16A (Methods)

The function is an even function, so every sine Fourier coefficient vanishes. Direct integration gives

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

Thus

∣ x ∣ = \frac{π}{2} - \frac{4}{π} k = 0 \sum \infty \frac{cos (( 2 k + 1 ) x )}{( 2 k + 1 ) ^{2}}

for the

2 π

-periodic extension.

b

(C_{1} + C_{2} x) e^{- x}

are not

2 π

-periodic unless

C_{1} = C_{2} = 0

, so there is a unique periodic solution. For a forcing term

a_{n} cos (n x)

, a trial solution

A_{n} cos (n x) + B_{n} sin (n x)

gives

A_{n} = \frac{a _{n} ( 1 - n ^{2} )}{( 1 + n ^{2} ) ^{2}}, B_{n} = \frac{2 n a _{n}}{( 1 + n ^{2} ) ^{2}} .

Using the Fourier series from part (a), for which only odd

n

occur and

a_{n} = - 4/ (π n^{2})

, yields

y (x) = \frac{π}{2} - \frac{4}{π} n \geq 1 n odd \sum \frac{( 1 - n ^{2} ) cos ( n x ) + 2 n sin ( n x )}{n ^{2} ( 1 + n ^{2} ) ^{2}} .

The coefficients decay fast enough to differentiate the series twice term by term, verifying the equation and periodicity.

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