Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 1 / c

Let $L=-(1-x^2)D^2+xD$, so $L{y}_n=n^2{y}_n$ and the forced equation is $({\gamma }^2-L)Z=e^{\epsilon x}$. The small-parameter Taylor expansion of the forcing isEach Chebyshev polynomial can be inverted separately under ${\gamma }^2-L$. None of the required denominators vanishes because $\gamma$ is a real noninteger. Thus the regular solution of a forced Chebyshev equation hasHere we select the same regular endpoint domain as in part (a), with finite function values and finite endpoint derivatives. This qualification is needed: the printed forced equation alone does not state endpoint conditions. Without this selection one can add $c_1\cos (\gamma \arccos x)+c_2\sin (\gamma \arccos x)$, which remain finite but generally have singular endpoint derivatives. With $x=\cos \theta$, regularity gives $Z_{\theta }(0)=Z_{\theta }(\pi )=0$; the corresponding homogeneous equation has only the zero solution for noninteger $\gamma$. This makes the displayed expansion unique.

Write the integral's multiplier in the Chebyshev polynomial basis:At ${\gamma }^2=5$, part (i) givesBy orthogonality, the $y_1$ term and all cross terms vanish. Using $I_0=\pi$ and $I_2=\pi /2$, the integral isThe remainder can be integrated against this integrable weight in the regular expansion, so it retains the stated order.