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- Jun 5th 2008, 11:28 AM #16

- Jun 5th 2008, 11:35 AM #17
Just define the given formula for all $\displaystyle n\geq 1$ and it works. The problem now is that this formula can show that 0!=1! without proving that $\displaystyle 1!=1$. Otherwise this formula would explain all. So using the fact that $\displaystyle 1!=1$ (since $\displaystyle n!=n(n-1)...1$) this works.

- Jun 5th 2008, 11:44 AM #18

- Jun 5th 2008, 12:27 PM #19

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- Jun 5th 2008, 12:39 PM #20

- Jun 5th 2008, 12:49 PM #21

- Jun 5th 2008, 01:07 PM #22

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Just look at how the Gamma function is defined $\displaystyle \Gamma (x) = \int_0^{\infty} e^{-t} t^{x-1} dt$. It can be proven that this converges when $\displaystyle x>0$. At $\displaystyle x=0$ the function becomes unbound as $\displaystyle x\to 0^+$. What about negative values? We cannot use the integral anymore. But we have a way around that. We use the property that $\displaystyle \Gamma (x+1) = x\Gamma (x)$. And so we can extend this to negative values also, for example,

*we define*$\displaystyle (-1/2)\Gamma (-1/2) = \Gamma (1/2)$. And so we can find $\displaystyle \Gamma (-1/2)$. The reason why we do that is to extend this property. The only problem is that at $\displaystyle x=0$ we have a problem and so we cannot redefine it at $\displaystyle x=-1,-2,-3,...$ as well.

(In complex analysis it is possible to extend the Gamma function everywhere on the complex plane except the non-positive integers).

- Jun 7th 2008, 03:37 PM #23

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Just out of curiosity... did the gamma function originate from considerations other than extensions of the factorial? I ask because I always wondered why $\displaystyle \Gamma(n+1)=n!$ rather than $\displaystyle \Gamma(n)=n!$ (the Pi function). It always trips we up because the latter seems more natural.

- Jun 7th 2008, 05:52 PM #24

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That is a good question. I think it comes from a historical mistake (this needs reference). I forgot who made this mistake, either Gauss or Euler. I think it was Euler. He used the Gamma function just defined about and therefore we have $\displaystyle \Gamma (n+1) = n!$ rather than $\displaystyle \Gamma (n) = n!$. There are many times in math were a mistake is kept for historical reasons.

- Jun 17th 2008, 08:17 AM #25
## Very basic

So why 0!=1?

this is an assumption. But now the question arise why this assumption was taken?

Its answer is very basic.

1) We know that number of ways of arranging r different things out n different things is =nPr=n!/(n-r)!

2)From fundamental principal of counting we know that number of ways of arranging n different things is n!

But number of ways of arranging n different things must also be equal to nPn (replacing r by n as all things are included)

therefor nPn=n! Or

n!/(n-n)!=n!

1/0! = 1

which is only possible if it is asrumed that 0!=1. Hence the assumption was taken.

- Jun 26th 2008, 02:52 AM #26

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- Jul 1st 2008, 06:22 AM #27

- Jul 1st 2008, 03:16 PM #28

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- Jul 1st 2008, 03:20 PM #29