What is the tenth superroot of $e$?

Solution 1:

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

Solution 2:

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.

Solution 3:

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).