SolomonZelman
  • SolomonZelman
trying to catch something about the gamma function.
Mathematics
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \Gamma(-x);~~{\rm where}~x\in {\bf Z},~x>0}\) this does not have a value, but rather an asymptote. (based on the graph and wolf) Why is that?
SolomonZelman
  • SolomonZelman
well, x≥0, really...
SolomonZelman
  • SolomonZelman
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?

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anonymous
  • anonymous
well, recall that \(\Gamma\) is analytically continued from the factorial: $$\Gamma(n+1)=n!$$and satisfies a similar function equation: $$\Gamma(t+1)=t\Gamma(t)$$
SolomonZelman
  • SolomonZelman
yes, i know these two facts....
ChillOut
  • ChillOut
http://www.iiste.org/Journals/index.php/MTM/article/viewFile/6988/7036 I think this article has some good points.
ChillOut
  • ChillOut
Page 3, equation (6).
SolomonZelman
  • SolomonZelman
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?
SolomonZelman
  • SolomonZelman
Chill Out, I wll read. tnx
SolomonZelman
  • SolomonZelman
oh it hits that idea, you end up dividing by 0! for negative x values, and this is why it diverges....
SolomonZelman
  • SolomonZelman
i mean by 0, and exclamation mark of mine is not to denote factorial
SolomonZelman
  • SolomonZelman
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....
ChillOut
  • ChillOut
I'm bad with analysis, just learning with this as much as I can :D
SolomonZelman
  • SolomonZelman
it always has references.... the author didn't want to make it easy, no, rather professional:O
anonymous
  • anonymous
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 \(\Gamma\)
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \(f(t)=et^n\):$$\int e^{-t}\cdot et^n\, dt=1+n+n(n-1)+\dots+n(n-1)(n-2)\cdots1$$
anonymous
  • anonymous
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 $$\int_0^\infty e^{-t} t^n\, dt=n!$$

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