Solution. \(|x|\) is even, so \(b_k = 0\); \(a_0 = \frac 2\pi \int _0^\pi x\,dx = \pi \), and \(a_k = \frac 2\pi \int _0^\pi x\cos kx\,dx = \frac {2((-1)^k - 1)}{\pi k^2}\), which is \(-\frac 4{\pi k^2}\) for odd \(k\) and \(0\) for even \(k\). So \[ |x| = \frac \pi 2 - \frac 4\pi \sum _{k\ \text {odd}}\frac {\cos kx}{k^2} , \] uniformly, by Theorem 7.2.3. At \(x = 0\): \(0 = \frac \pi 2 - \frac 4\pi \sum _{\text {odd}}\frac 1{k^2}\).

Solution. \(\frac 1\pi \int _{-\pi }^\pi x^2 = \frac {2\pi ^2}3 = \frac {\pi ^2}2 + \sum _{\text {odd}}\frac {16}{\pi ^2k^4}\), so \(\sum _{\text {odd}}k^{-4} = \frac {\pi ^2}{16}\cdot \frac {\pi ^2}6 = \frac {\pi ^4}{96}\). Since the even terms sum to \(\frac 1{16}\) of the whole, the whole is \(\frac {16}{15}\cdot \frac {\pi ^4}{96} = \frac {\pi ^4}{90}\).

Solution. The eigenfunctions are \(\cos k\pi x\), \(k \geq 0\). The coefficients are \(a_0 = 2\int _0^1x\,dx = 1\) and \(a_k = 2\int _0^1x\cos k\pi x\,dx = \frac {2((-1)^k - 1)}{k^2\pi ^2}\), so \[ u = \frac 12 - \frac 4{\pi ^2}\sum _{k\ \text {odd}}\frac {e^{-k^2\pi ^2t}\cos k\pi x}{k^2} \to \frac 12 , \] the mean initial temperature, as it must be, since an insulated rod conserves its heat.

Solution. With \(p = \frac 12\), \(b_k \propto \sin (k\pi /2)/k^2\), which vanishes for even \(k\). For odd \(k\), \(\omega _k^2b_k^2 \propto k^2\cdot k^{-4} = k^{-2}\), so the fraction in the fundamental is \(1/\sum _{\text {odd}}k^{-2} = 8/\pi ^2 = 0.811\).

Solution. Here \(a_k = 0\), \(\beta _1 = 1\), \(\beta _3 = \frac 12\), so \[ u = \frac {\sin \pi ct\,\sin \pi x}{\pi c} + \frac {\sin 3\pi ct\,\sin 3\pi x}{6\pi c} . \]

Solution. The partial sums are continuous and \(2\pi \)-periodic, so a uniform limit would be a continuous periodic function, but it would have to equal \(x\) on \((-\pi , \pi )\), which tends to \(\pi \) at the right end and \(-\pi \) at the left. At \(x = \pi \) every term \(\sin k\pi \) vanishes, so the sum is \(0\), the average of \(f(\pi -) = \pi \) and \(f(\pi +) = f(-\pi +) = -\pi \), as Theorem 7.2.2 says.