negative factorials

Apr 3, 2012 at 10:57pm
I was just reading that negative factorials are never defined, although they *can* be defined for all positive numbers based on the rationale that:


for all x's and y's such that x! = y
(x+1)!/y = x + 1


so for x = 2, y = 2, 2+1 = 3, 3! = 6, 6/2 = 3 = x+1

but this would not hold true for negative numbers:

for x = -1,,
x+1 = -1+1 = 0 / y = 0.

but i was thinking, that formula is silly, shouldn't it be a piecewise defined function so for negative domain the formula would be (x-1)!/y = x-1.

Apr 3, 2012 at 11:58pm
Take a look at
http://en.wikipedia.org/wiki/Gamma_function
The gamma function is a generalization of factorials, valid not only for integers.
Since gamma function has singularities for all negative integers, something similar can be said about factorials.
Last edited on Apr 3, 2012 at 11:58pm
Apr 4, 2012 at 12:11am
The factorial operator is just a special case of the gamma function.
In terms of this function, your formula
(x+1)/y = x+1

would be
G(z+1) = zG(z)


The complete definition of the gamma function is a complex integral over the half line (0, inf), and that can be extended to all complex numbers except the real negative integers.

Fun fact: (-0.5)! = 1.77245... (the square root of PI)
Apr 4, 2012 at 7:50pm
wow that really answers just about everything but my question. It says in the second sentence of the wikipedia article of Gamma functions that they are still ONLY defined for positive real numbers. My question was regarding a possible definition for negative numbers. I am not going to read the rest of the article.

@atropos ditto and interesting.
Last edited on Apr 4, 2012 at 8:02pm
Apr 4, 2012 at 10:10pm
The gamma function IS defined for all real numbers except the negative integers.

If you integrate by parts, you wind up with

G(x+n) = (x+n - 1)(x+n -2)(x+n -3).....(x+1)(x) G(x)

and that formula is valid for ALL x =/= {0, -1, -2, -3,....}.

EDIT: I just checked the wiki page; the second sentence you were referring to was talking about convergence of the integral.
Last edited on Apr 4, 2012 at 10:22pm
Topic archived. No new replies allowed.