Why is RH optimum?

Why is RH optimum?

Postby Guest » Wed Feb 13, 2019 2:25 am

FYI: RH (Riemann Hypothesis) means [tex]Re(z_{n}) = 1/2[/tex] for all n, positive integers.

https://en.m.wikipedia.org/wiki/Riemann_hypothesis

Keywords: Fundamental Theorem of Arithmetic

Hmm. A prime number, p, divides any integer, m,
if and only if 1) p = m or 2) [tex]2\le p \le m^{1/2}[/tex].

And of course, if 2) ... is false, then 1) p = m is true. Therefore, we shall consider only 2) ...

That exponent (e =1/2) of m in 2) ... is central because implies the truth of RH.

If we had 0 < e < 1/2, p will not divide all m according to 2) ...


And if we had 1/2 < e < 1, p will divide all m according to 2) ... but e is just too big for our needs here.

Hence, e = 1/2 is best or optimum which means RH is optimum.
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed Feb 13, 2019 9:59 am

Guest wrote:FYI: RH (Riemann Hypothesis) means [tex]Re([tex]z_{n}) = 1/2[/tex] for all n, positive integers.

Note: [tex]z_{n}[/tex] is the nth nontrivial zeta zero of the Riemann Zeta Function. And there is the unique prime number, [tex]p_{n}[/tex], associated with it.

https://en.m.wikipedia.org/wiki/Riemann_hypothesis

Keywords: Fundamental Theorem of Arithmetic

Hmm. A prime number, p, divides any integer, m,
if and only if 1) p = m or 2) [tex]2\le p \le m^{1/2}[/tex].

And of course, if 2) ... is false, then 1) p = m is true. Therefore, we shall consider only 2) ...

That exponent, e =1/2, of m in 2) ... is central because it implies the truth of RH.

If we had 0 < e < 1/2, then p does not divide all m according to 2) ...


And if we had 1/2 < e < 1, then p does divide all m according to 2) ... but e is just too big here.

Hence, e = 1/2 is best or optimum which means RH is optimum.


Please do not delete this post since it corrects the previous post.
Guest
 

Re: Why is RH optimum?

Postby Guest » Tue Mar 12, 2019 10:12 pm

FYI: RH (Riemann Hypothesis) means [tex]Re(z_{n}) = 1/2[/tex] for all n, positive integers.

Note: [tex]z_{n}[/tex] is the nth nontrivial zeta zero of the Riemann Zeta Function. And there is the unique prime number, [tex]p_{n}[/tex], associated with it.

https://en.m.wikipedia.org/wiki/Riemann_hypothesis

Keywords: Fundamental Theorem of Arithmetic

Hmm. A prime number, p, divides any integer, m,
if and only if 1) p = m or 2) [tex]2\le p \le m^{1/2}[/tex].

And of course, if 2) ... is false, then 1) p = m is true. Therefore, we shall consider only 2) ...

That exponent, e =1/2, of m^{1/2} in 2) ... is central because it implies the truth of RH.

If we had 0 < e < 1/2, then p does not divide all m according to 2) ...


And if we had 1/2 < e < 1, then p does divide all m according to 2) ... but e is just too big here.

Hence, e = 1/2 is best or optimum which means RH is optimum.
[/quote]

Please do not delete this post since it corrects the previous post.[/quote]
Guest
 

Re: Why is RH optimum?

Postby Guest » Tue Mar 12, 2019 10:15 pm

Guest wrote:FYI: RH (Riemann Hypothesis) means [tex]Re(z_{n}) = 1/2[/tex] for all n, positive integers.

Note: [tex]z_{n}[/tex] is the nth nontrivial zeta zero of the Riemann Zeta Function. And there is the unique prime number, [tex]p_{n}[/tex], associated with it.

https://en.m.wikipedia.org/wiki/Riemann_hypothesis

Keywords: Fundamental Theorem of Arithmetic

Hmm. A prime number, p, divides any integer, m,
if and only if 1) p = m or 2) [tex]2\le p \le m^{1/2}[/tex].

And of course, if 2) ... is false, then 1) p = m is true. Therefore, we shall consider only 2) ...

That exponent, e =1/2, of [tex]m^{1/2}[/tex] in 2) ... is central because it implies the truth of RH.

If we had 0 < e < 1/2, then p does not divide all m according to 2) ...


And if we had 1/2 < e < 1, then p does divide all m according to 2) ... but e is just too big here.

Hence, e = 1/2 is best or optimum which means RH is optimum.


