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

We are given the Legendre differential equation:

(1 − x²)y'' − 2xy' + n(n + 1)y = 0

Its solutions are the Legendre polynomials P_n(x).

---

Alternative Approach: Using Rodrigues’ Formula

Legendre polynomials can be defined by Rodrigues’ formula:

P_n(x) = (1 / 2ⁿn!) · dⁿ/dxⁿ [(x² − 1)ⁿ]

---

Step 1: Differentiate P_n(x)

Differentiating once,

P_n'(x) = (1 / 2ⁿn!) · dⁿ⁺¹/dxⁿ⁺¹ [(x² − 1)ⁿ]

---

Step 2: Evaluate at x = 1

At x = 1, the dominant contribution comes from the highest-order term of (x² − 1)ⁿ, which leads to the standard result:

P_n'(1) = n(n + 1) / 2

---

Step 3: Substitute n = 11

P₁₁'(1) = 11(11 + 1) / 2

P₁₁'(1) = 11 × 12 / 2

P₁₁'(1) = 66

---

Final Answer:

P₁₁'(1) = 66