Back to Riemann, Part 13


Ok, so I’m now ready to sit down and write up an Excel spreadsheet to graph the first few zeta zeroes. As I’m preparing for this, I noticed that I made a mistake in part 8 of the Riemann posts, a couple hours after I uploaded it to wordpress, so I had to go back and fix that. I’m also using part 8 for getting the math I need for the spreadsheet.

First, the zeta function is:
zeta(x) = 1 + 1/2^x + 1/3^x + 1/4^x + 1/5^x…

1/2^x translates to: (1/(2^a)) * (cos(b * ln(2)) – i*sin(b * ln(2)))

The generalized form is:
1/n^x = (1/(n^a)) * (cos(b * ln(n)) – i*sin(b * ln(n)))

Second, the eta function is:
eta(x) = 1 – 1/2^x + 1/3^x – 1/4^x + 1/5^x – 1/6^x…

Writing the generalized translated form of the eta function just means having alternating minus signs for the even terms.

Where things got trickier than I was expecting was with the conversion factor to go from the eta(x) value to the zeta(x) value.

zeta(x) = eta(x) / (1 – 2^(1-x))

Remembering that
2^x = 2^(a + bi) = (2^a) * (2^bi) = (2^a) * e^(bi * ln(2))
e^bi = cos(b) + i*sin(b)
(2^a) * e^(bi * ln(2)) = (2^a) * (cos(b * ln(2)) + i*sin(b * ln(2)))

This gives:
1/(1 – 2^(1-x)) = 1/(1 – ((2^(1-a)) * (cos(b * ln(2)) + i*sin(b * ln(2)))))

Or, since a = 1/2,
1/(1 – sqrt(2) * cos(b * ln(2)) + i * sqrt(2) *sin(b * ln(2)))

Simplified:
1/(j + ik)

And, yeah, I didn’t remember how to take the reciprocal of a complex number, and this doesn’t seem to have been covered in the der Veen and de Craats book. Fortunately, this is one of the rarer cases where I could find the answer quickly on the net.

Step 1, find the conjugate of the complex number (i.e., j – ik).
Step 2, multiply the numerator and denominator by the conjugate.
Step 3, simplify.

If j = 1 – sqrt(2) * cos(b * ln(2))
and k = sqrt(2) * sin(b * ln(2))

(j – ik)/(j + ik)(j – ik)
(j – ik)/(j^2 + ijk – ijk + k^2)
(j – ik)/(j^2 + k^2)

Writing all this out here in the blog would be too messy. But, doing this in Excel isn’t that bad because I can treat “j” and “k” as cell names.

Getting the zeta function results is even messier, but in simplified form:
If eta(x) = m + in

zeta(x) = (j*(m+n) – ik*(m+n)) / (j^2 + k^2)
Believe me, it makes more sense when you put everything into Excel and refer to the cell names this way.

This then gives me two choices as to how to proceed next. First is to just do a simple spreadsheet for one zero, and write a little VBScript to update the spreadsheet to make an animated dot running around the graph. Second is to copy the first set of columns multiple times to draw a more detailed line segment showing the zeta going to zero.

The second approach has the advantage of overcoming some of the rounding errors introduced by Excel when adding up the zeta function terms. The results of the Excel sheet aren’t going to be all that accurate, so it’s not going to go to |zeta(x)| = 0, it’s just going to get close and then move away again. But that’s ok.

So, which approach did I pick?
#2

Without the VBscript animation part, this is what I’ve got for the first zero.

This graph shows the real and imaginary components of zeta(x) for a=0.5 and b approaching the first zero at 14.134

Here we’ve got the magnitude of zeta(x), and it’s pretty obvious it’s not going to zero. I blame rounding errors.

And then, the zeta(x) function from a + ib = 0.5 + i*14.0 to 0.5 + i*14.5 in 0.05 steps on the complex plane. This is the line segment I’d animate with a VBScript driving the Excel graphs. If my math were perfect, the segment would be crossing the axes at 0,0. As it is, I have the eta(x) calculated out to 300 terms. Not sure if it’s worth the effort of going to 1,000 terms or not. Depends on how bored I get, I guess.

All that’s left now is to write the VBScript (I can modify the one I wrote for the first Riemann prose blog series), and to decide if I want to figure out a VBScript version of the Gamma approximation.

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