 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/18/c/solution

w

\int_{- 1}^{1} ∣ w (x) ∣^{2} d x = \frac{∣ a _{0} ∣ ^{2}}{2} + \sum_{n = 1}^{\infty} (∣ a_{n} ∣^{2} + ∣ b_{n} ∣^{2}) .

For

w = f - g

, write

s = sinh 1

a_{0} = 2 s

a_{n} = 2 (- 1)^{n} s / (1 + n^{2} π^{2})

and

b_{n} = nπ a_{n}

, because

g

is subtracted. Therefore

\int_{- 1}^{1} (f - g)^{2} d x = 2 s^{2} + 4 s^{2} n = 1 \sum \infty \frac{1 + n ^{2} π ^{2}}{( 1 + n ^{2} π ^{2} ) ^{2}} = 2 s^{2} [1 + 2 n = 1 \sum \infty \frac{1}{1 + n ^{2} π ^{2}}] = 2 sinh 1 cosh 1 = sinh 2 .

The sum was obtained in part (a). As a separate check,

f - g = e^{- x}

in the interval, whose square has integral

\int_{- 1}^{1} e^{- 2 x} d x = sinh 2

. Endpoint jumps have no effect on this integral.

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