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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?

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

$$ x_*^e=e $$

or

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

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


Update:

The real value seems to be

$$x\approx 1.46396$$

so the relative error using the infinte approximation

is

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

which is pretty awesome regarding the simplicity of this approximate solution

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  • $\begingroup$ @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
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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.

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  • $\begingroup$ 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
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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:

y=solve(x=1,1.5,x^x^x^x^x^x^x^x^x^x-exp(1))
 %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).

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  • $\begingroup$ @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
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    $\begingroup$ @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

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