Proof that the legendre polynomials are orthogonal polynomials

[center o bass]

I'm now studying the application of legendre polynomials to numerical integration in the so called gaussian quadrature. There one exploits the fact that an orthogonal polynomial of degree n is orthogonal to all other polynomials of degree less than n with respect to some weight function. For legendre polynomials that must mean that

$$\int_{-1}^{1} L_n(x) P_m(x) dx = 0$$

for all P(x) where m is less than n. How does one prove that the legendre polynomials are in the set of such orthogonal polynomials? It's okay that they are orthogonal among themselves, but I wonder how to show that they are orthogonal to everyone else with lower degree?

 

[mathman]

How does one prove that the legendre polynomials are in the set of such orthogonal polynomials?

This is unclear. What is the set you are referring to?

It's okay that they are orthogonal among themselves, but I wonder how to show that they are orthogonal to everyone else with lower degree?

Any polynomial of degree m can be represented as a linear combination of Legendre polynomials of degree at most m.

 

[Deveno]

show that the legendre polynomials of degree ≤ n, are linearly independent, and thus form a basis for all polynomials of degree ≤ n.

therefore, every polynomial of degree ≤ n can be written as a linear combination of the L_j (j = 0,1,2,...,n):

$$P_m(x) = a_0L_0(x) + \dots + a_nL_n(x)$$

which will make the only surviving term in the inner product the nth one:

$$\int_{-1}^1L_n(x)a_nL_n(x) dx$$

but if P_m is of degree < n, the coefficent a_n of L_n will be 0.

 

[center o bass]

Ah, thanks! That was the argument I was looking for.