 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/19c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

He_{n}^{'} = n He_{n - 1}, x He_{n} = He_{n + 1} + n He_{n - 1},

which follow by differentiating the Rodrigues formula. Thus

He_{n + 1} = x He_{n} - n He_{n - 1},

α_{n} = 0

and

β_{n} = n

For the squared norm, substitute the Rodrigues formula and integrate by parts

n

times. All boundary terms vanish under the Gaussian weight, and

He_{n}^{(n)} = n!

⟨ He_{n}, He_{n} ⟩ = (- 1)^{n} \int_{- \infty}^{\infty} He_{n} (x) \frac{d ^{n}}{d x ^{n}} e^{- x^{2} /2} d x = n! \int_{- \infty}^{\infty} e^{- x^{2} /2} d x = n! 2 π .

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