 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/17b/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

p_{n - 1}

gives

β_{n} = \frac{⟨ x p _{n} , p _{n - 1} ⟩}{⟨ p _{n - 1} , p _{n - 1} ⟩} = \frac{⟨ p _{n} , x p _{n - 1} ⟩}{⟨ p _{n - 1} , p _{n - 1} ⟩} = \frac{⟨ p _{n} , p _{n} ⟩}{⟨ p _{n - 1} , p _{n - 1} ⟩} .

β_{n} = n (n + \frac{1}{2}) .

The Rodrigues expansion also shows that the coefficient of

x^{n - 1}

p_{n}

- n (n + \frac{1}{2}) .

Comparing the coefficients of

x^{n}

p_{n + 1} = (x - α_{n}) p_{n} - β_{n} p_{n - 1}

therefore gives

- (n + 1) (n + \frac{3}{2}) = - n (n + \frac{1}{2}) - α_{n},

and hence

α_{n} = 2 n + \frac{3}{2} .

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