I'll be updating this as frequently as I can; but for the moment I thought I'd put down some graph's using Mike's graphing tool that Gottfried Linked me to.
Here is, beta_function(z,log(2),100) + tau_K(z,log(2),100,7) graphed over the region \( -1 \le \Re(z) \le 3.5 \) and \( 1\le\Im(z)\le 4 \)
And Here is, exp(beta_function(z-1,log(2),100) + tau_K(z-1,log(2),100,7)) graphed over the same region.
You'll notice they are virtually identical. Ergo; the functional equation is being satisfied. I've tried to edit some of my buggy code a bit. I'll update when I can make slightly better code. I'm trying to make more graphs at the moment, which are in the same flavour of this.
Wow, Mike's program is quite beautiful,
Here's my solution to the Abel equation with \( \log(2) \) multiplier over \( -1 \le \Re(z) \le 4 \) and \( 0.5 \le \Im(z) \le 2.5 \):
Here's the same solution, graphed over \( -1 \le \Re(z) \le 3.5 \) and \( -2.5 \le \Im(z) \le 2.5 \); which is more symmetrical:
And attached here is the solution with multiplier \( 1 +0.1i \) graphed over the region \( -1 \le \Re(z) \le 6 \) and \( 3.2 \le \Im(z) \le 9 \). You can see the singularities forming on the boundary; which the Riemann mapping will effectively remove, by pasting all the solutions together. This function has a period of \( \frac{2\pi i}{1+0.1i} \); so this graph repeats in the up/down direction at a slight angle.
Here is, beta_function(z,log(2),100) + tau_K(z,log(2),100,7) graphed over the region \( -1 \le \Re(z) \le 3.5 \) and \( 1\le\Im(z)\le 4 \)
And Here is, exp(beta_function(z-1,log(2),100) + tau_K(z-1,log(2),100,7)) graphed over the same region.
You'll notice they are virtually identical. Ergo; the functional equation is being satisfied. I've tried to edit some of my buggy code a bit. I'll update when I can make slightly better code. I'm trying to make more graphs at the moment, which are in the same flavour of this.
Wow, Mike's program is quite beautiful,
Here's my solution to the Abel equation with \( \log(2) \) multiplier over \( -1 \le \Re(z) \le 4 \) and \( 0.5 \le \Im(z) \le 2.5 \):
Here's the same solution, graphed over \( -1 \le \Re(z) \le 3.5 \) and \( -2.5 \le \Im(z) \le 2.5 \); which is more symmetrical:
And attached here is the solution with multiplier \( 1 +0.1i \) graphed over the region \( -1 \le \Re(z) \le 6 \) and \( 3.2 \le \Im(z) \le 9 \). You can see the singularities forming on the boundary; which the Riemann mapping will effectively remove, by pasting all the solutions together. This function has a period of \( \frac{2\pi i}{1+0.1i} \); so this graph repeats in the up/down direction at a slight angle.