# Tetration Forum

Full Version: Should tetration be a multivalued function?
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When we have a tetration $\vspace{15}{y_{x}=\,^xa}$ with his base "a" between $\vspace{25}{1< a \leq e^{\frac{1}{e}}}$, it tends to an asymptote value $\vspace{15}{y_{\infty}}$ such that $\vspace{25}{a^{y_{\infty}}={y_{\infty}}}$.

But that depends on the definition $\vspace{15}{^0a=1}$. If it were defined $\vspace{15}{^0a=y_{\infty}}$, we would get an horizontal line (drawn with red dashes in the graphic down).

The interesting part is when we define $\vspace{15}{^0a}$ a bit larger than $\vspace{15}{y_{\infty}}$, the tetration also converges to the same asymptote to the right, and another asymptote to the left, at the limit $\vspace{15}{y_{-\infty}}$, defined by $\vspace{25}{a^{y_{-\infty}}={y_{-\infty}}}$.

We get a Z-shaped tetration function (drawn in green), contained between $\vspace{15}{y_{-\infty}}$ and $\vspace{15}{y_{\infty}}$, which I call the Z curve.

The problem is that the Z curve is not uniquely defined. It depends on the value choosen for $\vspace{15}{^0a}$. Any value between $\vspace{20}{{y_{\infty}} < ^0a < {y_{-\infty}}}$ is valid, and the only difference this choice make, is the horizontal displacement on the curve. I drew the Z curve matching the origin with his inflection point (roughly).

The upper asymptote is also a non trivial function, and there is also another non trivial possible upper branch, obtained by choosing °a>-oo (drawn in blue line).

The question is, what would be convenient values for the definition of $\vspace{15}{^0a}$ for the green and blue branches?

If we take $\vspace{15}{a=\sqrt{2}}$, $\vspace{15}{^0a}$ may be multivalued at x=0: $\vspace{15}{^0(\sqrt{2})=(1,2,\; 2.
The inflection point in the Z curve is near to 3. Maybe that base has all integer values for °a? (°1,41421356 = (1,2,3,4,5) )

Note how the blue branch resembles the function $\vspace{15}{e^x}$.
So, I think that an expression for the derivative should also give (at least) 5 different results.
It's always been known that super-exponentials have an intricate and complicated Riemann surface, but aside from the bits we've seen in plots and graphics, I don't think there is a comprehensive enumeration of branch indexes or a system of labeling each branch. I can think of a few threads in this forum where such a branch index system has been theorized, but I don't remember if we came to any conclusions. It's easy to index some of the branches by adding an imaginary logarithm, but I'm not convinced that doing this enumerates all branches of super-exponentials.
(01/03/2016, 11:24 PM)marraco Wrote: [ -> ]The problem is that the Z curve is not uniquely defined. It depends on the value choosen for $\vspace{15}{^0a}$. Any value between $\vspace{20}{{y_{\infty}} < ^0a < {y_{-\infty}}}$ is valid, and the only difference this choice make, is the horizontal displacement on the curve. I drew the Z curve matching the origin with his inflection point (roughly).

It appears that these 3 branches are not connected in the real line, but I would suspect that they are connected in the complex plane (to make a Riemann surface), and that fixing ${}^0a = 1$ on the main branch might fix the value of the other branches, at least that's what I would think.
(01/04/2016, 12:01 AM)andydude Wrote: [ -> ]It appears that these 3 branches are not connected in the real line, but I would suspect that they are connected in the complex plane (to make a Riemann surface), and that fixing ${}^0a = 1$ on the main branch might fix the value of the other branches, at least that's what I would think.
[Note: the asymptotes make 5 branches]

If the transition is not singular, then the blue branch should be coincident with the black (main) branch for $\vspace{15}{a>e^{e^{-1}}}$. That may give a clue about his better value for °a.

[Edit: the transition for the blue branch must be singular, because it cannot smoothly change from an horizontal asymptote to a parallel line to the main branch, which has a vertical asymptote.
That unless tetration introduces a new kind of numbers which are continuous in the transition.]

They should be coincident, because for the blue branch, any value chosen for °a, is also somewhere in the main branch, so taking iterated logarithms/exponentiations should give the same result.
(01/03/2016, 11:24 PM)marraco Wrote: [ -> ]The question is, what would be convenient values for the definition of $\vspace{15}{^0a}$ for the green and blue branches?

