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Old 2005-10-15, 07:28   #34
akruppa
 
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Nowadays a complex variable as argument of a function is usually called "z", while "x" is typically used for functions in a real variable. The argument of Riemann's zeta function is complex, but he used "s" and the notation stuck.

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Old 2005-10-15, 17:48   #35
Citrix
 
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If I wanted to write a program to test the hypothesis, how would I calculate s? I do not understand how to get the value of s. Please help
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Old 2005-10-15, 18:13   #36
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Don't you understand the previous three posts?

Do you know what a function is?

Do you know what a variable is? If so, do you know what an independent variable is?

Do you know what the Riemann Hypothesis says?

Last fiddled with by jinydu on 2005-10-15 at 18:13
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Old 2005-10-15, 18:20   #37
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Don't you understand the previous three posts?
Yes

Do you know what a function is?
yes

Do you know what a variable is? If so, do you know what an independent variable is?
An independent variable is any integer? Any complex number? I do not know what values s can range from?

Do you know what the Riemann Hypothesis says?

I am not sure I understand this. That is why I am asking.
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Old 2005-10-15, 19:00   #38
xilman
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Quote:
Originally Posted by Citrix
Don't you understand the previous three posts?
Yes

Do you know what a function is?
yes

Do you know what a variable is? If so, do you know what an independent variable is?
An independent variable is any integer? Any complex number? I do not know what values s can range from?

Do you know what the Riemann Hypothesis says?

I am not sure I understand this. That is why I am asking.
In this particular case, "s" is the independent variable and it can take any complex value.

That was explained earlier in the thread but perhaps the statement wasn't explicit enough for you.

Paul
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Old 2005-10-15, 22:05   #39
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The Riemann Hypothesis is about the non-trivial zeros of the function zeta.

This means that we need to find the values s such that \large \zeta(s) = 0.

The trivial zeros are -2, -4, -6, ... and all negative even integers.

The non-trivial zeros are located (according to this Hypothesis) in complex values of s for which the real part equals 1/2.

Up to this moment we know that these zeros are located in the critical strip 0 < Re(s) < 1.

The infinite sum given by the original poster does not converge in the range where the RH applies.

Last fiddled with by alpertron on 2005-10-15 at 22:08 Reason: Corrected real part of non-trivial zeros
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Old 2005-10-16, 00:40   #40
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To DISPROVE the RH it would be sufficient to find any zero OFF the line.

Computers have been used to find many many zeroes.

But they cannot verify that infinite zeroes all lie on the line thus cannot alone PROVE the RH.

There are good shortcut tricks to locate zeroes with faster math.

Good treatment of all this in Crandall & Pomerance "Prime Numbers: A computational approach". Also try looking in wikipaedia.
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Old 2005-10-16, 02:00   #41
Citrix
 
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Everyone, Thank you very much. I have found the original manuscript and am trying to read it. If I have any more questions I will post here.

Thank you,
Citrix
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Old 2006-03-27, 17:52   #42
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Default Official Riemann Hypothesis thread

An intresting article about the Riemann Hypothesis:

Prime Numbers Get Hitched

Marcus du Sautoy, professor of mathematics at the University of Oxford, claims that mathematicians have discovered connections between zeros of Riemann's Zeta function and ... energy levels in the nucleus of heavy atoms. Moreover it turns out that physicists from well know facts could foretell the mathematicians unknown before properties of Zeta function.
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Old 2007-08-06, 03:58   #43
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Quote:
Originally Posted by alpertron View Post
The Riemann Hypothesis is about the non-trivial zeros of the function zeta.

This means that we need to find the values s such that \large \zeta(s) = 0.

The trivial zeros are -2, -4, -6, ... and all negative even integers.

The non-trivial zeros are located (according to this Hypothesis) in complex values of s for which the real part equals 1/2.

Up to this moment we know that these zeros are located in the critical strip 0 < Re(s) < 1.

The infinite sum given by the original poster does not converge in the range where the RH applies.
Can anyone please say how to compute Zeta for complex numbers?

For example, you say -2 is a zero of Zeta function
Zeta(n) = 1/1^n + 1/2^n + 1/3^n + ...

Thus
Zeta(-2) = 1 + 4 + 9 + ... The sum diverges
How do you say Zeta(-2) is zero?

Zeta(-1) = 1 + 1 + 1 + ... Also diverges
You say that -1 is not a zero.

How to evaluate Zeta off the critical line, ReP(z)=1/2?

What is the first zero of the Zeta Function in the critical line
(positive imaginary part)?

How do you evaluate that??
For example, if I take 1/2 + i

1 + 1/2^(1/2+i) + 1/3^(1/2+i) + ...

1 + 1/sqrt(2).2^i + 1/sqrt(3).3^i + ...

1 + 1/sqrt(2).(cos ln 2 + i sin ln 2) + 1/sqrt(3).(cos ln 3 + i sin ln 3) + ...

It is so sophisticated that the result doesn't seem to be real for any value of the imaginary part.

Give example of evaluation of first zero of Zeta function
and how to compute other zeros.

King
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Old 2007-08-06, 21:36   #44
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If I may ask a totally ignorant question:

In the very first post of this thread, someone said that if the Riemann(sp?) Hypothesis is proven, it would greatly speed up our search. My ignorant question is:

If the Riemann Hypothesis is wrong, what would be the consequences of using it? Note that I have no idea how it would be used, which is the reason I call this an ignorant question.

If the use of the Riemann Hypothesis causes the inclusion or exclusion of prime exponents in the Mersenne equation, perhaps the question of whether or not someone wants the Prime95 program to include the Hypothesis' findings in exponent selection could be an option.

If this is gibberish(I know some people like to use that word :) ) I apologize.
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