Welcome to MathOverflow

Question

15

2

How to find an integer part of $10^{10^{10^{10^{10^{-10^{10}}}}}}$? It looks like it is slightly above $10^{10^{10}}$.

flag	asked Oct 27 2011 at 0:18 Vladimir Reshetnikov 1,599●9●18

Either I am temorarily confused or I do not understand this question. (-10)^10 = 10^10, so? Or is there a typo? – quid Oct 27 2011 at 0:27

1

Only slightly related, but for amusement, check out math.stackexchange.com/questions/72646/… – Joel David Hamkins Oct 27 2011 at 0:27

@quid: It is -10^10, not (-10)^10. – GH Oct 27 2011 at 0:28

@GH: Thanks! So, it was me being confused. Sorry for the noise. – quid Oct 27 2011 at 0:35

1

Dear Vladimir, I dont think so. Evaluate using the approximation exp x is close to 1+x when x is small (so 10^x is close to 1+log_e 10 x, and also 10^{a+b}=10^a 10^b. Is there a research motivation behind this cute) question? – Gil Kalai Oct 27 2011 at 0:38

show 3 more comments

GH · Accepted Answer · 2011-10-27 01:09:51Z UTC

30

I think the number in question is $10^{10^{10}}+10^{11}\ln^4(10)$ plus a tiny positive number. That is, it starts with a digit $1$, followed by $10^{10}-13$ zeros, then by the string $2811012357389$, then a decimal point, and then some garbage (which starts like $4407116278\dots$).

To see this let $x:=10^{-10^{10}}$, a tiny positive number, and put $c:=\ln(10)$, an important constant. We have $$10^x=1+cx+O(x^2)$$ $$10^{10^x}=10^{1+cx+O(x^2)}=10+10c^2x+O(x^2)$$ $$10^{10^{10^x}}=10^{10+10c^2x+O(x^2)}=10^{10}+10^{11}c^3x+O(x^2)$$ $$10^{10^{10^{10^x}}}=10^{10^{10}+10^{11}c^3x+O(x^2)}=10^{10^{10}}+10^{10^{10}}10^{11}c^4x+O(x^2),$$ where $O(x^2)$ means something tiny all the way.

In the last expression we have $10^{10^{10}}10^{11}c^4x=10^{11}c^4$, which justifies my claim.

edited Oct 27 2011 at 1:09

answered Oct 27 2011 at 1:03

GH
15.1k●1●31●53

7

I'm not convinced the question belongs on MO, but I like the answer too much to vote the question down. – Gerry Myerson Oct 27 2011 at 4:45

1

@Gerry: I thought the same! :-) – GH Oct 27 2011 at 14:39

1

Robert Munafo has a "hypercalc", unfortunately not for the windows-environment, so I cannot use it (Actually it is a perl-script). mrob.com/pub/perl/index.html cite: "Hypercalc: An unusual calculator program. It represents numbers in a special way allowing the calculation of quantities much larger than tools such as bc, dc, MACSYMA/maxima, Mathematica and Maple, all of which use a bignum library. For example, you can use Hypercalc to determine whether 128^48^1024 is larger than 8^88^888. (...)" (from R. Munafo's site). Someone who could try it with this program? – Gottfried Helms Oct 27 2011 at 17:41

2

As far as I can tell, hypercalc just uses standard floating point representations of iterated logs. This suggests that it cannot handle the sort of precision necessary here. – S. Carnahan♦ Oct 28 2011 at 3:55

How to calculate [10^10^10^10^10^-10^10]?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.