Sunday, September 26, 2010

The gogolth digit of pi is 4.

"If this isn't true, I'll kill myself."

Is this a safe statement to make? Even with geometric expansion of computer power it seems quite likely that calculating the googolth digit of pi may not be done in my lifetime. Further, it might well be 4. Also the statement is not binding. And perhaps we can just keep upshifting the base until one of the bases actually is 4. In all likelihood any base greater than 5 should more or less have yet another set of odds up to base googol or so. I figure one could just keep trying.

So I think there's a way to make it true, and no real way to calculate it anyway (the best we've done is 5 trillion or ~10^12 so 1/10^88th as much as is needed). So I'd be safe with some sort of suicide pact based on pi. Or would I? Eh, doesn't matter either way. It's still a 4 in base-ten and suicide is contingent on non-4 in every base. If we assume it's normal and being unknown somewhat random, I have a 1/5th in base 5, 1/6th in base 6, 1/7th in base 7... or something I'd likely have to look up, let's just call it fifty fifty, but proving it would be an issue.

3 comments:

Tracy said...

If you look at each of the natural numbers in relation the number of times or rather the percentage in the number of times that they each appear in pi as pi is calculated out further and further, I wonder what these percentages would be or should be?

Tatarize said...

It depends entirely on whether pi is a normal number. It very likely is, but nobody has proved it yet. But are there say a 40% greater chance of getting a 2 after the 7 trillionth digit? Does the sequence 392393020392930 stop appearing ever after the 900 trillionth digit? Are the ratios equal?

Pi is very likely a normal number, in which case there would be as many sevens in pi as there are fours in pi, and as many fours in pi as there are digits of pi. All digits and sequences within pi should occur an infinite number of times. If you ran pi in base 4, and compared it to the base 4 nucleotides of my own entire genome, there should be as many zeroes in pi as there are copies of my entire genetic code.

Cantor's infinity cardinalities is strange and pi is a bit weird, but each of those statements is true and provable, assuming pi is a normal number which it has yet to be proven as such, then there really are as many zeroes as there are copies of my DNA.

And further, I can recite as many digits of pi as there are digits of pi, considering that I'll only recite the zeroes. But as there are as many zeroes as there are digits (assuming normalcy) then that's true.

The percentages should be equal for all sequences of equal length if pi is normal, and nobody has proven that it is.

PeterTheAble said...

You are safe. Consider the sum of 1/N for N=1..M. For M = 2^k the sum is greater than k/2 except for the trivial k=1 where the sum = 1/2.

Pi is irrational. Irrational numbers to not endlessly repeat digits like 1/3 or 1/7. This is true in any base. Although the density of 4s decrease as the base increases you are saved by fact that the sum of 1/N from 1 to M increases without bound as M approaches infinity.

Personal guess: well before base 100 (decimal) there will be a 4 in the 10^100th digit of pi.

Peter B