 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/18/c/solution

The function is even, so each product with a sine is an odd function and all the Fourier coefficients

b_{n}

vanish. Its constant coefficient is

a_{0} = \frac{2}{π} \int_{0}^{π} e^{x} d x = \frac{2 ( e ^{π} - 1 )}{π} .

For

n \geq 1

, part (a) gives the remaining Fourier coefficients:

a_{n} = \frac{2}{π} \int_{0}^{π} e^{x} cos (n x) d x = \frac{2 [ ( - 1 ) ^{n} e ^{π} - 1 ]}{π ( 1 + n ^{2} )} .

f (x) = \frac{e ^{π} - 1}{π} + \frac{2}{π} n = 1 \sum \infty \frac{( - 1 ) ^{n} e ^{π} - 1}{1 + n ^{2}} cos (n x) .

The periodic extension is continuous, including at

\pm π

, and piecewise continuously differentiable. Thus the Fourier series equals

f

everywhere. Its coefficients are also bounded in magnitude by a constant times

1/ n^{2}

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