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
$$
\begin{aligned}
&S_{0}(z)=a_{0}\left(z+\frac{1}{3 !} z^{3}+\frac{9}{5 !} z^{5}+\cdots\right) \\
&S_{1}(z)=a_{0}\left(1-\frac{1}{2 !} z^{2}-\frac{3}{4 !} z^{4}-\cdots\right) \\
&S_{2}(z)=a_{0}\left(z-\frac{3}{3 !} z^{3}-\frac{15}{5 !} z^{5}-\cdots\right) \\
&S_{3}(z)=a_{0}\left(1-\frac{9}{2 !} z^{2}+\frac{45}{4 !} z^{4}+\cdots\right)
\end{aligned}
$$