 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/a/solution

\frac{d}{d x} e^{- x^{2}} = - 2 x e^{- x^{2}}, \frac{d ^{2}}{d x ^{2}} e^{- x^{2}} = (4 x^{2} - 2) e^{- x^{2}} .

The defining prefactor then gives

H_{0} = 1, H_{1} = 2 x, H_{2} = 4 x^{2} - 2 .

The derivative-based definition uses nonnegative integers; negative orders are not part of this polynomial definition.

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