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anonymous
 5 years ago
why does is the integral form of n!(factorial)?
anonymous
 5 years ago
why does is the integral form of n!(factorial)?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The factorial function may be represented by a special function called, the gamma Function. You can't perform calculus on the factorial.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you heard of that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, but we are working with sequences/series, and are comparing the factorial with other integrals but he said that 0! = 1 and said to just believe him. Then he continued to write an integral form for n! which i believe is the gamma function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00! is 1 because we *define* it to be the case. It comes from permutations in combinatorics. If you have n objects and you're taking them r at a time, you have the total number of ways for doing this as,\[^nP_r=n \times (n1) \times (n2) \times ...\times (nr+1)\]If we now multiply this expression by 1, but in the form,\[1=\frac{(nr)!}{(nr)!}\]we get\[^nP_r=\frac{n \times (n1) \times (n2) \times ...\times (nr+1)(nr)!}{(nr)!}\]\[=\frac{n!}{(nr)!}\]Now, if we have n objects and we want to arrange things in order n ways, we'd have\[^nP_n=n!=\frac{n!}{(nn)!}=\frac{n!}{0!}\]For this to be true, we define\[0!=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so 1=0! rather than 0!=1. Think im gonna have to take some high level courses till i get to this but thanks, was interesting. You are very insightful.
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