wz1000 5 years 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^3The 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 black_knight 5 years 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). danharaj 5 years ago Right, that is the formula for multisets, or bags as they are also called. wz1000 5 years ago No, 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. zawerf 5 years ago How 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?
black_knight 5 years 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).
danharaj 5 years ago Right, that is the formula for multisets, or bags as they are also called. wz1000 5 years ago No, 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.
wz1000 5 years ago No, 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.
zawerf 5 years ago How 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?
> 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
This is also consistent with the fact that Set(A) ~ A -> Bool∑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).
Right, that is the formula for multisets, or bags as they are also called.
No, I believe the formula for multisets would be
since there are (n+k+1)c(k+1) possible unordered n-tuples over a set with cardinality k.How 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?
https://en.wikipedia.org/wiki/Combinatorial_species