Riemann was interested in understanding heat flow in a wire. As he was analyzing this, he developed a series of tools that could also be applied to the prime number distribution system.

What he came up with is the Riemann Functional Equation, which, according to der Veen and de Craats, is explained in more detail in chapter 9 of Jeffrey Stopple’s book “A Primer of Analytic Number Theory, From Pythagoras to Riemann.” This functional equation uses another mirror property of complex numbers to relate values to the left of the y-axis (x < 0) to those on the right side (x > 0), specifically as it applies to the zeta function.

z( -x ) = ((-2 * x!) / (2*pi)^(x+1)) * sin(pi*x/2) * z(x + 1)

(from the der Veen and de Craats book)

or,

(Image from the wiki article.)

That is, if you know zeta(x + 1) where x >= 0, then you can use this conversion function to get z( -x ).

Although both zeta(x) and eta(x) go to infinity for x < 0, using the functional equation extends the domain of zeta(x) to the entire complex plane, except for x=1, which still has a pole.

I’ll get to the idea of finding factorials of complex numbers in a minute.

Right now, if you look at the functional equation closely, you’ll see it has 4 parts:

-2 * x!

(2*pi)^(x+1)

sin(pi*x/2)

zeta(x+1)

We already know that zeta(x+1) doesn’t have zeroes for x > 1 across the complex plane, and that it converges to 1 for large x.

x! doesn’t have zeroes.

And (2*pi)^(x+1) doesn’t have zeroes.

However, sin(pi * x / 2) DOES go to 0 for integer multiples of pi.

0 = sin(0) = sin(pi) = sin(2*pi) = sin(3*pi) = sin(4*pi)…

And in the functional equation, that happens when x = -2, -4, -6, -8, etc.

These are the trivial zeroes, and they occur at all negative even integers of x.

Remember, the Riemann hypothesis says that the non-trivial zeroes for the zeta function are on the critical line, x=1/2.

This implies that there are trivial zeroes that don’t contribute to the hypothesis.

**And we now know where the trivial zeroes come from:
the sin(pi*x/2) component.**

—

Ok, factorials.

Traditionally, factorials are written as x!, which represents the product of all integers from 1 to x.

Example:

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

If we treat this pattern as a continuous line through all of the data points, it’s going to look something like this:

Meaning that technically we can have 3.3!, or 5.921!

It just means that the result is going to fall somewhere along the this line.

Now, trying to do something like this with imaginary numbers would be a little trickier, but it’s still something that can be visualized.

i! = i

2i! = i * 2i = -2

3i! = -2 * 3i = -6i

4i! = -6i * 4i = 24

5i! = 24 * 5i = 120i

which represents a rapidly growing spiral.

3.3i! would then be somewhere along that spiral between -6i and 24, which could be found with a variation of cos(x) + i*sin(x).

(Not really a proper spiral, but I think you get the point.)

Which is all really irrelevant when you mix the two, and try to get something like

(-6 + 4i)!

This is where the gamma function comes in. The gamma function is Riemann’s approach to complex factorials.

It’s also where I ran into a complete roadblock the first time I tried graphing the zeta function in my earlier blog series. The wiki entry was too difficult for me to follow at that time, and right now, der Veen and de Craats completely whiff out by referring to the wiki article.

However, there’s more than one way to approximate the gamma function, and the wiki article for the Lanczos approximation does have a code example in Python that I might try to steal from.

Briefly, the Lanczos approach uses a set of coefficients that can be pre-calculated and plugged into a table, then the gamma equation itself can be calculated as:

where “g” is an arbitrary constant with the real part > 1/2, and Ag(z) is:

If you pre-choose z, Ag(z) is just the sum:

Why go through all this work?

Well, Excel does have a gamma function built-in, but it’s only for real values of x, and doesn’t support complex values. Wolfram Alpha supports complex values for gamma, but it’s difficult to create the kind of animations I want to make.

On the other hand, the animations I’m most interested in are those for the zeta function in the critical strip (0 < x < 1), and that just requires the eta function, not the Riemann functional equation. So, this is “nice to know” material, but not essential. Then again, there’s a key part of the critical strip that I haven’t addressed yet that does depend on knowing what the functional equation is, and I’ll get to that next time.

The main reason for going through all this work, though, is to be able to prove to myself that I can write up the zeta function for the entire complex plane in one format or another. If not in Excel, then in VBScript (which I can then use to load values into Excel for graphing purposes).