How to perform factorial operation

1. Sep 29, 2003

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. Sep 29, 2003

StephenPrivitera

I believe that factorials aren't defined for nonintegers. Perhaps the intent is 5!/6!, which would be 1/6.

3. Sep 29, 2003

jcsd

Factorials only work for natural numbers, so 5/6 does not have a factorial.

4. Sep 29, 2003

Njorl

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

5. Sep 29, 2003

Staff Emeritus
The Gamma function is defined by &Gamma;(z) = &int;0oo tz-1dt

It has the property that n! = &Gamma;(n+1).

This isn't going to be a lot of help in solving the equation though.

6. Sep 29, 2003

jcsd

You can only take the factorial of a postive integer, thus &Gamma;(n+1) = n! is only true when n is a postive integer.

7. Sep 29, 2003

jcsd

Actually just done some research, using the gamma function you can find values for half-integrals.

8. Sep 29, 2003

ahrkron

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

&Gamma;(z) = &int;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)! = &Gamma;(5/6 + 1) = &int;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 &Gamma;(1+5/6) = &Gamma;(1.833) is slightly less than one.

9. Sep 29, 2003

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

In the half plane of complex numbers with real part > 1, it can be defined by &Gamma;(z-1) = &pi;/(&Gamma;(z)sin(&pi;z)) = &pi;z/(&Gamma;(1+z)sin(&pi;z)).

10. Sep 29, 2003