Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 2G (Topics in Analysis)

The $n$th Chebyshev polynomial is defined on $[-1,1]$ byThe cosine addition formula givesInduction therefore shows that every $T_n$ is a polynomial. Its defining cosine formula also gives

The assertion is clear for $T_1=x$. If $T_n$ has degree $n$ and leading coefficient $2^{n-1}$, the recurrence shows that the leading term of $T_{n+1}$ isthe lower-degree polynomial $T_{n-1}$ cannot cancel it. Hence

For $x=\cos \theta \in [-1,1]$,Both sides are polynomials in $x$. Since they agree on an interval, the polynomial identity theorem extends the equality to every real $x$. Thus $T_n$ is even for even $n$ and odd for odd $n$.

The $n$ roots of $T_n$ are distinct and lie in $(-1,1)$. Repeated use of Rolle's theorem shows that every root of $T_n^{(k)}$, for $0\le k<n$, also lies in $(-1,1)$. Its leading coefficient is positive, soTaking $k=r+1$ shows thatfor $0\le r\le n-1$; when $r=n-1$, the derivative is the positive constant $T_n^{(n)}$. This is the positivity of Chebyshev derivatives beyond the unit interval.

Writing $x=\cos \theta$ and taking the limit as $\theta \to 0$ givesPart (d), with $r=1$, shows that $T_n^{\prime }$ is increasing on $[1,\infty )$, soSince $T_n(1)=1$ and $n^2\ge n$,