 Solution

ID: 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

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

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