# Legendre Polynomials as an Orthogonal Basis

physconomic
Homework Statement:
If you had legendre polynomials defined in ##L^2([-1,1])##, with ##||Pn_2||^2=\frac{2}{2n+1}##, show that for any polynomial with p a set of ##L^2([-1,1])##, with degree less than n, we have the inner product of ##P_n## and p = 0. Find the polynomials ##P_0,... P_4##
Relevant Equations:
Integral form of inner product
If you had legendre polynomials defined in ##L^2([-1,1])##, with ##||Pn_2||^2=\frac{2}{2n+1}##, show that for any polynomial with p a set of ##L^2([-1,1])##, with degree less than n, we have the inner product of ##P_n## and p = 0. Find the polynomials ##P_0,... P_4##

Tried to use the integral form but getting no where with it

Gold Member
Your (source's) wording is confusing. Specifically: "show that for any polynomial with p a set of... less than ##n##". What is p here?
Does it mean: "with p a set of polynomials in ##L^2([-1,1])## with degree less than n"?
Does it mean" "with p in a set of Legendre polynomials in ..."?

Legendre polynomials are mutually orthogonal under the ##L^2([-1,1])## norm and indeed you can derive them fairly easily up to scalar multipliers by the Gram-Schmidt procedure applied to the power basis ##\{ 1,x,x^2,x^3, \ldots\}##.

Starting with ##P_0(x)=1## and ##P_1(x)=x## (already orthogonal), then ##p_2(x)=x^2## an
## \langle p_2,P_1\rangle = 0## but:
$$\langle p_2,P_0\rangle = \int_{-1}^1 1\cdot x^2 dx = 2/3$$
So we take:
## P_2 = p_2 -\frac{\langle P_0,p_2\rangle}{\langle P_0,P_0\rangle}\cdot P_0##
and that linear combination will be orthogonal to ##P_0##.
In this case with ##\langle P_0,P_0\rangle = 2## we get:
$$P_2 = p_2 - \frac{1}{3}P_0,\quad P_2(x) = x^2 -\frac{1}{3}$$
However with standard normalization the actual degree 2 Legrange polynomial is 3/2 times this one, namely ##P_2(x)=\frac{1}{2}(3x^2-1)##. (Standard normalization assures ##P_n(1)=1, P_n(-1)=(-1)^n##.)

That's all I can say without better context to understand the ?mis-worded? question.

physconomic