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The difference between them is that physical evolution takes one state to another state, whereas addition takes two numbers to another number.
by OscarCunningham 7y ago
The difference between them is that physical evolution takes one state to another state, whereas addition takes two numbers to another number.
- hyperpallium 7y ago|a| = 2 a^2 = 4
- hyperpallium 7y agoMost 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)...(1)). n!/n^n < 1/2 for all n>2 Does this converge to a fraction as n->infinity (i.e. for the naturals)? As a recurrence relation, we have: n!/n^n = (n-1)!/(n-1)^(n-1) x n(n-1)^(n-1)/n^n We could change variables to rewrite the factor as (n+1)n^n/(n+1)^(n+1) = n^n/(n+1)^n Since n+1 converges to n, I think this converges to 1. Meaning that the whole thing converges to something - but working out what it converges to is beyond my abiility. EDIT Wolfram says the limit is actually 0 https://www.wolframalpha.com/input/?i=n%21%2Fn%5En https://www.wolframalpha.com/input/?i=n%21%2Fn%5En lim_(n->∞) n^(-n) n! = 0 Reversibility is very rare indeed! At least, with the prior pr density implied from this comception of functions.
- OscarCunningham 7y agoStirling's approximation says that n! ~ sqrt(2 pi n) (n/e)^n. So n!/n^n ~ sqrt(2 pi n)e^-n.
- hyperpallium 7y agoCool! So the 1/e^n factor will bring it to 0, as n goes to infinity, I think?
- OscarCunningham 7y agoYeah, e^n grows much faster than sqrt(2 pi n), so sqrt(2 pi n)/e^n will tend to 0.