Proof. (i) \(\sum _tx_te^{-i\omega _jt} = \sum _tx_t\cos (\omega _jt) - i\sum _tx_t\sin (\omega _jt) = \frac n2(a_j - ib_j)\), whose squared modulus is \(\frac {n^2}4(a_j^2 + b_j^2)\); divide by \(2\pi n\). The harmonics at the Fourier frequencies are orthogonal over \(t = 1, \dots , n\), and for \(0 < j < n/2\), \[ \textstyle \sum _t\cos ^2(\omega _jt) = \sum _t\sin ^2(\omega _jt) = n/2 , \] so by the orthogonal-regression formula (each coefficient is \(\sum x_t\times \text {column}/\sum \text {column}^2\)) the least-squares coefficients are \(a_j\) and \(b_j\).

(ii) For \(j \ne 0\), \(\sum _te^{-i\omega _jt} = 0\) (a geometric sum over whole turns), so \(x_t\) may be replaced by \(x_t - \bar x\). Then \(\bigl |\sum _t(x_t - \bar x)e^{-i\omega _jt}\bigr |^2 = \sum _s\sum _t(x_s - \bar x)(x_t - \bar x)e^{-i\omega _j(t-s)}\), and grouping the pairs by \(k = t - s\) gives \(\sum _kn\,c_ke^{-i\omega _jk}\). □

Solution. Here \(n = 12\), \(\bar x = 0\) and \(\sum x_t^2 = 44\). The Fourier frequencies are \(\omega _j = 2\pi j/12\), periods \(12/j\), for \(j = 1, \dots , 6\). At \(j = 3\) (period \(4\)), \(\cos (\pi t/2)\) runs \(0, -1, 0, 1, \dots \) and \(\sin (\pi t/2)\) runs \(1, 0, -1, 0, \dots \), so \begin {align*} a_3 &= \tfrac 2{12}\bigl [(-1)(-3) + (1)(1) + (-1)(-3) + (1)(3) + (-1)(-1) + (1)(3)\bigr ] \\ &= \tfrac {14}6 = 2.333 , \\ b_3 &= \tfrac 2{12}\bigl [(1)(-1) + (-1)(-1) + (1)(-1) + (-1)(1) + (1)(1) + (-1)(1)\bigr ] \\ &= -\tfrac 26 = -0.333 , \end {align*}

and \(I(\omega _3) = \frac {12}{8\pi }(5.444 + 0.111) = 2.653\). The same computation at the other frequencies gives:

The sums of squares in the last column add to \(9.95 + 33.33 + 0.71 = 44.0 = \sum x_t^2\), as they should. The cycle of period \(4\) accounts for \(33.33/44 = 76\%\) of the variation, with most of the rest in a slow movement of period \(12\). The data look like quarterly observations with a strong seasonal pattern.