Now, Henryk will say: "But, Gianfranco, you are always late and .... approximated !"
This will happen concerning:
bo198214 Wrote:andydude Wrote:I just found the asymptotes of pentation, hexation, heptation, octation, and beyond! And they're fascinating:
 & = -\infty \\<br />
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) & = -2 \\<br />
\lim_{b \rightarrow -4}(a \begin{tabular}{|c|}\hline 6 \\\hline\end{tabular} b) & = -\infty \\<br />
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) & = -4 \\<br />
\lim_{b \rightarrow -6}(a \begin{tabular}{|c|}\hline 8 \\\hline\end{tabular} b) & = -\infty \\<br />
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 9 \\\hline\end{tabular} b) & = -6<br />
\end{blign}<br />
)
...
... these are really fascinating findings.
Indeed, they are! But, I think, they are not completely correct (Please, Andydude, correct me, if I'm wrong!! Please also remember my last msg to you "
Sure thing"!). In fact, take the first two lines:
 = -\infty \\<br />
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -2 \<br />
<br />
)
I agree that the first line is correct, for any
. In case of
, the limit is +oo. For
and
, we are in trouble. Concerning the second line, I should like to recall the following relations, the first of which valid for tetration:
a[3]b = a ==> h(a) = a[4]oo, depending on base a;
which also should imply the following, for pentation:
a[4]b = a ==> h(a) = a[5]oo, also depending on a.
I mean that the fixpoints (of exponentiation) determine the limit heights of tetration (we know that). Similarly, the fixpoints (of tetration) should determine the limit heights of pentation.
We had (I still have ...) a lot of problems in linking real and complex fixpoints with the tetration limit heights h and, I think, we are (I ... am) not yet completely prepared to study the fixpoints of tetration,
for any base a. Nevertheless, there is an area of tetration where the "plots" are more or less symmetrical, for various bases "
a", in the negative domains of the slog and of the sexp right (or second) operands, i.e.: [/a]sexp(x) and [/a]slog(x), for x < 0.
In fact, let us consider:
y = a[4]x and y = [/a]slog(x)
The two plots, always increasing "functions" for x < 0, have their two "tails" that cross themselves in one fixpoint, the coordinate(s) of which we may call "Sigma", with definitely: -2 < Sigma < 0.
(see:
http://forum.wolframscience.com/attachme...ostid=4192 ).
The values of Sigma are depending on base "a", i.e.:
[b --> -oo] lim a[5]b = Sigma(a).
We already found some approximated values of Sigma, in the following cases (always for operamd b < 0):
Sigma(e) = - 1.841..., obtained by the KAR/GFR linear approximation (Andydude can do better)
Sigma(Eta) = Sigma(e^(1/e)) = - Pi/2 = 1.570796327.. , graphically obtained by GFR, for b < 0
....
Sigma(oo) = -2 (is this true? I think it is ...).
We should then write:
 = -\infty \\<br />
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = Sigma(a))
And ... the rest should follow, accordingly. For base a > 1, I also agree on the limit value of a[s]n, for s -> oo and n -> -oo, which (by calling w the first infinite ordinal, or something like it ...) are indeed given by: a[w]n = n + 1
We should therefore write, for a > 1, n < 0:
[s --> +oo] lim a[s]n = a[w]n = n + 1
Quickfur would probably say: "
Much ado about nothing!" . We proposed to call "Omegation" such funny titanical operation. By the way, Henryk, well done, for your induction proof. Please check what I said and see if I didn't confuse issues, variables or signs.
My concern is what would probably happen in case of base
a < 1. I wonder if the entire sequence, mentioned by Andydude collapses, without even reaching n + 1. In fact, I have even doubts on the possibility of correctly defining the slog operation, for base a < 1.
But ... , Henryk, this is another story. If you allow me, I shall put an ad-hoc thread about that.
GFR
? Analytically?
for all
not -2 as I had first thought. Jay has found the more precise values in this thread which you should look to instead of this post, as I'm only accurate to 1 decimal place.