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I happened to ponder about the differentiation of the following function: $$f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$$ Now, while I do know how to manipulate power towers to a certain extent, and know the general formula to differentiate $g(x)$ wrt $x$, where $$g(x)=f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{.{^{.^{.}}}}}}}}}}$$ I'm still unable to figure out as to how I can adequately manipulate the function to differentiate it within its domain of convergence.


General formula: $$g'(x)=\frac{g^2(x)f'(x)}{f(x)\left[1-g(x)\ln(f(x))\right]}$$

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    Unless IM missing something obvious it will converge only for $[-1,1]$ – pancini Dec 29 '15 at 05:59
  • See if this question is of any help..... http://math.stackexchange.com/questions/1250381/derivative-of-the-power-tower – SchrodingersCat Dec 29 '15 at 06:10
  • @Brody: You can still interpret it as being right-associative, and as the limit of the finite partial towers (it just won't be the limit of an iterative process). I have deleted my answer because it was interpreting it as left-associative, which is almost certainly not the intended interpretation. – Eric Wofsey Dec 29 '15 at 06:34
  • @Brody Ya, Eric is right that this is right associative. Look up tetration for a little more information. The notation $a^{b^c}$ means $a^{(b^c)}$, not $(a^b)^c$ – Brevan Ellefsen Dec 29 '15 at 06:40
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    In any case, there appears to be no reason to believe that this sequence ever converges (except for $x=1$). If I had to guess at this point I would say the domain of $f$ is a finite set, so it doesn't make sense to differentiate it. – Eric Wofsey Dec 29 '15 at 06:45
  • @EricWofsey Depending on your definition, $f(0) = 1$... this is commonly how infinite tetration of $0$ is defined. Now, I agree with you that I can't think of more than a finite amount of converging values using naive methods. – Brevan Ellefsen Dec 29 '15 at 06:47
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    Can you confirm that by $x^{2x^{3x}}$ you mean $x^{2(x^{3x})}$, not $x^{(2x)^{3x}}$? In other words, do each of your exponents only apply to the previous "$x$" or the previous "$nx$"? – 2'5 9'2 Dec 29 '15 at 07:05
  • I intended $x^{2(x^{3x})}$... i.e. only to the previous "x". – Panglossian Oporopolist Dec 29 '15 at 07:46
  • So I suppose it's meaningless? (Is there a way to prove that the domain is only a finite set?) – Panglossian Oporopolist Dec 29 '15 at 07:51
  • Given my question and your answer, it might be nice to write this as $\left(x^2\right)^{\left(x^3\right)^{\left(x^4\right)^{\left(x^5\right)^{⋰}}}}$, which suggests that you also look at $\left(x^1\right)^{\left(x^2\right)^{\left(x^3\right)^{\left(x^4\right)^{\left(x^5\right)^{⋰}}}}}$. – 2'5 9'2 Dec 29 '15 at 08:23
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    @alex.jordan: It is unclear that your tower has the same limit, since its truncations are different. – Eric Wofsey Dec 29 '15 at 08:25
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    Numerically, the two potential definition of the function $$\lim\limits_{n\to\infty} x^{2(x^{3(x^{⋰^{(nx)})}})} \quad\text{ and }\quad \lim_{n\to\infty} x^{(x^2)^{(x^3)^{⋰^{(x^n)}}}}$$ seems equivalent. For $x \in (0,1)$. The limits of both sequences don't exist. However, the limits of their even sub-sequences do and seems to coincides to some smooth function. Same thing happen to their odd sub-sequences. The second form (corresponds to the one suggested by @alex.jordan) behaves better as the sub-sequences converge faster to their target. – achille hui Dec 29 '15 at 16:53
  • @achillehui interesting! Perhaps the second representation would be easier to analyze.... Now we just need to find a way to rigorously define a limit for values of these series – Brevan Ellefsen Dec 29 '15 at 17:56
  • That clears up a lot! I'll delete my previous comments as the thread seems to be getting cluttered. – Corellian Dec 29 '15 at 22:15
  • You may note that $$g(x)=f(x)^{g(x)}$$And so:$$g(x)=h^{-1}(x)$$where$$h(x)=\sqrt[x]{x}$$That is easier to differentiate using inverse rule and choosing the correct branches. – Simply Beautiful Art Feb 17 '16 at 21:22
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    glad to see that there is another bounty on this.... hopefully some more progress will be made with more exposure :/ this problem has occupied some of my free time, but I've found my mathematical knowledge too limited thus far... nothing spectacular has struck me. Really this question has boiled down to showing whether or not the function converges for more than $x=1$... I personally believe it does, but I don't know how to prove that and I would love to be proven wrong. Perhaps a more formal definition of the limit would work here than intuition? – Brevan Ellefsen Feb 19 '16 at 22:30
  • @SimpleArt What you say is true if $g(x) = f(x)^{g(x)}$, but that doesn't appear true to me here... obviously if the function converges from one height than the the function will converge from all heights, i.e. if $x^{4x^{5x\cdots}}$ converges than $g(x) = x^{2x^{3x\cdots}}$ does also, and the function satisfies a basic recursive definition using logarithms – Brevan Ellefsen Feb 19 '16 at 22:35

