Super root is one of the inverse functions of tetration defined as-


$\implies z=\sqrt[n]{y}_s$

We can easily get an infinite series representation of $\sqrt[n]{z}_s$ using the Lagrange inversion theorem:-



$\textstyle\displaystyle{s_n=\lim_{w\rightarrow 1}\frac{d^{n-1}}{dw^{n-1}}\left[\left(\frac{w-1}{^nw-1}\right)^n\right]}$

Those coefficients do not seem quite healthy.

Whatever I have written the super root in the form of an infinite sum but I want to know that if there is a way to write it in terms of the Lambert W Function. So what do I mean by that, I can clarify with an example-

For $n=1$ it's just the identity function whose inverse is itself.

For $n=2$ we have to find the inverse of $x^x$ which can find through Lambert W Function. The inverse is $e^{W(\ln(x))}$.

For $n\geq 3$, I have no idea, obviously we can just invent another function, but that ain't fun at all.

$\text{My Question}:-$

Is it possible to write $\sqrt[n]{z}_s$ as a finite combination of the elementary operators with the Lambert W Function included?

  • $\begingroup$ To capture the growth rate of tetration, we surely have to use the Lambert W function multiple times , if it is at all possible to find a closed form formula. Maybe , I am wrong, but I think noone has found such a formula for the super-root. It probably boils down to numerical methods. $\endgroup$
    – Peter
    Sep 27, 2021 at 13:36
  • $\begingroup$ @Tyma Gaidash. Yeah sure you can write it as an answer, but you have to note that it is a partial answer. And I can't accept(✔) an answer unless it's a complete one. $\endgroup$ Sep 29, 2021 at 4:03
  • $\begingroup$ @RounakSarkar I mean I can generalize the nesting to other power towers. Would this complete it? $\endgroup$ Sep 29, 2021 at 13:24
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    $\begingroup$ @Tyma Gaidash. If you can generalize then there is no problem. $\endgroup$ Sep 29, 2021 at 13:36
  • $\begingroup$ @Tyma Gaidash. You can say at the beginning of your answer that you don't have a finite representation of the super root. And also try to simplify those nestings instead of just nesting one at the top of another as they might show some patterns. $\endgroup$ Sep 29, 2021 at 14:06

1 Answer 1


The super root is just the inverse function which solves for $x$ in the $n$ height power tower:

$$\,^nx\mathop=^\text{def} n\big\{ x^{x^{x^…}}$$

Let’s find an inverse using logarithm nesting. I am going to use

$$\bigcirc^n f(y)=n\big\{f(\dots f(y))$$ to compactify notations for some function $f$

Here is an example using $n=6$. The nesting converges to the solution $x=y=y_0$ for an input of $y=y_0$. with $$\bigcirc^{n-1}\log_{x}\left(y\right)$$

meaning $n-1$ nestings of $\log_x$:

$$\,^nx\mathop=y\implies \,^{n-1}x =\log_x(y)\implies\bigcirc^{n-1}\log_x\left(y\right)=a_1(y)=a_0(y)\implies x=\lim_{k\to\infty} a_k(y)=a_\infty(y)=\bigcirc^{n-1}\log_{\bigcirc^{n-1}\log_{\ddots}}\left(y\right)=\{\bigcirc^\infty\bigcirc^{n-1}\log_x\}(y)\implies a_{k+1}(y)= \bigcirc^{n-1}\log_{a_k(y)}\left(y\right)$$

Let’s try a few simpler cases and see if there a pattern using a new way. Note that the domain might change, but the original function still stays the same in the same domain. The final function uses $y=y_0$ for a constant $y_0$ of which we want to find the value for which the power tower will evaluate to, or a recursive nested inverse. The function will converge to the graph of a line with zero or infinite slope like in the previous example:




$$x^x=y=y_0\implies \ln(y)=x\ln(x)\implies x=\log_x(y_0)= \log_{\log_{…}(y_0)}(y_0)=\left[\bigcirc^\infty \log_x \right](y_0)$$


$$x^{x^x}=y\implies \ln(\ln(y_0))= \ln\ln\left(x^{x^x}\right)=\ln\left(x^x\right)+\ln(\ln(x))) $$

Now we can solve for one of two branches and nesting because $x=f(x,y)=f(x=f(x,y),y)=f(f(x,y),y)$. Here is a graph showing the line segment to which the answer converges using the first method:

Let’s introduce a new notation with the subscript meaning recursion with respect to the variable:

$$\bigcirc_a^b (f(c,a))\mathop =^\text{def}n\big\{f(c,f(c,…f(c,a)))$$ $$\ln\left(x^x\right)+\ln(\ln(x)))=\ln(\ln(y))\implies x=e^{e^{\ln(\ln(y_0))-x\ln(x)}}= y_0^{x^{-x}}\implies x(y_0)= y_0^{\left(y_0^{…^{-…}} \right)^{-\left(y_0^{…^{-…}}\right)}}=\bigcirc_x ^\infty \left[y_0^{x^{-x}}\right]$$

Therefore we can do the following ignoring possible absolute value bars. Here is a graphical demonstration of the $n=4$ case showing the convergence and here is the rest of the demonstration:

$$\,^nx=n\big\{ x^{x^{…}}\ =y=y_0\implies x= y_0^\frac 1{(n-1)\big\{ x^{x^{…}} }=\sqrt[\left(\,^{n-1}x\right)]{y_0}\implies x(y_0)= \sqrt[\left(\,^{n-1}{\sqrt[\left(\,^{n-1}(…)\right)]{y_0}}\right)]{y_0}=\bigcirc_x^\infty \sqrt[\,^{n-1}x]{y_0} $$

Here is a graph of the second way using a combined method:

$$x^{x^x}=y=y_0\implies\ln\left(x^x\right)+\ln(\ln(x)))=\ln(\ln(y))\implies x=\sqrt[x]{\log_x(y_0)}\implies x(y_0)=\bigcirc_x^\infty \sqrt[x]{\log_x(y_0)} $$

So the best method would be to see what it best through testing.

Here is a “closed” form using the Fixed Point operator and Functional Root operator. The subscripts indicate the kth root. For convention, let the $k\in\Bbb N=1,2,3,…$ with the $1$st being the one closest to $0$. The other argument just tells the variable the operator is with respect to. Note that the fixed point has no index, so the result will be a set of all fixed points which can also be combined with the pre-recursion equation:

$$\,^n x=y\implies x=\left[x, ^n x=y \right]_k$$

$$^n x=y\implies ^n x-y+x=x\implies x=\{x\}=\text{FixedPoint}[x, ^n x-y+x]$$ Please correct me and give me some feedback.

  • $\begingroup$ If you write $\bigcirc^n$ in front of the $\log$'s, to show that there are $n$ of them, then that will compactify a lot of notation $\endgroup$ Sep 30, 2021 at 3:33
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    $\begingroup$ Don't worry I will edit your answer to add that. Then you can yourself just see how that will look. By the way what country do you live in? $\endgroup$ Sep 30, 2021 at 3:36
  • $\begingroup$ @RounakSarkar In the USA. You are from Bengal based on your profile. $\endgroup$ Sep 30, 2021 at 3:40
  • $\begingroup$ After a few days, I will do a bounty on this question. If no one is able to answer better then I will accept this answer and give you the bounty. $\endgroup$ Oct 2, 2021 at 10:49
  • $\begingroup$ This is way better, however do you have any other ideas? $\endgroup$ Oct 9, 2021 at 3:40

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