Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 3 / c

Solution

Complete multiplicativity and absolute convergence give the Euler product

D_{f} (s) = \prod_{p} (1 - \frac{f ( p )}{p ^{s}})^{- 1} (ℜ s > 1) .

σ = 1 + 1/ lo g x

. Taking logarithms of absolute values and expanding the local factors gives, uniformly in real

t

lo g ∣ D_{f} (σ + i t) ∣ = \sum_{p} \frac{ℜ ( f ( p ) p ^{- i t} )}{p ^{σ}} + O (1) = \sum_{p \leq x} \frac{ℜ ( f ( p ) p ^{- i t} )}{p} + O (1) .

The prime powers with exponent at least two contribute

O (1)

; changing

p^{- σ}

p^{- 1}

below

x

and estimating the tail above

x

also cost

O (1)

\sum_{p \leq x} \frac{ℜ ( f ( p ) p ^{- i t} )}{p} = lo g lo g x - D (f, n^{i t}; x)^{2} + O (1) .

Exponentiating yields

D_{f} (1 + \frac{1}{l o g x} + i t) ≍ (lo g x) exp (- D (f, n^{i t}; x)^{2}) .

Solved by gpt-5.6-sol high.

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