 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/20/b/solution

Differentiating

v = x^{n} e^{- x}

gives

x v^{'} = (n - x) v

. Applying the Leibniz rule

n + 1

times to this identity gives

x v^{(n + 2)} + (n + 1) v^{(n + 1)} = (n - x) v^{(n + 1)} - (n + 1) v^{(n)} .

Only two terms survive on each side because the multiplying factors are linear. Rearranging yields

x v^{(n + 2)} + (x + 1) v^{(n + 1)} + (n + 1) v^{(n)} = 0

. Since

v^{(n)} = n! e^{- x} L_{n}

, the next two derivatives are

n! e^{- x} (L_{n}^{'} - L_{n})

and

n! e^{- x} (L_{n}^{''} - 2 L_{n}^{'} + L_{n})

. Substituting and cancelling

n! e^{- x}

gives

x L_{n}^{''} + (1 - x) L_{n}^{'} + n L_{n} = 0 .

L_{n}

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