Just define the given formula for all and it works. The problem now is that this formula can show that 0!=1! without proving that . Otherwise this formula would explain all. So using the fact that (since ) this works.

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- Jun 5th 2008, 12:28 PM #16

- Jun 5th 2008, 12:35 PM #17
Just define the given formula for all and it works. The problem now is that this formula can show that 0!=1! without proving that . Otherwise this formula would explain all. So using the fact that (since ) this works.

- Jun 5th 2008, 12:44 PM #18

- Jun 5th 2008, 01:27 PM #19

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- Jun 5th 2008, 01:39 PM #20

- Jun 5th 2008, 01:49 PM #21

- Jun 5th 2008, 02:07 PM #22

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Just look at how the Gamma function is defined . It can be proven that this converges when . At the function becomes unbound as . What about negative values? We cannot use the integral anymore. But we have a way around that. We use the property that . And so we can extend this to negative values also, for example,

*we define*. And so we can find . The reason why we do that is to extend this property. The only problem is that at we have a problem and so we cannot redefine it at 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, 04:37 PM #23

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- Jun 7th 2008, 06:52 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 rather than . There are many times in math were a mistake is kept for historical reasons.

- Jun 17th 2008, 09:17 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, 03:52 AM #26

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- Jul 1st 2008, 07:22 AM #27

- Jul 1st 2008, 04:16 PM #28

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- Jul 1st 2008, 04:20 PM #29