Please do not delete this post since it corrects the previous post.
[/quote]
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed May 08, 2019 5:45 pm

Furthermore, a nontrivial zeta zero off the critical line (Re([tex]z_{n }[/tex]) = 1/2) does not make any sense. There’s no good reason for it. And it’s pure fiction or fake news too think otherwise.
Guest
 

Re: Why is RH optimum?

Postby Guest » Thu May 09, 2019 1:47 pm

Guest wrote:Furthermore, a nontrivial zeta zero off the critical line (Re([tex]z_{n }[/tex]) = 1/2) does not make any sense. There’s no good reason for it. And it’s pure fiction or fake news too think otherwise.


Yes! Any attempt to refute the famous and important Riemann Hypothesis, Re([tex]z_{n}[/tex]) = 1/2, is a very futile exercise since the hypothesis is true! Please do something worthwhile.

And I can give you n (from one to infinity) reasons why the Riemann Hypothesis is true! Amen!

Moreover, it is unsound (incomplete) to discuss the properties of the nontrivial and simple zeta zeros of the Riemann Zeta Function without discussing the properties of primes.

They are are interdependent or interrelated. Amen!

Note: The nth nontrivial and simple zeta zero exists if and only if the nth prime exists.
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed May 22, 2019 6:10 pm

Keywords: Fundamental Theorem of Arithmetic and the Riemann Hypothesis

Basic and Important Fact Confirming RH (Riemann Hypothesis or Right Hypothesis, Re(z) = 1/2) where z is any nontrivial and simple zero of the Riemann Zeta Function:

For all n > 1, where n is any positive composite number, there exists a prime number, p, such that p|n (p divides n)
where p [tex]≤ n^{1/2}[/tex] (p is less than or equal to the square root of n).

And thus, the exponent of of the expression, [tex]n^{1/2}[/tex], which is 1/2 represents the critical line (Re(z) = 1/2) and confirms RH. And in turn, RH confirms the above basic and important fact. Amen!

The Riemann Hypothesis or the Right Hypothesis is true! Please acknowledge that fact!

David Cole, https://www.researchgate.net/profile/David_Cole29

P.S. I cannot retired from mathematics in peace while there are ‘experts’ of RH who claimed RH is still a open problem. They are either in denial or they do not understand RH!
Guest
 

Re: Why is RH optimum?

Postby Guest » Thu May 23, 2019 7:14 pm

"Don't pay attention to "authorities," think for yourself."

-- Richard P. Feynman, a great scientist.
Guest
 

Re: Why is RH optimum?

Postby Guest » Fri May 24, 2019 1:42 pm

Guest wrote:Keywords: Fundamental Theorem of Arithmetic and the Riemann Hypothesis

Basic and Important Fact Confirming RH (Riemann Hypothesis or Right Hypothesis, Re(z) = 1/2) where z is any nontrivial and simple zero of the Riemann Zeta Function, [tex]\sum_{n=1}^{\infty }n^{-z} = 0[/tex]:

For all n > 1, where n is any positive composite number, there exists a prime number, p, such that p|n (p divides n)
where p [tex]≤ n^{1/2}[/tex] (p is less than or equal to the square root of n).

And thus, the exponent of of the expression, [tex]n^{1/2}[/tex], which is 1/2 represents the critical line (Re(z) = 1/2) and confirms RH. And in turn, RH confirms the above basic and important fact.
Amen!

The Riemann Hypothesis or the Right Hypothesis is true! Please acknowledge that fact!

Moreover, please remember that basic fact in bold font above when thinking about the meaning and the significance of RH.

David Cole, https://www.researchgate.net/profile/David_Cole29

P.S. I cannot retire from mathematics in peace while there are ‘experts’ of RH who claim RH is still a open problem. They are either in denial for personal or political reasons or they do not understand RH!
Guest
 

Re: Why is RH optimum?

Postby Guest » Sat Jun 01, 2019 2:09 pm

Guest wrote:"Don't pay attention to "authorities," think for yourself."

-- Richard P. Feynman, a great scientist.


I wonder why.

I wonder why.

I wonder why the Riemann Hypothesis is true!

I wonder why I wonder...

Now, I know why...

The Riemann Hypothesis is true! Amen!

-- Richard P. Feynman and David Cole.

Relevant Reference Links:

'Richard Feynman: A Life in Science'
by John Gribbin,

https://www.goodreads.com/book/show/56165.Richard_Feynman

https://en.m.wikipedia.org/wiki/Richard_Feynman
Guest
 

Re: Why is RH optimum?

