How to calculate extremely large exponential

Question

I would like to calculate the following nested exponential: $$ \Large{e^{e^{10^{72}}}} $$

In case that might be hard to read, in Mathematica functional notation it is:

Exp[Exp[10^72]]

My efforts fail due to computational overflow. I was hoping that Mathematica would recognize the very large size of the number and switch number format accordingly.

The number is and estimate the time it takes for a black hole to come to quantum equilibrium (per Leonard Susskind and ER=EPR lectures).

Answer 1

I think you underestimate how many digits this number has. Even if you used scientific notation, the exponent would overflow your PC's memory, so you'd need scientific notation for the scientific notation. I'm not sure which number format you expect to get, but I think the form it's already in is probably the most useful.

But let's try to get an idea for how big this number is (if that's even possible). Let's start with the inner exponent. $10^{72}$ is already a pretty big numbers. I just looked up some estimates for the number of atoms in the observable universe. It's somewhere around $10^{80}$. That's a factor of 100 million on top, but still $10^{72}$ is pretty impressive.

But wait, we've got $e^{10^{72}}$, that's $e$ multiplied by itself that many times. Can we use Mathematica to find out, how many digits this has? Yep:

Log[10., Exp[10^72]]
(* 4.34294*10^71 *)

So the number of digits in this number is only a few orders of magnitude short of the number of atoms in the universe. This is already incomprehensibly large.

But wait, we've got $e^{e^{10^{72}}}$. Yeah. That's how big it is.

But you were only asking for scientific notation. The problem is still that $e^{10^{72}}$ provides an estimate of the number of digits of the final result. But remember, even that number had an inconceivable amount of digits. So even if you had a bit of memory for each atom in the universe, you might just be able to store the exponent of your result. That's why even scientific notation doesn't cut it here.

You might be interested in arrow notation though.

Answer 2

You should take advantage of the fact that

a^b == c^(Log[c,a] b)

so that

E^(E^(10^72)) == 10^(Log[10,E] E^(10^72)) == 10^(Log[10,E] 10^(Log[10,E] 10^72))

Since Log[10,E] (that is, ln(10)) is about 2.3, the number of digits in your quantity (in base 10) is

~ 2.3 10^(2.3*10^72)

Answer 3

Wolfram alpha can deal with this number easily using repeated Power of 10 representation, though I don't know how to transfer this to mathematical without using the "Wolframalpha" command. https://www.wolframalpha.com/input/?i=E%5E%28E%5E%2810%5E72%29%29

Answer 4

Inspired by @三叶虫家族the193thDoctor’s answer about how Wolfram|Alpha represents large numbers:

ClearAll[represent];

represent[x_, Optional[base_?Positive, 10], 
 Optional[limit_?Positive, 100]] := 
 With[{calc = {#, Quiet@N[#]} &}, Which[
   Quiet@Reduce[x > 0],
      NestWhileList[
           ls \[Function] calc[Log[base, ls[[1]]] // FullSimplify],
           calc[x],
           #[[2]] === Overflow[] || #[[2]] > limit &] //
        Fold[
           {a, b} \[Function] Superscript[b, a],
           ConstantArray[base, Length[#]] // 
              ReplacePart[1 -> #[[-1, 2]] ] ] &,
   Quiet@Reduce[x < 0],
      -represent[-x, base, limit],
   True,
      0
   ]]

Test

Exp[Exp[10^72]] // represent // TeXForm

$10^{10^{10^{71.6378}}}$

Other parameters are supported:

represent[Exp[Exp[10^72]], 2, 1000] // TeXForm

$2^{2^{2^{239.708}}}$

represent[Exp[Exp[10^72]], Pi, 1000] // TeXForm

$ \pi ^{\pi ^{\pi ^{144.707}}} $

How much memory needed?

If an integer $n = 2^{2^x}$, then a minimum memory of $\lfloor{2^x}\rfloor+1$ bits is needed to store $n$.

In OP’s case, $x\approx 2^{239.708}$, so:

$$ \begin{align} \lfloor{2^x}\rfloor+1 \text{ bits} & \geqslant 2^x \text{ bits} \\ & = 2^{x-33} \text{ GB} \\ & > 2^{2^{239.7}} \text{ GB} \\ & > 10^{10^{71.63}} \text{ GB} \end{align} $$