Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 8

In normalized form, the coefficients are $p(x)=-x/(1-x^2)$ and $q(x)=n^2/(1-x^2)$. They are analytic near zero, so $\boxed{x=0\text{ is an ordinary point}}$.

At each of $x=\pm 1$, $p$ has a simple pole and $q$ at most a simple pole. The products $(x-x_0)p$ and $(x-x_0{)}^{2}q$ are analytic, so both are regular singular points. More precisely their limiting values are $1/2$ and zero, giving indicial equation $\rho (\rho -1)+\rho /2=0$, with exponents $0,1/2$.

To classify infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The transformed equation isAfter division, its coefficients satisfyBoth are analytic at $t=0$, while $P$ itself has a pole. Hence $\boxed{x=\infty \text{ is also a regular singular point}}$, with indicial exponents $\rho =\pm n$. These classifications remain valid at $n=0$, when the exponents at infinity coincide.

Because zero is an ordinary point, substitute $y={\sum }_{j\ge 0}{a}_j{x}^j$. Matching powers givesEven and odd coefficients evolve independently. Taking $(a_0,a_1)=(1,0)$ and $(0,1)$ constructs solutionsTheir initial values and derivatives give $W(y_{\mathrm{e}},y_{\mathrm{o}})(0)=1$, so they are linearly independent. These power series of the Chebyshev differential equation span the local solution space. A useful check is $y_{\mathrm{e}}=\cos (n\arcsin x)$ and $y_{\mathrm{o}}=\sin (n\arcsin x)/n$, with $y_{\mathrm{o}}=\arcsin x$ at $n=0$.

For a nonterminating parity series, consecutive nonzero terms satisfyThe ratio test therefore gives $\boxed{R=1}$ for each nonterminating series. The nearest singular coefficient points are $x=\pm 1$, both distance one from the expansion centre.

The polynomial exception is essential here. If $n$ is an integer, the series of the same parity as $|n|$ terminates and has $\boxed{R=\infty }$; the opposite-parity independent series remains nonterminating with radius one. For noninteger $n$, both have radius one. A regular singular point can be removable for one particular solution, so its presence does not force every individual solution to have finite convergence radius.

If $N=|n|$ is an integer, the recurrence has zero numerator at $j=N$. Choosing the parity containing $a_N$ therefore makes all later coefficients vanish. Conversely, for a nonzero polynomial of degree $N$, the coefficient of $x^N$ in the differential equation is $(n^2-N^2)a_N$, so $n^2=N^2$ is necessary.

Normalizing the terminating series by $y(1)=1$ gives the first Chebyshev polynomials:Substitution verifies each differential equation and endpoint value. More generally, $T_N(x)=\cos (N\arccos x)$; its recurrence $T_{N+1}=2x{T}_N-T_{N-1}$ also establishes polynomiality and the normalization.