Riemann prose, part 8


Some time ago, I was talking (writing) about finding the area of a backyard, and I said (wrote) something about taking unit steps, and adding up all of the lengths of the unit steps.

In my example, I used a yard 40’x100′. So,
40*1 + 40*1 + 40*1 … 100 times.

In taking narrower and narrower steps, we get an approximation closer to the actual area underneath a curve. If the curve is y = 5, for x from 0 to 5 (5 steps of unit 1),

area under the curve = 5+5+5+5+5 = 25

Generalized, the area under y = x, for n from 0 to x is

x * x = x^2

If the curve is: y = 3x, and we go from 0 to 4:

area(y) = 0+3+6+9+12 = 30 (because the steps are so big, we introduce an error of 4 * 3/2, or 6.)

Generalized, the area under a triangle y = ax, for n from 0 to b;

1/2 a*b^2

This is the start of integral calculus. If we take an infinite number of infinitely thin slices, we will get the precise value of the area under whatever curve we pick (ignoring the exceptions).

And, this where sigma comes in. Sigma represents the addition of a string of values from some lower limit up to the upper limit.

The above equation just tells us to add the positive integers from 1 to 100 (5050).

So, looking at the Riemann Zeta function one more time,

We’re going to sum a formula from 1 to infinity, where the formula uses a complex value, “s”, and “s” is in the form of “a+ib” in the complex plane. For each value of “n”, we will raise “n” to the complex exponent and take the reciprocal (divide 1 by that number). “n^s” is the same as:

n^a * (cos(ln(n^b)) + i*sin(ln(n^b)))

Yeah, so what? How do we draw this?

Basically, we’re going to look at the entire complex plane as one big sheet, then make it three dimensional by adding a z-axis, where z is the total of all the sums for each point “a+ib”.

What this is going to do is to give us a ripply, wavy sheet that sometimes goes to negative z, sometimes holds at a fixed value (converges) and sometimes goes to positive infinity on the z-axis.

Why? Why do we care?

What got me started is the Riemann Hypothesis, which says that the Riemann Zeta function equals 0 when, and only when, the real part of s = 1/2 (a is the square root).

Actually, there are two sets of places where the zeta function equals 0, the “trivial zeros” and the “non-trivial zeros”. The trivial zeros are at s = -2, -4, -6, etc. (“a” is a negative even integer, and “b” is 0).

The non-trivial zeros are all in the “critical strip” where “a” is close to 1/2, but no one’s been able to prove mathematically that they’re all actually on the “critical line” of a=1/2. The first trillion non-trivial zeros have been calculated with a computer, and they’re all on the critical line, but calculating them is not the same as proving the hypothesis.

Now, if only it were that easy.

If we add up a series of numbers and the total goes to infinity, then the series is said to “not converge”. Conversely, if the total keeps getting closer to some value without exceeding that, the series is said to converge at that value.

x = 1 + 2 + 3… etc. goes to infinity.
x = 1/1 + 1/2 + 1/3 + 1/4… still goes to infinity, only slower
x = 1/1 + 1/4 + 1/9 + 1/16 * 1/25… converges to about 1.64

The faster the farther values get smaller, the more likely the total is to be convergent.

The problem is that the Riemann Zeta function is not convergent when “a” is < 1.

Let me repeat that – the Zeta function as given above goes to infinity when the real part of s is less than 1. (And, 1/2 is less than 1, the last time I checked.)

This is why mathematicians will use other forms of the equation to find the Riemann Zeros, specifically the analytic continuation. Unfortunately, I haven’t wrapped my head around that yet.

The formula I came up with:

f(x) = sum(1/(n^a * cos(ln(n^b) + n^a*i*sin(ln(n^b))))

doesn’t work. (sob) But it does make pretty pictures.

For x = (0.5 + i* 14.13):

At the first non-trivial zero, this version of the Riemann Zeta just keeps getting bigger as you continue summing up the n^s parts. It doesn’t converge to (0, 0).

(Graph of the zeta function, with a = 1/2, and i*b from 1.8 to 15. Nope, zeros here.)

I kind of thought I was doing it wrong, since we usually see formulas using sines and cosines as cos(2 * PI * x). So, I changed the spreadsheet to use 2*PI. The sum converges, but not to (0, 0).

The real and imaginary parts just don’t both go to 0 simultaneously, which is a bad thing for the home team.

 

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2 Comments

  1. Anonymous

     /  May 12, 2016

    Great plot convincing me that series does not converge.

    Reply
  1. Back to Riemann | threestepsoverjapan

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