How do black holes destroy information and why is that a problem? • Hacker News Board

Re: How do black holes destroy information and why is that a problem?

#91

Post by OscarCunningham » Tue, Aug 27, 2019, 2:52 PM UTC

hyperpallium wrote: post #90

hyperpallium wrote: post #56

Earlier quoted context omitted.

|a| = 2 a^2 = 4

Most functions are not invertible. In modulo arithmetic (numbers wrap around like days of the week, months, hours in a day, or the units column, e.g. mod 10 has numbers 0 to 9, and 9+1=0) we can count them. There are n^n possible functions (each of the n inputs could map to any n values), but only n! invertible functions (the first input can be to any of n , the second can be to any but the first's, etc (n)(n-1)(n-2)…

Stirling's approximation says that n! ~ sqrt(2 pi n) (n/e)^n.

So n!/n^n ~ sqrt(2 pi n)e^-n.

Re: How do black holes destroy information and why is that a problem?

#92

Post by hyperpallium » Wed, Aug 28, 2019, 7:10 AM UTC

OscarCunningham wrote: post #91

hyperpallium wrote: post #90

Earlier quoted context omitted.

Most functions are not invertible. In modulo arithmetic (numbers wrap around like days of the week, months, hours in a day, or the units column, e.g. mod 10 has numbers 0 to 9, and 9+1=0) we can count them. There are n^n possible functions (each of the n inputs could map to any n values), but only n! invertible functions (the first input can be to any of n , the second can be to any but the first's, etc (n)(n-1)(n-2)…

Stirling's approximation says that n! ~ sqrt(2 pi n) (n/e)^n. So n!/n^n ~ sqrt(2 pi n)e^-n.

Cool! So the 1/e^n factor will bring it to 0, as n goes to infinity, I think?

Re: How do black holes destroy information and why is that a problem?

#93

Post by OscarCunningham » Wed, Aug 28, 2019, 10:42 AM UTC

hyperpallium wrote: post #92

OscarCunningham wrote: post #91

Earlier quoted context omitted.

Stirling's approximation says that n! ~ sqrt(2 pi n) (n/e)^n. So n!/n^n ~ sqrt(2 pi n)e^-n.

Cool! So the 1/e^n factor will bring it to 0, as n goes to infinity, I think?

Yeah, e^n grows much faster than sqrt(2 pi n), so sqrt(2 pi n)/e^n will tend to 0.