7 Answers7

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It is not completed answer but I thought it can be an approach for such kind of questions. Thus I decided to post it. Let's define

$$f_n(x)=nx^{(n+1)x^{(n+2)x^{(n+3)x^{(n+4)x^{(n+5)x^{(n+6)x^{.{^{.^{.}}}}}}}}}}$$

Your function can be found by $$f_1(x)=f(x)$$ and you are looking for $$\frac {\partial f_n(x)}{\partial x} \bigg|_{n=1}=f'(x)$$

We can easily see a relation for $f_n(x)$ $$f_n(x)=nx^{f_{n+1}(x)}$$ $$\ln(f_n(x))=\ln(n) +f_{n+1}(x)\ln(x)$$

$$\frac {\partial \ln(f_n(x))}{\partial x}=\frac {\partial (f_{n+1}(x)\ln(x))}{\partial x}$$

$$\frac {\partial \ln(f_n(x))}{\partial x}=\frac {\partial (f_{n+1}(x)\ln(x))}{\partial x}$$

$$\frac{\partial f_n(x)}{\partial x} = f_n(x) \frac{\partial f_{n+1}(x)}{\partial x}\ln(x)+\frac {f_{n+1}(x)f_n(x)}{x}$$

----Let's put $n=1,2,3,....$

$$\frac{\partial f_1(x)}{\partial x}= \frac{\partial f_{2}(x)}{\partial x}f_1(x)\ln(x)+\frac {f_{2}(x)f_1(x)}{x}$$

$$\frac{\partial f_2(x)}{\partial x} = \frac{\partial f_{3}(x)}{\partial x}f_2(x)\ln(x)+\frac {f_{3}(x)f_2(x)}{x}$$

$$\frac{\partial f_3(x)}{\partial x} = \frac{\partial f_{4}(x)}{\partial x}f_3(x)\ln(x)+\frac {f_{4}(x)f_3(x)}{x}$$

$$.$$ $$.$$ $$.$$

$$\frac{\partial f_1(x)}{\partial x} = U(x)+\frac {f_{1}(x)f_2(x)}{x}+\frac {f_{1}(x)f_2(x)f_3(x) \ln(x) }{x}+\frac {f_{1}(x)f_2(x)f_3(x) f_4(x) \ln^2(x) }{x}+.... \tag{1}$$

Where $$U(x)=\lim\limits_{ n\to \infty } \frac{\partial f_n(x)}{\partial x} \ln^n(x)\prod_{k=1}^{n-1} f_n(x) $$

Finally we can express the derivative as

$$f'(x)=\frac{\partial f_1(x)}{\partial x} = U(x)+ \sum_{k=1}^{\infty} \frac {\ln^{k-1}(x)f_1(x)\prod_{n=2}^{k+1} f_n(x)}{x}$$

Note: Yet I have not found what $U(x)$ is. I estimate that $U(x)$ will vanish but I have not proved it yet. Maybe someone can help with some numerical values that if $U(x)=0$ or not. Thanks a lot for contributions and advice

An observation : I noticed a similar pattern with general formula $g'(x)$ in the question and $f'(x)$ that I wrote in (1). $$g(x)=h(x)^{h(x)^{h(x)^{h(x)^{h(x)^{h(x)^{h(x)^{.{^{.^{.}}}}}}}}}}$$ $$g'(x)=\frac{g^2(x)h'(x)}{h(x)\left[1-g(x)\ln(h(x))\right]}=$$

