 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/1/d/solution

p_{n} (x) = \sum_{j = 0}^{n} \frac{( - 1 ) ^{j}}{j !} (j n) x^{j} = \frac{e ^{x}}{n !} \frac{d ^{n}}{d x ^{n}} (e^{- x} x^{n}) .

The leading coefficient is

(- 1)^{n} / n!

, so

p_{n}^{(n)} = (- 1)^{n}

. Substitute this differentiated expression into one factor in the squared weighted norm and apply integration by parts

n

∥ p_{n} ∥_{w}^{2} = \frac{1}{n !} \int_{0}^{\infty} p_{n} (x) \frac{d ^{n}}{d x ^{n}} (e^{- x} x^{n}) d x = \frac{( - 1 ) ^{n}}{n !} \int_{0}^{\infty} e^{- x} x^{n} p_{n}^{(n)} (x) d x = \frac{1}{n !} \int_{0}^{\infty} e^{- x} x^{n} d x = 1 .

Every endpoint term contains a derivative of

e^{- x} x^{n}

of order less than

n

n

, including

n = 0

without any integrations by parts.

 New to topics? Read the docs here!