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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?

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Long ago I answered a similar question detailing these things, see Legendre Equation Properties .

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