Question

Hoping someone can make sense of this:
(negative decimal factorial)

Answer 1

http://www.wolframalpha.com/input/?i=%28-0.999%29%21

Answer 2

question: why is it 999?

Answer 3

well you aren't allowed to take the factorial of a decimal, so wolfram has used the gamma function which is ok for decimals,
\[\Gamma(z+1)=z!\]

Answer 4

so wolfram has done this

\[(-0.999)!=\Gamma (-0.999+1)=\Gamma(0.001)\]

Answer 5

and the gamma function is defined differently, so that Γ(0.001) can be computed?

Answer 6

now a property of the Gamma function is that
\[\Gamma(z)=\frac{\Gamma(z+1)}z\]

so we get \[\Gamma(0.001)=\frac{\Gamma(1.001)}{0.001}=1000\times\Gamma(1.001)\]

Answer 7

thats not the only property of the gamma function though?

Answer 8

now we can use a table of graph get a numeric approximation

Answer 9

\[1.000<1.001<1.100\]
\[\Gamma(1.000)<\Gamma(1.001)<\Gamma(1.100)\]
\[1<\Gamma(1.001)<0.9514\]

Answer 10

and 1.001 is much closer to 1.000 , than 1.100

so we can approximate \[\Gamma(1.001)\approx0.99\].


so

\[1000×Γ(1.001)\approx1000\times0.99=990\] which is about right

Answer 11

Technically I think wolfram is wrong, because the factorial can only be taken of a non negative integer

Answer 12

\[\large{\ddot\smile}\]

Answer 13

that helps alot, thank you :)

Answer 14

one crazy thing about the gamma functions is that\[\Gamma(\tfrac12)=\sqrt\pi\]

Answer 15

does that mean that (1/2 - 1)! = root(pi)?

Answer 16

on wiki it says that gamma(x)=(x-1)!

Answer 17

wolfram would agree that (1/2 - 1)! = √π
but i would not because i would say the factorial can only be taken of nonnegative integers

Answer 18

lol, this is weird stuff

Answer 19

\[Γ(z+1)=z!\]
let \(z+1=x\)
\(\qquad \qquad z=x-1\)

\[Γ(x)=(x-1)!\]

Answer 20

weird and true,

i think the gamma function is related to growth

Answer 21

great information, thanks again