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This \(\exp(-x)\) you are seeing, is slightly inaccurate while at the same time very accurate. Let's look at:
\[
\vartheta(x,-1)\\
\]
And let's plot it over \(x \in [0,70]\)--using my program, so slightly inaccurate.
You can notice it goes all the way to the millions at about \(x \approx10,20,30\). RIGHT AFTER THOUGH, it decays to zero just like \(\exp(-x)\)...
Now the exact decay is more wonky, and the best I can prove is \(O(1/x)\), but if I choose \(z = -1+i\delta\) then this decay goes even faster. And the exponential decay term you have sussed out appears over and over again. But there's a lot of error, remember that, and these errors can add growth we don't like.
The main argument I make, which is purely a mathematical statement: Is that:
\[
\int_0^\infty|\vartheta(x,-1)|x^{-\sigma}\,dx < \infty\\
\]
In true honest form though, I expect \(\vartheta\) to have decay like \(\exp(-x)\). I just have no real idea how to prove this. So I settled with the above,
EDIT: Also \(\vartheta\) has nothing to do with the \(\theta\) that Kneser or Sheldon use; what Trappman calls the crescent iteration. \(\vartheta\) is a function I have used to describe fractional calculus (fancy mellin transforms) and its relation to iteration theory. \(\vartheta\)/fractional calculus/mellin transforms has worked perfectly for geometric fixed points. Geometric fixed points are pretty easy in the long run. The real trouble is: The same formulas work for neutral fixed points, but I could never prove this. I could never solidify why the above code even worked. Now I can...
Thanks to you and Trappman.... And all the stuff we clarified, talked about, looked at, ran through. Fucking everything.
This \(\exp(-x)\) you are seeing, is slightly inaccurate while at the same time very accurate. Let's look at:
\[
\vartheta(x,-1)\\
\]
And let's plot it over \(x \in [0,70]\)--using my program, so slightly inaccurate.
You can notice it goes all the way to the millions at about \(x \approx10,20,30\). RIGHT AFTER THOUGH, it decays to zero just like \(\exp(-x)\)...
Now the exact decay is more wonky, and the best I can prove is \(O(1/x)\), but if I choose \(z = -1+i\delta\) then this decay goes even faster. And the exponential decay term you have sussed out appears over and over again. But there's a lot of error, remember that, and these errors can add growth we don't like.
The main argument I make, which is purely a mathematical statement: Is that:
\[
\int_0^\infty|\vartheta(x,-1)|x^{-\sigma}\,dx < \infty\\
\]
In true honest form though, I expect \(\vartheta\) to have decay like \(\exp(-x)\). I just have no real idea how to prove this. So I settled with the above,
EDIT: Also \(\vartheta\) has nothing to do with the \(\theta\) that Kneser or Sheldon use; what Trappman calls the crescent iteration. \(\vartheta\) is a function I have used to describe fractional calculus (fancy mellin transforms) and its relation to iteration theory. \(\vartheta\)/fractional calculus/mellin transforms has worked perfectly for geometric fixed points. Geometric fixed points are pretty easy in the long run. The real trouble is: The same formulas work for neutral fixed points, but I could never prove this. I could never solidify why the above code even worked. Now I can...
Thanks to you and Trappman.... And all the stuff we clarified, talked about, looked at, ran through. Fucking everything.