The New Riemann Hypothesis, Part 2

Where Part 1 left off showed an alternate form of the Riemann Zeta Function is

\zeta(\alpha + \beta{i}) = \sum \limits_{n=1}^\infty \frac{e^{\omega_n{i}}}{n^\alpha}

This is not the final form of the equation. The next step is applying Euler’s formula. Euler came up with a proof to show that any imaginary exponent is a rotation in the complex plane. While I could take a massive detour here and explain to you exactly what Euler’s formula is and how to interpret and understand it, Kalid over at has done a great job with it already. If you want to learn more about it, click here.

The mathematical statement of Euler’s law is this:

e^{xi} = \cos{x} + i\sin{x}

Taking Euler’s formula and applying x = \omega_n as seen in the numerator of the Zeta Function the following form is found.

\zeta(\alpha + \beta{i}) = \sum \limits_{n=1}^\infty \frac{\cos{\omega_n} + i\sin{\omega_n}}{n^\alpha}

Interpret this as a spiral placement of marbles on a balancing board as previously mentioned in the last post. How though? \cos{\omega_n} is an entirely real term. It determines how much of the distance from the center is in the real direction. The {i}\sin{\omega_n} term is purely imaginary because of the {i} being multiplied. It determines how much of the distance from the center is in the imaginary direction. \omega_n = \beta\ln{n}, which calculates the angle between the horizontal and a line drawn between each point created for each value of n. 1/{n^\alpha} represents the distance from the center. That’s how the interpretation in the image below is found.

Riemann Spiral Explained.001

Since \cos{\omega_n} is entirely real while i\sin{\omega_n} is entirely imaginary, it’s possible to split the Riemann Zeta Function into two different sums. Those sums would look like this

\sum \limits_{n=1}^\infty \frac{\cos{\omega_n}}{n^\alpha} = 0

\sum \limits_{n=1}^\infty \frac{i\sin{\omega_n}}{n^\alpha} = 0

It should be noted that i is in every term in the second sum, which means it can be factored out. After that it can be divided out since the sum is equal to zero. That would look something like this is mathematical notation.

\sum \limits_{n=1}^\infty \frac{i\sin{\omega_n}}{n^\alpha} = i\sum \limits_{n=1}^\infty \frac{\sin{\omega_n}}{n^\alpha}=0

Dividing through by i yields

\sum \limits_{n=1}^\infty \frac{\sin{\omega_n}}{n^\alpha}=0

From here, both the \sin{} terms and the \cos{} terms are treated as separately or put back together. For brevity’s sake, I will use the \sin{} terms only to show how the proof unfolds, then I will show how it is extended to both sets.

I’m not sure exactly how many posts are left in this blog until the proof in these posts are explained and the more formal proof is posted at the end, but if I ventured a guess, I would say it’s near the halfway point. I have to explain a few things about reducing the number of terms, the nature of calculating \ln{0} and adding in something which provides the necessary conditions for calculating \alpha. I look forward to sharing with those with you soon. Thanks for reading this far!


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