It looks like this paper written by fellow forum members, also uses 1, 3, 5 as the points which fix the branches.
(01/04/2016, 12:35 AM)andydude Wrote: [ -> ]
(01/03/2016, 11:24 PM)marraco Wrote: [ -> ]The question is, what would be convenient values for the definition of $\vspace{15}{^0a}$ for the green and blue branches?

It looks like this paper written by fellow forum members, also uses 1, 3, 5 as the points which fix the branches.

Still, it looks like 3 and 5 had been chosen arbitrarily.
Let's call c₁ to the superior asymptotic limit, and c₂ to the inferior. There are interesting relationships between those numbers.

They can be deduced from the known relation that defines those numbers: The base of the function Y(x)=ˣa is $\vspace{27}{a=c_1\,^{\frac{1}{c_1}}}$, and also $\vspace{27}{a=c_2\,^{\frac{1}{c_2}}}$

then:

$\vspace{25}{\frac{c_2}{c_1}=\frac{ln(c_2)}{ln(c_1)}}=a^{c_2-c_1}$

$\vspace{25}{{\color{Red}c_1\,^{c_2}=c_2\,^{c_1}=a^{c_1.c_2}}}$

The last one is analogous to the relationship between the asymptotes for bases between $\vspace{15}{ 0

(04/13/2015, 08:01 PM)marraco Wrote: [ -> ]for bases between $\vspace{15}{0 it remains bounded between 0 and 1 (for x>0), and converges to c, were c is the solution of $\vspace{15}{a^{a^c}}\,=\,c$, which seems to have 2 roots, c₁ and c₂, with
$\vspace{15}{a^{c_1}\,=\,c_2}$
$\vspace{15}{a^{c_2}\,=\,c_1}$
$\vspace{15}{{ \color{Red} {c_1}^{c_1}\,=\,c_2^{c_2}\,=\,{a^{c_1.c_2}} }}$
$\vspace{15}{a\,=\,{c_1}^{\frac{1}{c_2}}\,={c_2}^{\frac{1}{c_1}}}$

I suspect that this relation is the key to solve tetration equations:

$\vspace{15}{a^{c_1}\,=\,a^{a^{c_2}}\,=\,a^{a^{a^{c_1}}}\,=\,...$

Here is tetration base a=0.01:
c₁ = 0,941488369
c₂ = 0,013092521

c₁ is the limit to the left, and c₂ to the right.
Let r be a real number 0<r<1

We can numerically calculate c₍ᵣ₎=ʳa and c₍₋ᵣ₎=⁻ʳa with Sheldonison's knesser-gp.

No proof, but numerically:
$\vspace{15}{(^{r+1}a)^{^{-(r+1)}a}=(^{-r}a)^{^{r}a}}$

and in general

$\vspace{15}{(^{r+(n+1)}a)^{^{-(r+(n+1))}a}=(^{-(r+n)}a)^{^{r+n}a}}$

This remains valid for -(r+n)<-2, where the iterated logarithms are complex numbers.

This gives a natural way to define °a for the Z branch. °a should be the value that makes:

$\vspace{15}{^{-1}a\,^{^1a}\,=\,^1a\,\,^{^{-1}a}\,=\,^{0}a\,^{^0a}\, \; or \; (log_a{^0a})\,^{(a^{(^0a))}=(a^{(^0a)})\,^{(log_a{^0a})}$

then it also satisfies:

$\vspace{15}{(^{r+(n+1)}a)^{^{-(r+(n+1))}a}=(^{-(r+n)}a)^{^{r+n}a}}$

Example (for the main branch)

For the Z branch
(01/08/2016, 06:26 PM)marraco Wrote: [ -> ]This gives a natural way to define °a for the Z branch. °a should be the value that makes:

$\vspace{15}{^{-1}a\,^{^1a}\,=\,^1a\,\,^{^{-1}a}\,=\,^{0}a\,^{^0a}\, \; or \; (log_a{^0a})\,^{(a^{(^0a))}=(a^{(^0a)})\,^{(log_a{^0a})}$

then it also satisfies:

$\vspace{15}{(^{r+(n+1)}a)^{^{-(r+(n+1))}a}=(^{-(r+n)}a)^{^{r+n}a}}$
I retract that. The last equation is satisfied no matter what the value of °a is.
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