It can be rewritten as

$$g'(x)=\frac{g^2(x)h'(x)}{h(x)}(1+g(x)\ln(h(x))+g^2(x)\ln^2(h(x))+g^3(x)\ln^3(h(x))+....)$$

If $h(x)=x$ then we can obtain

$$g'(x)=\frac{g^2(x)}{x\left[1-g(x)\ln(x)\right]}=$$ $$g'(x)=\frac{g^2(x)}{x}(1+g(x)\ln(x)+g^2(x)\ln^2(x)+g^3(x)\ln^3(x)+....)$$ $$g'(x)=\frac{g^2(x)}{x}+\frac{g^3(x)}{x}\ln(x)+\frac{g^4(x)}{x}\ln^2(x)+....$$

It has similarities with my formula for $f'(x)$ that I wrote above. This result supports my idea that $U(x)$ may vanish.

V.G
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Mathlover
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  • Nice observations! A couple things I would point out are that your comparisons to $g(x)$ should be taken lightly, as $g(x)$ is tetration, while $f(x)$ is not. Also, you are implicitly assuming that there are more than a finite amount of points that satisfy the function... See the comments above and my observations for more detail. Perhaps if we can show that an infinite amount of values exist within some domain then we can get leaps ahead with your answer, and maybe even have more ideas for $U(x)$... – Brevan Ellefsen Dec 29 '15 at 17:42
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Partial Answer / Observation
So it is clear that this function can be defined for $f(0)$ and $f(1)$, and is definitely defined only within some portion of this domain... accordingly, I chose to evaluate what happens with $0.5$ as you increase the tower height. Numerically, it appears that two limits are approached, one for even heights and one for odd heights (this is the nature of power towers evaluated between $0$ and $1$ for any I have ever calculated... there is probably a proof of this somewhere, at least for any sequence of heights that reach some limit). Regardless, it appears that the function is simply periodic between these two limits, and I would say that this should hold as you continue to increase the tower height. Now, this isn't a proof in any sense, (even for the value $0.5$), as numerical analysis alone won't cut this, but I think it provides some interesting insight, and is the only thing that got me any sort of result after hours of looking into this. (Note that I am not referencing iterated functions with the superscript, but the height of the tower.) As is pointed out in the comments, this function is probably only defined at a finite amount of points... One could probably find a way to show that a point such as $0.5$ diverges by analyzing the property that causes this dual limit (I am fairly certain it is due to the domain $[0,1]$), but I am not sure that such observations would be sufficient to prove this across the entire domain. $$f^1(0.5) \approx x = 0.5$$ $$f^2(0.5)\approx x^{2x} = 0.5$$ $$f^3(0.5)\approx x^{2x^{3x}} = 0.612...$$ $$f^4(0.5)\approx x^{2x^{3x^{4x}}} = 0.439...$$ $$f^5(0.5)\approx 0.679...$$ $$f^6(0.5)\approx 0.374...$$ $$f^9(0.5)\approx 0.804...$$ $$f^{10}(0.5)\approx 0.305...$$ $$f^{14}(0.5)\approx 0.3040559...$$ $$f^{18}(0.5)\approx 0.3040557...$$ $$f^{15}(0.5)\approx 0.81045144...$$ $$f^{19}(0.5)\approx 0.81045145968867...$$ $$f^{23}(0.5)\approx 0.81045145968869...$$

