Assuming this could be done how long would it take to show everything? We ran in to the roadblock when we tried to calculate how many frames there were total in a non-exponential form. We were looking at the grayscale version, so there are 2[sup]1310719[/sup] frames, but that is not too useful for our purposes. We also assumed that the monitor this was displaying on was refreshing 100 times per second.
2[sup]1310719[/sup]/100 gives us the total time in seconds, but how do you calculate that? None of the digital means we tried could work it. Please help assuage our curiosity.
log[sub]10[/sub]2 = 0.301
y = 1310719
So, y * log[sub]10[/sub]2 = 394565.7.
A = 10[sup]394565.7[/sup] = 5 x 10[sup]394565[/sup].
Divide that by 100, and we get 5 x 10[sup]394563[/sup] seconds.
(This would be 1.6 x 10[sup]394556[/sup] years, which is a little longer than most of us are willing to wait around for.)
First of all, the leap from greater than the number of particles in the universe to all knowledge is just plain idiotic. (But see the note at the end.)
Secondly, 2^131079 is about 10^394565.7349, so subtact 2 (dividing by 100) from the exp. gives 10^394563.7349 seconds. Now as we all know “there are (about) Pi seconds in a nano-century”. So subtract about 9.5 (losing accuracy here) to get roughly 10^394554.2 centuries. Since the universe is about 10^8.2 centuries old, the two numbers are noways comparable.
Back to a small but meaninful number: Chaintin defined a number, Omega, with a property that is
"embodies an enormous amount of wisdom in a very small space … inasmuch as its first few thousands digits, which could be written on a small piece of paper, contain the answers to more mathematical questions than could be written down in the entire universe. " (Charles Bennett)
Wolfram’s new book is suppose to also cover a similar issue.
(Called away before final submit., so a little redundancy here.)