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How to perform factorial operation

  1. Sep 29, 2003 #1
    I don't know if this is the right place for this question, but here it goes.

    Could someone explain how I would go about solving for x of x=(5/6)!

    Thanks
     
  2. jcsd
  3. Sep 29, 2003 #2
    I believe that factorials aren't defined for nonintegers. Perhaps the intent is 5!/6!, which would be 1/6.
     
  4. Sep 29, 2003 #3

    jcsd

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    Factorials only work for natural numbers, so 5/6 does not have a factorial.
     
  5. Sep 29, 2003 #4

    Njorl

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    There is a function that is equivalent to taking factorials that works on real numbers (maybe just positive reals, it's been a while) as well. I believe it is the Gamma function.
    Njorl
     
  6. Sep 29, 2003 #5

    selfAdjoint

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    The Gamma function is defined by Γ(z) = ∫0oo tz-1dt

    It has the property that n! = Γ(n+1).

    This isn't going to be a lot of help in solving the equation though.
     
  7. Sep 29, 2003 #6

    jcsd

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    You can only take the factorial of a postive integer, thus Γ(n+1) = n! is only true when n is a postive integer.
     
  8. Sep 29, 2003 #7

    jcsd

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    Actually just done some research, using the gamma function you can find values for half-integrals.
     
  9. Sep 29, 2003 #8

    ahrkron

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    There is a small, yet important omission here. The definition includes a negative exponential:

    Γ(z) = ∫0oo tz-1e-tdt

    The exponential is important because it makes the integral converge for almost all values of z (the exponential goes to zero much faster than the growth of tz-1).

    In particular, for what you want, you can obtain the value as

    (5/6)! = Γ(5/6 + 1) = ∫0oot(5/6+1)-1e-tdt

    Mathworld has a nice entry for the Gamma function. In the plot, you can see that the value for Γ(1+5/6) = Γ(1.833) is slightly less than one.
     
  10. Sep 29, 2003 #9

    selfAdjoint

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    Woops! Sorry. You are actually right. BTW the z in the definition is a complex variable, and the Γ function is meromorphic (it has poles at 0 and negative integers, but is otherwise analytic). And with that proviso, we have the recurrence relation Γ(z+1) = zΓ(z).

    In the half plane of complex numbers with real part > 1, it can be defined by Γ(z-1) = π/(Γ(z)sin(πz)) = πz/(Γ(1+z)sin(πz)).
     
  11. Sep 29, 2003 #10
    Oops, my bad.

    I've looked at the mathworld site, many time actually.

    It turns out this problem stems back a few years, and I just realised its not factorial I'm actually wondering about (tried it on my TI-89)

    Okay, here is there real problem then. My calculator will give sums for negative numbers and fractions, and I was confused about that.

    This is actually a lot less confusing than i thought. Thanks for the input everyone. I remembered wrong:frown: but hey its been a while.
     
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