Dear Jaydfox!
Concerning:
I am very happy to see that you are approaching this problem exactly like (KAR and myself) we did (perhaps with less precision) in a progress report we posted to the NKS Forum on 25-07-2006, copy attached (NKS Forum III - Final).
In fact, in the case of pentation (to the base b), we may indicate it as:
y = b-penta-x, or y = b ยง x , or y = b [5] x (GFR-KAR conventions), or:
y = b ||| x, according to your conventions (sorry for the arrows, they are .... up!).
As a matter of fact, in that occasion and for the particular case of of b = e, we have shown that pentation is definable also for (integer) hyperexponents x < 0 and that this generates an asymptotic behaviour for x -> -oo. See the attachment, Section 4, pages 8 and 9, formulas 13 to 15.
We noticed that two near successive points of the plot in that area must always be linked by y(x-1) = sln y(x), where sln is the "natural" slog (base e), and by y(x+1) = sexpn y(x), where sexpn is the "natural" tetration operator (base e). We then concluded that the asymptotic value of y could be immediately found by putting:
sexpn(y) = slog(y)
This means that we can find the asymptotic value of y, for x -> -oo, at the intersection of sln(x) with sexpn(x) and we called "Sigma" this numerical value. We got, with our first approximations:
Sigma = -1.84140566043697..
But, very probably, your numerical value is better.
I understand that this was also your conjecture, which was verified through your calculations. Could I have confirmation of the most precise value of Sigma obtainable via more formal and precise calculations, for example using the Andrew's excellent slog approximation? Or... else? I (together with KAR) would be very happy if you could kindly produce that.
Thank you very much in advance.
GFR
(Annex attached on 2-02-2008. Previous attachment missing. Sorry)
Concerning:
jaydfox Wrote:We know that \( e\uparrow\uparrow\uparrow0=1 \), and \( e\uparrow\uparrow\uparrow-1=0 \). But we can find \( e\uparrow\uparrow\uparrow-2 \) by finding \( \mathrm{slog}_e(0) \), which is -1. Then we can find \( e\uparrow\uparrow\uparrow-3 \) by finding \( \mathrm{slog}_e(-1) \). This will quickly take us outside the radius of convergence, so in order to get maximum accuracy, we'll find \( \mathrm{slog}_e\left(\exp_e(-1)\right)-1 \).
Using my 1200-term accelerated solution, the first few iterations give us the following:
\( e\uparrow\uparrow\uparrow0=1 \)
\( e\uparrow\uparrow\uparrow-1=\mathrm{slog}_e(1)=0 \)
\( e\uparrow\uparrow\uparrow-2=\mathrm{slog}_e(0)=-1 \)
\( e\uparrow\uparrow\uparrow-3=\mathrm{slog}_e(-1)=-1.636358354286028979629049436 \)
\( e\uparrow\uparrow\uparrow-4=\mathrm{slog}_e(-1.636358354286028979629049436)=-1.813170483098635639971748853 \)
And so on... Taken to similar precision, the fixed point is -1.850354529027181418483437788.
Going in the forward direction for iteration:
\( e\uparrow\uparrow\uparrow1=\mathrm{sexp}_e(1)=2.718281828459045235360287471 \)
\( e\uparrow\uparrow\uparrow2=\mathrm{sexp}_e(2.718281828459045235360287471)=2075.968335058065833574141757 \)
And so on... Obviously, the next iteration is beyond the scope of scientific notation.
....................................
I am very happy to see that you are approaching this problem exactly like (KAR and myself) we did (perhaps with less precision) in a progress report we posted to the NKS Forum on 25-07-2006, copy attached (NKS Forum III - Final).
In fact, in the case of pentation (to the base b), we may indicate it as:
y = b-penta-x, or y = b ยง x , or y = b [5] x (GFR-KAR conventions), or:
y = b ||| x, according to your conventions (sorry for the arrows, they are .... up!).
As a matter of fact, in that occasion and for the particular case of of b = e, we have shown that pentation is definable also for (integer) hyperexponents x < 0 and that this generates an asymptotic behaviour for x -> -oo. See the attachment, Section 4, pages 8 and 9, formulas 13 to 15.
We noticed that two near successive points of the plot in that area must always be linked by y(x-1) = sln y(x), where sln is the "natural" slog (base e), and by y(x+1) = sexpn y(x), where sexpn is the "natural" tetration operator (base e). We then concluded that the asymptotic value of y could be immediately found by putting:
sexpn(y) = slog(y)
This means that we can find the asymptotic value of y, for x -> -oo, at the intersection of sln(x) with sexpn(x) and we called "Sigma" this numerical value. We got, with our first approximations:
Sigma = -1.84140566043697..
But, very probably, your numerical value is better.
I understand that this was also your conjecture, which was verified through your calculations. Could I have confirmation of the most precise value of Sigma obtainable via more formal and precise calculations, for example using the Andrew's excellent slog approximation? Or... else? I (together with KAR) would be very happy if you could kindly produce that.
Thank you very much in advance.
GFR
(Annex attached on 2-02-2008. Previous attachment missing. Sorry)
