# Using Orthogonality of Legendre Polynomials to determine a Recurrence Relation

This is something that I've been struggling on for a few hours now and would appreciate any help:

Rodrigue's formula:

$$P_n(x)$$ = $$\frac{1}{2^nn!} \frac{d^n}{dx^n}(x^2-1)^n$$

The Legendre polynomials (given by the above formula) $$\{P_0,...,P_n\}$$ form an orthogonal basis of the space of all polynomials of degree at most $$n$$ (integer).

Let $$P_n(x) = a_{n,n}x^n +a_{n-1,n}x^{n-1} + ...+ a_{0,n}$$

We are given that:

1) The leading coefficient $$a_{n,n} = \frac{(2n)!}{2^n(n!)^2}$$

2) The polynomial $$xP_n(x)$$ of degree $$n+1$$ can be written as: $$xP_n(x) = \frac{n+1}{2n+1}P_{n+1}(x) + Q(x)$$

Question: Show that $$Q(x)$$ is orthogonal to $$P_j(x)$$ for all $$j\geq n-2$$.

Conclude that there exists constants A and B such that

$$Q(x) = AP_n(x) +BP_{n-1}(x)$$

Using the normalisation condition relation $$P_n(1)=1$$ for all $$n \geq 0$$, and the fact that $$\int_{-1}^{1} x|P_n(x)|^2dx=0$$, show that $$A=0$$ and $$B=\frac{n}{2n+1}$$. Deduce from this recurrence relation:

$$(n+1)P_{n+1}(x) = (2n+1)xP_n(x)-nP_{n-1}(x)$$

My attempt:

For orthogonality:

I have considered the inner product of $$Q(x)$$ and $$P_j(x)$$ w.r.t weight function $$w(x)=1$$. However I'm having some issues with the integration and not sure how to proceed with the rest of the question.

$$\langle{Q(x)}, {P_j(x)} \rangle$$ = $$\int_{-1}^{1} Q(x)P_j(x) dx$$

=$$\int_{-1}^{1} (xP_n(x)-\frac{n+1}{2n+1}P_{n+1}(x))P_j(x) dx$$

=$$\int_{-1}^{1} xP_n(x)P_j(x) dx - \frac{n+1}{2n+1}\int_{-1}^{1}P_{n+1}(x)P_j(x) dx$$

I know that this must equal zero, but how exactly? And also how can we conclude that constants $$A$$ and $$B$$ exist such that $$Q(x)$$ takes the second form?