# Derivative of x

• climbhi

#### climbhi

Is there a way to find the derivative, or antiderivative for that matter, of x!. Or is there a special function for that or something?

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. in fact, all of calculus rests on certain axioms that only the real numbers satisfy.

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.

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.

In that case you can take the discrete-space derivative.

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 occurring at every integral step.

eNtRopY

There is a continuous extension to x! called the gamma function. Using G for gamma, it has the property that for integers, G(x+1)=x!.

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

&Gamma(n) = &int(0, &infin) xn-1 e-x dx

Hurkyl

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Just wondering my calculator will give a numerical value for things such as 2.5! and 3.5! For the ! command to work on my calculator you have to use either 1,2,3... or any 1/2 value of those and strangely also -1/2. How is it getting these numbers? Is it perhaps just picking the values off the Gamma function?

Originally posted by climbhi
Is it perhaps just picking the values off the Gamma function?

yup. check 1/2! it should be &radic;&pi;

Originally posted by lethe
yup. check 1/2! it should be &radic;&pi;

Well actually according to my calculator it is (&radic;&pi;)/2, is this actually what it should be?

Originally posted by climbhi
Well actually according to my calculator it is (&radic;&pi;)/2, is this actually what it should be?

umm... yeah, i guess so. oops. i guess i meant to say that &Gamma;(1/2) = &radic;&pi; which implies that (1/2)! = &Gamma;(3/2) = (1/2)&Gamma;(1/2) = 1/2&radic;&pi;. OK

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

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So I wonder why my calculator will only give values for half integer factorals and only for one negative number (-.5) would it really have been that hard for them to program in the rest of the gammma function if they were going to include all the half integers? Is there any real common use for half integer factorals that would incline them to include this feature?

Oh also, Loren, just thinking that for a few of those infinite sums and products you were trying to work out, many of them involved factorals. Could you perhaps use this Gamma function in the place and see if you could get anything to fall out?

Originally posted by lethe
&Gamma;(3/2) = (1/2)&Gamma;(1/2)

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?

Try integration by parts.

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

&Gamma;(x+1) = x!

&Gamma;(x) = &int;t=0..infinitytx-1e-t dt = (x-1)!

so

x! = &int;t=0..infinitytxe-t dt

so

d/dx(x!) = &int;t=0..infinityln(t)*txe-t dt

good luck evaluating that.