# Tetration Forum

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andydude Wrote:You're right, I'm wrong. $\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -1.85$ 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.

Andrew Robbins

So what are the exact values for :

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) = ?$

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 9\\\hline\end{tabular} b) = ?$

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 11\\\hline\end{tabular} b) = ?$

etc?

I am looking for them to prove or disprove the fine structure constant approximation which seems to improve with each next value above [5], but the accuracy is bad as I took them from Andrew's graph.

Ivars
Ivars Wrote:So what are the exact values for :

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) = ?$

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 9\\\hline\end{tabular} b) = ?$

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 11\\\hline\end{tabular} b) = ?$

etc?
Well, always for $b < 0$ let us start from (Hey, boys, I am starting using TeX ... Waaaoow!):

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) = c$

For an estimation of $c$, I think we may proceed like this, assuming the new more precise asymptotic value obtained by Andydude (step 0):

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -1.85..$, and:

$\lim_{b\rightarrow-1.85..}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -\infty$

Then:

step 1 - Calculate $y = e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b$ vor various values of $b < 0$ and produce a graphical continuous and smooth plot, if possible;

step 2 - Produce the graphical inversion of the previous plot, always in the $b < 0$ domain, so obtaining the pentalog of $b < 0$ ;

step 3 - Estimate the intersection between the two pentation/pentalog diagrams (always in $b < 0$), which should coincide with one the pentation fixpoints for $y = b$. The coordinate of this point should be what we may call $h$. This $h$ is the coordinate of the penta-fixpoint and, therefore of the hexation asymptote. We can see thay it must be $h < -1.85..$.

step - Proceed again as in step 3 with the hexation/hexalog plots and the result will be $c < h$, such that:

$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) = c$, and:

$\lim_{b\rightarrow c}(e \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) = -\infty$

And ... so on !!!!
However, ... it is a very long way ... Perghaps there are shortcuts.

GFR
[Sorry, Administrator, these comments of mine, apart the first 5 lines are completely wrong. I shall correct them asap. Perhaps, I was tired! - GFR]
bo198214 Wrote:May I ask if anyone has an idea how much this result depends on the actually chosen tetration extension to real numbers? I mean the b[n]-k=-k+1 for 0$\le$k$\le$n-3 does not depend on the extension.

Surely. Just as $b[4](-\infty)$ is the fixed point obtained by iterating $\log_b(x)$, so is $b[5](-\infty)$ the fixed point obtained by iterating $\text{slog}_b(x)$ which also corresponds to the fixed point of ${}^{x}b$. This works well when the fixed point is an attracting fixed point, but poorly for a repelling fixed point.

For $1 < b < \eta = e^{1/e}$ tetration has 2 fixed points, and for $b > B$ (does $B = \eta$?) tetration has only the lower fixed point. Since the lower fixed point of tetration falls between two integers is does depend on which extension is used.

To summarize, one thing we can know for sure regardless of which extension is used, is that even hyper-exponentials follow: $b [N] (-Y) = (-\infty)$ and odd hyper-exponentials follow: $b [N] (-\infty) = (-X)$. All odd hyper-exponentials should be real-valued over all reals. All even hyper-logarithms should be real-valued over all reals. All hyper-exponentials above 3 map [-1,0] -> [0,1] and all hyper-logarithms above 3 map [0,1] -> [-1,0]. Another common property is that the range of odd hyper-exponentials is bounded below, just as the domain of even hyper-logarithms is bounded below.

Since these findings are all using negative hyper-exponents, then they are essentially specifying the number of times to iterate the appropriate hyper-(N-1)-logarithm. All odd hyper-exponentials are real-valued over all reals BECAUSE the hyper-(N-1)-logarithm is also real-valued over all reals. One conclusion we can draw from the statements above is that the domain of even hyper-exponentials is bounded below, because the domain of odd hyper-(N-1)-logarithms is bounded below. Accordingly, the range of odd hyper-exponentials is bounded below, because the range of even hyper-(N-1)-logarithms is bounded below.

I hope that made sense. I will try and make this more formal and more clear in a further post.

Andrew Robbins
andydude Wrote:For $1 < b < \eta = e^{1/e}$ tetration has 2 fixed points, and for $b > B$ (does $B = \eta$?) tetration has only the lower fixed point. Since the lower fixed point of tetration falls between two integers is does depend on which extension is used.

Thanks for the explanation. Maybe also interesting for Ivars asking for *the* value of each $\lim_{x\to -\infty}$b[2n+1]x.
GFR Wrote:$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -1.85..$, and:

$\lim_{b\rightarrow-1.85..}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -\infty$

Wah, the second one is wrong! As Andydude wrote, the odd hyperexponentials are bounded below. The even hyper exponentials b[2n]x are not bounded below but only defined on -2n+2=-(2n-3)-1<x.

