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

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