Riemann Zeta Function Implies the Harmonic Series...

[Guest]

Student: "Hmm. Okay, I understand that the sum of the Harmonic Series (HS) is infinite. But how does one infer the Harmonic Series from the Riemann Zeta Function (RZF)?

Professor: "Great question! If we think very big and connect the Fundamental Theorem of Arithmetic (FTA) with RZF via the Riemann Hypothesis (RH), we can discover an affirmative answer to your question.

Student: "No one knows if RH is true! Right?"

Professor: "Fair statement! Let's assume RH is 'almost true' for now. Is that acceptable?"

Student: "That's a fair assumption! And I find that quite acceptable since extraordinary exceptions may prove RH is not sound!"

Professor: "Let's begin with RZF: [tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex]

where [tex]z = a \pm i * b[/tex] where a and b are real numbers.

Next, we equate RZF to zero: [tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex] = 0. And we know that [tex]0 \le a \le 1[/tex] is true. Right?"

Student: Yes! Your inequality, [tex]0 \le a \le 1[/tex], is an important Riemann Theorem.

Professor: Correct! Let's take a short break.

Student: Okay.

[Guest]

Student: "Hmm. Okay, I understand that the sum of the Harmonic Series (HS) is infinite. But how does one infer the Harmonic Series from the Riemann Zeta Function (RZF)?

Professor: "Great question! If we think very big and connect the Fundamental Theorem of Arithmetic (FTA) with RZF via the Riemann Hypothesis (RH), we can discover an affirmative answer to your question.

Student: "No one knows if RH is true! Right?"

Professor: "Fair statement! Let's assume RH is 'almost true' for now. Is that acceptable?"

Student: "That's a fair assumption! And I find that quite acceptable since extraordinary exceptions may prove RH is not sound!"

Professor: "Let's begin with RZF: [tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex]

where [tex]z = a \pm i * b[/tex] where a and b are real numbers.

Next, we equate RZF to zero: [tex]\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}[/tex] = 0. And we know that [tex]0 \le a \le 1[/tex] is true. Right?"

Student: Yes! Your inequality, [tex]0 \le a \le 1[/tex], is an important Riemann Theorem.

Professor: Correct! Let's take a short break.

Student: Okay.

Professor: Furthermore, our Riemann Zeta Equation (RZE) becomes [tex]\zeta(z) = \sum_{k=2}^{\infty }\frac{1}{k^{z}} = -1[/tex].


Moreover, [tex]k^{z} = k^{ a \pm i * b} = k^{ a} * k^{ \pm i * b} = k^{ a} *( cos(b * log(k)) \pm sin(b * log(k)) )[/tex] via Euler's equation.

Therefore, [tex]\zeta(z) = \sum_{k=2}^{\infty }\frac{1}{k^{z}} = \sum_{k=2}^{\infty }\frac{1}{ k^{ a} *( cos(b * log(k)) \pm sin(b * log(k))} = -1[/tex].

Next, we multiply the expression, [tex]cos(b * log(k)) \pm sin(b * log(k))[/tex], with its complex conjugate, [tex]cos(b * log(k)) \mp sin(b * log(k))[/tex].

And we obtain, [tex]( cos(b * log(k)) \pm sin(b * log(k)) )* ( cos(b * log(k)) \mp sin(b * log(k)) ) = cos^{2}(b * log(k)) + sin^{2}(b * log(k)) = 1[/tex].

Now, our RZE has been transformed.

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

Student: Professor, well done! But I suspect there's much work to come.

Professor: Right! And now, we must begin to think very big! And we must also begin to think about FTA.

Next, we multiply the right side and left side of our current RZE by [tex]N^{a} = \prod_{k=2}^{\infty } {k^{a}}[/tex].

And we have:

(RZE): [tex]\sum_{k=2}^{\infty }N^{a}_{k^{a}} * ( cos(b * log(k)) \mp sin(b * log(k)) ) = -N^{a}[/tex] where [tex]N^{a}_{k^{a}} = \frac{N^{a}}{k^{a}}[/tex].

Student, please ponder the significance of our most current RZE with respect to FTA.

Moreover, what is the largest prime factor of [tex]N[/tex]?

Of course, [tex]N = \infty[/tex] and [tex]N^{a} = \infty[/tex]. And what can we say about [tex]N^{a}_{k^{a}}[/tex]?

