Back to Riemann



(Image from Amazon)

I mentioned about a week ago that I received this book for Christmas. It’s a pretty thin volume, at only 144 pages, with 54 pages dedicated to exercise solutions and the index, but it lays out the math for the Riemann hypothesis clearly enough that I’m getting pretty close to understanding how to put the formulas into Excel this time. So, I’m going to try to write up my thoughts as I go in an attempt to understand everything myself.

This book, compiled by der Veen and de Craats, started out as an online internet course at the University of Amsterdam, aimed at advanced secondary school students to get them interested in university-level math. It includes URLs for a couple free web-based math applications that can be used to help solve some of the exercises, or draw plots of functions of interest. These are: Wolfram Alpha and Sage. The authors make extensive use of Alpha for the book exercises.

When I left off at the end of my Riemann prose series I’d gotten stuck with the fact that the standard representation of Riemann’s zeta function doesn’t work for the region we need it for.

Where “s” is a complex number in the form of s = a + bi. (Later, written as z instead of s.)

If b=0, that is we only look at numbers on the real number line, we get something like the following graph.

This just shows the results of summing “n” from 1 to 11, while calculating “a” from -5 to plus 5. It’s pretty clear from this graph that f(x) is going to infinity fast as “a” goes more negative (a < 0). Even for a=0, summing “n” from 1 to 11, the total reaches 11. Limiting the graph to the portion for a >= 1, we get the next chart.

If a=1, f(1) reaches 3.02. If n goes to infinity, so does f(1). That is, when f(1) = 1/1 + 1/2 + 1/3 + 1/4…, f(1) doesn’t converge. However, when a = 2 or greater, the function does converge. f(2) = 1/1 + 1/4 + 1/9 + 1/16… goes to about 1.6, and larger values of “a” converge closer and closer to 1.

What’s important to take away from this graph at the moment is that f(a) never crosses the x-axis, and therefore there’s never any point at which f(a) = 0. Which is the part we want.

If z (or, “s”) = a + bi for non-zero values of b, what happens? Well, in the previous blog series I’d shown that “i” represents a phase shift, where we can move from the x-axis to the y-axis in the complex plane, and then back.

1 * i = i
i * i = -1
-1 + i = -i
-1 * i = 1

And, x^bi represents a rotation around the origin in the form of x*cos(b) + x*i*sin(b).

So, if z = a + bi, and f(z) = x^z = x^(a + bi),
this is functionally the same as (x^a) * x*(cos(b) + i*sin(b)).

Summing 1 + 1/2^z + 1/3^z + 1/4^z …

where z = a + bi

Just means that we keep using the function n^a * n(cos(b) + i*sin(b)) to get each individual term of f(z), and then add that to the current running total.

If a > 1, we’re going to get a spiral converging on a non-zero point:

And if a < 1, the function goes to infinity.

This is where I’d given up, because the wiki article kept using notation that I had trouble rewriting to fit into Excel. However, the der Veen and de Craats book is very promising. To go any further, I need to cover some of the basics a bit more.

Questions:
How did Riemann decide on the zeta function in the first place?
What was he trying to prove with the zeta function?
What is the correlation between the zeta function and prime numbers?
If the important part of the zeta curve is where f(z) = 0, and the existing version of the zeta function never equals 0, what do we do to fix things?
What are logs, what is e, and how do they relate to the zeta function?

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