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climbhi

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climbhi

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however, there is a function defined on the reals that agrees with the factorial at integer values. it is called the gamma function. it is integrable and derivable.

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eNtRopY

In that case you can take the discrete-space derivative.Originally posted by lethe

well, the factorial is only defined on the natural numbers, and there is no sensible way to take a derivative of any function on the naturals.

d(x!)/dx = [(x+1)! - x!]

If you want to get formal, you can consider the domain of x values the set of all real numbers superimposed with a series of delta functions occuring at every integral step.

eNtRopY

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mathman

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(In order to display the formula for G, I need integral sign, exponent, and integral limits.)

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Hurkyl

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&Gamma(n) = &int

Hurkyl

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climbhi

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yup. check 1/2! it should be √πOriginally posted by climbhi

Is it perhaps just picking the values off the Gamma function?

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climbhi

Well actually according to my calculator it is (√π)/2, is this actually what it should be?Originally posted by lethe

yup. check 1/2! it should be √π

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umm... yeah, i guess so. oops. i guess i meant to say that Γ(1/2) = √π which implies that (1/2)! = Γ(3/2) = (1/2)Γ(1/2) = 1/2√π. OKOriginally posted by climbhi

Well actually according to my calculator it is (√π)/2, is this actually what it should be?

i m always making that mistake. all the time, i forget and say that Γ(n) = n!, when actually, it should be that Γ(n+1) = n!

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climbhi

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climbhi

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climbhi

I don't follow this step. By Hurkyl's given definition of what the Gamma function is I don't see how you can take the 1/2 outside. Am I missing something, how does that work?Originally posted by lethe

Γ(3/2) = (1/2)Γ(1/2)

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Hurkyl

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The reason your calculator only gives results for positive integers and any half integer is because those are the only values for which an exact solution is known. Well, technically the exact solution is known for nonpositive integers; +/-&infin.

The values for the gamma function can be found from the recursive relationship:

&Gamma(1) = 1

&Gamma(1/2) = sqrt(&pi)

n &Gamma(n) = &Gamma(n+1)

Hurkyl

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Γ(x) = ∫

so

x! = ∫

so

d/dx(x!) = ∫

good luck evaluating that.

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Hurkyl

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&Gamma'(x) = &Gamma(x) &Psi(x)

Where &Psi is the digamma function!

http://mathworld.wolfram.com/DigammaFunction.html

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