Let's take a break.

[Guest]

Student: Professor, before we break. I have a question or observation to make. I agree [tex]N = \infty[/tex] but the value of [tex]N^{a}[/tex] depends on the value of a. Right?

Professor: Student, well done! If a = 0, then [tex]N^{a} = 1[/tex], and if [tex]0 < a \le 1[/tex], then [tex]N^{a} = \infty[/tex]. Why?

We will take a break now.

____________________________________________________________
Author of our story is David Cole,

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

P.S. Please ponder RH!

[Guest]

The previous post needed some editing. Here's an update.

Student: Professor, before we break. I have a question or observation to make. I agree [tex]N = \infty[/tex] but the value of [tex]N^{a}[/tex] depends on the value of a. Right?

Professor: Student, well done! If a = 0, then [tex]N^{a} = 1[/tex], and if [tex]0 < a \le 1[/tex], then [tex]1 < N^{a} \le \infty[/tex]. Why?

We will take a break now.

____________________________________________________________
Author of our story is David Cole,

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

P.S. Please ponder RH!

[Guest]

\frac{a}{b}\begin{cases} x + y = 4 \\ x - y = 0 \end{cases}

Professor: Furthermore, our Riemann Zeta Equation (RZE) becomes [tex]\zeta(z) = \sum_{k=2}^{\infty }\frac{1}{k^{z}} = -1[/tex].

Moreover, [tex]k^{z} = k^{ a \pm i * b} = k^{ a} * k^{ \pm i * b} = k^{ a} *( cos(b * log(k)) \pm sin(b * log(k)) )[/tex] via Euler's equation.

Therefore, [tex]\zeta(z) = \sum_{k=2}^{\infty }\frac{1}{k^{z}} = \sum_{k=2}^{\infty }\frac{1}{ k^{ a} *( cos(b * log(k)) \pm sin(b * log(k))} = -1[/tex].

Next, we multiply the expression, [tex]cos(b * log(k)) \pm sin(b * log(k))[/tex], with its complex conjugate, [tex]cos(b * log(k)) \mp sin(b * log(k))[/tex].

And we obtain, [tex]( cos(b * log(k)) \pm sin(b * log(k)) )* ( cos(b * log(k)) \mp sin(b * log(k)) ) = cos^{2}(b * log(k)) + sin^{2}(b * log(k)) = 1[/tex].

Now, our RZE has been transformed.

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

Student: Professor, well done! But I suspect there's much work to come.

Professor: Right! And now, we must begin to think very big! And we must also begin to think about FTA.

Next, we multiply the right side and left side of our current RZE by [tex]N^{a} = \prod_{k=2}^{\infty } {k^{a}}[/tex].

And we have:

(RZE): [tex]\sum_{k=2}^{\infty }N^{a}_{k^{a}} * ( cos(b * log(k)) \mp sin(b * log(k)) ) = -N^{a}[/tex] where [tex]N^{a}_{k^{a}} = \frac{N^{a}}{k^{a}}[/tex].

Student, please ponder the significance of our most current RZE with respect to FTA.

Moreover, what is the largest prime factor of [tex]N[/tex]?

Of course, [tex]N = \infty[/tex] and [tex]N^{a} = \infty[/tex]. And what can we say about [tex]N^{a}_{k^{a}}[/tex]?

Let's take a break.

Student: Professor, before we break. I have a question or observation to make. I agree [tex]N = \infty[/tex], but the value of [tex]N^{a}[/tex] depends on the value of a. Right?

Professor: Student, well done! If a = 0, then [tex]N^{a} = 1[/tex], and if [tex]0 < a \le 1[/tex], then [tex]1 < N^{a} \le \infty[/tex]. Why?

We will take a break now.

____________________________________________________________
Author of our story is David Cole,

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

P.S. Please ponder RH!

_____________________________________________________________

Student: Professor, the largest prime of N is [tex]p_{\infty } = \infty[/tex]. And the values of [tex]N^{a}_{k^{a}}[/tex] depend on k and a? For example,

if [tex]\frac{1}{2} \le a \le 1[/tex], then [tex]N^{a}_{k^{a}}[/tex] goes from [tex]\infty[/tex] to [tex]1[/tex] as [tex]k[/tex] goes from [tex]2[/tex] to [tex]\infty[/tex], respectively.

