# Tetration Forum

Full Version: Generalized recursive operators
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2 3 4
Cool.

That raises another question. Let x=0 and y=infinity. Then we have a R(x) b = a R(y) b. Which x's have y's such that a R(x) b is the same as a R(y) b for this interval?

Potentially, if we can find a function that will generate y's from x's, and if some of the x's are real numbers, then we may be able to define a R(n) b for real n such that 0 < n < 1--at least, when b < 0 and a > 1.

Alex
andydude Wrote:Wow! I can't believe I gave that away for free! I would pay hundreds to get my hands on this code! Good thing I didn't have to... I just found the asymptotes of pentation, hexation, heptation, octation, and beyond! And they're fascinating:

\begin{align}
\lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) & = -2 \\
\lim_{b \rightarrow -4}(a \begin{tabular}{|c|}\hline 6 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) & = -4 \\
\lim_{b \rightarrow -6}(a \begin{tabular}{|c|}\hline 8 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 9 \\\hline\end{tabular} b) & = -6
\end{blign}

where the box is GFR's box notation for hyperops, and a is sufficiently close to e, because thats the number I used.

I suppose you could see this from the integer versions of these operators, but I think the continuous (or if not continuous, mostly real-valued) versions make it easier to see.

Andrew Robbins

This is fantastic.

Ivars
Hi, Andydude!

Ref.:
andydude Wrote:... in the limit, all hyper-operators return to the successor operation, like the circle of life...
Andrew Robbins.
Please see also the KAR-GFR Thread posted to the NKS Forum on 25-07-2006:
http://forum.wolframscience.com/showthre...eadid=1168
Particularly the annex "Notes on Hyperoperations - Second progress report - NKS Forum III" . Page 12, formula 21.

We called "Omegation" that limit hyper-operation. Indeed it is sending us back to the successor operation and, therefore, to "Zeration" (see a forthcoming thread about that, in this Forum).

GFR
Hello,

This has a fascinating intutive appeal, especially the appearance of even negative integers-just like trivial 0 of Riemann zeta function.

What are the few next values on other axis (- 3, ..., - 5, ... ) and how accurate they seem to be? Meaning is e.g. -1,85.. really close to asymptotic value or it can be - 1,9..

I am asking because of this sum:

=1/1,85035452902718^8-1/(2*1,8503545290271^7+1/(3*1,8503545290271^6-1/(4*1,8503545290271^5+1/(5*1,8503545290271^4=0,007297583=1/137,0316766

which makes sense to me as long as we include only (operations up to) pentation in its approximation.

So long as pentation is faster than tetration, it means that processes (phase transitions) described by tetration need pentation to run processes inside them, - and so on. Infinitely nested phase transitions and not only phase transitions- nested time scales in general.

Ivars Fabriciuss
We have to check all the asymptotic values of y = b[s]x for x -> -oo, for any b and for "s" odd integer. There is a pattern.

GFR
Is it possible to have analytic values y = b[s]x for x -> -oo, [s] odd integer in any of the bases?

Put other way, do there ( does there) exist a base ( an algorithm to construct a base) which would lead to analytic values for all odd [s] as x-> - oo?

I have a gut feeling this relates to Euler's old problem of finding sum of alternating sign terms of :

1/1^n -1/2^n+1/3^n-1/4^n+1/5^n-1/6^n+1/7^n - where n- odd integer.

Well , Euler new only one hyperoperation-tetration, so he could not go above n=3 in his search.

Or, may be this sum is already found?

Ivars
Well, it's a long way... !

Let's start considering that, at rank s=4, tetration, we have that :
y = b^y = b[3]y <---> b = yth-rt(b) = selfrt(y) <---> y = b[4](+oo) = h <---> b = (+oo)th-srt(h).

We may then suppose that, very probably, we should also have that:
y = b§y = b[4]y <---> b = yth-srt(b) = selfsrt(y) <---> y = b[5](+oo) = g <---> b = (+oo)th-ssrt(g).

