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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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    $\begingroup$ Unless IM missing something obvious it will converge only for $[-1,1]$ $\endgroup$ – Elliot G Dec 29 '15 at 5:59
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    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Dec 29 '15 at 6:45
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    $\begingroup$ 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 "$nx$"? $\endgroup$ – alex.jordan Dec 29 '15 at 7:05
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    $\begingroup$ @alex.jordan: It is unclear that your tower has the same limit, since its truncations are different. $\endgroup$ – Eric Wofsey Dec 29 '15 at 8:25
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    $\begingroup$ 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. $\endgroup$ – achille hui Dec 29 '15 at 16:53
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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. $$f^1(0.5) \approx x = 0.5$$ $$f^2(0.5)\approx x^{2x} = 0.5$$ $$f^3(0.5)\approx x^{2x^{3x}} = 0.612...$$ $$f^4(0.5)\approx x^{2x^{3x^{4x}}} = 0.439...$$ $$f^5(0.5)\approx 0.679...$$ $$f^6(0.5)\approx 0.374...$$ $$f^9(0.5)\approx 0.804...$$ $$f^{10}(0.5)\approx 0.305...$$ $$f^{14}(0.5)\approx 0.3040559...$$ $$f^{18}(0.5)\approx 0.3040557...$$ $$f^{15}(0.5)\approx 0.81045144...$$ $$f^{19}(0.5)\approx 0.81045145968867...$$ $$f^{23}(0.5)\approx 0.81045145968869...$$

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  • $\begingroup$ I see. Indeed interesting... Have you encountered oscillating limits for other functions before? $\endgroup$ – Kugelblitz Dec 29 '15 at 8:07
  • $\begingroup$ @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! $\endgroup$ – Brevan Ellefsen Dec 29 '15 at 8:11
  • $\begingroup$ Nice... I haven't come across oscillating limits in power towers before... there doesn't seem to be any literature devoted to it either. $\endgroup$ – Kugelblitz Dec 29 '15 at 8:14
  • $\begingroup$ @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!) $\endgroup$ – Brevan Ellefsen Dec 29 '15 at 8:19
  • $\begingroup$ How do you define $f(0)$?. We don't define $0^0$ $\endgroup$ – Ross Millikan Dec 29 '15 at 15:26
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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.

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  • $\begingroup$ The OP explained in comments (in response to alex.jordan) that the intended interpretation is $x^{2(x^{3x})}$, not $x^{(2x)^{(3x)}}$ -- i.e., each exponent applies only to the $x$, not the $kx$. $\endgroup$ – Barry Cipra Dec 29 '15 at 16:11
  • $\begingroup$ @BarryCipra: I see that now. I had fun with this version. I'll put a note to that effect. Thanks. $\endgroup$ – Ross Millikan Dec 29 '15 at 16:19
  • $\begingroup$ Good stuff @RossMillikan! $\endgroup$ – Kugelblitz Dec 29 '15 at 16:24
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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 $$\frac {\partial f_n(x)}{\partial x} |_{n=1}=f'(x)$$

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.

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  • $\begingroup$ 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 $U(x)$... $\endgroup$ – Brevan Ellefsen Dec 29 '15 at 17:42
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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 $X=-\ln|f(x)|\quad \to \quad g'(x)=\frac{d}{dX}\left(\frac{W(X)}{X} \right) \frac{dX}{dx}$

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

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

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  • $\begingroup$ Nice Observations +1 $\endgroup$ – Kevin Feb 23 '16 at 9:00
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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 $0.8<x<1$.

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  • $\begingroup$ 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. $\endgroup$ – Brevan Ellefsen Feb 24 '16 at 2:22
  • $\begingroup$ @BrevanEllefsen Yes, I see. Good to see people working together on a question that feels forgotten. $\endgroup$ – 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.

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  • $\begingroup$ 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!! $\endgroup$ – Brevan Ellefsen Feb 24 '16 at 2:20
  • $\begingroup$ 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 $\endgroup$ – Yiannis Galidakis Feb 24 '16 at 6:42
  • $\begingroup$ 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 $\endgroup$ – Brevan Ellefsen Feb 24 '16 at 20:57
  • $\begingroup$ 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 $\endgroup$ – Yiannis Galidakis Feb 24 '16 at 22:33

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