a tetration limit for base $a > e^{1/e}$

Question

Let $a$ be a real number with $a > e^{1/e}$ and $a <> e$.

$slog$ means superlog base $e$ and $sexp$ means superexp base $e$.

$sloga$ means superlog with base $a$ and $sexpa$ means superexp base $a$.

$k$ is a real number with $k>0$ and $x$ is a real number with $x>0$.

Conjecture

There exists a real constant $Q > 0$ for each k (as a function of k) such that as $x$ goes to +infinity :

$Q = \dfrac{sexp( slog(sexp(slog(x)+k)/sexpa(sloga(x)+k)) -k)}{x}$

or equivalently

$sexp(slog(Qx) + k) = sexp(slog(x)+k) / sexpa(sloga(x)+k)$

Now this is a conjecture about when $x$ goes to $oo$. If $k$ goes to OO we can show there is a $Q$ for that $k$. To see that we use the famous change of base :

lim $k$ (change of base)

$Ax = sexp( slog( sexpa(sloga(x)+k) ) -k)$

Where $A$ is a nonzero real number.

or equivalently

$sexp(slog(Ax) + k) = sexpa(sloga(x)+k)$

Now if we plug this in the previous equations we get

$=> sexp(slog(Qx) + k) = sexp(slog(x)+k) / sexp(slog(Ax)+k)$

And thus $Q$ and $A$ are trivially linked.

If $A > 1 \implies Q < 1$. If $A < 1 \implies Q > 1$ And $0<QA<oo$

Thus if $k$ goes to oo we have solved the Conjecture , but if $k$ is finite the use of the base change is more dubious.

How to prove the Conjecture ?

Answer 1

Not a full answer or intended as an answer but related and too long for comment.

Concerning that related base change idea.

My apologies if this is already posted on some math forum , MSE or MO.

Also Im not 100 % sure.

More apologetics ; this is a sketch from 'tommy1729' from an email so credit goes to him.

Attempted proof (sketch)

title : change of base 'wobble' proof.

sexp(slog( sexpa(sloga(x)+k) ) -k) = A x + o(1)

( for lim k = oo POSITIVE INTEGER ! )

sexp(slog( sexpa(sloga(x+y)+k) ) -k) = A (x+y) + o(1)

sexp(slog( sexpa(sloga(x)+k+1/2) ) -k) = A (x+y) + o(1)

--

sloga(x+y) = sloga(x)+1/2

y = sexpa(sloga(x)+1/2) - x

A(x+y) = A sexpa(sloga(x)+1/2)

--

sexp(slog( sexpa(sloga(x)+k+1/2) ) -k) = A sexpa(sloga(x)+1/2) + o(1)

sexp(slog( sexpa(sloga(x)+k+1/2) ) -k-1/2) = sexp( slog( A sexpa(sloga(x)+1/2) + o(1) ) -1/2)

= ? = A x + o(1)

reverse ?

?<=> slog(A x +o(1)) +1/2 = slog( A sexpa(sloga(x)+1/2) + o(1) )

=> assume o(1) = 0 (?)

=> y >> x

?<=> slog(A x) + 1/2 = slog( A (x+y) )

AND

  sloga(x) + 1/2 = sloga(x+y)

<= slog(A z) = sloga(z)

? }=> A z = sexp(sloga(z))

( = <=> A sexpa(z) = sexp(z) )

<=> sexp(sloga(z)) = sexp(slog( sexpa(sloga(z)+k) ) -k)

=> sloga(z) = slog( sexpa(sloga(z)+k) ) -k use x = sloga(z) <=> x = slog( sexpa(x+k) ) -k

statement false , proof by contradiction. QED.

end proof

=> argument against your OP on MSE.

It is surprising that this is INDEPENDANT of the type of abel or inverse abel (super) we use.

Im am still not feeling well so forgive If i made a mistake.

end quote tommy1729.

I did not have time to format sorry.

Maybe this is what Shel wanted partially.

Answer 2

This is meant as a comment to make the question more precise before I can think of an answer at all, but doesn't fit in a comment-box

Did I get your formulae correctly with this, where the tiny superscript circle means iteration

$ \tag1 \qquad \displaystyle Q = {\exp°^{-k} \left({ \exp°^k(x) \over \exp_a°^k(x) }\right) \over x}$

$ \tag2 \qquad \displaystyle \exp°^k(Qx) = { \exp°^k(x) \over \exp_a°^k(x) }$

I read the other formula concerning "base-change"
$ \tag3 \qquad \displaystyle Ax = \exp°^{-k}(\exp_b°^{k}(x)) $

If this is all correct, then fine so far; but I'm still missing some understanding of your arguments. Clearly if the rhs in (3) converges to some real number when k goes to infinity, then this number can be related to x by some real cofactor A -but that's simply trivial... so I do not see any information in this.

