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 Incoming links: Hermite polynomial derivative recurrence

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 20 / c / Solution 2026-10-18

Differentiate the product in the definition of

H_{n}

. Using the identity in (b),

H_{n}^{'} = (- 1)^{n} e^{x^{2}} (2 x u^{(n)} + u^{(n + 1)}) = - 2 n (- 1)^{n} e^{x^{2}} u^{(n - 1)} = 2 n (- 1)^{n - 1} e^{x^{2}} u^{(n - 1)} .

The last expression is

2 n H_{n - 1}

, so

H_{n}^{'} = 2 n H_{n - 1} (n \geq 1) .

Also

H_{0}^{'} = 0

separately. This is the Hermite polynomial derivative recurrence in the physicists' normalization.

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