10/23/2021, 11:47 PM
Hey,
I made some big graphs (more are on the way, they just compile really slow). I've stuck with the convention that \(e^{bz}\) is the exponential and not \(b^z\)--this makes it a bit easier to evaluate.
Here is, \(\text{tet}_{1,\sqrt{2}}(z)\) over a large domain. This is tetration base \(b = \log(2)/2\) with period \(2 \pi i\). I kept it to 100 iterations, which gives about 37 digit accuracy. Again, I've coded overflows to zero.
Everything is very regular and smooth; interestingly enough, the essential singularities are very quiet; they don't really contribute much. I think it's because we are using a bounded base.
I thought I'd take a crack at a complex base too. So this is tetration with base \(b = \log(2)/2 + 0.3i\)--which still resides in the Shell-Thron region; and \(2\pi i\) period again. I'm betting we get something just as nice for all values in the Shell-Thron region. I kept it to 100 iterations, which gives about 28 digit accuracy. Again, I've coded overflows to zero:
I'm currently drawing up \(b = \log(1/2)\) and \(b= 1/e\) at the moment (recalling I treat the base as \(e^{bz}\) and not \(b^z\)), but I couldn't make all these graphs synchronously because more than two graphs at once slows my computer too much.
This seems to imply that the infinite composition/beta method is leagues more effective in the Shell-Thron region.
I made some big graphs (more are on the way, they just compile really slow). I've stuck with the convention that \(e^{bz}\) is the exponential and not \(b^z\)--this makes it a bit easier to evaluate.
Here is, \(\text{tet}_{1,\sqrt{2}}(z)\) over a large domain. This is tetration base \(b = \log(2)/2\) with period \(2 \pi i\). I kept it to 100 iterations, which gives about 37 digit accuracy. Again, I've coded overflows to zero.
Everything is very regular and smooth; interestingly enough, the essential singularities are very quiet; they don't really contribute much. I think it's because we are using a bounded base.
I thought I'd take a crack at a complex base too. So this is tetration with base \(b = \log(2)/2 + 0.3i\)--which still resides in the Shell-Thron region; and \(2\pi i\) period again. I'm betting we get something just as nice for all values in the Shell-Thron region. I kept it to 100 iterations, which gives about 28 digit accuracy. Again, I've coded overflows to zero:
I'm currently drawing up \(b = \log(1/2)\) and \(b= 1/e\) at the moment (recalling I treat the base as \(e^{bz}\) and not \(b^z\)), but I couldn't make all these graphs synchronously because more than two graphs at once slows my computer too much.
This seems to imply that the infinite composition/beta method is leagues more effective in the Shell-Thron region.