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kirbykirby

  • 3 years ago

Verifying the property of the Gamma function: Gamma(n) = (n-1)!

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  1. kirbykirby
    • 3 years ago
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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

  2. anonymous
    • 3 years ago
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    really? you have done all the hard work!

  3. kirbykirby
    • 3 years ago
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    I dunno why I can't see this.. my brain is fried probably from doing the first part LOL

  4. anonymous
    • 3 years ago
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    yeah i guess so

  5. anonymous
    • 3 years ago
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    \(n!=n(n-1)!\)

  6. anonymous
    • 3 years ago
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    that is \(n!\) can be defined as the recursion \(1!=1\) and \(n!=n(n-1)!\)

  7. kirbykirby
    • 3 years ago
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    OH i see right!! Oh thank you. Haha oh my that was fairly simple :P

  8. kirbykirby
    • 3 years ago
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    Thanks :)!!

  9. anonymous
    • 3 years ago
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    yeah way way more simple than showing \(\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha -1)\)

  10. kirbykirby
    • 3 years ago
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    For sure!!!

  11. anonymous
    • 3 years ago
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    yw

  12. anonymous
    • 3 years ago
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    like opening the jar after someone already loosened it

  13. kirbykirby
    • 3 years ago
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    @satellite73 I couldn't help but wonder how do you keep your equations on the same line?

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