Difference between revisions of "Riemann Hypothesis"

Revision as of 09:28, 21 August 2006

The Riemann Hypothesis is a well-known conjecture in analytic number theory that states that all nontrivial zeros of the Riemann zeta function have real part $1/2$ . From the functional equation for the zeta function, it is easy to see that $\zeta(s)=0$ when $s=-2,-4,-6,\ldots$ 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 $\pi(x)$ denote the number of primes less than or equal to x, and let $\mathrm{Li}(x)=\int_2^x \frac{1}{\ln t}\; dt$ . Then an equivalent statement of the Riemann hypothesis is that $\pi(x)=\mathrm{Li}(x)+O(x^{1/2}\ln(x))$ .

One fairly obvious try to prove the Riemann Hypothesis (which unfortunately doesn't work) is to consider the reciprocal of the zeta function, $\frac{1}{\zeta(s)}{=}\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$ , where $\mu(n)$ refers to the Möbius function. Then one might try to show that $\frac{1}{\zeta(s)}$ admits an analytic continuation to $\Re(s)>\frac{1}{2}$ . Let $M(n)=\sum_{i=1}^n \mu(i)$ be the Mertens function. It is easy to show that if $M(n)\le\sqrt(n)$ for sufficiently large $n$ , then the Riemann Hypothesis would hold. However, A. M. Odlyzko and H. J. J. te Riele showed that this conjecture is in fact false. The Riemann Hypothesis would also follow if $M(n)\le C\sqrt{n}$ for any constant $C$ ; however, this is believed to be false as well.

Some equivalent statements of the Riemann Hypothesis are

The zeta function has no zeros with real part between $\frac{1}{2}$ and 1
$\zeta_a(s)$ has all nontrivial zeros on the line $Re(s)=\frac{1}{2}$
All nontrivial zeros of all L-series have real part one half where an L-series is of the form $\displaystyle\sum_{n=1}^\infty \frac{a_n}{n^s}$ . This is the generalized Riemann Hypothesis because in the Riemann Hypothesis, $a_n$ is 1 for all n
$|M(x)|\le cx^{1/2+\epsilon}$ for a constant c and where $M(x)=\displaystyle\sum_{n\le x}\mu(n)$

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Disproof of the Mertens Conjecture

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 One fairly obvious try to prove the Riemann Hypothesis (which unfortunately doesn't work) is to consider the reciprocal of the zeta function, <math>\frac{1}{\zeta(s)}{=}\sum_{n=1}^\infty \frac{\mu(n)}{n^s}</math>, where <math>\mu(n)</math> refers to the [[Möbius function]]. Then one might try to show that <math>\frac{1}{\zeta(s)}</math> admits an [[analytic continuation]] to <math>\Re(s)>\frac{1}{2}</math>. Let <math>M(n)=\sum_{i=1}^n \mu(i)</math> be the [[Mertens function]]. It is easy to show that if <math>M(n)\le\sqrt(n)</math> for sufficiently large <math>n</math>, then the Riemann Hypothesis would hold. However, A. M. Odlyzko and H. J. J. te Riele showed that this conjecture is in fact false. The Riemann Hypothesis would also follow if <math>M(n)\le C\sqrt{n}</math> for any constant <math>C</math>; however, this is believed to be false as well.
+Some equivalent statements of the Riemann Hypothesis are
+* The zeta function has no zeros with real part between <math>\frac{1}{2}</math> and 1
+* <math>\zeta_a(s)</math> has all nontrivial zeros on the line <math>Re(s)=\frac{1}{2}</math>
+* All nontrivial zeros of all L-series have real part one half where an L-series is of the form <math>\displaystyle\sum_{n=1}^\infty \frac{a_n}{n^s}</math>.  This is the generalized Riemann Hypothesis because in the Riemann Hypothesis, <math>a_n</math> is 1 for all n
+* <math>|M(x)|\le cx^{1/2+\epsilon}</math> for a constant c and where <math>M(x)=\displaystyle\sum_{n\le x}\mu(n)</math>
 ==Links==
 [http://www.dtc.umn.edu/~odlyzko/doc/arch/mertens.disproof.pdf Disproof of the Mertens Conjecture]

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Difference between revisions of "Riemann Hypothesis"

Revision as of 09:28, 21 August 2006

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