5 ms·
> Set(A) = ∑n A^n/n! = e^A since there are n! permutations to quotient by. But this is wrong. The number of sets over A={0,1,2} is 2^3 = 8 /= e^3 The actual s
by wz1000 8y ago
> Set(A) = ∑n A^n/n! = e^A since there are n! permutations to quotient by.
But this is wrong. The number of sets over A={0,1,2} is 2^3 = 8 /= e^3
The actual series should be
Set(A) = ∑n (A*(A-1)*(A-2)..(A-n+1))/n! = 2^A
This is also consistent with the fact that Set(A) ~ A -> Bool
- zawerf 8y agoHow do you interpret the derivative of that? It's now log(2)*2^A. What does "log(2)" mean as a type here? Do you expand it back out as a taylor series like in the circular list example?
- danharaj 8y agoRight, that is the formula for multisets, or bags as they are also called.
- wz1000 8y agoNo, I believe the formula for multisets would be M(x) = ∑n (A*(A+1)*(A+2)..(A+n-1))/n! since there are (n+k+1)c(k+1) possible unordered n-tuples over a set with cardinality k.
- black_knight 8y ago∑Aⁿ/n! actually represents the multisets (aka bags) of elements from A. A multiset is a list, but we quotient out by permutations (hence n!) The division here can be seen as an action of the permutation group (which has n! element).