Hmm, but for the form that was given (euler integral form) it's only defined for x>0.
But, when you add the relation:
gf(x)=gf(x+1)/x

Then it's defined for negative values also (except integer negatives).

Another way to see that 0!=1 is:
(n+k)! / n! = (n+k)(n+k-1)(n+k-2)...(n+2)(n+1)

Set n = 0.

k! / 0! = (k)(k-1)(k-2)...(2)(1) = k!

That is true only when 0! = 1.


/Mankind gave birth to God.


[edited by - silvren on March 23, 2003 7:37:19 AM]

but for the form that was given (euler integral form) it''s only defined for x>0

What makes you say that?

And just to be more pedantic, the gamma function is defined everywhere, however it''s not analytic on the negative integers.

quote:

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What makes you say that?

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I mean for positive x values the gamma is defined like this:
gf(x) = int[0,inf] (t^(x-1)*e^-t dt) , x>0

BUT for negative values it's defined recursively like this:
gf(x)=gf(x+1)/x x<0

And the gammafunction that somebody posted was for positive values not negative.


/Mankind gave birth to God.

[edited by - silvren on March 24, 2003 1:26:05 PM]

quote:

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Original post by silvren
I mean for positive x values the gamma is defined like this:
gf(x) = int[0,inf] (t^(x-1)*e^-t dt) , x>0

BUT for negative values it''s defined recursively like this:
gf(x)=gf(x+1)/x x<0

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Sorry, you''re right. I didn''t think about the integral not being defined at t = 0.

In any case, it''s definately defined for 0! = Γ(1) = 1.

ok. little off topic anyway.

/Mankind gave birth to God.