I thought I'd take a crack at base = 1/2
#7
So I've switched over to 64-bit pari-gp in hopes this may solve some of my problems. And I decided to take a whack at \( b = 10^{-5} \). We're going to do the exact same procedure. So I thought I'd add some new notation. We're going to switch from \( b^z \) to \( e^{bz} \) here; so let's take the logarithm of the normal idea of a base.

\(
\varphi_\lambda(b,s) = \Omega_{j=1}^\infty \frac{e^{bz}}{e^{\lambda(j- s)} +1}\,\bullet z\\
\)

So that I don't have to keep defining new function names. And this function satisfies the functional equation,

\(
\varphi_\lambda(b,s+1) = e^{b\varphi_\lambda(b,s)}/(e^{-\lambda s} + 1)\\
\)

It is \( 2\pi i / \lambda \) periodic in \( s \), and is real valued for all variables real. And decays to zero at \( \Re(s) = -\infty \). Additionally it is holomorphic for \( b\neq 0,\,\lambda(j-s)\neq (2k+1)\pi i \).

We're going to focus on \( b_0 = \log(0.00001) \) and for simplicity keep \( \lambda = 1 \)--so it's 2 pi i periodic. And here's a graph of \( \varphi_1(b_0,x) \) for \( x \in [-1,3] \).

   

We define the iteration similarly,

\(
\mu^0 = 0\\
\mu^{n+1}(s) = \log(1+e^{-s})/b_0 - \log(1+\frac{\mu^{n}(s+1)}{\varphi_1(b_0,s+1)})/b_0\\
\)


Now, this method works fine, but it's really really reallly weird. It almost just looks like a sinusoid times 1E-10. The functional equation is defined upto 20 digits or so atm. I am very confused by all this. But the equation is being satisfied. I think it's because the normalization constant is about \( x_0 = -1000 \) or something; not sure tbh. But the superfunction equations being satisfied.

It is definitely not working perfectly; it worked really well with \( b = -e \), but \( b = \log(0.00001) \) is very very very weird. It basically just oscillates about the fixed point.

I thought I'd just post the code (it's 64 bit this time, so there shouldn't be as much of a problem)--rather than dealing with all the nuances. I won't make as many graphs as before. But I do believe this still works. Again, the math says this should still work, but coding it is crazy weird. I'll leave it to you to make your own judgements.


Remember to run beta_init(100) before doing anything. This takes about 30 mins on my pc to compile--the files too big to attach the initialization file. After that you'll get Init_tiny.dat which you can include in any future compilations. This just initializes \( \varphi \) for \( b = \log(0.00001) \).

You can also reinitialize for arbitrary b by changing the base variable at the beginning of the file. So you can run all the tests you want on weird values, ember. Just change the base value, reinitialize the beta function; and have at it. I have stuck to 100 iterations as the default, but recall, for b = e, about 10 iterations is a max (plus you need a limiter). But for values \( b < e^{1/e} \), 100 iterations should work fine. Complex plane, not sure, but prolly won't work.


.gp   abel_tiny_t.gp (Size: 7.52 KB / Downloads: 693)


Here's a graph of \( \varphi_1(b_0,x) + \mu^{100}(x) \) over \( x \in [-1002,1000] \) which shows the points where it hits 1 and 0. 


   


It's boxy as hell, and further out it just looks constant with a small wobble of 1E-10. I'm not too sure about all of this in the complex plane. it gets really wonky; I think that singularities are popping up surreptitiously so the complex plane is a little off. You can still grab taylor series to about \ps 100 though. And the functional equation is capping at about E-27 precision; when you use the taylor series, that is.

Regards, James




Well this is even wackier than I expected. This function is bananas! The above graph doesn't do it justice at all; this jumps up and down between 0 and 1, and eventually settles to a line. But other than that, it looks a lot like a square wave for a large portion of its domain. The above graph has sampling errors, so it looks like lines, but this is really a square wave looking thing. But taylor series converge fairly damn well; I'm not sure what to make of all this o.o

over \( x \in [500,505] \):

   
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RE: I thought I'd take a crack at base = 1/2 - by JmsNxn - 09/30/2021, 05:38 AM

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