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How would one find: $$\frac{\mathrm d}{\mathrm dx}{}^xx?$$ where ${}^ba$ is defined by $${}^ba\stackrel{\mathrm{def}}{=}\underbrace{ a^{a^{\cdot^{\cdot^{\cdot^a}}}}}_{\text{$b$ times}}$$

Work so far

The interval that I am working in is $(0, \infty)$. It doesn't make much sense to consider negative numbers. Although there exists no extension to the reals for tetration I am going to assume that it exists. My theory is that it shouldn't change the algebra involved; (correct me if I am wrong).

Some visual analysis on the curve and you can see that it diverges to $+\infty$ extremely rapidly. This means that the derivative is going to have similar properties as well.

Let $f(x, y):={}^yx$ so we can rewrite our tetration as $f(x, x)$. Now using the definition of the total derivative: $D\;g(x, y)=\partial_xg(x, y)+\partial_yg(x, y)$. This should allow us to differentiate $f$.

$$D\;f(x,y)=\frac{\partial}{\partial x}{}^yx+\frac{\partial}{\partial y}{}^yx$$

Let's focus on the first partial derivative $\partial_x{}^yx$. This is just the case of differentiating a finite power tower as $y$ is treated constant.

Firstly looking at some examples do derive a general formula for $D\;\;{}^nx$: $$ \begin{array}{c|c} n & D\;\;{}^nx\\ \hline 0 & 0\\ 1 & 1\\ 2 & {}^2x(\log x + 1)\\ 3 & {}^3x\times {}^2x\times x^{-1}(x\log x(\log x + 1)+1) \end{array} $$

It is easy to see that there is some pattern emerging however because of it's recursive nature I could not form a formula to describe it.

Edit

$$\dfrac{d}{dx}\left(e^{{}^nx \log(x)}\right)={}^{n+1}x\dfrac{d}{dx}\left({}^nx \log(x)\right)={}^{n+1}x\left(({}^nx)' \log(x)+\frac{{}^nx}{x}\right)$$ The recursive formula for the partial was pointed out in comments however an explicit formula would be more useful for this purpose.

The second partial derivative is interesting and relies on properties of tetration.

I was hoping for it to be similar to exponention such that $$D_y \;x^y=D_y\;e^{y\log x}=e^{y \log x}\log x\;D_y\;y=x^y\log x$$ However I am not sure of an '$e$ for tetration' but I hope it would be something like this: $$D_y \;{}^yx=D_y\;{}^{y\;\text{slog} x}t={}^{y\;\text{slog} x}t\;\text{slog} x\;D_y\;y={}^yx\;\text{slog}\; x$$ Where $\text{slog}$ denotes the super logarithm (slogorithm), an inverse of tetration.

Edit

This may as well be a possible identity which can easily be applied to the above: $$\text{slog}\;\left({}^yx\right)=\text{slog}^y\;(x)$$

I am unsure about using slogorithms and tetration in this way and I feel I might just be abusing notation.

Work on Tetration

I will update this section with more rigorous definitions and properties of tetration. I cannot prove all of them now.

For $x\in \Bbb R$ and $n \in \Bbb N$, $${}^nx:=\underbrace{ x^{x^{\cdot^{\cdot^{\cdot^x}}}}}_{\text{$n$ times}}\tag{1}$$

For $x\in\Bbb R$ and $a,b\in\Bbb N$?, $${{}^b({}^ax)={}^{a^b}x\tag{$\not2$}}$$ Through simply algebra you can find that the above is not the case.

Update:

This is just differentiating the pentation function.

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    $\begingroup$ For the pattern in $\dfrac{d}{dx}\left({}^nx\right)$, use induction: the derivative of ${}^{n+1}x$ is $$\dfrac{d}{dx}\left(e^{{}^nx \log(x)}\right)={}^{n+1}x\dfrac{d}{dx}\left({}^nx \log(x)\right)={}^{n+1}x\left(({}^nx)' \log(x)+\frac{{}^nx}{x}\right).$$ $\endgroup$
    – Ian Mateus
    Dec 29, 2013 at 20:15
  • $\begingroup$ $a^{{}^b a}={}^{b+1}a$ and monotonicity might well be enough to have a unique interpolation. $\endgroup$
    – Phira
    Dec 29, 2013 at 20:18
  • $\begingroup$ Although not in the interval you require, it seems that: $$\frac{\text{d}}{\text{d}x}{}^nx = (x^{-1})({}^{n-2}x)({}^{n-1}x)(1+\log({}^{n-1}x)(1+\log({}^{n-2}x)(\cdots(1+\log(x))))),$$ for integer $n>0$. $\endgroup$
    – pathfinder
    Dec 29, 2013 at 23:10
  • $\begingroup$ This is similar: math.stackexchange.com/questions/616014/… $\endgroup$
    – Simply Beautiful Art
    Feb 2, 2016 at 23:55
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    $\begingroup$ I'm fairly certain that $^b(^ax)\ne^{a^b}x$... try $a=1$ and $b=x=2$. $\endgroup$
    – Simply Beautiful Art
    Apr 30, 2017 at 13:46

2 Answers 2

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It seems to me that before much more progress can be made in the calculus of ${}^xy$, more fundamental questions have to be answereed, such as, how to define ${}^xy$ for rational $x$? It's clear how the OP's definition works if $x$ is a non-negative integer; but how do we define ${}^xy$ if, say, $x = 7/2$? What then is "one-half" of an occurrance of $x$ in the exponential "tower" which is supposed to be ${}^xy$?

