= 27d567bc2d48d9f40e9ccc20ea370b15)
gjm comments on Open thread, 9-15 June 2014 - Less Wrong Discussion
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (239)
In what follows, all logs are to base 3.
Definition: [a,b,c,d] := a^(b^(c^d))), etc.
Lemma: log [a,b,...,z] = log a^[b,...,z] = (log a) [b,...,z].
Definition: {n} := [3,...,3] with n 3's.
Lemma: log {n} = {n-1}.
Definition: G := [10,10,100].
OK. So we want to know when {n} > [G,G]. Taking logs, this is the same as {n-1} > G log G = [10,10,100] log [10,10,100]. Taking logs again, it's the same as {n-2} > log log [10,10,100] + log [10,10,100] which (unless it comes out amazingly close) is the same as {n-2} > log [10,10,100] = (log 10) [10,100]. Taking logs again, this is the same as {n-3} > log log 10 + log [10,100] = log log 10 + 100 log 10, which again is basically the same as {n-3} > 100 log 10, and now we're in the realm of small numbers and we find that [3,3] is too small but [3,3,3] is way more than enough.
So I'm pretty sure the answer is that you need six 3s in your tower.
Since it's easy to get this kind of thing wrong -- which is why I introduced notation and lemmas intended to make the manipulations as simple as possible -- I just did it again a different way (a couple of steps of algebra on paper, plus some ordinary floating-point arithmetic on a computer) and without the informal throwing away of much smaller bits. I can confirm that five 3s aren't nearly enough and six 3s are way more than you need.
[EDITED to remove a definition and a lemma that I never actually used, and to add a little clarification.]
By the way, here's the handwavy heuristic version that gives the right answer.
When you are comparing two exponentials with really big exponents, only the exponent really matters even if the difference in bases is huge (unless it's really huge or the exponents are really close). So the following should all be the same:
and now we can see that 3^3 < 100 but 3^3^3 > 100, so n-3=3 and n=6.
If the detailed algebra in the parent isn't enough to make "look at the exponent and ignore the bases" plausible, here's another way to see it that happens to work neatly in this case:
(10^10^100) ^ (10^10^100) = 10^(10^100 . 10^10^100) = 10^10^(100 . 10^100) = 10^10^10^102
so the difference between 10 and 10^10^100 on the base is the same as the difference between 100 and 102 three levels up!
... EDIT, on happening to reread this days later: No, I slipped up in the calculation above and the result is actually a lot closer.
(10^10^100) ^ (10^10^100) = 10^(10^100 . 10^10^100) = 10^10^(100 + 10^100) = 10^10^10^(100+teenytiny)
because 10^100 + 100 is barely bigger than 10^100 at all. In fact it turns out that "teenytiny" is about 4 x 10^-99. So: the difference between 10 and 10^10^100 on the base is the same as the difference between 100 and 100 + 4e-99 three levels up.