Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-25/3/c/solution
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 25 3 c Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
Put and , for sufficiently large . Apply the Landau zero-free-region theorem withwhere is fixed and small enough that . On its two discs, the real part is at least and the imaginary part is comparable to . The given Richert bound for the Riemann zeta function therefore gives, on the part left of one,On the part right of one, the separately given bound gives the same conclusion. We may thus choose for one fixed sufficiently large . Also , so the logarithmic term in the Landau zero-free-region theorem is . Its conclusion isfor large and a sufficiently small positive . Complex conjugation supplies negative heights. This is the Vinogradov-Korobov zero-free region. Only the stated Richert upper bounds, the Euler product, the pole at one and the proved Landau zero-free-region theorem were used; no prior zero-free-region theorem was assumed.
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