Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 6

Divide the Chebyshev differential equation by $1-x^2$. Its coefficients $-x/(1-x^2)$ and $m^2/(1-x^2)$ are analytic at zero, so $\boxed{x=0\text{ is an ordinary point}.}$ The finite singular points of a differential equation are $x=\pm 1$.

Insert $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$, $a_0\ne 0$. The coefficient of the lowest power $x^{\sigma -2}$ is $\sigma (\sigma -1)a_0$, hence $\sigma =0$ or $1$. The next power gives $\sigma (\sigma +1)a_1=0$, and the remaining powers yieldFor $\sigma =0$, $a_1$ is free andChoosing $a_1=0$ selects the even power series, with $y(0)=a_0,y^{\prime }(0)=0$. For $\sigma =1$, the next-power equation forces $a_1=0$ andThis selects the odd power series, with $y(0)=0,y^{\prime }(0)=a_0$. Their different initial values prove linear independence. Although the Frobenius method produces two possible leading exponents, zero is an ordinary point, and a general $\sigma =0$ series already includes both initial constants.

Let $M=|m|$ for integer $m$. In the even or odd power series whose parity matches $M$, the numerator in the recurrence relation vanishes at the coefficient of $x^M$. All coefficients of lower powers in that parity are nonzero, and all higher ones vanish. Therefore a polynomial solution exists, of polynomial degree $M$, containingFor nonnegative $m$ these are the Chebyshev polynomials $T_m$. The normalized low-degree cases follow directly from the recurrence relation:Enforcing $T_m(1)=1$ givesTo see that normalization at one is always possible, set $x=\cos \theta$ in the Chebyshev differential equation. For $Y(\theta )=y(\cos \theta )$, the differential expression $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$ becomes $Y^{\prime \prime }$. Thus $Y=\cos (M\theta )$ gives $T_M(x)=\cos (M\arccos x)$ on $[-1,1]$ and has value one at $x=1$. It is a polynomial because the cosine addition formula gives $T_{M+1}=2x{T}_M-T_{M-1}$. Negative integer $m$ produces the same solution as $M=|m|$.

Write a nonterminating parity series as $y=x^r{\sum }_{j\ge 0}{b}_j{x}^{2j}$, where $r=0$ or $1$. Its recurrence relation isThe ratio test applied to its nonzero successive terms has limiting absolute ratio $|x{|}^2$. Hence the series converges absolutely for $|x|<1$ and diverges for $|x|>1$, withAt $x=\pm 1$, the limiting ratio is one, so the basic ratio test alone is inconclusive. Retain the next term of the ratio instead:For large $j$ the ratio is positive. Taking logarithms and summing gives $\log |b_j|=-\frac{3}{2}\log j+O(1)$; the $O(j^{-2})$ remainders have a finite sum. Thus $|b_j|\le C{j}^{-3/2}$ eventually. Comparison with the convergent p-series proves absolute convergence at both endpoints as well:The radius of convergence one is the distance from the expansion point to the nearest singular points of a differential equation, $x=\pm 1$. Convergence of function values at those points does not make the solutions analytic there; endpoint derivatives can diverge. The terminating Chebyshev polynomial is an exception with infinite radius of convergence. For example, at $m=0$ the other solution is $\arcsin x$, whose endpoint values are finite while its first derivative diverges.