Brevan Ellefsen
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  • I see. Indeed interesting... Have you encountered oscillating limits for other functions before? – Panglossian Oporopolist Dec 29 '15 at 08:07
  • @Kugelblitz Yes, I have seen this happen when working with power towers before. Forgive me if I can't think of any specific cases off the top of my head (I've messed with a LOT of power towers), but I remember this being used to indicate divergence of towers before. If I can remember where I have seen this before I will let you know! – Brevan Ellefsen Dec 29 '15 at 08:11
  • Nice... I haven't come across oscillating limits in power towers before... there doesn't seem to be any literature devoted to it either. – Panglossian Oporopolist Dec 29 '15 at 08:14
  • @Kugelblitz No, unfortunately there isn't much in general to go off of for power towers. The Tetration Forum has some stuff on it, but is mostly focused on tetration (thus the name). Nevertheless, I find that there is a treasure trove of info on the site, and you could probably get a question like this answered far faster (although most of these mathematicians are far above you and I, and are quite specialized, so the responses you get might be a bit harsher than a downvote!) – Brevan Ellefsen Dec 29 '15 at 08:19
  • How do you define $f(0)$?. We don't define $0^0$ – Ross Millikan Dec 29 '15 at 15:26
  • @RossMillikan generally tetration of $0$ is *defined* to be $1$, and most attempts to extend tetration assume this. (I think Wikipedia briefly mentions this on the tetration page). Nevertheless, you are right... If you don't define it this way you get an indeterminate form. – Brevan Ellefsen Dec 29 '15 at 17:52
  • @Kugelblitz It appears to me that this converges for at least one value between $0.7$ and $0.8$, likely much closer to $0.7$. For $x \geq 0.8$ the partial sums of the even towers grow and the sums of the odd towers shrink. For $x \leq 0.7$ the even towers shrink and the odd towers grow. Even better, it appears that for all $x \geq 0.8$ the even sums start below their limit and the odd sums above their limit, so the limits seem to be converging. As long as the limits can be shown to never cross each other in finite time but to monotonically increase for all $x$ then we have a convergent domain! – Brevan Ellefsen Dec 30 '15 at 07:55
  • @Kugelblitz In other words, I might have numerically found a domain where this is actually valid. Now, everything analytical I have tried has failed, but perhaps I just need to focus on the domain where convergence is possible. I hope for a proof soon! – Brevan Ellefsen Dec 30 '15 at 07:56
  • @Kugelblitz One problem is that as I move in from $0.7$ upward the two limits seem to get closer and closer until they then seem to diverge, although this takes much longer the more I approach $0.8$ and higher... Thus, I can (with great difficulty) "constrain" the domain I have to check, but I can't numerically confirm a single point works other than $1$... For all I know, $0.9$'s limits diverge, but only after using a tower of height $20000$... – Brevan Ellefsen Dec 30 '15 at 08:10
  • @Kugelblitz Crap... the problem is even worse than I imagined. Apparently the even/odd sums will continue decreasing for a while and then start increasing again, so throw monotonic increase/decrease out the door.... Best we could hope for is only one of these switches at a given point and then for monotonic increase/decrease after said point (or at least for finite points... if this is periodic we're screwed) – Brevan Ellefsen Dec 30 '15 at 08:25
  • @Kugelblitz Since I lack the resources to continue checking past a height of 50ish in depth using excel we will either need a better method or to show some point analytically... if anyone can show that 0.9 converges/diverges it would mean the world to me. – Brevan Ellefsen Dec 30 '15 at 08:29
  • Sorry I couldn't respond before; I'm currently very busy with some banal exam prep. I see... great effort so far! I suggest you ask a question regarding the doubts you have before I become free (I'll be somewhat free in 3 hours or so). Let me have a go at analytical methods then and see if we can write a paper (seeing as it would be a terrible waste not to make your valuable effort known) on this quizzacious function. – Panglossian Oporopolist Dec 30 '15 at 09:21
  • @Kugelblitz it's fine, I completely understand... I'm very busy today as well so I won't be able to work on the problem much for the next 12 hours or so. I agree that I should post by doubts... Assuming you don't have an analytical breakthrough soon I will do so as soon as I am available. – Brevan Ellefsen Dec 30 '15 at 13:42
  • @RossMillikan, I agree that $f(0)$ is not defined, but for a somewhat different reason. If you accept $0^0=1$, then *finite* tetration toggles back and forth between $1$ and $0$: i.e., $0^{0^0}=0^1=0$, $0^{0^{0^0}}=0^0=1$, etc. – Barry Cipra Dec 30 '15 at 13:47
  • I believe the behavior you have seen, alternating between two values as the tower height increases, will be very general and has a simple explanation. When we get high, the top number will be $x$ to some high power, which is approximately $0$ for $x \in (0,1)$. That will make the next layer down approximately $1$. As you add layers up high, the $0$'s and $1$'s will alternate. Your values of $0.304$ and $0.810$ depend on the parity of the number of layers because it changes whether the tenth layer up (say) is $0$ or $1$. – Ross Millikan Jan 01 '16 at 22:24
  • @RossMillikan yes, I too have noticed this as well, however there is a problem. When messing with finite towers I find that the values generally alternate between being greater than the value of x used and below it. However, there are times where it is either above or below for multiple iterations. If it was uniform then I could easily show divergence, but I'm not sure how to show that the values don't flip enough to achieve convergence... Reminds me of the prime counting function and the log integral related to it, and hoe they flip every once in a while – Brevan Ellefsen Jan 01 '16 at 22:57
  • @RossMillikan perhaps it would suffice to show that regardless of the effect I point out, that each level approaches 1 or 0 as you mention in your comment, and so I need only show that finitely many values could have these two limits converge... How to prove this has stumped me though. Any ideas? I haven't looked at the problem for a couple days due to other tasks. – Brevan Ellefsen Jan 01 '16 at 23:00
  • I find the plot of [$(x^{10})^{x^{11}}$](http://www.wolframalpha.com/input/?i=plot+%28x%5E10%29%5E%28x%5E11%29+from+0+to+1) interesting. It is solidly $1$ up to $x=0.6$, then starts to fall because $x^{11}$ isn't close enough to $0$. As you get low in the stack the small exponents take you away from the powers being $0$ or $1$. Then if you look at [$(x^{10})^{{(x^{11})}^{x^{12}}}$](http://www.wolframalpha.com/input/?i=plot+%28x%5E10%29%5E%28x%5E11%29%5E%28x%5E12%29+from+0+to+1) it is solidly zero up to $x=0.6$ I know the alternation is true, but don't know how to show it. – Ross Millikan Jan 02 '16 at 02:55
  • @RossMillikan same. I've written some code in Excel to check points as I mention in my comments above, but, just as with the graphs, I can only *predict* an interval of covergance (if any), not *prove* it. I considered showing the limit of f(h,x) - f(h-1,x) (where h is the height) which has roots 1/h and 1, and alternates between being greater than and lesser than 0 for even/odd tower height, but this isn't leading me anywhere so far. – Brevan Ellefsen Jan 02 '16 at 03:04
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Note: The below applies to a different function than OP intended. I was studying $$f(x)=f(x)=x^{(2x)^{(3x)^{(4x)^{(5x)^{(6x)^{(7x)^{.{^{.^{.}}}}}}}}}}$$ but don't want to delete this.

