Since $5^k\equiv1\pmod4$,
$$
g\left(\frac\pi2+\frac1n\right)-g\left(\frac\pi2\right)
=-\sum_{k=0}^\infty5^{-k}\sin(5^k/n).
$$
Choose $5^m\leq n<5^{m+1}$. For $k\leq m$, $\sin(5^k/n)\geq c5^k/n$, so each of these $m+1$ same-sign terms has magnitude at least $c/n$. The remaining tail is $O(5^{-m})=O(n^{-1})$. Hence
$$
\boxed{\left|g\left(\frac\pi2+\frac1n\right)-g\left(\frac\pi2\right)\right|
\geq c\frac{\log n}{n}},\qquad
\boxed{\omega(g,n^{-1})\geq c\frac{\log n}{n}}.
$$
Thus $E_n(f)=O(n^{-1})$ 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-318/1/c/solution