Character large sieve 2026-10-06
For coefficients supported on consecutive integers,The star restricts to primitive Dirichlet characters. Their finite Fourier transform of a primitive Dirichlet character has normalization of absolute value . Orthogonality of Dirichlet characters therefore bounds the weighted contribution for one modulus by . Sum over and use the exponential-sum large sieve on the distinct reduced fractions of denominators at most .
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 27 1 c Solution Created 2026-10-03 Updated 2026-10-06
For coefficients supported on an interval of length , write and . The variance form of the large sieve statesThe constant is absolute. One may take the explicit right side , by orthogonality of roots of unity and the exponential-sum large sieve proved in Question 2.
Take , , and the indicator function of the -smooth numbers up to . Applying the given smooth-number density with parameter giveswith a harmless adjustment of the constant for integer endpoints. If an odd prime has least quadratic nonresidue , then : a quadratic nonresidue always occurs among . Every prime factor of every selected smooth number is thus a nonzero quadratic residue modulo . By the multiplicativity of the Legendre symbol, every selected number is a nonzero quadratic residue modulo .
There are nonzero quadratic nonresidue classes, and on all of them. Their contribution to the variance is at leastIf denotes the number of exceptional primes, the variance form of the large sieve, with , yields . ConsequentlyThis is the bounded exceptional primes for least quadratic nonresidues argument. Using an interval of length is what matches the term; an interval of length would not give a bounded exceptional set.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 27 2 a Solution Created 2026-10-03 Updated 2026-10-06
The points are -spaced if their circular spacing satisfies for , where is distance to the nearest integer. Ordinary distance on the real line would be insufficient because the complex exponential is periodic.
Let and . Multiplication by this unit-modulus factor leaves unchanged and places the frequencies of in . Put . The permitted Sobolev–Gallagher inequality, in the form needed here, isFor , the arcs about the have disjoint interiors on the circle group. Summing and applying the Cauchy-Schwarz inequality givesThe Cauchy-Schwarz inequality here follows by expanding and minimizing over . For completeness, the finite-interval Parseval identities follow by expanding the squares: is one at and zero at every other integer . Thus and . We obtain the exponential-sum large sieve boundIf , there is at most one point, and the direct Cauchy-Schwarz inequality bound proves the same assertion.
Variance form of the large sieve 2026-10-06
For coefficients supported on an interval of consecutive integers, set and . ThenThe orthogonality of roots of unity gives . The distinct fractions have circular spacing at least ; now apply the exponential-sum large sieve.