Equivalence of Roots on the Exponential Level - $a^a$ ($n$ $times$) $ = b,$ $find$ $a$

Question

My question is how would the concept of roots, on the multiplication level, be expanded to the exponential level?

For example, the equivalence of roots on the addition level is division as to find $a$ in "$a + a$ ($n$ $times$) $ = b$" you divide $b$ by $n$ ($a = \frac{b}{n}$).

Moving up a level, roots find $a$ in "$a\times{a}$ ($n$ $times$) $=b$" when you take the root of $b$ to the index of $n$ ($a = \sqrt[n]{b}$).

Finally moving up to exponents, the equivalence of roots finds $a$ in "$a^a$ ($n$ $times$) $= b$" which leads me to my question of what this exactly is and if it exists? Also, looking at the pattern of how these equivalences work the "root" for exponents would have to involve the reverse of tetration (aka. division for tetration). If you know anything about this or can lead me to any external sources that would be greatly appreciated.

exponentiation is not associative so one needs to be careful when one defines the iterated exponential, whenever one defines we atleast have one root for every $b>1$ — Aditya Dwivedi, Dec 16 '22 at 04:34
There are 2 answers on https://math.stackexchange.com/questions/1415029/solve-xxx-1 (for the specific case of 3'rd superroot). — Gottfried Helms, Dec 16 '22 at 14:33

Gottfried Helms · Answer 1 · 2022-12-18T00:43:35.267

A short example that it might be solvable. And to get not only one root, but infinitely many... Let's say, $b=3+4î$, and let's assume, we look for the 4-fold "superroot" $b$ such that $b^{b^{b^b}}=a$.

One solution can often be found, when we go for logarithms,such that $a^a \cdot \log(a) + \log(\log(a)) = \log(\log(b)) $
In Pari/GP I used the following loop:

 a1= 2+ I 
 llb = log(log(b))
 \\ and then the loop
 for(k=1,50,a1=a1+ ( a1^a1*log(a1)+log(log(a1) )-llb)/5)
 \\ perhaps you run that loop much more often... 
 print([a1,a1^a1^a1^a1 - b])

Here the $5$ in the denominator smoothes the iterations a bit; sometimes/often you get oscillating or divergent inbetween results when this were not needed; the division by some small number (like $2$ or $5$) smoothes such oscillations a bit out.

And the protocol:

a1=2+I
%1858 = 2 + I
for(k=1,50,a1=a1+ ( a1^a1log(a1)+log(log(a1) )-llb)/5);[a1,a1^a1^a1^a1 - b]
%1860 = [0.783810452536 + 4.91321170105I, -2.15880810921 + 0.862152498110*I]
for(k=1,50,a1=a1+ ( a1^a1log(a1)+log(log(a1) )-llb)/5);[a1,a1^a1^a1^a1 - b]
%1862 = [2.40191290725 + 6.25202990786I, -0.848936804266 + 0.794139691466*I]
for(k=1,50,a1=a1+ ( a1^a1log(a1)+log(log(a1) )-llb)/5);[a1,a1^a1^a1^a1 - b]
%1864 = [2.61793681780 + 6.34004308898I, 0.00147753555577 - 0.00641784928147*I]
for(k=1,50,a1=a1+ ( a1^a1log(a1)+log(log(a1) )-llb)/5);[a1,a1^a1^a1^a1 - b]
%1866 = [2.61688549860 + 6.34029457755I, 1.41457341857 E-5 + 1.80101968639 E-5*I]
for(k=1,50,a1=a1+ ( a1^a1log(a1)+log(log(a1) )-llb)/5);[a1,a1^a1^a1^a1 - b]
%1868 = [2.61688711244 + 6.34029117923I, -7.98341102288 E-8 - 1.00455595947 E-9*I]
\ .... repeating the loop converges .....

Newton-iteration on a sufficiently-approximated value for $a_1$ may better improve the result to arbitrary precision.

Additional note: in my older answer (see link) I arrived then at the possibility that there is an infinite number of solutions, which can be found, when appropriately multiples of $2 \pi î$ are added to the logarithms in the loop. (But this is not my scope here)

Extending the example to the $5$'th "superroot" (awful name...) of $b$, using moreover the 3-fold logarithm (instead the 2-fold as before),converges even faster:

llb=(log(log(b))) \\ %1894 = 0.619211291069 + 0.522706299508*I
a1=2+I            \\ %1896 = 2 + I
for(k=1,50,a1=a1+ (log( a1^a1^a1log(a1)+log(log(a1) ))-log(llb))/5);[a1,a1^a1^a1^a1^a1 - b]
%1898 = [1.80323294737 + 1.88023731125I, 1.84013147894 E-10 + 4.86457525906 E-11*I]
for(k=1,50,a1=a1+ (log( a1^a1^a1log(a1)+log(log(a1) ))-log(llb))/5);[a1,a1^a1^a1^a1^a1 - b]
%1900 = [1.80323294737 + 1.88023731124I, -2.23667218308 E-22 + 8.47505332430 E-22*I]
\ very good convergence, 10 digits precision per 50 iterations

score 0 · Answer 2 · answered Dec 16 '22 at 23:01

Such a tetration is denoted by $^na$ (Mathjax: $^na$ or ${}^na$ ). For a fixed $n$, it is a strictly increasing function of $a$ on $[1,\infty)$. It is also continuous. As $^n1 = 1$ and $\lim_{a\to \infty} {}^na = \infty$, this is enough to ensure that for any $b \ge 1$, there is a unique $a \ge 1$ with $^na = b$.

The case for $b < 1$ is more challenging.

Equivalence of Roots on the Exponential Level - $a^a$ ($n$ $times$) $ = b,$ $find$ $a$

2 Answers2