anonymous
  • anonymous
why does is the integral form of n!(factorial)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
wups, what is*
anonymous
  • anonymous
The factorial function may be represented by a special function called, the gamma Function. You can't perform calculus on the factorial.
anonymous
  • anonymous
Have you heard of that?

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anonymous
  • anonymous
Yes, 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
  • anonymous
0! 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,\[^nP_r=n \times (n-1) \times (n-2) \times ...\times (n-r+1)\]If we now multiply this expression by 1, but in the form,\[1=\frac{(n-r)!}{(n-r)!}\]we get\[^nP_r=\frac{n \times (n-1) \times (n-2) \times ...\times (n-r+1)(n-r)!}{(n-r)!}\]\[=\frac{n!}{(n-r)!}\]Now, if we have n objects and we want to arrange things in order n ways, we'd have\[^nP_n=n!=\frac{n!}{(n-n)!}=\frac{n!}{0!}\]For this to be true, we define\[0!=1\]
anonymous
  • anonymous
Ok, 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.
anonymous
  • anonymous
No probs.

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