Or do you want to say that the change of the rhs becomes zero, when k increases, so you refer to the derivative with respect to k ? And do you then assume some relation between A and the base b (or x) ?

Answer 3

The "k" is being used as a fractional exponentiation indication, as x goes to infinity. For example, substituting k=1 into the Mick's equations yields the following equations. $$sexp(slog(x)+1)=e^x$$ $$sexpa(sloga(x)+1)=a^x$$ $$sexp(slog(Qx)+1)=e^{Qx}$$

So then, for k=1, Q=1-log(a), by taking the logarithm of the following equation, after substituting the above into Mick's equation. new comment: I now realize Mick has already generated this result, and posted it yesterday. $$e^{Qx}=\frac{e^x}{a^x}$$ $$Qx=x-x*log(a)$$ $$Q=1-log(a)$$

Continuing on, assuming that Q(k,a) is meant to be a function of both k and a, lets generate the equations for k=0.5 $$sexp(slog(x)+0.5)=\exp^{o 0.5}(x)$$ $$sexpa(sloga(x)+0.5)=\exp_a^{o 0.5}(x)$$ $$sexp(slog(Qx)+0.5)=\exp^{o 0.5}(Qx) = \frac{\exp^{o 0.5}(x)}{\exp_a^{o 0.5}(x)}$$

Here are some numerical calculations for k=1/2 to help show that a similar limit probably does not exist for fractional iterations of k. Fractional iterations for different bases tend to behave differently for different values of x. For example, for x=10^100, the half iterate for 2^x is larger than the half iterate of e^x, which is not intuitive. For x=10^10, the half iterate of 2^x is smaller than the half iterate of e^x, as would intuitively be expected. For these calculations, I used the pari-gp code, which is online at http://math.eretrandre.org/tetrationforum/showthread.php?tid=486 $$sexp(slog(10^{100})+0.5)=5.342915750*10^{6890248}$$ $$sexp2(slog2(10^{100})+0.5)=2.041342098*10^{6928081}$$ $$sexp(slog(10^{10})+0.5)=5.588492691*10^{266}$$ $$sexp2(slog2(10^{10})+0.5)=1.258585157*10^{253}$$

There is a 1-cyclic relationship, as x gets larger, governing $$\theta(x)=sloga(sexp(x))-x$$ As x gets larger, the value for theta(x) quickly converges, so that theta(x) is arbitrarily close to theta(x+1), theta(x+n). But theta(x) converges to a 1-cyclic real valued function, as opposed to converging to a constant, which might be expected. The 1-cyclic function that theta(z) converges to has an amplitude varying by about +/-0.001, is infinitely differentiable, and surprisingly, is nowhere analytic. I don't know if this has been published outside of the tetration forum, http://math.eretrandre.org/tetrationforum/showthread.php?tid=14&page=5, but it is a very interesting phenomena, originally discovered in looking for a "basechange" version of tetration, which turns out to be different than the unique tetration solution I used to generate results here, which is defined everywhere in the complex plane except for singularities at integers<=-2.
- Sheldon

Answer 4

So we want to show that for each $k \ge 1$ [I assume that $k$ is an integer], $x \simeq \log^k(\exp^k(x)/\exp_a^k(x)) = \log^{k-1}(\exp^{k-1}(x) - \log(a) \exp^{k-1}_a(x))$, for sufficiently large $x$. If $k = 1$, we have $\log(\exp(x)/a^x) = x(1 - \log(a))$, and we are done. So assume $k \ge 2$.

Assuming $e^{1/e} < a < e$, we need only pick $x$ large enough that $\exp^{k-1}_a(x) < a^{\exp^{k-2}(x)} = \exp^{k-1}(x)^{\log(a)} < \frac{\delta}{\log(a)} \exp^{k-1}(x)$. For then $\log^{k-1}(\exp^{k-1}(x) - \log(a) \exp^{k-1}_a(x)) > \log^{k-2}(\exp^{k-2}(x) + \log(1 - \delta)) $, and the latter quantity can clearly be made arbitrarily close to $x$ by picking $\delta$ small.

On the other hand, suppose that $a > e$; then we can pick $x$ large enough that $\delta > |\log(a) \exp_a^{k-2}(x) + \log(\log(a)) - \log(\log(a) \exp_a^{k-1}(x) - \exp^{k-1} x)|$ (since $\exp^{k-1} x$ is small compared to $\exp_a^{k-1}(x)$). Then we can pick $\delta$ small enough that $\log^k(\exp^k(x)/\exp_a^k(x))$ is within $\epsilon$ of $\log^{k-2}(\log(a) \exp_a^{k-2}(x) + \log(-\log(a)))$. But this is asymptotically equivalent to either $\log(a) x$ or $x$, so we are done.