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

jcsd

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

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
 

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.
 

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.
 

jcsd

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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 Γ(z) = ∫0oo tz-1dt
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.
 

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)).
 
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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