 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v/4/b/i/solution

u = x / L

and

q = nπ

. The required integrals follow from

\int e^{u} cos (q u) d u = \frac{e ^{u} ( c o s ( q u ) + q s i n ( q u ))}{1 + q ^{2}}, \int e^{u} sin (q u) d u = \frac{e ^{u} ( s i n ( q u ) - q c o s ( q u ))}{1 + q ^{2}} .

Consequently the Fourier coefficients are

a_{0} = 2 (e - 1), a_{n} = \frac{2 [( - 1 ) ^{n} e - 1 ]}{1 + n ^{2} π ^{2}}, b_{n} = \frac{2 nπ [ 1 - ( - 1 ) ^{n} e ]}{1 + n ^{2} π ^{2}} (n \geq 1) .

e - 1 + 2 n = 1 \sum \infty \frac{( - 1 ) ^{n} e - 1}{1 + n ^{2} π ^{2}} cos \frac{nπ x}{L},

and its sum is

e^{x / L}

everywhere on

[0, L]

2 n = 1 \sum \infty \frac{nπ [ 1 - ( - 1 ) ^{n} e ]}{1 + n ^{2} π ^{2}} sin \frac{nπ x}{L},

and its sum is

e^{x / L}

(0, L)

and zero at the two endpoints. The open and filled circles in the sketches distinguish these endpoint limits from the actual values of the sum.

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