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

 

[jambaugh]

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.

 

[jambaugh]

Immediate followup. Here's a point to remember which may help understand the question and its solution:
The Legendre polynomials up to degree n form a basis for the space of all polynomials up to degree n.

Note in my prior post that while $p_2: p_2(x)=x^2$ was not orthogonal to $P_0$, the constructed $P_2$ was orthogonal to both $P_0$ and $P_1$ (and thus to any linear combination of them >wink wink<).