Before we worry about derivatives, we need to find where the function is defined and continuous. $f(x)$ has to be seen as the limit of the sequence $x, x^{2x},x^{(2x)^{(3x)}}\dots $ If $x \gt 1$ this clearly diverges to infinity, if $x=0$ it is undefined. It might turn out to be defined at some negative integers, but will not be defined at other negative $x$, so we can concentrate on $0 \lt x \le 1$. If $x=1$, the sequence is just $1$ to higher and higher powers, so $f(1)=1$. If $\frac 12 \lt x \lt 1$ the sequence is a number less than $1$ to higher and higher powers, so the limit is $0$. If $x=\frac 12,$ the sequence is $\frac 12$ to ($1$ to higher and higher powers), so $f(\frac 12)=\frac 12$. If $\frac 13 \lt x \lt \frac 12,$ we have $2x \lt 1$, so $x^{(2x)^{\text { lots }}}$ goes to $0$. In general, if $x = \frac 1k$ we just evaluate up to $(k-1)x$. If $\frac 1{k+1} \lt x \lt \frac 1k$ we evaluate up to $(k-2)x$ because the powers of $kx$ will go to zero, so the powers of $(k-1)x$ go to $1$. Summing up $$f(x)=\begin {cases} 1&x=1\\0&\frac 12 \lt x \lt 1\\\frac 12&x=\frac 12\\1&\frac 13 \lt x \lt \frac 12\\ \frac 13^{\frac 23} &x=\frac 13 \\x&\frac 14 \lt x \lt \frac 13 \\\text {tower up to }(k-1)x&x=\frac 1k\\\text{tower up to }(k-2)x&\frac 1{(k+1)x} \lt x \lt \frac 1k\end {cases}$$

Now it is clear that $f(x)$ is differentiable at all points in $(0,1)$ that are not of the form $\frac 1k$ but the derivative will get very messy as $x$ gets small.