We have the general relation
$\lim_{x\to -2n+2}$ b[2n]x = $-\infty$.
in this case:
$\lim_{x\to -4}$ b[6]x = $-\infty$

though that is probably not right either. But the true value $-2n+2
$\lim_{x\to X}$ b[2n]x = $-\infty$.
Yes! My comments were completely wrong. Sorry about that. I shall amend them asap.

GFR
Concerning:
bo198214 Wrote:
GFR Wrote:$\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -1.85..$, and:
$\lim_{b\rightarrow-1.85..}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -\infty$
Wah, the second one is wrong! As Andydude wrote, the odd hyperexponentials are bounded below. The even hyper exponentials b[2n]x are not bounded below but only defined on -2n+2=-(2n-3)-1<x.
We have the general relation

$\lim_{x\to -2n+2}$ b[2n]x = $-\infty$.
in this case:

$\lim_{x\to -4}$ b[6]x = $-\infty$
Actually, believe or not, what I wanted to say was:

[b -> -00] lim (e[5]b) = - 1.85...

[b -> -1.85..] lim (penta-ln b) = - oo

The problem, for me, was again produced by the TeX animals. After the fifth slash, I fall asleep and I try to compensate using the copy-and-glue technology. Unfortunately, it was late and I was very tired. I just wanted to show that the tails of the sln and sexp (base e) plots cross themselves in a point "sigma", which has real, negative but not integer coordinates. That's all. But, this is the past.

@Henryk -
I agree on the Andydude overall scheme, concerning the odd/even hyperations. I just wished to draw the attention to the need of finding the right numerical values. It is also clear (... for me, I hope not to be wrong) that in the odd (3, 5, 7, ..) ranks y = b[2n+1]x have horizontal asymptotes (for x -> -oo), say for y = - k, and, therefore, their corresponding hyperlogs have vertical asymptotes for x -> - k (with y -> -oo). It is also clear that, for even ranks (2, 4, 6, ..) the contrary is true (horizontal asymptotes for the hyperlogs and vertical asymptotes for the hyperops). However, the problem remains of finding (exactly) such numbers. I think that we must completely solve the "tetration" business, before going further.

Probably, you are also right, even formula [x -> -2n+1]lim(b[2n]x) = - oo may be wrong.

@Ivars -
The role of the selfroot is essential. Nevertheless, it is important as a solution of the y = b[s]y functional equation. But, unfortunately (so to say) it is not its unique "functional root". Another solution can be found by considering y = b[s](b[s]y), or [/b]s-log(y) = y = b[s]y.

E.g. for rank 3 (exponentiation), we should consider (supposing base b) an expression such: [/b]log x = x = b^x. I. e. also other intersection points between the exp and the log, which will give other "branches" of the functional roots, are relevant. And, the same should be valid for other hyperops ranks.

GFR
GFR Wrote:The problem, for me, was again produced by the TeX animals. After the fifth slash, I fall asleep and I try to compensate using the copy-and-glue technology.

*Gives the poor TeX-haunted GFR a hug*

Quote:I agree on the Andydude overall scheme, concerning the odd/even hyperations. I just wished to draw the attention to the need of finding the right numerical values.

Ya, which however depend on the extension method.

Quote:I think that we must completely solve the "tetration" business, before going further.
Whatever "solving the 'tetration' business" means ... perhaps designate "the" solution and spread around the world that it is the only true solution
Yeah ..., we must try harder. Truth and unicity, this is the question.
It would appear that we have our work cut out for us, in defineing higher operators, addition, multiplication, exponentiation, tetration, pentation, hexation....ultration.

What I am interested in in is generalized triadic operators, for higher operators, you know y=x[w]z.

I mean, could you graph this formula/function, for {y, x, w, z}eR?

I could only assumme that you'd map it to 3-space, and 1 time space, making 4 dimensions, or perhaps you could do this in 2D space, with one axis a color spectrum.

Please be aware that there are 100^3 definitions (speaking generally...) for this triadic operator, and I call this the "ultra-operator", or a "wild operator".

Why 100^3?

Well, because:

+/-{0, N, Q, R, T}+/-{0, iN, iQ, iR, iT} gives us 100 possibilitys, where:

N=Integers, Q=Rationals, R=Irrationals, T=Transendentals.

The formula above may even be 8-dimensional if you were to bring in complex numbers, and also, did you consider inverse operations? -Here:

f^-1(a, b)-->(a[b]c)=c, f^-1(b, c)-->(a[b]c)=a, f^-1(a, c)-->(a[b]c)=b.

Remember, also, todays mathematics is tomorrows physics, so perhaps one day these higher operators will have a practical real world application.
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