Why is RH optimum?

Re: Why is RH optimum?

Postby Guest » Sun Jun 09, 2019 10:23 pm

Some Food for Thought:

'Prime-counting function', by
https://en.m.wikipedia.org/wiki/Prime-counting_function;

'Computing [tex]\pi(x)[/tex] Analytically' by
Prof. David J. Platt.

Please see attached file for details.
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Guest
 

Re: Why is RH optimum?

Postby Guest » Sun Jun 09, 2019 11:14 pm

Guest wrote: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} \frac{1}{m^{z_n}}[/tex], \Leftrightarrow

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.


The derivation question is not difficult to answer as we shall soon see.

[tex]\zeta(z_{n} ) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}}[/tex]
=
Guest
 

Re: Why is RH optimum?

Postby Guest » Sun Jun 09, 2019 11:20 pm

We are having some technical difficulties cutting and pasting... We shall try again tomorrow.
Guest
 

Re: Why is RH optimum?

Postby Guest » Mon Jun 10, 2019 10:39 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} \frac{1}{m^{z_n}}[/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 positive real value, x.


The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and [tex]p_n[/tex]), Simple Nontrivial Zeros (z and [tex]z_n[/tex]) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p [tex]≤ k^{1/2}[/tex].


HS (v1): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n}[/tex];

RZF (v1): [tex]\ \ \zeta(z_{n} ) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}}[/tex].

HS (v2): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n}[/tex];

RZF (v2): [tex]\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}}[/tex].

We need to check our derivations. And we have more work...
Guest
 

Re: Why is RH optimum?

Postby Guest » Mon Jun 10, 2019 10:42 pm

Guest wrote: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} \frac{1}{m^{z_n}}[/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 positive real value, x.


The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and [tex]p_n[/tex]), Simple Nontrivial Zeros (z and [tex]z_n[/tex]) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p [tex]≤ k^{1/2}[/tex].


HS (v1): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

RZF (v1): [tex]\ \ \zeta(z_{n} ) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex].

HS (v2): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty[/tex];

RZF (v2): [tex]\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = 0[/tex].

We need to check our derivations. And we have more work...
Guest
 

Re: Why is RH optimum?

Postby Guest » Mon Jun 10, 2019 11:26 pm

Guest wrote:
Guest wrote: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} \frac{1}{m^{z_n}}[/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 positive real value, x.


The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and [tex]p_n[/tex]), Simple Nontrivial Zeros (z and [tex]z_n[/tex]) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p [tex]≤ k^{1/2}[/tex].


HS (v1): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

RZF (v1): [tex]\ \ \zeta(z_{n} ) = \ \ \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex].

HS (v2): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty[/tex];

RZF (v2): [tex]\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = 0[/tex].

We need to check our derivations. And we have more work...
Guest
 

Re: Why is RH optimum?

Postby Guest » Tue Jun 11, 2019 3:58 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} \frac{1}{m^{z_n}}[/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 positive real value, x.


The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and [tex]p_n[/tex]), Simple Nontrivial Zeros (z and [tex]z_n[/tex]) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p [tex]≤ k^{1/2}[/tex].


HS (v1): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

RZF (v1): [tex]\ \ \zeta(z_{n} ) = \ \ \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex].

HS (v2): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty[/tex];

RZF (v2): [tex]\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = 0[/tex].

We need to check our derivations. And we have more work...

Confirmed Observations:

The Riemann Hypothesis is extraordinarily exceptional since all nontrivial and simple zeros of RZF are accounted for. And there are no exceptions or 'stray' (off the critical line, Re(z) = 1/2) nontrivial zeros possible.

Furthermore, we have the following equations too:

[tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ = \infty[/tex] and [tex]\ \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

[tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ = \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex];

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty[/tex] and [tex]\ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty[/tex];

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = -1[/tex].

Dave.
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed Jun 12, 2019 5:58 pm

Notes:

From the equations,

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty[/tex] (Euler's Proof) and [tex]\ \ \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty[/tex],

we compute,

[tex]\lim_{n \to \infty} \frac{\sum_{n=1}^{\infty }\frac{1}{p_n}}{\sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n}} \rightarrow 0[/tex].

Relevant Reference Link:

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed Jun 12, 2019 10:40 pm

Oops!

The equation,

[tex]\sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty[/tex], is wrong!

The above equation is adding too many (infinitely many) identical composite numbers... We shall fix the problem and post a revision soon.
Guest
 

Re: Why is RH optimum?

Postby Guest » Wed Jun 12, 2019 11:20 pm

Guest wrote:Oops!

