Orthogonality in Legendre polynomials

[Wingeer]

Homework Statement

There is a recursion relation between the Legendre polynomial. To see this, show that
the polynomial $$x p_k$$ is orthogonal to all the polynomials of degree less than or equal k-2.

Homework Equations

<p,q>=0 if and only if p and q are orthogonal.

The Attempt at a Solution

I used the Gram-Schmidt algorithm to find an expression for $$p_k$$. I came up with the following:
$$p_k = q_k - \left[ \sum_{n=0}^{k-1} proj_{p_n}(q_k) \right]$$
where $$q_k=x^k$$ for $$k=0,1,2, \cdots$$
and $$proj_{p_n}(q_k)=\frac{\langle q_k,p_n \rangle}{\langle p_n,p_n \rangle} \cdot p_n$$.
However I am stuck. At first I thought induction with k=3 as the first case, but in the case k=m I am stuck. Is there a theorem that can guarantee that if a vector is not orthogonal to one of the vector in the basis, it is not orthogonal to none of the preceding vectors in the basis?

 

[vela]

Use

$$\int_{-1}^1 [xP_k(x)]P_n(x)\,dx = \int_{-1}^1 P_k(x)[xP_n(x)]\,dx$$

and what you know about the degree of xP_n(x).

 

[Wingeer]

Hm. But if k is an odd integer, then for n=k-1
$$\int_{-1}^{1} xP_k(x) P_n(x) dx = 0$$ since the integrand is odd?

 

[vela]

Actually, if n=k-1, P_k and P_n will have opposite parity, so their product will be odd. Since x is odd, xP_kP_n will be even. In any case, the problem with using parity is it only gets you half of what you want.

Think about the fact that the Legendre polynomials form a basis.

 

[Wingeer]

Okey. So if the Legendre polynomials forms a basis they are by definition of a basis linearly independent. I.e. they are all orthogonal. But xP_i when i=k-1 is of the same degree as P_n when and they need not be orthogonal. Is this all?

 

[vela]

Independent isn't synonymous with orthogonal. Moreover, linear independence isn't really the characteristic of a basis you need for this problem.

Also, why do you keep looking at the case where i=k-1? The problem asked you to show xP_k is orthogonal to all P_i where i≤k-2.

 

[Wingeer]

No, you're right; it is not synonymous. Then what? Hints are appreciated.
I thought it was a natural step following from the premise, that you showed wrong.

 

[vela]

What's the second requirement for being a basis?

 

[Wingeer]

That the basis spans the given vector space. Alas, I can't see the connection.

 

[vela]

Hint: xP_i(x) is a polynomial of degree k-1 or less.

 

[Wingeer]

So P_k(x) is not in the span of xP_i(x) as it is of a higher degree in the vector space F[x] if you consider xP_i(x) as your basis?

 

[vela]

That's true, but it's not enough to conclude P_k and xP_i are orthogonal. In R², for example, (1,0) isn't in the span of {(1,1)}, but the two vectors aren't orthogonal.

There is a recursion relation between the Legendre polynomial. To see this, show that
the polynomial is orthogonal to all the polynomials of degree less than or equal k-2.

I think I might have misunderstood the problem slightly. Because you said "the polynomials," I read this to mean the Legendre polynomials of degree less than or equal to k-2. If you meant polynomials of degree less than or equal to k-2 in general, you want to show

$$\int_{-1}^1 [xP_k(x)]q(x)\,dx = \int_{-1}^1 P_k(x)[xq(x)]\,dx$$

You use exactly the same reasoning, but you might find the ambiguity of q(x) makes it easier to figure the problem out.

 

[Wingeer]

No, you're right in that they need not be orthogonal. What more do I need?

Hm. I just copied and pasted the problem text. Now I am unsure if "the polynomials" are referring to the Legendre polynomials or polynomials in general. Is the solution different regarding which type of polynomials we are looking at?

 

[vela]

Hm. I just copied and pasted the problem text. Now I am unsure if "the polynomials" are referring to the Legendre polynomials or polynomials in general. Is the solution different regarding which type of polynomials we are looking at?

No, it's the same.

Hint: Think about the set {P₀(x), P₁(x), …, P_k-1(x)}.

 

[Wingeer]

Okey. My thoughts: The set is a basis for the polynomials up to degree k-1. This means they are linearly independent and that they span the polynomials of lower or equal degree. I know that we can use the Gram Schmidt process to construct a set of orthonormal vectors from this set, but I do not think that's what you're hinting at.