Postby Guest » Tue Jun 04, 2019 7:22 pm

Guest wrote:Keywords: Fundamental Theorem of Arithmetic and the Riemann Hypothesis

Basic and Important Fact Confirming RH (Riemann Hypothesis or Right Hypothesis, Re(z) = 1/2) where the complex variable, z, is any nontrivial and simple zero of the Riemann Zeta Function, [tex]\zeta(z) =\sum_{n=1}^{\infty }\frac{1}{n^{z}} = \sum_{n \ is \ prime}^{\infty }\frac{1}{n^{z}} + \sum_{n \ is \ not \ prime}^{\infty }\frac{1}{n^{z}} = 0[/tex]:

For all n > 1, where n is any positive composite number, there exists a prime number, p, such that p|n (p divides n)
where p [tex]≤ n^{1/2}[/tex] (p is less than or equal to the square root of n).

And thus, the exponent of of the expression, [tex]n^{1/2}[/tex], which is 1/2 represents the critical line (Re(z) = 1/2) and confirms RH. And in turn, RH confirms the above basic and important fact.
Amen!

The Riemann Hypothesis or the Right Hypothesis is true! Please acknowledge that fact!

Moreover, please remember that basic fact in bold font above when thinking about the meaning and the significance of RH.

David Cole, https://www.researchgate.net/profile/David_Cole29

P.S. I cannot retire from mathematics in peace while there are ‘experts’ of RH who claim RH is still a open problem. They are either in denial for personal or political reasons or they do not understand RH!
Guest
 

Re: Why is RH optimum?

Postby Guest » Tue Jun 04, 2019 10:09 pm

FYI:

Rieman's Zeta Equation is equivalent to Euler's Equation:

[tex]\zeta(z) = \sum_{n=1}^{\infty }\frac{1}{n^{z}}[/tex]
[tex]= 1 \\\ + \sum_{\forall n \ that's \ prime}^{\infty }\frac{1}{n^{z}} + \sum_{\forall n \ that's \ not \ prime}^{\infty }\frac{1}{n^{z}} = 1 + e^{iπ} = 0[/tex].
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed Jun 05, 2019 1:35 am

Guest wrote:FYI:

Rieman's Zeta Equation (RZE) is equivalent to Euler's Equation (ER):

[tex]\zeta(z) = \sum_{n=1}^{\infty }\frac{1}{n^{z}}[/tex]
[tex]\\\ = 1 + \sum_{\forall n \ that's \ prime}^{\infty }\frac{1}{n^{z}} \ + \
\sum_{\forall n \ that's \ not \ prime}^{\infty }\frac{1}{n^{z}} \ \\ = 1 + e^{iπ} \\ = 1 - 1 = 0[/tex].


Hmm. Since RZE has infinitely many unique solutions, we must adapt ER accordingly:

[tex]\zeta(z) = \sum_{n=1}^{\infty }\frac{1}{n^{z}}[/tex]
[tex]\\\ = 1 + \sum_{\forall n \ that's \ prime}^{\infty }\frac{1}{n^{z}} \ + \
\sum_{\forall n \ that's \ not \ prime}^{\infty }\frac{1}{n^{z}} \\ = 1 + e^{i(2k + 1)π} \\ = 1 - 1 = 0[/tex] for all integers, [tex]k \ge 0[/tex].
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed Jun 05, 2019 1:46 am

Guest wrote:
Guest wrote:FYI:

Rieman's Zeta Equation (RZE) is equivalent to Euler's Equation (EE):

[tex]\zeta(z) = \sum_{n=1}^{\infty }\frac{1}{n^{z}}[/tex]
[tex]\\\ = 1 + \sum_{\forall n \ that's \ prime}^{\infty }\frac{1}{n^{z}} \ + \
\sum_{\forall n \ that's \ not \ prime}^{\infty }\frac{1}{n^{z}} \ \\ = 1 + e^{iπ} \\ = 1 - 1 = 0[/tex].


Hmm. Since RZE has infinitely many unique solutions, we must adapt EE accordingly:

[tex]\zeta(z) = \sum_{n=1}^{\infty }\frac{1}{n^{z}}[/tex]
[tex]\\\ = 1 + \sum_{\forall n \ that's \ prime}^{\infty }\frac{1}{n^{z}} \ + \
\sum_{\forall n \ that's \ not \ prime}^{\infty }\frac{1}{n^{z}} \\ = 1 + e^{i(2k + 1)π} \\ = 1 - 1 = 0[/tex] for all integers, [tex]k \ge 0[/tex].
Guest
 

Re: Why is RH optimum?

