# Is there a way to write the $n^\text{th}$ super root in terms of the lambert W function?

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

$$y=^nz$$

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

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

$$\textstyle\displaystyle{\sqrt[n]{z}_s=1+\sum_{k=1}^{\infty}s_n\frac{(z-1)^n}{n!}}$$

Where,

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

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?

• 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. Sep 27, 2021 at 13:36
• @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. Sep 29, 2021 at 4:03
• @RounakSarkar I mean I can generalize the nesting to other power towers. Would this complete it? Sep 29, 2021 at 13:24
• @Tyma Gaidash. If you can generalize then there is no problem. Sep 29, 2021 at 13:36
• @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. Sep 29, 2021 at 14:06

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:

$$n=1:$$

$$x=y$$

$$n=2$$:

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

$$n=3:$$

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

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.

• 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 Sep 30, 2021 at 3:33
• 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? Sep 30, 2021 at 3:36
• @RounakSarkar In the USA. You are from Bengal based on your profile. Sep 30, 2021 at 3:40
• 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. Oct 2, 2021 at 10:49
• This is way better, however do you have any other ideas? Oct 9, 2021 at 3:40