Why is RH optimum?

[Guest]

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?

Don't worry! Let go of doubts! Equations, 1A and 1B, are correct for all integers, [tex]j\ge 0[/tex].

[Guest]

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?

Don't worry! Let go of doubts! Equations, 1A and 1B, are correct for all integers, [tex]j\ge 0[/tex].

Hmm. However, your equations, 1A and 1B, are redundant!
They do not generate any new solutions beyond the first solution, [tex]b = b_{1 }= 14.1[/tex].

Maybe, the following equations will work when [tex]b = b_{1 }= 14.1[/tex] with [tex]\ \ a = \frac{1}{2}[/tex]:

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

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

[Guest]

Note: We must know [tex]b_{n }[/tex] before we can compute [tex]b_{n+1 }[/tex] from the following equations, 1A and 1B:

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

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

And hopefully, our equations are correct!

Warning: We are not sure that equations, 1A and 1B, are correct!

[Guest]

Here's a tentative fix:

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

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

where [tex]b_{n+1} = b_{n} \pm \triangle b_{n}[/tex] for some real value, [tex]\triangle b_{n} \ne 0[/tex].

[Guest]

We conclude the Riemann Hypothesis is still true!

And we apologize for any sloppy (mistaken) mathematics posted earlier...

David Cole.

P.S. I am done with the Riemann Hypothesis.

[Guest]

We conclude the Riemann Hypothesis is still true!

And we apologize for any sloppy (mistaken) mathematics posted earlier...

David Cole.

P.S. I am done with the Riemann Hypothesis.

Moreover, I am also done with David Hilbert's Eighth Problem and with prime number theory. And hopefully, my work is significant...

I want to advance the current theory of energy in the context of David Hilbert's Sixth Problem and the Lyapunov's Second Method for Stability...

David Cole,

Relevant Reference Links:

https://www.researchgate.net/project/Hilberts-Sixth-Problem-https-enwikipediaorg-wiki-Hilberts-sixth-problem;

https://www.researchgate.net/project/Important-Question-about-the-Lyapunovs-Second-Method-for-Stability.

[Guest]

Relevant Reference Link:

'Riemann Zeta Function Implies the Harmonic Series...',

https://www.math10.com/forum/viewtopic.php?f=63&t=8577.

[Guest]

FYI: "The Optimization Principle for the Riemann Hypothesis" by H. Saidane, Ph.D.,

https://vixra.org/pdf/1806.0444v2.pdf

[Guest]

FYI: "The Optimization Principle for the Riemann Hypothesis" by Prof. H. Saidane, Ph.D.,

https://vixra.org/pdf/1806.0444v2.pdf

Reference Link:

'Lectures on The Riemann Zeta–Function' by Prof. K. Chandrasekharan,

https://julianoliver.com/share/free-science-books/tifr01.pdf.

Enjoy!

[Guest]

“Nothing takes place in the world whose meaning is not that of some maximum or minimum.” ― Leonhard Euler.

[Guest]

“Nothing takes place in the world whose meaning is not that of some maximum or minimum.” ― Leonhard Euler.

Moreover, Nature is a grand optimizer (minimizer or maximizer), and therefore, one and only one critical line, Re(z) = a = 1/2 (the Riemann Hypothesis), is perfect for detecting all primes in the Riemann Zeta Function,

[tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex] = 0 (zero indicates the minimum or ground state or the fundamental basis level [existence of primes]) for all simple nontrivial zeros, z = a ± bi ...

Dave.

[Guest]

Key Ideas: Fundamental Theorem of Arithmetic, integer k is king

Fundamental Rule/Law of Prime Number Theory:

For ALL integers, [tex]k > 1[/tex], there exists a prime number, [tex]p[/tex], which DIVIDES [tex]k[/tex] or [tex]p | k[/tex] such that either [tex]p = k[/tex] or [tex]p \le k^\frac{1}{2}[/tex].


BIG FACT: The OPTIMAL exponent, [tex]\frac{1}{2}[/tex] , of [tex]k^\frac{1}{2}[/tex] is confirmed by the Riemann Hypothesis.

Remark: We hope you are convinced that the Riemann Hypothesis is true! Godspeed! Go Blue!

Dave.

[Guest]

...

P.S. I am done with the Riemann Hypothesis.

Oops! If there are "authorities" (so-called experts) who doubt the Riemann Hypothesis, then my work is not done. And I am still learning too.

Dave.

[Guest]

Important Remark: The Fundamental Theorem of Arithmetic, integer [tex]k \ge 1[/tex], [tex]\sum_{k=1}^{\infty }\frac{1}{k^z} = 0[/tex], and the truth of the Riemann Hypothesis imply the divergent Harmonic Series, [tex]\sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex].

[Guest]

Important Remark: The divergent Harmonic Series, [tex]\sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex] implies the Fundamental Theorem of Arithmetic, integer [tex]k \ge 1[/tex], [tex]\sum_{k=1}^{\infty }\frac{1}{k^z} = 0[/tex], and the truth of the Riemann Hypothesis.

[Guest]

Final Remark: Never, never, never,..., never doubt the Riemann Hypothesis! Go Blue!

[Guest]

Important Remark: The divergent Harmonic Series, [tex]\sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex] implies the Fundamental Theorem of Arithmetic, integer [tex]k \ge 1[/tex], [tex]\sum_{k=1}^{\infty }\frac{1}{k^z} = 0[/tex], and the truth of the Riemann Hypothesis.

FYI:

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} \ge 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... = 1 + 1 + 1 + 1 + ... = 1 + 2 + 3 + 4 + 5 + ...
= 2 + 3 + 5 + 7 + 11 + 13 + ... = 2 + 4 + 6 + 8 + ... = 3 + 6 + 9 + 12 + 15 + ... = 5 + 10 + 15 + 20 + 25 + ... = p + 2p + 3p + 4p + 5p + ... = \infty[/tex] where p is any positive prime number.

Relevant Reference Link: Prove the Harmonic Series diverges to infinity.

[Guest]

“Nothing takes place in the world whose meaning is not that of some maximum or minimum.” ― Leonhard Euler.

Moreover, Nature is a grand optimizer (minimizer or maximizer), and therefore, one and only one critical line, Re(z) = a = 1/2 (the Riemann Hypothesis), is perfect for detecting all primes in the Riemann Zeta Function,

[tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex] = 0 (zero indicates the minimum or ground state or the fundamental basis level [existence of primes]) for all simple nontrivial zeros, z = a ± bi ...

Dave.

FYI: If a = 1 and b = 0, then z = 1 ± 0i = 1, then we have [tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex], the divergent Harmonic Series.

[Guest]

Update:
...

FYI: If a = 1 and b = 0, then z = 1 ± 0i = 1 and we have [tex]\zeta(z = 1) = \sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex], the divergent Harmonic Series.

[Guest]

In a nutshell, [tex]\zeta(z = 1) = \sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex] if and only if [tex]\zeta(z = \frac{1}{2} ± bi ) = \sum_{k=1}^{\infty }\frac{1}{k^{ \frac{1}{2} ± bi}} = 0.[/tex]