Question

trying to catch something about the gamma function.

Answer 1

\(\large\color{black}{ \displaystyle \Gamma(-x);~~{\rm where}~x\in {\bf Z},~x>0}\)
this does not have a value, but rather an asymptote. (based on the graph and wolf)

Why is that?

Answer 2

well, x≥0, really...

Answer 3

is there anything i can try to dig, like to write a power series representation for it and see what happens at particular values of x, or what should i try?

Answer 4

well, recall that \(\Gamma\) is analytically continued from the factorial: $$\Gamma(n+1)=n!$$and satisfies a similar function equation: $$\Gamma(t+1)=t\Gamma(t)$$

Answer 5

yes, i know these two facts....

Answer 6

http://www.iiste.org/Journals/index.php/MTM/article/viewFile/6988/7036 I think this article has some good points.

Answer 7

Page 3, equation (6).

Answer 8

I have a little thought.....

Factorial, working forward
1!=1
2!=1•2=2
3!=1•2•3=6
4!=1•2•3•4=24

Now working backwards
3!=(1•2•3•4)÷4=24÷4=6
2!=(1•2•3)÷3=6÷3=2
1!=(1•2)÷2=2÷2=1

going back further
0!=1÷1=1
(-1)! (really a gamma of -2, but....) = 1÷0 ... bingo?

Answer 9

Chill Out, I wll read. tnx

Answer 10

oh it hits that idea, you end up dividing by 0! for negative x values, and this is why it diverges....

Answer 11

i mean by 0, and exclamation mark of mine is not to denote factorial

Answer 12

will see how much i can understand from that..... tnx for googling for me.....

If any sugestions, everyone is welcome to say anything.... (even to study more before posting this question), of course....

Answer 13

I'm bad with analysis, just learning with this as much as I can :D

Answer 14

it always has references.... the author didn't want to make it easy, no, rather professional:O

Answer 15

lol, that identity is all you need:
$$\Gamma(t+1)=t\Gamma(t)\implies \Gamma(t)=\frac1t\Gamma(t+1)\\\implies \Gamma(t)=\frac1{t(t+1)(t+2)\dots (t+k)}\Gamma(t+k+1)$$
so if are interested in extending \(\Gamma\) to negative numbers \(t\), we just need to pick a \(k\) big enough so that \(t+k+1\) is in the domain of the already defined \(\Gamma\)

Answer 16

now you'll notice that if we try to do this for negative integers, we'll have trouble -- the only way to make \(t+k+1>0\) for negative integer \(t\), positive integer \(k\) will necessarily mean that one of those factors in the denominator *will* be zero, so our definition will blow up and does not apply to negative integers

Answer 17

and the integral definition of the gamma function just follows from observations of integration by parts with sufficiently 'small' growth: $$\begin{align*}\int_1^\infty e^{-t} f(t)\, dt&=-e^{-t}f(t)\big|_1^\infty-\int_1^\infty (-e^{-t}) f'(t)\,dt\\&=\frac1e f(1)+\int_1^\infty e^{-t} f'(t)\,dt\\&=\frac1e\left(f(1)+f'(1)+f''(1)+\dots \right)\end{align*}$$now imagine if we took \(f(t)=t^n\) we know the derivatives look like: $$f(1)=1\\f'(1)=n\\f'(2)=n(n-1)\\\dots\\f^{(n)}(1)=n(n-1)(n-2)\cdots 1\\f^{(n+1)}(1)=0\\f^{(n+2)}(1)=0\\\dots$$ so consider \(f(t)=et^n\):$$\int e^{-t}\cdot et^n\, dt=1+n+n(n-1)+\dots+n(n-1)(n-2)\cdots1$$

Answer 18

oops, that last integral should be on \((1,\infty)\). anyways, now that is more or less like the Laplace transform: $$\int_0^\infty e^{-st} t^n\, dt =\frac{n!}{s^{n+1}}$$let \(s=1\) so $$\int_0^\infty e^{-t} t^n\, dt=n!$$