The equation,

[tex]\sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty[/tex], is wrong!

The above equation is adding too many (infinitely many) identical composite numbers... We shall fix the problem and post a revision soon.


The correct equation is

[tex]\sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} = \infty[/tex].
Guest
 

Re: Why is RH optimum?

Postby Guest » Thu Jun 13, 2019 1:10 am

An Update:

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} \frac{1}{m^{z_n}}[/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 positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and [tex]p_n[/tex]), Simple Nontrivial Zeros (z and [tex]z_n[/tex]) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p [tex]≤ k^{1/2}[/tex].


HS (v1): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

RZF (v1): [tex]\ \ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex].

HS (v2): [tex]\sum_{k=1}^{\infty }\frac{1}{k} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ \ = \infty[/tex];

RZF (v2): [tex]\ \ \zeta(z) = \ \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{{p_n}^{z}} \ \ \ = 0[/tex].

We need to check our derivations. And we have more work...

Confirmed Observations:

The Riemann Hypothesis is extraordinarily exceptional since all nontrivial and simple zeros of RZF are accounted for. And there are no exceptions or 'stray' (off the critical line, Re(z) = 1/2) nontrivial zeros possible.

Furthermore, we have the following equations too:

[tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ = \infty[/tex] and [tex]\ \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

[tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ = \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex];

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty[/tex] and [tex]\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty[/tex];

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n^{z}} \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ = -1[/tex].

Notes:

From the equations,

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty[/tex] (Euler's Theorem) and [tex]\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty[/tex],

we compute as [tex]n \rightarrow \infty[/tex],

[tex]\ \ \ \frac{\sum_{n=1}^{\infty }\frac{1}{p_n}}{ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}}} \ \ \rightarrow 0[/tex].

Relevant Reference Link:

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
Guest
 

Re: Why is RH optimum?

Postby Guest » Thu Jun 13, 2019 1:15 am

Guest wrote:An Update:

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} \frac{1}{m^{z_n}}[/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 positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and [tex]p_n[/tex]), Simple Nontrivial Zeros (z and [tex]z_n[/tex]) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p [tex]≤ k^{1/2}[/tex].


HS (v1): [tex]\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

RZF (v1): [tex]\ \ \ \zeta(z_n) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex].

HS (v2): [tex]\sum_{k=1}^{\infty }\frac{1}{k} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ \ = \infty[/tex];

RZF (v2): [tex]\ \ \zeta(z) = \ \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{{p_n}^{z}} \ \ \ = 0[/tex].

We need to check our derivations. And we have more work...

Confirmed Observations:

The Riemann Hypothesis is extraordinarily exceptional since all nontrivial and simple zeros of RZF are accounted for. And there are no exceptions or 'stray' (off the critical line, Re(z) = 1/2) nontrivial zeros possible.

Furthermore, we have the following equations too:

[tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ = \infty[/tex] and [tex]\ \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty[/tex];

[tex]\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ = \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0[/tex];

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty[/tex] and [tex]\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty[/tex];

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n^{z}} \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ = -1[/tex].

Notes:

From the equations,

[tex]\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty[/tex] (Euler's Theorem) and [tex]\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty[/tex],

we compute as [tex]n \rightarrow \infty[/tex],

[tex]\ \ \ \frac{\sum_{n=1}^{\infty }\frac{1}{p_n}}{ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}}} \ \ \rightarrow 0[/tex].

Relevant Reference Link:

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
Guest
 

Re: Why is RH optimum?

Postby Guest » Thu Jun 13, 2019 2:06 am

I apologize for the sloppy (mistaken) mathematics posted here. As we know, mathematical research is hard work. So please continue to help me find my flaws whenever possible. Thank you!

The proof of Riemann Hypothesis is the cumulative work of many outstanding or great mathematicians, Euler, Gauss, Riemann, Hardy, Littlewood, Siegel, Erdős, Montgomery (Go Blue!), Lagarias (Go Blue!), Odlyzko, and others. And I am trying my best to make a significant contribution too.

Thank you for your patience! And Go Blue!

David Cole.
Guest
 

Re: Why is RH optimum?

Postby Guest » Fri Jun 14, 2019 5:20 pm

We conclude that the Riemann Hypothesis is true! Do you agree?

David Cole,

https://www.researchgate.net/profile/David_Cole29.
Guest
 

Re: Why is RH optimum?

Postby Guest » Fri Jun 14, 2019 9:26 pm

Guest wrote:We conclude that the Riemann Hypothesis (RH) is true! Do you agree?

