The New Riemann Hypothesis, Part 1

Focus is a strange thing. Too far away and things blend together, too close and we can’t see the picture. Sometimes it’s necessary to look at things from multiple distances to get a better understanding.

While others may use different variable names then I have used, the statements are all equivalent when stating the Riemann Hypothesis. The Riemann Hypothesis says for The Riemann Zeta Function to equal zero, alpha must equal \frac{1}{2} if it is between but not equal to 0 or 1. In mathematical terms it looks like this:

\zeta(s) = \sum\limits_{n=1}^\infty \frac{1}{n^s}=0   only when
s = \frac{1}{2} + \beta{i}    for    0< \alpha<1

If you stick with me for a bit, I’ll explain this in a physical sense.

Originally, in my proof for The Riemann Hypothesis, I thought of it as a statement about “wave heights” adding to zero. To put a picture into your head, the blue lines in the image below shows where the combinations of these “wave heights” were calculated. I thought The Riemann Hypothesis was a special statement about these “wave heights” adding to zero for certain values of s. This was a bad representation. I went through a couple dozen before I settled. Life lesson, question everything, even something you see. You may have the wrong perspective!


An easier to understand interpretation of the Riemann Hypothesis would be a set of points in a spiral-like pattern. Imagine each of the points being a marble sitting on a rectangular board. While doing this, take some liberties with your imagination because there would be infinite marbles. Placing the center of that board on a pin, if the board stayed balanced, it would represent The Riemann Zeta Function equaling zero. In proper math, that looks like this:

\zeta(s) = 0.

How does this relate to the variable s?

The variable s describes each marble’s placement with relationship to the center of the board. The real part of s tells us how far away from the center of the board, while the imaginary term tells us the angle that distance is at with respect to the horizontal axis. The image below may explain it.

Riemann Spiral Explained.001

Again, this board remaining perfectly balanced would represent The Riemann Zeta Function equaling zero, so the Riemann Hypothesis says that this board won’t balance unless each marble is a distance of \sqrt{Number Of Marble Placement} away from the center. Even in that case, there are only certain values of the angles between each marble placement that cause the board to balance, this would be determined by the imaginary part of s.

Disclaimer: We’re about to get into some math.

This math comes from my formal proof, which will be posted at the end of this blog series. Restating the Riemann Zeta Function and the definition of s:

\zeta(s) = \sum \limits_{n=1}^\infty \frac{1}{n^s} = 0
s= \alpha + \beta{i}

Replacing s in the Riemann Zeta Function with \alpha + \beta{i}. This is an equivalent substitution.

\zeta(\alpha + \beta{i}) = \sum \limits_{n=1}^\infty \frac{1}{n^{\alpha + \beta{i}}} = 0

From here, apply the laws of exponentiation that state equivalent definitions for manipulating variables with exponents. Two steps to illustrate each law being applied.

First Step

 \zeta(\alpha + \beta{i}) = \sum \limits_{n=1}^\infty \frac{1}{n^\alpha{n^{\beta{i}}}} = 0

Second Step

\zeta(\alpha + \beta{i}) = \sum \limits_{n=1}^\infty \frac{n^{-\beta{i}}}{n^\alpha} = 0

In this case, the first step was expanding the definition of the exponent, the second step was moving the imaginary portion to the numerator. More substitutions are coming. First, by the definition of natural logarithms, e^{\ln(n)} = n. Substituting this in for n in the numerator gives:

\zeta(\alpha + \beta{i}) = \sum \limits_{n=1}^\infty \frac{(e^{\ln{n}})^{-\beta{i}}}{n^\alpha}= \sum \limits_{n=1}^\infty \frac{e^{-\ln{n}\beta{i}}}{n^\alpha} = 0

Like all mathematicians, I don’t write more than necessary, so for simplicity sake, the substitution \omega_n = -\ln{n}*{\beta} is made.

\zeta(\alpha + \beta{i}) = \sum \limits_{n=1}^\infty \frac{e^{-\ln{n}\beta{i}}}{n^\alpha} = \sum \limits_{n=1}^\infty \frac{e^{\omega_n{i}}}{n^\alpha} = 0

This is almost the final form of my version of the Riemann Hypothesis, but it’s time to stop for now. I’ll finish this up in my next post and show you how this form comes back to the spiral form I showed at the beginning.


One thought on “The New Riemann Hypothesis, Part 1

  1. The sum 1/n^s for n=1 to infinity does not evaluate to be zero at any point. In fact it isn’t even defined for s=1/2+ti. I haven’t read your ‘proof’ in depth but if this is the function you used for your final result, you have ran into a serious error. Perhaps you need to read more into analytic continuation before dedicating so much time into this.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s