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2006-06-18 06:23:23 · 3 answers · asked by cj k 4 in Science & Mathematics Mathematics

3 answers

You'd have to define x! for arbitrary x. Normally, it is only defined for x a non-negative integer.

Luckily, there is a standard function, the gamma function, such that gamma(n) = (n-1)! such that:

gamma(x)=(x-1)*gamma(x-1)

so gamma(x+1) is taken to be the usual extension of the factorial function to the real line.

The derivative of gamma is fairly complicated - see the source below.

2006-06-18 06:32:13 · answer #1 · answered by thomasoa 5 · 2 0

The "factorial" is only defined for whole numbers, not for all real numbers. ( What does (0.5)! mean? ) The factorial, as it is usually defined, does not have a derivative...

HOWEVER! There is a nice function ( The Gamma function ) which is like a generalization of the factorial. Gamma is differentiable.

For more details see http://en.wikipedia.org/wiki/Gamma_function

2006-06-18 06:36:11 · answer #2 · answered by AnyMouse 3 · 0 0

x! normally only works for integer numbers, a discrete, and hence non-continuous series, and those things doesn't have derivatives.

You must first define a continuous function f such that f(x) = x! for all integer x, and then (f(x)/f(x-1) = x) for all other numbers. That can be tricky!

2006-06-18 09:43:11 · answer #3 · answered by User1 2 · 0 0

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