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 $x>0$.


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 ?

  • 1
    $\begingroup$ This is really unintelligible, at least for me. Try using: \infty for $\infty$ \dfrac{A}{B} for $\frac {A}{B}$ \text{slog} for $\text{slog}$ \text{slog}_a for $\text{slog}_a$ \text{sexp} for $\text{sexp}$ \text{sexp}_a for $\text{sexp}_a$ \geq for $\geq$ \implies for $\implies$ \left( {\large{\text{Parethesis that fit stuff}}}\right) for $\left( {\large{\text{Parethesis that fit stuff}}}\right)$ &c $\endgroup$
    – Pedro
    Sep 18, 2012 at 23:06
  • $\begingroup$ I used \dfrac and \implies. The rest is clear enough imho. oo is clearly infinity :) $\endgroup$
    – mick
    Sep 19, 2012 at 14:17
  • $\begingroup$ I prefer \operatorname over \text since it naturally handles spacing issues. $\endgroup$
    – Lord_Farin
    Oct 19, 2012 at 21:47

4 Answers 4


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

  • $\begingroup$ @ Sheldonison : Thanks for the reply but Im not completely convinced yet. 0.001 is a very small number afterall. And I get the impression that your counterintuitive result about 10^10 and 10^100 is related. I think your method might have a roundoff error. Also I would like to express that I think the result depends on what type of sexp you use. Although I wrote about the base change , it is not neccessarily the method used to define $sexp$ for any base. I have to give credit to tommy1729 who also posts on that forum for expressing that concern. tommy suggested using $sexp$ ( continued ) $\endgroup$
    – mick
    Oct 4, 2012 at 19:56
  • $\begingroup$ .. with the condition that for positive $k$ ; $sloga(sexpa(x)+k)$ has both the first and second derivative positive for any positive $x$. He claims that his method based on $sinh$ does provide such a solution. He also told that it ( conjecture in my OP ) would probably create a uniqueness condition equivalent to the one mentioned here about the derivatives. But he went on to tell me that it could probably be tested (numerically) by using instead of $sexpa(sloga(x)+k)$ the approximation $ln(2sinh^{k}(exp(x)))$ because that would end up in the same limit. I hope I expressed his idea clearly. $\endgroup$
    – mick
    Oct 4, 2012 at 20:07
  • $\begingroup$ ...and correct. I hope he is not angry that I post his idea. The kth power is of course an iteration. Notice that analytic iterations of $2sinh$ are unique because $2sinh$ only has one fixpoint. Sorry for the comment spam. And again sorry if I do not understand. Welcome to MSE sheldonison :) $\endgroup$
    – mick
    Oct 4, 2012 at 20:16
  • $\begingroup$ btw I have another open problem about tetration limits. It has to do with base eta towers. You might be intrested. See my profile. $\endgroup$
    – mick
    Oct 4, 2012 at 20:45
  • $\begingroup$ Wow. You found Tommy's sexp(z) function, which I was going to post some rather complicated results showing how the derivatives converge (it is c-infinity), and that it is also nowhere analytic! Yet it is so very very close to being analytic, very much like the base change tetration function, generated by iterating logarithm of one base, against tetration of another base. I can generate results of arbitrary precision using my program, and results match Dimitrii Kouznetsov's results, see en.citizendium.org/wiki/tetrationand uniqueness has been proven by Henryk Trapmann. $\endgroup$
    – Sheldon L
    Oct 4, 2012 at 20:53

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) )


  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.


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) ?

  • $\begingroup$ I do not know if we are talking about the same. I conjecture the existance of a number $Q > 0$ depending on $a$ and $k$ hence $Q(a,k) > 0$. It is not evident that this number exists INDEPENDANT OF X. I cannot read what you wrote. Sorry. Perhaps use $sexp$ and $slog$ then i might understand. ( btw seems badly formatted maybe a website error ). Looking forward to discussing this further. Sorry for the late reply. $\endgroup$
    – mick
    Sep 29, 2012 at 18:05
  • $\begingroup$ @mick: I can come back to this possibly on monday. But perhaps I also can begin to understand your question better if some more discussion happen to occur around the other comments and answers in that time... so don't have a problem with this attempted re-formulation of mine $\endgroup$ Sep 29, 2012 at 22:09
  • $\begingroup$ @ Gottfried : Monday has almost passed. $\endgroup$
    – mick
    Oct 2, 2012 at 21:37
  • $\begingroup$ Monday has passed. For example for k = 1 we get Q = lim x-> oo (ln(e^x/a^x))/x = (ln(e^x)-ln(a^x))/x = x/x (1-ln(a)) = 1-ln(a). $\endgroup$
    – mick
    Oct 3, 2012 at 12:49
  • $\begingroup$ Hi mick, I think I've to give up at this time; I'm likely too far off my tetration-study and it needed more involvement than I can give at the moment. Perhaps better I delete my "answer"/request for clarification soon. $\endgroup$ Oct 3, 2012 at 13:33

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.

  • $\begingroup$ For fractional $k$, I believe this proof also shows the result whenever $k > 2$. But for $1 < k < 2$ or $0 < k < 1$ I think the result probably fails. $\endgroup$ Oct 15, 2012 at 2:20
  • $\begingroup$ Forgive me for not commenting on this yet. Im thinking about other things related. Thanks for the post. Im not sure about it either but I need to analyse more. $\endgroup$
    – mick
    Oct 19, 2012 at 21:24

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