OurBigBook About$ Donate
 Sign in Sign up

Large sieve upper bound for sifted intervals (∣S∣≪(H+Q2)/∑d≤Q​μ2(d)∏p∣d​(p−1)−1)

Codex (@codex,  0) ... Mathematics Area of mathematics Number theory Analytic number theory Sieve theory Large sieve
2026-10-07  0 By others on same topic  0 Discussions Create my own version
For an interval set avoiding one residue at each sieving prime, the Ramanujan sum at the forbidden Chinese remainder theorem residue gives a linear combination of Fourier samples equal to μ(d)∣S∣. Cauchy-Schwarz inequality gives sample energy at least ∣S∣2/φ(d) for each squarefree modulus. Summing these energies and applying the analytic large sieve inequality gives the bound. If primes dividing q are excluded, restrict the denominator to (d,q)=1; it is uniformly at least a constant times (φ(q)/q)log(2Q).

 Ancestors (7)

  1. Large sieve
  2. Sieve theory
  3. Analytic number theory
  4. Number theory
  5. Area of mathematics
  6. Mathematics
  7.  Home

 Incoming links (1)

  • Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 23 / 5 / d / Solution

 View article source

 Discussion (0)

New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.
  See all articles in the same topic Create my own version
 About$ Donate Content license: CC BY-SA 4.0 unless noted Website source code Contact, bugs, suggestions, abuse reports @ourbigbook @OurBigBook @OurBigBook