We introduce some notation for writing really big (but finite) numbers. A googol, denoted g, is defined by $g = 10^{100}$. A googolplex, denoted G, is defined by $G = 10^g$. A MathPatharoo, denoted M, is defined by $M = G^G$. Next, for positive integers m, n define m ↑ n (spoken “em uparrow en”) as follows: $$m ↑ 1 = m$$ $$m ↑ (n+1) = m^{m↑n}$$

And, find the smallest positive integer value of n for 2↑n, so that $2↑n \ge M$

After trying some values for n in 2↑n, the value is always 2 raised to itself $n-1$ times. For example, 2↑2 has a value of $2^2$, 2↑3 has a value of $2^{2^2}$, 2↑4 has a value of $2^{2^{2^2}}$, and so on. I do not know how to approach finding a value of n for 2↑n, so that $2↑n \ge M$ because the numbers are all way too big. Thanks for the help!

NOTE: The ↑ is not any other mathematical term and doesnt relate to any other mathematical expression. It is just an expression made up for this specific problem. Also, I must show exactly on paper how this answer was derived.

  • $\begingroup$ $$2↑M \ge M$$.. $\endgroup$ – N. S. Dec 29 '16 at 17:43
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    $\begingroup$ Your $\uparrow$ is Knuth's $\uparrow\uparrow$. Is the difference in notation deliberate? $\endgroup$ – hmakholm left over Monica Dec 29 '16 at 17:51
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    $\begingroup$ There is a very nice discussion of questions like this at Robert Munafo's website, and he even made a "hypercalc"-calculator implementing arithmetic and value-comparision with expressions like this. see for a start mrob.com/pub/perl/hypercalc.html $\endgroup$ – Gottfried Helms Dec 29 '16 at 21:59
  • $\begingroup$ This is question 2 on the ongoing MathPath 2017 Qualifying Test. $\endgroup$ – hmakholm left over Monica Jan 13 '17 at 9:40
  • $\begingroup$ And this by the same OP is the start of question 3, but may scrape by by virtue of not really getting to the meat of the question. $\endgroup$ – hmakholm left over Monica Jan 13 '17 at 9:53

We have $$M = (10^{10^{100}})^{10^{10^{100}}} = 10^{10^{100}10^{10^{100}}} = 10^{10^{100+10^{100}}} \lt 10^{10^{10^{101}}} \\ \lt 2^{4\cdot 10^{10^{101}}} \lt 2^{64^{10^{101}}} \lt 2^{2^{6\cdot 10^{101}}} \lt 2^{2^{10^{102}}} \lt 2^{2^{2^{4\cdot 102}}} \lt 2^{2^{2^{2^{512}}}} \le 2^{2^{2^{2^{9}}}} \le 2^{2^{2^{2^{2^4}}}} \le 2^{2^{2^{2^{2^{2^2}}}}} = 2\uparrow\uparrow 7 $$

We can also approach it the other way:

$2\uparrow\uparrow 2 = 4$

$2\uparrow\uparrow 3 = 16$

$2\uparrow\uparrow 4 = 65536$

$2\uparrow\uparrow 5$ is astronomically larger than a googol. In particular it is much larger than $\log_2(10)$ googols. But it is less than a googolplex.

$2\uparrow\uparrow 6$ is therefore much larger than a googolplex -- in fact, easily larger than a googol googolplexes.

Since your $M$ is approximately $10^{\text{googolplex}}$, $2\uparrow\uparrow 7$ is therefore larger than $M$.

  • $\begingroup$ You're a genius!!!!! Oh my god! I'm speechles! Thanks!!!!!! $\endgroup$ – mathtron999 Dec 29 '16 at 18:03
  • $\begingroup$ Can you supplement this with an estimate showing that $M > 2 \uparrow\uparrow 6$? $\endgroup$ – Erick Wong Dec 29 '16 at 18:18
  • $\begingroup$ @mathtron999: No, but I have some experience in solving this kind of puzzles. The most important thing to remember here is that (a) whether the base is $2$ or $10$ doesn't matter much in the big scale of things, an (b) one can afford really outrageous overapproximations in the early steps, because their effect soon vanishes anyway. $\endgroup$ – hmakholm left over Monica Dec 29 '16 at 18:24
  • $\begingroup$ @ErickWong: Does the added material satisfy you of that? $2\uparrow\uparrow 5$ is less than a googolplex, so $2\uparrow\uparrow 6$ must be less than $2^{\text{googolplex}}$ and therefore also less than $M$. $\endgroup$ – hmakholm left over Monica Dec 29 '16 at 18:27
  • $\begingroup$ @HenningMakholm Yup, thanks! +1 $\endgroup$ – Erick Wong Dec 29 '16 at 19:59

$M = G^G = 2^{G\log_2 G} = 2^{G^2 \log_2 g} = 2^{100 G^2 \log_2 10}\approx 2^{332.19 G^2}$.

$2 ↑ k = 2^{...}\ge 2^{332.19 G^2}$

$k-1 \ge 332.19 G^2$

$k-2 = \log_2 332.19 G^2= 2*8.375 \log_2 G = g*2*8.375 \log_2 g = 100*g*2*8.375 \log_2 10 = 100*g*2*8.375 *3.3219$

$k-3 = 2*3.3219 * 100*3.3219*\log_2 8.375*\log_2*3.3219$

So $k = \lfloor 2*\log_2 10*100*\log_2 10*\log_2(\log_2 10*100)*\log_2 10 + 3 \rfloor$


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