So, with only using the value \(\text{slog}'(0)\) from Sheldon, I've managed to draw \(\chi(z)\) to about 2-3 digit accuracy. THIS IS MONUMENTAL. I do not need Kneser's riemann mapping, or Sheldon's version of it--other than the value \(\text{slog}'(0)\). We don't even need sheldon to find this value; it is precisely \(\lim_{z\to 0} \text{slog}'(\log(z))/z = \lim_{z \to - \infty} \text{slog}'(z)/\exp(z)\). This value is findable using Kouznetsov, and any good mathematician worth their salt can find this value. I'm just using sheldon's program because it's quick and easy to calculate this value. But you don't need Kneser or Sheldon, per se.
Here is a graph of the beta-chi function, which I remind the reader:
\[
\chi(z) = \beta(\text{slog}_K(z))\\
\]
Without constructing \(\text{slog}_K(z)\) using any pre-existing function. So all we need is \(\beta\), and no mention of Sheldon's code.
As a by-product of constructing \(\chi\) we actually end up constructing Kneser's slog. This isn't that hard to see if we think about this more abstractly. \(P = \text{slog}_K(z)\) is the unique function such that:
\[
\chi(e^z) = \frac{e^{\chi(z)}}{1+e^{-P(z)}}\\
\]
So if you construct one accurately, you construct the other! So I already have some code for Kneser's slog; which again uses none of sheldon's code, other than the acquisition of \(\text{slog}'(0)\). For now my code is only safe for 2-3 digits. But I'm considering this a milestone that this method will work. And if you have \(\chi\), then functionally you've constructed kneser from \(\beta\). Where we get:
\[
\chi(\text{tet}_K(z)) = \beta(z)\\
\]
Which, I've confirmed for more digits than the slogarithm; about 4-5 digits.
I should also add, that when I say "confirmed to 2-3 digits"--I mean the Taylor series are correct upto to 2-3 significant digits. So if \(\frac{d^j}{dx^j} \text{slog}_K^{(j)}(x)/j! = A\). Then my code finds \(A\) upto the first 2-3 significant digits. I'm still having a few mistakes; but otherwise everything works flawlessly.
The math makes perfect sense too. We don't actually need \(\text{slog}_K\) to construct \(\chi\)'s exponential series. We just need it's behaviour near \(-\infty\) which has exponential decay to \(-2\); which makes it fairly easy--even easier when all we need is the first order expansion.
My goal is a little blurry at this point; but I think I'm getting close to the right answer! I believe; the independent construction of \(\chi\) attached with \(\beta\) produces a novel construction style for Kneser. And it's actually fucking working this time
. This makes much more sense than my earlier attempts. We're legitimately just writing:
\[
\text{tet}_K(z) = \chi^{-1}\left(\beta(z)\right)\\
\]
And since \(\beta\) already looks so much like tetration this should be possible, so long as \(|\Im(z)| < \pi\). Getting it to work outside of the period case, will be difficult, but should be doable. But I'll cross that bridge after I construct \(\chi\) to desired accuracy... I will cross that bridge when I get there.
Here is a graph of the beta-chi function, which I remind the reader:
\[
\chi(z) = \beta(\text{slog}_K(z))\\
\]
Without constructing \(\text{slog}_K(z)\) using any pre-existing function. So all we need is \(\beta\), and no mention of Sheldon's code.
As a by-product of constructing \(\chi\) we actually end up constructing Kneser's slog. This isn't that hard to see if we think about this more abstractly. \(P = \text{slog}_K(z)\) is the unique function such that:
\[
\chi(e^z) = \frac{e^{\chi(z)}}{1+e^{-P(z)}}\\
\]
So if you construct one accurately, you construct the other! So I already have some code for Kneser's slog; which again uses none of sheldon's code, other than the acquisition of \(\text{slog}'(0)\). For now my code is only safe for 2-3 digits. But I'm considering this a milestone that this method will work. And if you have \(\chi\), then functionally you've constructed kneser from \(\beta\). Where we get:
\[
\chi(\text{tet}_K(z)) = \beta(z)\\
\]
Which, I've confirmed for more digits than the slogarithm; about 4-5 digits.
I should also add, that when I say "confirmed to 2-3 digits"--I mean the Taylor series are correct upto to 2-3 significant digits. So if \(\frac{d^j}{dx^j} \text{slog}_K^{(j)}(x)/j! = A\). Then my code finds \(A\) upto the first 2-3 significant digits. I'm still having a few mistakes; but otherwise everything works flawlessly.
The math makes perfect sense too. We don't actually need \(\text{slog}_K\) to construct \(\chi\)'s exponential series. We just need it's behaviour near \(-\infty\) which has exponential decay to \(-2\); which makes it fairly easy--even easier when all we need is the first order expansion.
My goal is a little blurry at this point; but I think I'm getting close to the right answer! I believe; the independent construction of \(\chi\) attached with \(\beta\) produces a novel construction style for Kneser. And it's actually fucking working this time
. This makes much more sense than my earlier attempts. We're legitimately just writing:\[
\text{tet}_K(z) = \chi^{-1}\left(\beta(z)\right)\\
\]
And since \(\beta\) already looks so much like tetration this should be possible, so long as \(|\Im(z)| < \pi\). Getting it to work outside of the period case, will be difficult, but should be doable. But I'll cross that bridge after I construct \(\chi\) to desired accuracy... I will cross that bridge when I get there.