Ross Millikan
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Comment about $g(x)=f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{.{^{.^{.}}}}}}}}}}$

The comment is too long to be posted in the comments section. That is why it is posted in the answer section. But this is not an answer.

If the power tower is convergent $g(x)=f(x)^{g(x)}=e^{g\ln|f|}$

The derivative of $g(x)$ can be explicitly expressed thanks to the Lambert W function :

$ge^{-g\ln|f|}=1 \quad \to \quad -g\ln|f|e^{-g\ln|f|}= -\ln|f|$

$-g\ln|f|=W\left( -\ln|f|\right) \quad$ where $W$ is the Lambert W function.

$$g(x)=-\frac{1}{\ln|f(x)|}W\left( -\ln|f(x)|\right)$$

With condition $\quad 0<|f(x)|<e^{1/e}\quad$

We know that $\frac{dW(X)}{dX}=\frac{W(X)}{X\left(W(X)+1 \right)}$

With $X=-\ln|f(x)|\quad \to \quad g'(x)=\frac{d}{dX}\left(\frac{W(X)}{X} \right) \frac{dX}{dx}$

$g'(x)= \left( -\frac{W(X)}{X^2}+\frac{W(X)}{X^2\left(W(X)+1 \right)} \right) \left(-\frac{f'}{f} \right) = \left( -\frac{W(X)^2}{X^2\left(W(X)+1 \right)} \right) \left(-\frac{f'}{f} \right)$

$$g'(x)=\left(\frac{ W\left(-\ln|f(x)|\right) }{-\ln|f(x)| } \right)^2 \frac{1}{W\left(-\ln|f(x)|\right)+1}\:\frac{f'(x)}{f(x)}\quad \text{in} \quad 0<|f(x)|<e^{1/e}$$

JJacquelin
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1

Looking at this graph, it doesn't appear to converge for values near $x=0$ and it may converge for values close and larger than $x=1$. Around $x=0.2$, it oscillates a lot and doesn't appear to converge to a single value around $x<0.8$.

It also appears to have 'good' convergence for $0.8<x<1$.

Simply Beautiful Art
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  • I came to the same conclusion months ago... See my comments. I even calculated out to Heights of 40 or so, but couldn't find a solid proof... The oscillation just seems to get smaller and smaller, and I don't know how to prove whether or not it decreases to 0 for an infinite height. – Brevan Ellefsen Feb 24 '16 at 02:22
  • @BrevanEllefsen Yes, I see. Good to see people working together on a question that feels forgotten. – Simply Beautiful Art Feb 24 '16 at 20:28
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It can be proved by induction that the iteration of $\lambda\cdot e^{x}$ is equivalent to the iteration of $e^{\lambda\cdot x}$ modulo the base. Therefore $g(x)$ is equivalent to:

$$g(x)=(x^2)^{(x^3)^{(x^4)^{\cdots}}}$$

For the above to make sense, using Barrow's Theorem (here), we must have $\forall n>1\in\mathbb{N}$:

$$e^{-e}\le x^n\le e^{1/e}\Leftrightarrow$$

$$e^{-\frac{e}{n}}\le x \le e^{\frac{1}{ne}}$$

Precisely because the above must hold for all $n>1$, the applicable constraint reduces to:

$$\lim_{n\to\infty}e^{-\frac{e}{n}}\le x \le \lim_{n\to\infty}e^{\frac{1}{ne}}\Leftrightarrow$$

$$x=1$$

If $x=1$ however, then your expression is identically equal to 1, hence the derivative of your expression vanishes everywhere on its domain.

  • What does this say about $x<0$ though?? For that matter, what are you saying about $x>1$?? The tower clearly diverges for all $x>1$, so I'm not sure how to interpret your result. Looks really promising though!! – Brevan Ellefsen Feb 24 '16 at 02:20
  • It says nothing about $x\neq 1$ because the expression doesn't make sense for $x\neq 1$. The domain of the expression is just $x=1$. That's why the derivative is singular there @BrevanEllefsen –  Feb 24 '16 at 06:42
  • ah, I see. So your result shows that the function diverges for $x < 1$ indirectly as well?? Interesting... I'm now just curious which points diverge to alternating limits and which diverge to infinity – Brevan Ellefsen Feb 24 '16 at 20:57
  • No, this result doesn't show that the function diverges for $x<1$, indirectly (or otherwise). It just shows that the expression for $g(x)$ remains afortiori undefined for all $x\neq 1$ (and as such, questions about divergence or convergence for values $x\neq 1$ are really inconsequential (to the fact that the derivative is singular at $x=1$)). @Brevan Ellefsen –  Feb 24 '16 at 22:33
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First of all, we need to see how to define this function and if it exists. The truth is that there are two functions behind it, not one.

