# Difference between revisions of "Riemann Hypothesis"

m |
|||

Line 11: | Line 11: | ||

* <math>|M(x)|\le cx^{1/2+\epsilon}</math> for a constant c and where <math>M(x)=\sum_{n\le x}\mu(n)</math> | * <math>|M(x)|\le cx^{1/2+\epsilon}</math> for a constant c and where <math>M(x)=\sum_{n\le x}\mu(n)</math> | ||

− | * <math>\sigma(n) < e^\gamma n\ | + | * <math>\sigma(n) < e^\gamma n\ log_{log_n}</math> for all <math>n > 5040</math>. This being equivalent to the Riemann Hypothesis is Robin's Theorem, proved in 1984 by Guy Robin. |

==Links== | ==Links== |

## Latest revision as of 12:01, 20 October 2016

The **Riemann Hypothesis** is a famous conjecture in analytic number theory that states that all nontrivial zeros of the Riemann zeta function have real part . From the functional equation for the zeta function, it is easy to see that when . These are called the trivial zeros. This hypothesis is one of the seven millenium questions.

The Riemann Hypothesis is an important problem in the study of prime numbers. Let denote the number of primes less than or equal to *x*, and let . Then an equivalent statement of the Riemann hypothesis is that .

One fairly obvious method to prove the Riemann Hypothesis is to consider the reciprocal of the zeta function, , where refers to the Möbius function. Then one might try to show that admits an analytic continuation to . Let be the Mertens function. It is easy to show that if for sufficiently large , then the Riemann Hypothesis would hold. The Riemann Hypothesis would also follow if for any constant .

Some equivalent statements of the Riemann Hypothesis are

- The zeta function has no zeros with real part between and 1
- has all nontrivial zeros on the line
- All nontrivial zeros of all L-series have real part one half where an L-series is of the form . This is the generalized Riemann Hypothesis because in the Riemann Hypothesis, is 1 for all n
- for a constant c and where

- for all . This being equivalent to the Riemann Hypothesis is Robin's Theorem, proved in 1984 by Guy Robin.