Alright, so I did a bunch of coding at the cottage this easter weekend. It was shitty horrible weather, so I spent most of the time inside on the laptop grinding out the code. The program is nowhere near done, but I got a first glimpse of what to expect.
Essentially I've successfully coded in:
\[
\exp_{y^{1/y}}^{\circ s}(\log^{\circ s}_{y^{1/y}}(a) + y)\\
\]
In limited scenarios... but it should be good enough for semi-operators. I've managed to make a close first order approximation to the solution:
\[
0 <s> 2\\
\]
So, this isn't the exact graph, but expect it to look something like this over \(0 \le s \le 2\):
The protocol I've written is still very beta beta beta, so I'm not going to post it. And it mostly works for \(\alpha \approx 0\), so I can for example do:
\[
0.5 <s> 2\\
\]
Again, we need to slightly perturb these solutions to get the correct solution. But it should look something like this.
The trouble I'm having now, is figuring out how to solve the root of a three variable equation. I have to read more pari-gp protocols. Beware, that when I release this program it's going to be slow as fuck. This is going to be a very slow program. And I really can't think of a work around
As a better explanation of these graphs; they are precisely:
\[
\alpha <s> 2 = \text{these graphs} + \mathcal{O}(s)\,\,\text{for}\,\,s \approx 0\\
\]
And similarly for \(s \approx 1,2\); but now it's \(\mathcal{O}(s-1),\mathcal{O}(s-2)\).
FUCK YA! I figured it out for more anomalous values. Here are some more graphs.
As we know:
\[
2 <s> 2 =4
\]
I was having trouble accessing the repelling fixed point to do this. But I can do it now. It produces a straight line at 4.
If I take something weird like:
\[
3.2 <s> 2\\
\]
We get this graph:
Again, we have to perturb these graphs slightly to get the right answer... but my code's almost there!!!!! I'm having trouble with \(3.2 <s> 4\) at the moment; so dealing with the repelling fixed point when it's in the s-exponent is a little tricky. But that's probably enough for tonight. This is definitely just a dumb coding error though, nothing mathematical. I just need to call it a night and look at it tomorrow ,lol. But here is the attracting fixed point of \(b = \sqrt[3]{3}\), let's call it \(u_0\).
\[
2 <s> u_0\\
\]
EDIT!!!!
Alright, I figured out how to add repelling fixed points much better. Now I can have repelling fixed points for all the variables. Here is:
\[
3<s>3\\
\]
Such that this is just the first order expansion in \(s\) of the \(\varphi\) argument.
I've done all the preliminary math at this point. Now I just have to efficiently solve for a 3 dimensional root equation. This will definitely take me a while.
As an italian Mphlee, I'm sure you can appreciate this. Romans had a perfect way of building bridges. But people could never build bridges as well as the romans. No one could make a bridge like the romans. And it all hinged on the roman keystone. They used to make bridges, with a strange brick in the center that just held everything together. And that was the legacy of roman thought. The \(\varphi\) I'm looking for, is the keystone of a roman bridge... I've got a shoddy looking bridge here. I just need the italian roman keystone and it's perfect! lmao!
Essentially I've successfully coded in:
\[
\exp_{y^{1/y}}^{\circ s}(\log^{\circ s}_{y^{1/y}}(a) + y)\\
\]
In limited scenarios... but it should be good enough for semi-operators. I've managed to make a close first order approximation to the solution:
\[
0 <s> 2\\
\]
So, this isn't the exact graph, but expect it to look something like this over \(0 \le s \le 2\):
The protocol I've written is still very beta beta beta, so I'm not going to post it. And it mostly works for \(\alpha \approx 0\), so I can for example do:
\[
0.5 <s> 2\\
\]
Again, we need to slightly perturb these solutions to get the correct solution. But it should look something like this.
The trouble I'm having now, is figuring out how to solve the root of a three variable equation. I have to read more pari-gp protocols. Beware, that when I release this program it's going to be slow as fuck. This is going to be a very slow program. And I really can't think of a work around
As a better explanation of these graphs; they are precisely:
\[
\alpha <s> 2 = \text{these graphs} + \mathcal{O}(s)\,\,\text{for}\,\,s \approx 0\\
\]
And similarly for \(s \approx 1,2\); but now it's \(\mathcal{O}(s-1),\mathcal{O}(s-2)\).
FUCK YA! I figured it out for more anomalous values. Here are some more graphs.
As we know:
\[
2 <s> 2 =4
\]
I was having trouble accessing the repelling fixed point to do this. But I can do it now. It produces a straight line at 4.
If I take something weird like:
\[
3.2 <s> 2\\
\]
We get this graph:
Again, we have to perturb these graphs slightly to get the right answer... but my code's almost there!!!!! I'm having trouble with \(3.2 <s> 4\) at the moment; so dealing with the repelling fixed point when it's in the s-exponent is a little tricky. But that's probably enough for tonight. This is definitely just a dumb coding error though, nothing mathematical. I just need to call it a night and look at it tomorrow ,lol. But here is the attracting fixed point of \(b = \sqrt[3]{3}\), let's call it \(u_0\).
\[
2 <s> u_0\\
\]
EDIT!!!!
Alright, I figured out how to add repelling fixed points much better. Now I can have repelling fixed points for all the variables. Here is:
\[
3<s>3\\
\]
Such that this is just the first order expansion in \(s\) of the \(\varphi\) argument.
I've done all the preliminary math at this point. Now I just have to efficiently solve for a 3 dimensional root equation. This will definitely take me a while.
As an italian Mphlee, I'm sure you can appreciate this. Romans had a perfect way of building bridges. But people could never build bridges as well as the romans. No one could make a bridge like the romans. And it all hinged on the roman keystone. They used to make bridges, with a strange brick in the center that just held everything together. And that was the legacy of roman thought. The \(\varphi\) I'm looking for, is the keystone of a roman bridge... I've got a shoddy looking bridge here. I just need the italian roman keystone and it's perfect! lmao!