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.
