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

Table of contents
- Solution c

Solution

Since

5^{k} \equiv 1 (mod 4)

,

g (\frac{π}{2} + \frac{1}{n}) - g (\frac{π}{2}) = - \sum_{k = 0}^{\infty} 5^{- k} sin (5^{k} / n) .

Choose

5^{m} \leq n < 5^{m + 1}

. For

k \leq m

,

sin (5^{k} / n) \geq 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})

. Hence

g (\frac{π}{2} + \frac{1}{n}) - g (\frac{π}{2}) \geq c \frac{lo g n}{n}, ω (g, n^{- 1}) \geq c \frac{lo g n}{n} .

Thus

E_{n} (f) = O (n^{- 1})

does not imply

ω (f, n^{- 1}) = O (n^{- 1})

; the logarithmic gap prevents a characterization of that approximation class by this first modulus alone.

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