And I accept that if [tex]0 < a \le 1[/tex], then [tex]1 < N^{a} \le \infty[/tex]. That statement makes sense since the value of a dominates the value of N in the term, [tex]N^{a}[/tex].

For example, if a is almost zero, then [tex]N^{a} > 1[/tex]...

Professor, your RZE is not right!

(RZE): [tex]\sum_{k=2}^{\infty }N^{a}_{k^{a}} * ( cos(b * log(k)) \mp i *sin(b * log(k)) ) \ne -N^{a}[/tex] where [tex]N^{a}_{k^{a}} = \frac{N^{a}}{k^{a}}[/tex]

Corrections:

(RZE_A): [tex]\sum_{k=2}^{\infty }N^{a}_{k^{a}} * cos(b * log(k)) = -N^{a}[/tex] where RZE_A is the real part of RZE;

(RZE_B): [tex]\sum_{k=2}^{\infty }N^{a}_{k^{a}} * \mp i * sin(b * log(k)) = 0 * i = 0[/tex] where RZE_B is the imaginary part of RZE

where [tex]N^{a}_{k^{a}} = \frac{N^{a}}{k^{a}}[/tex].

Professor: Good work!!

According to FTA, all prime factors of N are all the prime numbers, [tex]2, 3, 5,7, ..., p_{\infty } = \infty[/tex]. And we can achieve this result only when [tex]a \ge \frac{1}{2}[/tex].

Hmm. So on the left side of RZE_A, we have an infinite series of numbers that sum to a transfinite number, [tex]-N^{a}[/tex], on the right side of RZE_A. Moreover,

[tex]N^{a} = ( 2^{\infty} * 3^{\infty} * 5^{\infty} * 7^{\infty} ... *p_{\infty }^{\infty} )^{a} =\infty[/tex].

The Harmonic Series (HS) states [tex]\frac{1}{N} *\sum_{k=1}^{\infty }N_{k} = N = \infty[/tex] where [tex]N_{k} = \frac{N}{k}[/tex].

Therefore, [tex]N^{a} * N^{a} = N^{2a} = N[/tex] implies HS if and only if [tex]a =\frac{1}{2}[/tex].

Together, FTA and HS, confirm the truth of RH or [tex]a = \frac{1}{2}[/tex].

Student: Professor, your argument is interesting, but I am not convinced RH is true! Can you explain RZE_B?

Professor: Okay, I accept your doubts. And I apologize for my sloppy mathematics too. RZE_B means there is harmony/unity, and the infinite series of the imaginary parts of RZE_B sum to zero for some values of [tex]a \ge \frac{1}{2 }[/tex] if you doubt RH.

[Guest]

RH is true! And that's the end of our story.

Dave.

[Guest]

"Please the ponder the wonderful Harmonic Series for it is central to the understanding of the Riemann Hypothesis and the Prime Number Theorem..." -- Dave.

Relevant Reference Link:

'The connection between the Riemann hypothesis and the harmonic series',

https://math.stackexchange.com/questions/3272146/the-connection-between-the-riemann-hypothesis-and-the-harmonic-series.

[Guest]

"Please ponder the wonderful Harmonic Series for it is central to the understanding of the Riemann Hypothesis and the Prime Number Theorem..." -- Dave.

Relevant Reference Link:

'The connection between the Riemann hypothesis and the harmonic series',

https://math.stackexchange.com/questions/3272146/the-connection-between-the-riemann-hypothesis-and-the-harmonic-series.

[Guest]

Recap: For all positive integers, [tex]1 < k \le \infty[/tex], there exists a prime number, [tex]p[/tex], such that p [tex]\le k^{\frac{1}{2}}[/tex] or [tex]p = k[/tex]. Furthermore,

[tex]k = k^{\frac{1}{2} - i * b} *k^{\frac{1}{2} + i * b}.[/tex]... The Riemann Hypothesis (RH) is best! And the Riemann Hypothesis (RH) is also true!

[Guest]

Relevant Reference Links:

'Proof of Riemann Hypothesis',

https://www.math10.com/forum/viewtopic.php?f=63&t=1549;

'Why is RH optimum?',

https://www.math10.com/forum/viewtopic.php?f=63&t=8042.