Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 1 / iii / Solution

Set $x=\cos \theta$ and $Y(\theta )=y(\cos \theta )$, with $0\le \theta \le \pi$. The chain rule converts $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$ to $Y^{\prime \prime }$, so $Y=\cos (n\theta )$ solves $Y^{\prime \prime }+n^{2}Y=0$. Hence $y_n(x)=\cos (n\arccos x)$ is a solution in the interior.

For nonnegative integer $n$, De Moivre's theorem makes its polynomial nature explicit:Every term is a polynomial of degree at most $n$. For $n\ge 1$, the coefficient of $x^n$ is ${\sum }_j\left(\genfrac{}{}{0ex}{}{n}{2j}\right)=2^{n-1}$, so its polynomial degree is exactly $n$. These are the Chebyshev polynomials $T_n$. Their polynomial form also proves the required boundedness of the function and derivative at both endpoints. Negative labels reproduce $T_{|n|}$ and have degree $|n|$.

The first three areUnder $wdx=-d\theta$, their cross inner products become integrals of $1$, $\cos \theta$ and $\cos 2\theta$ on $(0,\pi )$. The constant-cosine integrals vanish, and $\cos \theta \cos 2\theta =(\cos \theta +\cos 3\theta )/2$ also integrates to zero. Thus all three pairs are orthogonal. Their squared weighted norms are $\pi ,\pi /2,\pi /2$, respectively; more generally