By $^nx$, I mean $x$ tetrated to $n$. So, basically, I'm looking for the solution of the equation $$\large x^{x^{x^{x^{x^{x^{x^{x{^{x^x}}}}}}}}}=e$$. Is there some way to find the approximate value by using some infinite series or anything? I can only figure out that the value should be between $1$ and $2$. And, it should be far away from $2$ because for $x=2$, $^{10}x$ is a very very large number. Is there some way to approximate it?

  • 2
    $\begingroup$ You can use the knuth notation $x\uparrow\uparrow 10=e$. $\endgroup$ – zwim Jan 30 '17 at 10:40
  • 1
    $\begingroup$ @zwim: I've never seen that notation before. Wkipedia uses $^nx$. $\endgroup$ – Dove Jan 30 '17 at 10:42
  • 2
    $\begingroup$ 1.46395824687941822939451822947123 looks like a reasonable starting approximation... $\endgroup$ – PM 2Ring Jan 30 '17 at 10:51
  • 1
    $\begingroup$ $~^nx=y\require{cancel}\cancel\implies x=~^{1/n}y$ $\endgroup$ – Simply Beautiful Art Jun 12 '17 at 1:07

Assuming that the infinte tetraetion

$$ x^{x^{x^{....}}}=e $$

exists (which is indeed the case in the sense of a limit, see here for more detail), the limiting value is given by

$$ x_*^e=e $$


$$ x_*=e^{1/e}\approx1.44466786 $$

which should serve as an extremly good approximation for the solution of $$x\uparrow\uparrow10=e$$


The real value seems to be

$$x\approx 1.46396$$

so the relative error using the infinte approximation


$$ \frac{|x-x_*|}{x}\approx0.0131 $$

which is pretty awesome regarding the simplicity of this approximate solution

  • $\begingroup$ @GottfriedHelms i have no cas at hand today, i will check tomorrow. thanks for your patience! $\endgroup$ – tired Jan 31 '17 at 17:44
  • $\begingroup$ @GottfriedHelms you are indeed correct...somehow i used $x\uparrow\uparrow11$. i correct my answer now $\endgroup$ – tired Feb 1 '17 at 10:52

Considering that even $x^x=a$ doesn't come with a formula for $x$ either, I think it is not unreasonable to search for the result by traditionnal means like dichotomy, especially since the tetration is an increasing function in the considered interval.

Also since $x\uparrow\uparrow n=e$ imposes strong bounds on $x$ in $[1,2]$ as you stated else it would diverge quickly, we are somehow in the ideal range for the $pow$ function accuracy.

For large $n$, the infinite approximation given by tired would work fine, and for small $n$, it would not be a big deal for a computer to calculate $x\uparrow\uparrow n$ with required accuracy.

  • $\begingroup$ the infinite approximation is pretty accurate in this case, i wouldn't have expected that $\endgroup$ – tired Jan 30 '17 at 11:39
  • $\begingroup$ If one allows the Lambert W function, $$x^x=a\implies x=\frac{\ln(a)}{W(\ln(a))}$$ $\endgroup$ – Simply Beautiful Art Jun 12 '17 at 1:14

What you asking for seems to me to be the 10'th "superroot" (or possibly one should call this and introduce such a term like "tetroot of order 10") .

Using Pari/GP we can do the following:

 %97 = 1.46395824688 \\ lines with %<number>= ... is output of the interpreter

y^y^y^y^y ^y^y^y^y^y
 %98 = 2.71828182846

y^y^y^y^y ^y^y^y^y^y-exp(1)
 %99 = 0.E-201

I've some time ago done a little analysis of this problem and a general path to find a power series.
In further generalization I think we can even interpolate to superroots of fractional order (in your case, you ask for the (integer) 10'th superroot)

(However, I wouldn't interpret tetration this way).

  • $\begingroup$ @Gottdried Helmea: I don't know about generalizing super-roots, but I may have found a way to generalize super-logarithms: math.stackexchange.com/questions/2134589/…. Maybe, the generalization of super-roots follows from that. $\endgroup$ – Dove Feb 8 '17 at 12:02
  • 1
    $\begingroup$ @Dove: in your link - a nice approach. I still like some fresh ideas and creativity here. However, I've another definition for tetration, in a sense as function $f: x_0 \to x_1 $ and then $g : x_0 \to x_{0.5} $ and with that same *function* $g$ we have $g : x_{0.5} \to x_1 $ such that we interpret tetration as $ f(x) = g(g(x))$ or as $f^{°h}(x) = f(f^{°h-1}(x)) $ which is a 3-argument operation between a starting value $x_0$, the function itself and the number of iterations of the function $h$ producing the result $x_h$ and not a 2-argument operation like $\;^h x $ . $\endgroup$ – Gottfried Helms Feb 8 '17 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.