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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).

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    $\begingroup$ 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. $\endgroup$ – Martin Ender Jan 12 '15 at 17:29
  • $\begingroup$ I agree with @MartinBüttner, you might change the e's to 10's or vice versa, but that would probably obscure the physics behind. $\endgroup$ – mikuszefski Jan 12 '15 at 17:32
  • $\begingroup$ I knew it was very large but probably did not guess that large for my Mac's 16 GB memory. I was hoping to find a way merely to represent the number powers of 10. But, powers of some larger numbers work. Given this is Mathematica, I guessed there might be some easier way like Log, Log or something. $\endgroup$ – K7PEH Jan 12 '15 at 17:32
  • $\begingroup$ @Szabolcs -- yes, I can do that but I am not actually interested in the actual number -- just a rough estimate of the size of the number such as total number of digits. I have the same overflow problem with using nested Log functions to determine digit count -- actually, just noticed the comment above about changing e's to 10's and that works for me -- the physics is obscure to begin with! $\endgroup$ – K7PEH Jan 12 '15 at 17:41
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    $\begingroup$ @Mr.Wizard -- this number comes from a lecture by Professor Leonard Susskind of Stanford University on the connection between Einstein-Rosen Bridge (wormholes) and the EPR correlation (quantum entanglement) within Black Holes. It is a measure of the time for a black hole to achieve some kind of quantum state equilibrium after the creation of a black hole. A much better explanation is from the lecture itself available here: youtube.com/watch?v=OBPpRqxY8Uw (this is part 1, and part 2 is easily located on Youtube). $\endgroup$ – K7PEH Jan 12 '15 at 18:35
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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.

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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)
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  • $\begingroup$ Which is a lot of digits ;P $\endgroup$ – evanb Jan 12 '15 at 19:45
  • $\begingroup$ Yes, I did all that this morning (well, at least, this morning where I am). I got sidetracked in assuming that this was something that Mathematica should be able to do before, as others pointed out, I realized just how big this number actually is. $\endgroup$ – K7PEH Jan 12 '15 at 22:15

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