4 ms·

Almost, but not quite. You could say "n! is the product of the natural numbers <= n." By this definition 0! is an empty product because there are no natural num

Almost, but not quite. You could say "n! is the product of the natural numbers <= n." By this definition 0! is an empty product because there are no natural numbers <= 0. However, this same reasoning would say (-1)! = 1, (-2)! = 1, etc. And while you could define negative factorials that way, it's better in practice to leave negative factorials undefined.

itsjareds 12y ago
Could you then modify the definition to "for all n >= 0, n! is the product of natural numbers <= n"? That sounds kind of like the definition "f(0) = 1; f(n) = (n-1)! * n, n > 0".
matchu 12y ago
That's just because we need to define the set more gracefully: "n! is the product of the first n natural numbers". The negative factorial issue arises from a slightly awkward specification of the set, not from jameshart's core idea.
rbehrends 12y ago
This is really a different issue. You'd be extending the domain of factorial beyond what it's traditionally been used for (same as if you extended it to include other numbers outside the natural). That you can do it for some sets (gamma function) doesn't matter, because it also stands on its own. And if you extend it, it's not the factorial function anymore, which is defined to have the non-negative integers as its domain. But conceptually the primary use case in combinatorics is for permutations, i.e. $n!=|S_n|$, which only makes sense for $n \ge 0$. You could even make an argument for n!= because $|S_0|$ should be 1 on its own (because there's exactly one permutation of the empty set) without saying that it makes formulas easier. (This, I note again, matters regardless of what the gamma function does.)
jameshart 12y ago
That same reasoning would also say that π! is 6, which is not really a useful result, so we are probably best looking for a different definition. There's also some controversy about whether "there are no natural numbers <= 0"...

itsjareds 12y ago

Could you then modify the definition to "for all n >= 0, n! is the product of natural numbers <= n"? That sounds kind of like the definition "f(0) = 1; f(n) = (n-1)! * n, n > 0".

matchu 12y ago

That's just because we need to define the set more gracefully: "n! is the product of the first n natural numbers". The negative factorial issue arises from a slightly awkward specification of the set, not from jameshart's core idea.

rbehrends 12y ago

This is really a different issue. You'd be extending the domain of factorial beyond what it's traditionally been used for (same as if you extended it to include other numbers outside the natural). That you can do it for some sets (gamma function) doesn't matter, because it also stands on its own. And if you extend it, it's not the factorial function anymore, which is defined to have the non-negative integers as its domain. But conceptually the primary use case in combinatorics is for permutations, i.e. $n!=|S_n|$, which only makes sense for $n \ge 0$. You could even make an argument for n!= because $|S_0|$ should be 1 on its own (because there's exactly one permutation of the empty set) without saying that it makes formulas easier. (This, I note again, matters regardless of what the gamma function does.)

jameshart 12y ago

That same reasoning would also say that π! is 6, which is not really a useful result, so we are probably best looking for a different definition. There's also some controversy about whether "there are no natural numbers <= 0"...