Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 2

The Chebyshev alternation theorem says that $p^{\ast }\in {\mathcal{P}}_n$ is the unique best uniform approximation to $f\in C[-1,1]$ if and only if there are at least $n+2$ ordered pointsand a sign $\epsilon \in \{-1,1\}$ such thatThus the extremal error has common magnitude and alternating sign at $n+2$ points.

Suppose for contradiction that some $q\in {\mathcal{P}}_n$ satisfiesAt $t_i$,Because $|f(t_i)-q(t_i)|<a_i$, the values $p(t_i)-q(t_i)$ have strictly alternating signs. The intermediate value theorem therefore gives at least one root of $p-q$ in each of the $n+1$ intervals $(t_i,t_{i+1})$. A nonzero polynomial of degree at most $n$ cannot have $n+1$ distinct roots. If $p-q$ were identically zero, its error at $t_i$ would be $a_i\ge a_{\ast }$, also a contradiction. Hence

For any $n\ge 0$, take $f_n=T_{n+1}$, the Chebyshev polynomial of degree $n+1$. It has $n+2$ alternating extrema of magnitude one on $[-1,1]$. The Chebyshev alternation theorem shows that the zero polynomial is best from ${\mathcal{P}}_n$, with error one. Since it also lies in ${\mathcal{P}}_{n-1}$ when $n\ge 1$,

Now suppose $f^{(n)}(x)>0$ throughout $[-1,1]$ and, contrary to the claim, $E_{n-1}(f)=E_n(f)$. A best $p\in {\mathcal{P}}_{n-1}$ would then also be best in ${\mathcal{P}}_n$. Its error $e=f-p$ would have $n+2$ alternating extrema by the alternation theorem, hence at least $n+1$ distinct zeros. Applying the Rolle theorem $n$ times gives a zero ofcontradicting positivity. Therefore

For $n\ge 1$, choose the integer $m$ withand define the partial sumIt belongs to ${\mathcal{P}}_n$, and the triangle inequality givesAt the $3^{m+1}+1$ pointsevery tail term has the same alternating sign becauseThusThere are at least $n+2$ such points because $n<3^{m+1}$. The Chebyshev alternation theorem provesFor $n=0$, the partial sum is empty and the same argument at $x=\pm 1$ gives $p_0=0$ and $E_0(f_0)={\sum }_{k=0}^{\infty }{a}_k$.