In other words, by calling "g" the infinite pentation (infinite super-tower .. !) to the base "b", we may discover that "g", at rank s=5 (like "h" for rank s=4) can also converge in some domains of "b". But this needs a strict mathematical demonstration, itself requiring the continuity (analyticity?) of both slog and sexp, at rank s=4. First, we must solve and consolidate this. Then, we shall see. The consequence of that would be that the infinite pentation "g(b)" would be depending on "b" and that g(b) should be the inverse of b(g), like h(b) is the inverse of b(h).

Nevertheless (...), even before any accurate demonstration, it can be easily seen that y = b[5]x = b§x = b-penta-x, at lest for bases near the value b = e, has an asymptotic behaviour, for x -> -oo (minus infinity). By synthetically examining the situation (please, ... try to remain cool, bo!), we must admit that, for x < 0, we indeed have that:
x = b#x <---> [ b]slog(x) = x <---> b#x = [ b]slog(x).

Now, both with the (GFR/KAR) "linear approximation" and with the Robbins (at present, more precise) "smoothings", we can easily obtain the value of "x", for which the before-mentioned relation is satisfied. We (KAR/GFR) checked it for base b = e and it worked nicely.

We also have the surprise to discover, for the "supertowers" the possibility of having "negative heights". How about the "push-down", Gottfried ?

As a (provisional) conclusion, I must say that the infinite supertowers (pentations) correspond to fixpoints of sexp(x) and slog(x), like the infinite towers (tetrations) were determined by the fixpoints of exp(x) and log(x). This "pattern" is probably repeated for all the hyperops hierarchy.

For base b = e, the sexp(x) and slog(x) admit one very clear fixpoint for x < 0, (we may call "sigma" the value of x satisfying that). Well, this "sigma" (if ... correctly calculated) is the ordinate of the horizontal asymptote of y = e[5]x.

In the other cases of b, the fixpoint are, maybe, two, three, ... oder ... ?

Here we are, for the moment.

GFR
I think that the ordinates of the x -> -oo asymptotes of y = b[5]x are depending on the values of base b.

In particular, it can be shown that, for b = Eta = e^(1/e) = 1.444667861.., we have two fixpoints in y = b#x and that they probably (...) are:
x = {-Pi/2, +Pi/2} = {-1.570796327..., +1.570796327..}, giving two symmetrical horizontal asymptotes to the base-Eta pentation, y = Eta[5]x, with the same (positive, for x -> +oo, and negative, for x -> -oo) values.

So, infinite pentations may also be finite, but I presume that, for b slightly greater than Eta, the pentation will easily ... explode!

GFR
GFR,

See? That sounds rather nice to have such a link between convergence limit of tetration and ordinate of asymtotic value of pentation. That would mean as noticed earlier that there may be an infinite pahtway from Tetration to Pentation etc. Going into infinity and back( my be wrong comment, please do not take into account).

The double values returned are also interesting, the positive asymptote value should correspond to some very slow operation- may be inverse pentation, what ever it means-is there a definition?.

How did You show (it can be shown...) that what You have shown ? Analytically?

Ivars
Another thought-just forgive me please:

-i was the result of infinite tetration of e^pi/2
-pi/2 is a result of infinite pentation of e^(1/e)

-pi/2 is also lni/i which in my outragoues formulation was dI= -pi/2 .

dI was considered by me a hypersurface of hypervolume -i . Now it seems that by moving one operation up we get that surface value if we change base from e^pi/2 to e^(1/e).
That would be fine as long as hypersurfaces are 1 dimension away from hypervolumes by definition. But why did we need to change base to obtain it? On other hand, why not?

Also, see how it looks nested:

-i=h(i^-i) = e^(-i*(pi/2)) = e^ (h(i^(1/i)) * pentation (e^(1/e))) with appropriate signs (infinite operations.)

Now it seems interesting to know which exponentation operation infinitely applied gives e?

(1+1/n)^n =e ?

And which multiplication applied infintely gives 1?

What else should follow in both directions?

Ivars
Pages: 1 2 3 4