Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 318 / 1

The trigonometric Chebyshev alternation theorem says that $p_n^{\ast }\in {\mathcal{T}}_n$ is best exactly when its error has at least $2n+2$ cyclically ordered extrema of equal magnitude and alternating sign. Let $5^m\le n<5^{m+1}$ and set $x_j=j\pi /5^{m+1}$. For every $k\ge m+1$,ThusThere are $2\cdot 5^{m+1}\ge 2n+2$ such extrema, while the triangle inequality bounds the tail by their common magnitude. Hence

The inverse theorem for trigonometric approximation givesSumming ${\nu }^{-\alpha }$ provesFor $c_k=a^{-k}$, part (a) gives $E_n(g)\asymp a^{-m}\asymp n^{-{\log }_{5}a}$. If $a<5$, an increment $h=\pi /5^{m+2}\le 1/n$ at $x=0$ has nonnegative summands and its $k=m+2$ term is $\asymp n^{-{\log }_{5}a}$. Part (c) handles $a=5$. Therefore

Since $5^k\equiv 1(\mathrm{mod}4)$,Choose $5^m\le n<5^{m+1}$. For $k\le m$, $\sin (5^k/n)\ge c{5}^k/n$, so each of these $m+1$ same-sign terms has magnitude at least $c/n$. The remaining tail is $O(5^{-m})=O(n^{-1})$. HenceThus $E_n(f)=O(n^{-1})$ does not imply $\omega (f,n^{-1})=O(n^{-1})$; the logarithmic gap prevents a characterization of that approximation class by this first modulus alone.