Solution. \(f(\omega ) = (1.25 + \cos \omega )/(2\pi )\), so \(f(0) = 2.25/(2\pi ) = 0.358\), \(f(\pi /2) = 1.25/(2\pi ) = 0.199\), \(f(\pi ) = 0.25/(2\pi ) = 0.040\). The integral is \(\frac 1{2\pi }\int _{-\pi }^\pi (1.25 + \cos \omega )\,d\omega = \frac 1{2\pi }(2.5\pi + 0) = 1.25\), which is \(\Var (X_t) = 1 + 0.25\).

Solution. \(f(0)/f(\pi ) = (1 + 2\alpha + \alpha ^2)/(1 - 2\alpha + \alpha ^2) = (1.6/0.4)^2 = 16\). Slow variation carries sixteen times the spectral density of the fastest alternation: the series moves in smooth swings rather than jumping about.

Solution. The values are \(0, -3, 0, 3, 0, -3, 0, 3\). At \(j = 2\), \(\sum x_t\cos (\pi t/2) = 3\sum \cos ^2(\pi t/2) = 3 \times 4 = 12\), so \(a_2 = \frac 28 \times 12 = 3\), the amplitude of the cosine, and \(b_2 = \frac 28\sum 3\cos (\pi t/2)\sin (\pi t/2) = 0\). So \(I(\omega _2) = \frac {8}{8\pi }(9 + 0) = 9/\pi = 2.86\). At \(j = 1\) and \(j = 3\), \(a_j\) and \(b_j\) are sums of products of different harmonics, which are orthogonal over \(t = 1, \dots , 8\), so they are \(0\) and the periodogram vanishes. All the variance, \(\sum x_t^2 = 36 = \frac 82 \times 9\), is at the one frequency.