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SolomonZelman
 one year ago
trying to catch something about the gamma function.
SolomonZelman
 one year ago
trying to catch something about the gamma function.

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle \Gamma(x);~~{\rm where}~x\in {\bf Z},~x>0}\) this does not have a value, but rather an asymptote. (based on the graph and wolf) Why is that?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0well, x≥0, really...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0is 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, recall that \(\Gamma\) is analytically continued from the factorial: $$\Gamma(n+1)=n!$$and satisfies a similar function equation: $$\Gamma(t+1)=t\Gamma(t)$$

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0yes, i know these two facts....

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2http://www.iiste.org/Journals/index.php/MTM/article/viewFile/6988/7036 I think this article has some good points.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Page 3, equation (6).

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I 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?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Chill Out, I wll read. tnx

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0oh it hits that idea, you end up dividing by 0! for negative x values, and this is why it diverges....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0i mean by 0, and exclamation mark of mine is not to denote factorial

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0will 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....

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2I'm bad with analysis, just learning with this as much as I can :D

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0it always has references.... the author didn't want to make it easy, no, rather professional:O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, 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 \(\Gamma\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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: $$f(1)=1\\f'(1)=n\\f'(2)=n(n1)\\\dots\\f^{(n)}(1)=n(n1)(n2)\cdots 1\\f^{(n+1)}(1)=0\\f^{(n+2)}(1)=0\\\dots$$ so consider \(f(t)=et^n\):$$\int e^{t}\cdot et^n\, dt=1+n+n(n1)+\dots+n(n1)(n2)\cdots1$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, 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 $$\int_0^\infty e^{t} t^n\, dt=n!$$
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