Define the family of functions:

$$f_n(x)=x^{(n+1)f_{n+1}(x)}$$

But, there is an instant trouble. First of all we have

$$f_1(1)=1$$ and $$f_1(1^{+})=\infty$$

The trouble is how to define $f(0)$. Let us define a partial function

$$ f_{m,n}(x)=\begin{cases} & x^{f_{m,n+1}(x)},\text{ if } n \leq m \\ & 1,\text{ if } n>m \end{cases} $$

and then try to find the limiting function. And here is the problem.

If $m$ is even then the function starts at $1$ and has a jump at $\frac{1}{m}$ towards $0$. If $m$ is odd then the the function starts at $0$ and has a jump at $\frac{1}{m}$ towards $1$.

Going towards infinity, we will obtain two versions of the functions: even that starts at $0$ and is monotonically increasing and odd that has a minimum between $0$ and $1$ and starts at $1$.

(We can agree that neither of two is continuous at $0$ or we need to extend the function at $0$ ourselves, because the limiting process is never going to reach $0$ or $1$ value at $f_1(0)$ by itself.)

odd and even case

odd and even case

Now regarding first derivative, there is no closed form of it. We can simply state it the same way the function is defined and then use an infinite chain to define it in full.

$$f_n'(x)=(n+1)x^{(n+1)f_{n+1}(x)-1} \left (x\ln(x)f'_{n+1}(x)+f_{n+1}(x) \right )$$

This makes it obvious that $f_1'(1)=2 \cdot 3 \cdot 4 \cdot 5 ... = \infty$ while $f_1'(0) = 0$

Now one thing is certain and that is that the difference between $f_{2n}$ and $f_{2n+2}$ is all smaller and smaller as $n$ tends infinity. The same for $f_{2n+1}$ and $f_{2n+3}$.

so we can try to use that and define that at $n \gg 1$

$$f_n(x)=y \sim x^{(n+1)x^{(n+2)y}}$$

Now, here comes the story. First, since $x$ is between $0$ and $1$ $x^n$ for very large $n$ tends to $0$. So we arrive at the expression

$$y = 0^{0^y}$$

which can lead only to $y=0$ or $y=1$

So at infinity we have $f_n(x)$ reaching flat $1$. (We could take $0$ as well but that is the same as saying that $f_{n-1}(x)$ reaches $1$. So $1$ is more universal.)

For this reason, our analysis of starting with the above definition for $f_{m,n}(x)$ is just right.

Other than this it is not likely that it is possible to write this function in any known and simple closed form even with all other special functions available. It is simply a beast on its own starting from its dual nature.

Another curious question is how to represent these exponential functions. It is difficult to find a fine cut and the difference between

$$x^{2^{x^{3}}}$$

and

$$x^{2x^{3}}$$

and

$$x^{(2x)^{3}}$$

This notation should reduce ambiguity and do notice that extended exponential is no longer commutative.

$$x^{2 \cdot x^{3 \cdot x^{4 \cdot x^{5 \cdot x...}}}}=\operatorname*{E}\limits_{n=1}^{\infty}x^{n+1}$$

$$x^{(2x)^{(3x)^{(4x)^{(5x)...}}}}=\operatorname*{E}\limits_{n=1}^{\infty}nx$$

$x^{2^{x^{3}}}$ is obviously a special case with alteration $$ (x,n) = \begin{cases} & x, n \mod 2=1\\ & \frac{n}{2}+1, n \mod 2=0 \end{cases} $$

$$\operatorname*{E}\limits_{n=1}^{\infty}(x,n)$$

A general problem with this extended exponentiation is that you technically need to read it right to left in order to clear the confusion.

Alex Peter
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