Postby Guest » Fri Jun 07, 2019 9:33 pm

Minor Update:

[tex]\zeta(z) = \sum_{n=1}^{\infty }\frac{1}{n^{z}}[/tex]
[tex]\\\ = 1 + \sum_{\forall n \ that's \ prime}^{\infty }\frac{1}{n^{z}} \ + \
\sum_{\forall n > 1\ that's \ not \ prime}^{\infty }\frac{1}{n^{z}} \\ = 1 + e^{i(2k -1)π} \\ = 1 - 1 = 0[/tex] for all integers, [tex]k \ge 1[/tex].
Guest
 

Re: Why is RH optimum?

Postby Guest » Fri Jun 07, 2019 10:39 pm

A Recap On Why The Riemann Hypothesis Is True:

There are infinitely many nontrivial and simple zeros of the complex Riemann Zeta Function,

[tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex],

whose part equals one-half or Re(z)= 1/2, and there are infinitely many primes. Every nontrivial and simple zero has a unique simple prime associated with it according to the following important equation:

[tex]f(n, p_n, z_n) =[/tex] [tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} m^{-z_n} = 0[/tex]

where [tex]z_n[/tex] is the nth nontrivial and simple zero of the Riemann Zeta Function, and [tex]p_n[/tex] is the nth simple prime.

Moreover, the wonderful Harmonic Series is the prime source of analysis for the great results, the Riemann Hypothesis and the Prime Number Theoreom...

David Cole.
Guest
 

Re: Why is RH optimum?

Postby Guest » Sat Jun 08, 2019 12:36 am

Guest wrote:A Recap On Why The Riemann Hypothesis Is True:

There are infinitely many nontrivial and simple zeros of the complex Riemann Zeta Function,

[tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex],

whose part equals one-half or Re(z)= 1/2, and there are infinitely many primes. Every nontrivial and simple zero has a unique simple prime associated with it according to the following important equation:

[tex]f(n, p_n, z_n) =[/tex] [tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} m^{-z_n} = 0[/tex]

where [tex]z_n[/tex] is the nth nontrivial and simple zero of the Riemann Zeta Function, and [tex]p_n[/tex] is the nth simple prime.

Moreover, the wonderful Harmonic Series is the prime source of analysis for the great results, the Riemann Hypothesis and the Prime Number Theoreom...

David Cole.


Note: The important complex equation,

[tex]f(n, p_n, z_n) =[/tex] [tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} m^{-z_n} = 0[/tex]

is derived from the complex Riemann Zeta Function.
Guest
 

Re: Why is RH optimum?

Postby Guest » Sat Jun 08, 2019 12:48 am

Guest wrote:
Guest wrote:A Recap On Why The Riemann Hypothesis Is True:

There are infinitely many nontrivial and simple zeros of the complex Riemann Zeta Function,

[tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex],

whose part equals one-half or Re(z)= 1/2, and there are infinitely many primes. Every nontrivial and simple zero has a unique simple prime associated with it according to the following important equation:

[tex]f(n, p_n, z_n) =[/tex] [tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} m^{-z_n} = 0[/tex]

where [tex]z_n[/tex] is the nth nontrivial and simple zero of the Riemann Zeta Function, and [tex]p_n[/tex] is the nth simple prime.

Moreover, the wonderful Harmonic Series is the prime source of analysis for the great results, the Riemann Hypothesis and the Prime Number Theorem...

David Cole.


Note: The important complex equation,

[tex]f(n, p_n, z_n) =[/tex] [tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} m^{-z_n} = 0[/tex]

is derived from the complex Riemann Zeta Function.
Guest
 

Re: Why is RH optimum?

Postby Guest » Sat Jun 08, 2019 2:17 pm

There is a great deal of important mathematical literature which depends on the assumption of the Riemann Hypothesis. And that is the right assumption to make since the Riemann Hypothesis is true!

In Lord GOD and in the Riemann Hypothesis, we trust! Amen!
Guest
 

Re: Why is RH optimum?

Postby Guest » Sun Jun 09, 2019 9:40 pm

Given the values, [tex]z_{n}[/tex] and [tex]p_{n}[/tex], how do we correctly derive equation two,

2. [tex]f(n, p_n, z_n) =[/tex] [tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} m^{-z_n} = 0[/tex],

from equation one,

1. [tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex] = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate real value, x.
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