Solution. The recursion is \((k + 2)(k + 1)a_{k+2} = -a_k\). From \(a_0\): \(a_{2k} = (-1)^ka_0/(2k)!\); from \(a_1\): \(a_{2k+1} = (-1)^ka_1/(2k + 1)!\). So \(y = a_0\cos t + a_1\sin t\), with infinite radius of convergence, as Theorem 8.1.1 predicts for constant coefficients.

Solution. \(p = t/(t^2 + 4)\) and \(q = 1/(t^2 + 4)\) are analytic except at \(t = \pm 2i\). The distance from \(1\) to \(\pm 2i\) is \(\sqrt {1 + 4} = \sqrt 5\), so the series converge at least for \(|t - 1| < \sqrt 5 = 2.24\).

Solution. For \(r = \frac 12\): \(a_1 = a_0/(1 \cdot 5) = \frac 15\), \(a_2 = a_1/(2 \cdot 7) = \frac {1}{70}\), so \(y_1 = t^{1/2}\bigl (1 + \frac t5 + \frac {t^2}{70} + \cdots \bigr )\). For \(r = -1\): \(a_1 = a_0/(1 \cdot (-1)) = -1\), \(a_2 = a_1/(2 \cdot 1) = -\frac 12\), so \(y_2 = t^{-1}\bigl (1 - t - \frac {t^2}{2} - \cdots \bigr )\).

Solution. By Theorem 8.2.2(ii), \(y_2 = J_0(t)\log t + \sum _{k\geq 1}c_kt^k\), which tends to \(-\infty \) as \(t \to 0^+\) because \(J_0(0) = 1\). A general solution is \(AJ_0 + By_2\), bounded at \(0\) only if \(B = 0\). This is why the displacement of a drum, which must be finite at the centre, involves only \(J_0\) and not \(Y_0\).

Solution. With \(u = (t^2 - 1)^n\), \(\int _{-1}^1P_n^2 = \frac {1}{4^n(n!)^2}\int _{-1}^1u^{(n)}u^{(n)}\). Integrating by parts \(n\) times moves all derivatives onto one factor, the boundary terms vanishing because \(u\) and its first \(n - 1\) derivatives vanish at \(\pm 1\): \(\int u^{(n)}u^{(n)} = (-1)^n\int u\,u^{(2n)} = (-1)^n(2n)!\int _{-1}^1(t^2 - 1)^n\,dt\), since \(u^{(2n)} = (2n)!\). With \(\int _{-1}^1(1 - t^2)^n\,dt = \frac {2^{2n+1}(n!)^2}{(2n + 1)!}\), this gives \(\frac {(2n)!\,2^{2n+1}(n!)^2}{4^n(n!)^2(2n + 1)!} = \frac {2}{2n + 1}\). For \(n = 2\), \(\int _{-1}^1\bigl (\frac 32t^2 - \frac 12\bigr )^2dt = \int _{-1}^1\bigl (\frac 94t^4 - \frac 32t^2 + \frac 14\bigr )dt = \frac 9{10} - 1 + \frac 12 = \frac 25\).

Solution. With \(\nu = \frac 12\), \(J_{3/2} = \frac 1tJ_{1/2} - J_{-1/2} = \sqrt {\frac 2{\pi t}}\Bigl (\frac {\sin t}{t} - \cos t\Bigr )\). Every Bessel function of half-integer order is elementary in this way; they are the spherical Bessel functions of the wave equation in three dimensions.