Proof. (i) With \(P_k = 0\) for \(k \geq 1\) and \(Q_2 = 1\), the recursion of Theorem 8.2.2 becomes \(F(\nu + k)a_k = -a_{k-2}\), that is \(k(k + 2\nu )a_k = -a_{k-2}\), with \(a_1 = 0\). So only even coefficients survive, \(a_{2k} = -a_{2k-2}/(4k(k + \nu ))\), and with \(a_0 = 1/(2^\nu \Gamma (\nu + 1))\) and \(\Gamma (z + 1) = z\Gamma (z)\) this gives the series. The ratio of successive terms tends to zero, so it converges everywhere. (ii) For non-integer \(\nu \) the same calculation with \(-\nu \) works, since \(k(k - 2\nu ) \neq 0\), and the two solutions behave like \(t^\nu \) and \(t^{-\nu }\) at \(0\), so they are independent. For \(\nu = n\), \(1/\Gamma (k - n + 1) = 0\) for \(k < n\), the series for \(J_{-n}\) starts at \(k = n\), and re-indexing \(k = j + n\) gives \((-1)^nJ_n\); the second solution comes from case (ii) or (iii) of Theorem 8.2.2, with a logarithm. (iii) Differentiate the series of \(t^\nu J_\nu = \sum _k\frac {(-1)^kt^{2k+2\nu }}{2^{2k+\nu }k!\,\Gamma (k + \nu + 1)}\) term by term and use \((2k + 2\nu )/\Gamma (k + \nu + 1) = 2/\Gamma (k + \nu )\); the second identity is similar, and expanding both derivatives and adding gives the recurrence. (iv) With \(\Gamma (k + \frac 32) = \frac {(2k + 1)!}{4^kk!}\cdot \frac {\sqrt \pi }{2}\), the series becomes \(\sqrt {\frac 2{\pi t}}\sum _k\frac {(-1)^kt^{2k+1}}{(2k + 1)!}\); the other is similar. (v) Multiply \(e^{tz/2} = \sum _j\frac {(tz/2)^j}{j!}\) by \(e^{-t/(2z)} = \sum _k\frac {(-t/(2z))^k}{k!}\), and collect the coefficient of \(z^n\), \(n \geq 0\): it is \(\sum _k\frac {(-1)^k(t/2)^{n+2k}}{(n + k)!\,k!} = J_n(t)\). □