How do you simplify #(3/2)! #?

2 Answers
Nov 11, 2015

It is undefined since #3/2 !in ZZ_0^+#

Explanation:

Factorial is only defined for positive integres and 0, ie for #ZZ_0^+##.

Nov 11, 2015

#(3/2)! = 1/2Gamma(1/2) = sqrt(pi)/2#

Explanation:

Factorials were traditionally defined only for Whole Numbers #(0,1, 2, 3, ...)#, also called non-negative integers #ZZ^+#, by the relation,

#n! = n(n-1)(n-2)(n-3)...(3)(2)(1)#

and

#0! = 1#

This did not take into account non-integral and negative numbers.

To the rescue came the Gamma Function #Gamma(z)#, which could handle all numbers in the complex plane except non-positive integers #(-1, -2, -3, ...)#. If you do not know what the complex plane is, don't worry. It is just like a regular 2-dimensional plane of numbers (like a graph sheet), with a few exceptional and beautiful properties. You will learn about it eventually in higher math classes.

The Gamma Function is defined as a definite integral,

#Gamma(z+1) = int_0^oot^ze^-tdt#

How is the Gamma Function related to the Factorial Function? Well,

#z! = Gamma(z+1)#

The advantage of representing the Factorial in terms of the Gamma function was that we could make use of several of the neat little properties of the Gamma function to evaluate non-integral factorials. One such property states that,

#Gamma(z+1) = nGamma(n)#...(eqn. 1)

Another property is,

#Gamma(z)Gamma(1-z) = pi/sin(piz)# (for all #0 < z < 1#) ...(eqn. 2)

If you use #z = 1/2# in Eqn. 2, you simply get,

#Gamma(1/2) = sqrt(pi)#

The number whose factorial we wish to evaluate does not lie between 0 and 1, so we'll make use of Eqn. 1,

#Gamma(3/2) = Gamma(1/2 + 1) = 1/2Gamma(1/2) = 1/2sqrt(pi)#

This is how you evaluate non-integral values under a factorial. Keep in mind that the Gamma Function explodes for negative-integral values of z.