Using the recurrence relationprove that

Question

Using the recurrence relation

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

prove that

$(2n+1)^2\int_{-1}^{1}x^2$ $P_{n}^{2}(x)dx=\frac{2(n+1)^2}{(2n+3)}+\frac{2n^2}{(2n-1)}$

22 Mar 2023

Answer :

Word Count : 418

We start with the given recurrence relation:

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

Multiplying both sides by x and rearranging, we get:

$x(n+1)P_{n+1}(x) = (2n+1)x^2P_{n}(x) - nxP_{n-1}(x)$

Taking the derivative of both sides with respect to x, we get:

$(n+1)P_{n+1}'(x) + x(n+1)P_{n+1}''(x) = (2n+1)(2xP_{n}(x) + x^2P_{n}'(x)) - nP_{n-1}'(x)$

Multiplying both sides by $x^2$ and integrating from -1 to 1, we get:

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