree at most $n+1$), and the coefficient of $P_m$ in this expansion if $(xP_,P_m)/(P_m,P_m)$. Therefore $xP_n$ is a linear combination of $P_{n-1}$, $P_n$ and $P_{n+1}$.
Finally note that this works for any family of orthogonal polynomials, not just Legendre polynomials.
Consider the following diagram, courtesy of mathinsight.org:
You can think of $(a \cdot u) u$ as the piece of $a$ that is in the direction of $u$. The part that is left over, $a - (a \cdot u) u$, must naturally be the missing side of the triangle, and hence is perpendicular to $u$. So at each step of the Gram-Schmidt process, the formula
$$ v_{n+1} = a - \sum_{j=1}^n \langle a, u_j \rangle u_j, \quad u_{n+1} = v_{n+1}/ \|v_{n+1} \|$$
does the following: it first subtracts all the pieces of $a$ that are in the same direction as all the $u_j$, then it renormalizes. The resulting vector must be orthogonal to all the $u_j$'s since you just subtracted out all the pieces that were not perpendicular.
Best Answer
Gram-Schmidt for the case of the Legendre polynomials can be naively performed by defining $p_n$ to be the $L^2$ coprojection of $x^n$ onto the span of $1,x,\dots,x^{n-1}$.
Computing this coprojection using Gram-Schmidt gives us the recursion:
$$p_n(x)=x^n-\sum_{k=0}^{n-1} \frac{\int_{-1}^1 p_k(y) y^n dy}{\int_{-1}^1 p_k(y)^2 dy} p_k (x).$$
This works. But we can do better by choosing to project $x p_{n-1}(x)$ instead of $x^n$. Now $x p_{n-1}(x)$ still has degree $n$ and the leading coefficients are the same, so this will give us the same as projecting $x^n$ would. Thus we write:
$$p_n(x)=x p_{n-1}(x)-\sum_{k=0}^{n-1} \frac{\int_{-1}^1 y p_k(y) p_{n-1}(y) dy}{\int_{-1}^1 p_k(y)^2 dy} p_k(x).$$
Why is this any better? Because actually only two terms of the sum are nonzero! This is because $p_k$ is orthogonal to $p_j$ for all $j<k$, and is therefore orthogonal to all polynomials of degree $d<k$ as well. Therefore the inner product $(yp_k,p_{n-1})$ is zero as soon as $k+1<n-1$. (It is precisely the trick $(p_k,xp_{n-1})=(xp_k,p_{n-1})$ that is specific to the world of orthogonal polynomials. Essentially everything else we did here would work in an abstract inner product space.) Thus we get the "three term recurrence relation":
$$p_n(x)=x p_{n-1}(x)-\frac{\int_{-1}^1 y p_{n-1}(y)^2 dy}{\int_{-1}^1 p_{n-1}(y)^2 dy} p_{n-1}(x) - \frac{\int_{-1}^1 y p_{n-2}(y) p_{n-1}(y) dy}{\int_{-1}^1 p_{n-2}(y)^2} p_{n-2}(x).$$
By proving formulae for these integrals by induction, you can derive a general form for this three term recurrence relation. Standardizing by taking $P_n(x)=\frac{p_n(x)}{p_n(1)}$ gives the standard Legendre polynomials. (Note that the Legendre polynomials are not orthonormal in the sense of $L^2$.)
Now given the resulting recurrence for the Legendre polynomials, which turns out to be
$$(n+1) P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$$
you can prove this recurrence given another expression for the Legendre polynomials, e.g. a Rodriguez formula like the one you posted. It then follows that the two are the same once you check that they satisfy the same initial conditions i.e. $P_0=1,P_1=x$.