Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 8 / Solution

In monic form the Chebyshev differential equation has $P(x)=-x/(1-x^2)$ and $Q(x)=p^2/(1-x^2)$. Both are analytic near zero, so it is an ordinary point. At $x=1$, $(x-1)P=x/(x+1)$ and $(x-1{)}^{2}Q=-p^2(x-1)/(x+1)$ are analytic; at $x=-1$, $(x+1)P=x/(x-1)$ and $(x+1{)}^{2}Q=-p^2(x+1)/(x-1)$ are analytic. Thus both endpoints are regular singular points.

Insert $y_q={\sum }_{n\ge 0}{a}_n{x}^{n+q}$ and equate coefficients. The recurrence relation isExplicitly the factors are $(n^2-p^2)/[(n+2)(n+1)]$ for $q=0$ and $[(n+1{)}^2-p^2]/[(n+3)(n+2)]$ for $q=1$. The lowest equation for $q=1$ forces $a_1=0$. For $q=0$, one may choose $a_1=0$ to select an even solution, independent of the odd $q=1$ solution. The leading coefficient $a_0$ is free in either chosen parity branch.

For a nonterminating branch the consecutive parity-term ratio tends to $x^2$. Hence the ratio test gives convergence for $|x|<1$, and radius of convergence one. Terminating branches are polynomials with infinite radius. For $p=m\ge 0$ an integer, choose $q$ of the same parity as $m$: the recurrence stops once $n+q=m$, producing a degree-$m$ Chebyshev polynomial. Normalization at $1$ gives all nonzero coefficients through degree three:

Set $Y(\theta )=y(\cos \theta )$. Direct differentiation gives $Y^{\prime \prime }={\sin }^2\theta y^{\prime \prime }-\cos \theta y^{\prime }$, so the equation becomes $Y^{\prime \prime }+m^{2}Y=0$. For $m>0$, $Y=A\cos (m\theta )+B\sin (m\theta )$. A polynomial $y$ has finite derivative at $x=1$, and therefore $Y^{\prime }(0)=-\sin 0y^{\prime }(1)=0$, forcing $B=0$. Also $Y(0)=T_m(1)=1$, so $A=1$. For $m=0$, the transformed solution is affine in $\theta$ and the same derivative condition removes its linear term. ThusThe endpoint values follow by continuity. The finite-derivative condition is what excludes the independent nonpolynomial branch, rather than normalization alone.