This blog is about the nature of things including ourselves.


“We must know;  we shall know.”  — David Hilbert.

“When the solution is simple, GOD is answering.”  — Albert Einstein.

Riemann Hypothesis states that the real part of all nontrivial zeros,

s = a ± b * i,

of the Riemann zeta function equals one-half (a = 1/2)  where 

Re(s = a ± b * i) = a,

and where the Riemann zeta function is  Σ 1/k^s  (from k = 1 to k = ∞).


Reference links:

Riemann Hypothesis;

Why is the Riemann Hypothesis true?;

Subtle relations: prime numbers, complex functions, energy levels and Riemann;

150 Years of Riemann Hypothesis;

Turing and the Computational Tradition in Pure Mathematics (http://www.computing-conference.ugent.be/file/16).


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“… deepest interrelationships in analysis are of an arithmetical nature.”  — Hermann Minkowski.

In the simplest sense, the Riemann Hypothesis (RH) confirms for all positive integers,  k > 1,

there exists a prime, p, which divides k such that either

p = k

or

p ≤ k^( 1/2 = a = Re(s = a ± b * i) )

where s is the nontrivial zero of the Riemann zeta function.

Furthermore, only s + s’ = 1  over 0 ≤ a ≤ 1 satisfies the  divergent Euler zeta function (Harmonic Series) for which there exists infinitely prime numbers (see Euler’s proof)  where a = 1/2.  

Hence, s = 1/2 + b * i and s’ = 1/2 – b* i are the only nontrivial zeta zeros of the Riemann zeta function in the interval, 0 ≤ a ≤ 1.

Note:  s = a + b * i and s’ = a – b* i are the nontrivial zeta zeros of the Riemann zeta function in the interval, 0 ≤ a ≤ 1.

Therefore, the answer is affirmative.

The Riemann Hypothesis is true!   — David Cole.

“If you can’t explain simply, you don’t understand it well enough.” — Albert Einstein.


Reference Links:

What is the correct proof of the famous and important Polignac Conjecture?.

Romans 8:28

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Comments on: "Why is the Riemann Hypothesis true?" (3)

  1. Try solving the following system of two equations derived from the equation, Σ 1/k^s (from k = 1 to k = ∞) = 0 + 0 *i = 0:

    1. Σ cos (b * log (k) ) / k^a (from k = 2 to k = ∞) = -1

    and

    2. Σ sin (b * log (k) ) / k^a (from k = 2 to k = ∞) = 0

    where 0 ≤ a ≤ 1
    and where s = a ± b*i is a complex number, and s is also the nontrivial zero of the Riemann zeta function.

    You have two equations, one and two, with two unknowns, a and b.

    What is the solution?

    Bonne chance!!

    David Cole

    Like

    • Notes:

      I. Looking at equations one and two above, we see periodicity associated with the variable b indirectly and not with the variable a. Why?

      II. Both variables, a and b, are of degree one.

      What does this information tell us about the solution of equations, one and two, in terms of variables, a and b?

      Hint. We expect all simple zeros, and there’s more … What more can we state factually?

      Like

      • There’s uniqueness associated with the variable a which is of degree one, and there’s periodicity associated with the variable b indirectly which is also of degree one. So, whatever is the true value of variable a, it is unique and the only one. And we know what the true value of variable a is. Variable a equals one-half (or a = 1/2).

        And because of the periodicity of the trigonometric functions, cosine and sine, there are infinitely many b values. Each b value is unique, and each one is associated with a = 1/2. Thus, s = 1/2 +/- b *i solves the Riemann zeta function.

        We conclude these observations confirm my proof of the Riemann Hypothesis and all empirical evidence which supports the Riemann Hypothesis. Furthermore, all nontrivial zeros of the Riemann zeta function are simple.

        Like

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