kirbykirby
Verifying the property of the Gamma function:
Gamma(n) = (n-1)!
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kirbykirby
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\[\Gamma (\alpha)=\int\limits_{0}^{\infty}x^{\alpha-1}e^{-x}dx\]
I understand how to get to the relationship \[\Gamma (\alpha)=(\alpha-1)\Gamma (\alpha-1)\] for any alpha >1
But I'm not sure how t get to :\[\Gamma (n)=(n-1)!\] for n is a positive integer
anonymous
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really? you have done all the hard work!
kirbykirby
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I dunno why I can't see this.. my brain is fried probably from doing the first part LOL
anonymous
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yeah i guess so
anonymous
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\(n!=n(n-1)!\)
anonymous
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that is \(n!\) can be defined as the recursion \(1!=1\) and \(n!=n(n-1)!\)
kirbykirby
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OH i see right!! Oh thank you. Haha oh my that was fairly simple :P
kirbykirby
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Thanks :)!!
anonymous
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yeah way way more simple than showing \(\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha -1)\)
kirbykirby
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For sure!!!
anonymous
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yw
anonymous
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like opening the jar after someone already loosened it
kirbykirby
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@satellite73 I couldn't help but wonder how do you keep your equations on the same line?