Chapter 16: Problem 22
The origin is an ordinary point of the Chebyshev equation, $$ \left(1-z^{2}\right) y^{\prime \prime}-z y^{\prime}+m^{2} y=0 $$ which therefore has series solutions of the form \(z^{\sigma} \sum_{0}^{\infty} a_{n} z^{n}\) for \(\sigma=0\) and \(\sigma=1\). (a) Find the recurrence relationships for the \(a_{n}\) in the two cases and show that there exist polynomial solutions \(T_{m}(z)\) : (i) for \(\sigma=0\), when \(m\) is an even integer, the polynomial having \(\frac{1}{2}(m+2)\) terms; (ii) for \(\sigma=1\), when \(m\) is an odd integer, the polynomial having \(\frac{1}{2}(m+1)\) terms. (b) \(T_{m}(z)\) is normalised so as to have \(T_{m}(1)=1 .\) Find explicit forms for \(T_{m}(z)\) for \(m=0,1,2,3\). (c) Show that the corresponding non-terminating series solutions \(S_{m}(z)\) have as their first few terms $$ \begin{aligned} &S_{0}(z)=a_{0}\left(z+\frac{1}{3 !} z^{3}+\frac{9}{5 !} z^{5}+\cdots\right) \\\ &S_{1}(z)=a_{0}\left(1-\frac{1}{2 !} z^{2}-\frac{3}{4 !} z^{4}-\cdots\right) \\\ &S_{2}(z)=a_{0}\left(z-\frac{3}{3 !} z^{3}-\frac{15}{5 !} z^{5}-\cdots\right) \\\ &S_{3}(z)=a_{0}\left(1-\frac{9}{2 !} z^{2}+\frac{45}{4 !} z^{4}+\cdots\right) \end{aligned} $$
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