The <inverse theorem for trigonometric approximation> gives
$$
\omega(f,n^{-1})\leq\frac Cn\sum_{\nu=0}^nE_\nu(f).
$$
Summing $\nu^{-\alpha}$ proves
$$
\boxed{\omega(f,n^{-1})=
\begin{cases}O(n^{-\alpha}),&0<\alpha<1,\\O(n^{-1}\log n),&\alpha=1.\end{cases}}
$$
For $c_k=a^{-k}$, part (a) gives $E_n(g)\asymp a^{-m}\asymp n^{-\log_5a}$. If $a<5$, an increment $h=\pi/5^{m+2}\leq1/n$ at $x=0$ has nonnegative summands and its $k=m+2$ term is $\asymp n^{-\log_5a}$. Part (c) handles $a=5$. Therefore
$$
\boxed{\omega(g,n^{-1})\asymp
\begin{cases}n^{-\log_5a},&1<a<5,\\n^{-1}\log n,&a=5.\end{cases}}
$$

Solution

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

 Solution

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