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

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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 '11 at 0:27
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Only slightly related, but for amusement, check out math.stackexchange.com/questions/72646/… –  Joel David Hamkins Oct 27 '11 at 0:27
    
@quid: It is -10^10, not (-10)^10. –  GH from MO Oct 27 '11 at 0:28
    
@GH: Thanks! So, it was me being confused. Sorry for the noise. –  quid Oct 27 '11 at 0:35
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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 '11 at 0:38

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up vote 44 down vote accepted

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.

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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 '11 at 4:45
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@Gerry: I thought the same! :-) –  GH from MO Oct 27 '11 at 14:39
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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 '11 at 17:41
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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 '11 at 3:55

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