Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 3 / a

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- Solution a

Solution

With

μ_{\leq U} (d) = μ (d) 1_{d \leq U}

and analogous notation for

Λ_{\leq V}

, Vaughan identity is

Λ = Λ_{\leq V} + μ_{\leq U} * lo g - μ_{\leq U} * Λ_{\leq V} * 1 + μ_{> U} * Λ_{> V} * 1.

Thus, when

U V \leq N

,

n \leq N \sum Λ (n) e (α n^{2}) = n \leq V \sum Λ (n) e (α n^{2}) + d \leq U \sum μ (d) m \leq N / d \sum (lo g m) e (α d^{2} m^{2}) - d \leq U \sum μ (d) m \leq V \sum Λ (m) r \leq N / (d m) \sum e (α d^{2} m^{2} r^{2}) + d > U, m > V, r \geq 1 d m r \leq N \sum μ (d) Λ (m) e (α d^{2} m^{2} r^{2}) .

The first term is short, the next two are Type I sums, and the last becomes a Type II bilinear sum after grouping variables and applying a dyadic decomposition.

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