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Derivative of x!

  1. Apr 18, 2003 #1
    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?
     
  2. jcsd
  3. Apr 19, 2003 #2
    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.
     
  4. Apr 19, 2003 #3
    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 occuring at every integral step.

    eNtRopY
     
  5. Apr 19, 2003 #4

    mathman

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    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.)
     
  6. Apr 19, 2003 #5

    Hurkyl

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

    Hurkyl

     
    Last edited: Apr 19, 2003
  7. Apr 19, 2003 #6
    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?
     
  8. Apr 20, 2003 #7
    yup. check 1/2! it should be √π
     
  9. Apr 20, 2003 #8
    Well actually according to my calculator it is (√π)/2, is this actually what it should be?
     
  10. Apr 20, 2003 #9
    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. OK


    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!
     
    Last edited: Nov 20, 2003
  11. Apr 20, 2003 #10
    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?
     
  12. Apr 20, 2003 #11
    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?
     
  13. Apr 20, 2003 #12
    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?
     
  14. Apr 20, 2003 #13

    Hurkyl

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    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
     
  15. Apr 20, 2003 #14
    Γ(x+1) = x!

    Γ(x) = ∫t=0..infinitytx-1e-t dt = (x-1)!

    so

    x! = ∫t=0..infinitytxe-t dt

    so

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

    good luck evaluating that.
     
  16. Apr 20, 2003 #15

    Hurkyl

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