How to find the integer part of big number?

Question

How to calculate the integer part $$\left \lfloor10^{10^{10^{10^{10^{-10^{10}}}}}} \right \rfloor ?$$ Does this equal $$10^{10^{10}}? $$ Both Maple and Mathematica fail with it.

PS. Unmotivated votes down are not a constructive answer.

Answer 1

${-10^{10}}$ is very negative, hence $10^{-10^{10}}$ is very close to $0$, hence $\alpha:=10^{10^{-10^{10}}}$ is very close to $1$. Hence we expect the number in question to be somewhat close to $10^{10^{10}}$ and it is no wonder that CAS fail (or would produce a 10 billion digit output if they did not fail). It looks like our only chance to obtain a definite answer seems to be that we are lucky enough to show that $10^{10^{10^{\alpha}}}$ is in a suitable interval of length at most $1$

Lemma. If $a<b<a+\log \frac{3}{\ln 10}$, then $10^a+\ln 10\cdot (b-a)10^a<10^b<10^a+3(b-a)10^a$.

Proof. Let $f(x)=10^x$. Then $f'(x)=\ln 10\cdot 10^x>0$. From the MVT we have $$\frac{10^b-10^a}{b-a}=f'(\xi)$$ for some intermediate $\xi$. The claim follows from $\ln10\cdot10^a=f'(a)<f'(\xi)<f'(a+\log\frac{3}{\ln 10})=3\cdot 10^a$. $_\square$

For the applicability of the lemma in what follows, note that $\log\frac 3{\ln 10}\approx 0.1149$.

First we find $$\tag2 1+\ln 10\cdot 10^{-10^{10}}<\alpha<1+3\cdot 10^{-10^{10}}.$$ From this, $$\tag3 10+10(\ln 10)^2\cdot 10^{-10^{10}}<10^\alpha<10+90\cdot10^{-10^{10}}.$$ Next $$ \tag4 10^{10}+10^{10}(\ln 10)^3\cdot 10^{-10^{10}}<10^{10^\alpha}<10^{10}+270000000000\cdot 10^{-10^{10}}.$$ Finally, $$ \tag5 10^{10^{10}}+10000000000(\ln 10)^4<10^{10^{10^\alpha}}<10^{10^{10}}+810000000000.$$ Considering the size of the numbers involved, this is very close to the desired result - and yet so far away. Can this be imrpoved? Well, it seems that we need to replace $3$ in the lemma with a number $c$ much closer to $\ln 10$ - in fact so close that $10^{10}c^4<\lceil10^{10}(\ln10)^4\rceil$. To acchieve this, we need something like $c<\ln10+10^{-13}$. Then $\log\frac{c}{\ln 10}\approx 1.88\cdot 10^{-14}$.

So we try the Improved lemma. Let $c=\ln 10+10^{-13}$. If $a<b<a+\log\frac{c}{\ln 10}$, then $$10^a+\ln 10 \cdot (b-a)10^a<10^b<10^a+c(b-a)10^a. $$

(Proof as above). We can repeat the steos above, noting that even $10^{10}c^310^{-10^{10}}$ is small enough (namely $\ll 10^{-14}$), so that this time we obtain $$ 10^{10^{10}}+10^{10}(\ln10)^4<10^{10^{10^\alpha}}<10^{10^{10}}+10^{10}c^4<\left\lceil10^{10^{10}}+10^{10}(\ln10)^4\right\rceil.$$ We conclude that $$ \left\lfloor10^{10^{10^ {10^{10^{-10^{10}}}}}}\right\rfloor=10^{10^{10}}+\left\lfloor10^{10}(\ln 10)^4\right\rfloor =10^{10^{10}}+281101235738.$$