SOLUTION: gamma of n where n>0. Why n is positive?

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Question 1062420: gamma of n where n>0.
Why n is positive?

Found 2 solutions by ikleyn, rothauserc:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
This post is nonsense.


Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
I am taking n to be an element of the natural numbers, then
:
G is the gamma function which is defined for all numbers except negatives
:
There are four properties(P) of G
:
(P1) G[1] = 1
(P2) G[n+1] = n * G[n]
(P3) G[n] = (n-1)!
(P4) G[1/2] = square root(Pi)
:
We have by P2 and P3
:
(1/2)! = G[3/2] = (1/2) * G[1/2]= (square root(Pi) / 2)
:
(n+k)! / n! = ((n+k)(n+k-1)(n+k-2)...(n+2)(n+1)(n)(n-1)....(2)(1)) / ((n)(n-1)...(2)(1)) =
:
= (n+k)(n+k-1)(n+k-2)...(n+2)(n+1)
:
let n = 0, then
:
k! / 0! = (k)(k-1)(k-2)...(2)(1) = k! makes sense only if 0! = 1
: