How to perform factorial operation

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

StephenPrivitera

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

jcsd

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

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

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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.

jcsd

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

jcsd

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

ahrkron

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Originally posted by selfAdjoint
The Gamma function is defined by &Gamma;(z) = &int;0oo tz-1dt
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.

Staff Emeritus
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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)).

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 but hey its been a while.

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