David Cole,

https://www.researchgate.net/profile/David_Cole29.


If we are still in doubt about the validity of RH, let's do some computations.

Let z = a + i * b, and let [tex]\bar{z}[/tex] = a - i * b. Note: [tex]\bar{z}[/tex] is the complex conjugate of z.

Therefore, z + [tex]\bar{z}[/tex] = 1 which implies 2a = 1 which implies a = Re(z) = Re([tex]\bar{z})[/tex] = 1/2 which implies RH, and we are done.

Note: z + [tex]\bar{z}[/tex] = 1 implies the Harmonic Series, the prime source of our previous analysis from past posts.


However, if we are still in doubt about the validity of RH, let's do the harder computations.

From equation one,

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

the RH doubters need to solve the following system of two derived equations of equation one:

1A. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * cos(b * log(k))[/tex] = -1;

1B. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * sin(b * log(k))[/tex] = 0;


And the result is again, a = Re(z) = 1/2 for all the different values of b. And there are infinitely many distinct values of b. Thus, RH is true!
Guest
 

Re: Why is RH optimum?

Postby Guest » Sun Jun 16, 2019 11:22 am

\le
Guest wrote:
Guest wrote:We conclude that the Riemann Hypothesis (RH) is true! Do you agree?

David Cole,

https://www.researchgate.net/profile/David_Cole29.


If we are still in doubt about the validity of RH, let's do some computations.

Let z = a + i * b, and let [tex]\bar{z}[/tex] = a - i * b. Note: [tex]\bar{z}[/tex] is the complex conjugate of z.

Therefore, z + [tex]\bar{z}[/tex] = 1 which implies 2a = 1 which implies a = Re(z) = Re([tex]\bar{z})[/tex] = 1/2 which implies RH, and we are done.

Note: z + [tex]\bar{z}[/tex] = 1 implies the Harmonic Series, the prime source of our previous analysis from past posts.


However, if we are still in doubt about the validity of RH, let's do the harder computations.

From equation one,

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

the RH doubters need to solve the following system of two derived equations of equation one:

1A. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * cos(b * log(k))[/tex] = -1;

1B. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * sin(b * log(k))[/tex] = 0.


And the result is again, a = Re(z) = 1/2 for all the different values of b. And there are infinitely many distinct values of b. Thus, RH is true!


An Update:

1A. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2j\pi)}{k^{a}}= -1[/tex] ;

1B. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2j\pi)}{k^{a}}= 0[/tex]

where [tex]0 \le a \le 1[/tex] (Riemann's Theorem) for all integers, [tex]j\ge 0[/tex].
Guest
 

Re: Why is RH optimum?

Postby Guest » Tue Jun 25, 2019 8:45 am

"Nature is a grand optimizer, and therefore, one and only one critical line, Re(z) = 1/2 (the Riemann Hypothesis), is enough to define all primes and all nontrivial zeros of the Riemann Zeta Function." Hmm. Do you agree?
Attachments
RH Critical Line.png
RH Critical Line.png (11.77 KiB) Viewed 80 times
Guest
 

Re: Why is RH optimum?

Postby Guest » Mon Jul 01, 2019 12:09 pm

Hmm.

1A. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2\pi j)}{k^{a}}= -1[/tex] ;

1B. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2\pi j)}{k^{a}}= 0[/tex]

where [tex]0 \le a \le 1[/tex] (Riemann's Theorem) for all integers, [tex]j\ge 0[/tex].

Are there some or infinitely many values of j which make equations, 1A and 1B, false?
Guest
 

Re: Why is RH optimum?

Postby Guest » Mon Jul 01, 2019 12:12 pm

Guest wrote:Hmm.

1A. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2\pi j)}{k^{a}}= -1[/tex] ;

1B. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2\pi j)}{k^{a}}= 0[/tex]

where [tex]0 \le a \le 1[/tex] (Riemann's Theorem) for all integers, [tex]j\ge 0[/tex].

Are there some or infinitely many values of j which make equations, 1A and/or 1B, false?
Guest
 

Re: Why is RH optimum?

Postby Guest » Mon Jul 01, 2019 12:24 pm

Hmm. We assume [tex]\ \ a = \frac{1}{2}[/tex], and we consider the solutions of the following equations:

1A. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2\pi j)}{k^{a}}= -1[/tex] ;

1B. [tex]\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2\pi j)}{k^{a}}= 0[/tex]

for all integers, [tex]j\ge 0[/tex].

Are there some or infinitely many values of j which make equations, 1A and/or 1B, false?
Guest
 

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