I am reminded here of the way $x^y$ is extended from integers through the reals, by starting with a careful, consistent and believable definition of $(p / q)^{(r / s)}$ for integral $p, q, r, s$; once we have that, a simple, consistent and believable continuity argument allows us to accept a definition of $x^y$ for real $x, y > 0$. We know what $(p / q)^r = (p^r / q^r)$ means; we know what it means for a positive real $z$ to satisfy $z^s = (p / q)^r$, so we can get a handle on $(p / q)^{(r / s)}$ from which, by continuity, we can generalize to $x^y$. I think an analogous method is needed here, but I don't know what it is. But I think my question of the preceding paragraph might be worth considering early on in this game.

Of course, perhaps there is a (reasonably) simple, consistent and believable argument to contruct ${}^xy$ using $\exp()$, $\log()$, etc., or some sort of differential or similar equation ${}^xy$ must satisfy, or perhaps one could learn something from the $\Gamma$ function and factorials here which would bypass, at least temporarily, the need to address how ${}^{(p / q)}(r / s)$ is supposed to work, but sooner or later the question will have to be faced, I'll warrant.

This is an interesting, though speculative, arena and I am glad to have participated. But until I can answer my own questions to my better satisfaction, I will refrain from further remarks, except to bid those who are ready to climb such unknown heights, "Excelsior!

Hope this helps, at least with the spirit of the adventure if not with the direction. Happy New Year,

and as always,

Fiat Lux!!!

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    $\begingroup$ This will not work, since exponentiation is not associative. The extension of exponentiation with $a^(1/b) = \sqrt[b]{a}$, $b \in \mathbb{N}$ and $b \ne 0$ only works because we define $a^{1/b} a^{1/b} ... a^{1/b}$ ($b$ times) to equal $a^{1/b + 1/b + ... + 1/b}$, which is a natural definition due to the law $a^b a^c = a^{b + c}$. But no analogous law holds for tetration, because exponentiation is not associative, while multiplication is. So trying to use a similar definition there would probably not work, or give an ill-behaved function. $\endgroup$
    – The_Sympathizer
    Dec 29, 2013 at 23:33
  • $\begingroup$ @ mike4ty4: what exactly will not work? $\endgroup$
    – Robert Lewis
    Dec 29, 2013 at 23:35
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    $\begingroup$ Eh, maybe that wasn't the right way to say it. But it seems like you're coming at this from the angle of "find a law analogous to that of exponentiation to extend tetration". But there's a very good reason the law $a^b a^c = a^{b + c}$ exists for even just integer $a$, $b$, and $c$, and that's because the multiplication operation is associative. That is not the case for exponentiation. So there does not seem any reason to reasonably suppose such a law must exist. $\endgroup$
    – The_Sympathizer
    Dec 29, 2013 at 23:40
  • $\begingroup$ One should never forget that exponentiation involving complex numbers is clock-arithmetic (with modulus $2 i \pi $). The problems occuring when iterating exponentiation and involve its inverse and even when you attempt to involve fractional iterations are similar to that to find, say, the $1/3$ of $6$ o'clock using the direct division down to $2$ o'clock versus taking $4*6=24=0$, then $0*1/3=0$ and then $0/4=0$ or something similar. (This is surely an ugly example, well, I'm just lazy today, but I simply want to point out that rational arithmetic cannot really give a really useful clue here) $\endgroup$
    – Gottfried Helms
    Dec 30, 2013 at 13:03
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Your "$e$ for tetration" cannot exist. Indeed let $T(x,y)={}^yx$ be the extended tetration and let $t$ be the analogue to $e$ for tetration, in the sense that $$\partial_yT(x,y)=\partial_yT(t,y\operatorname{slog}x) = T(t,y\operatorname{slog}x)\operatorname{slog}x$$ Here $\operatorname{slog}x$ is the (unique, if $T$ is strictly increasing in its first argument) positive real number such that $x=T(x,1)=T(t,\operatorname{slog}x)$. Then $\operatorname{slog}t=1$ and thus by the above we have $$\partial_yT(t,y)=T(t,y)$$ which implies $T(t,y)=T(t,0)e^y=e^y$, which is obviously a contradiction.

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  • $\begingroup$ I think I made a mistake when talking about slogorithms. Maybe it is $$\text{slog}\;\left({}^yx\right)=\text{slog}^y\;(x)$$ $\endgroup$
    – Ali Caglayan
    Dec 29, 2013 at 20:30
  • $\begingroup$ @Alizter It doesn't matter for my answer. The point is that if if you assume that there is some $t$ satisfying what I wrote for my definition of $\operatorname{slog}$, then you get a contradiction. $\endgroup$
    – Daniel Robert-Nicoud
    Dec 29, 2013 at 20:32
  • $\begingroup$ Making the intuitive $\text{multiplication}\mapsto\text{exponentiation}$, I would expect the result to be $T^{\mathrm{slog} x}$. Are you aware of a neat contradiction in this case? $\endgroup$
    – Ian Mateus
    Dec 29, 2013 at 20:40
  • $\begingroup$ @IanMateus Sorry, but I don't understand what you're trying to say. Could you elaborate? $\endgroup$
    – Daniel Robert-Nicoud
    Dec 29, 2013 at 20:50
  • $\begingroup$ I'll reply soon, tetration gives me a huge mind bending. $\endgroup$
    – Ian Mateus
    Dec 29, 2013 at 21:12

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