Question

Let \( y = P_n(x) \) be the unique polynomial of degree \( n \) satisfying the Legendre differential equation \[ (1 - x^2)y'' - 2xy' + n(n + 1)y = 0 \quad {and} \quad y(1) = 1. \] Then, the value of \( P_{11}'(1) \) is equal to _________ (in integer).

Hint:

For Legendre polynomials, the derivative at \( x = 1 \) can be computed using the formula \( P_n'(1) = \frac{n(n+1)}{2} \).

Answer 1

Correct Answer: 66

We are given the Legendre differential equation: \[ (1 - x^2)y'' - 2xy' + n(n + 1)y = 0. \] The solutions to this equation are the Legendre polynomials \( P_n(x) \), where \( n \) is the degree of the polynomial. The Legendre polynomials satisfy the following recurrence relation: \[ P_0(x) = 1, \quad P_1(x) = x, \] \[ (n+1)P_{n+1}(x) = (2n+1)x P_n(x) - nP_{n-1}(x). \] For this problem, we need to find \( P_{11}'(1) \). It is known that the derivative of the Legendre polynomial at \( x = 1 \) for any \( P_n(x) \) is related to the formula: \[ P_n'(1) = \frac{n(n+1)}{2}. \] Thus, for \( n = 11 \): \[ P_{11}'(1) = \frac{11(11 + 1)}{2} = \frac{11 \times 12}{2} = 66. \] Therefore, the value of \( P_{11}'(1) \) is: \[ \boxed{66}. \]