Difference between revisions of "Riemann Hypothesis"

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The '''Riemannn Hypothesis''' is a yet unproven conjecture stating that all interesting solutions to <math>\zeta (x) = 0</math> lie on the vertical line denoting <math>\frac{1}{2}</math> in the four-dimensional graph of the [[harmonic series|zeta function]] with a [[complex number]] as input. This hypothesis is one of the [http://www.claymath.org/millennium/ millenium questions].
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The '''Riemannn Hypothesis''' is a well-known [[conjecture]] in [[analytic number theory]] that states that all nontrivial [[zero]]s of the [[Riemann zeta function]] have real part <math>1/2</math>. From the [[functional equation]] for the zeta function, it is easy to see that <math>\zeta(s)=0</math> when <math>s=-2,-4,-6,\ldots</math> These are called the trivial zeros. This hypothesis is one of the seven [http://www.claymath.org/millennium/ millenium questions].
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The Riemann Hypothesis is an important problem in the study of prime numbers. Let <math>\pi(x)</math> denote the number of primes less than or equal to ''x'', and let <math>\mathrm{Li}(x)=\int_2^x \frac{1}{\ln t}\; dt</math>. Then an equivalent statement of the Riemann hypothesis is that <math>\pi(x)=\mathrm{Li}(x)+O(x^{1/2}\ln(x))</math>.
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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 [[Mobius function]]. (My laptop is acting up, so I'm having trouble getting the umlaut on the o. Can someone help me? Thanks.) 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.
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==Links==
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[http://www.dtc.umn.edu/~odlyzko/doc/arch/mertens.disproof.pdf Disproof of the Mertens Conjecture]

Revision as of 11:53, 24 June 2006

The Riemannn 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 Mobius function. (My laptop is acting up, so I'm having trouble getting the umlaut on the o. Can someone help me? Thanks.) 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.

Links

Disproof of the Mertens Conjecture