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anonymous

  • 4 years ago

a!+b! = a^b

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  1. anonymous
    • 4 years ago
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    I can be rewritten using the gamma function as: \[\Gamma(a+1) + \Gamma(b+1) = a^{b}\]

  2. anonymous
    • 4 years ago
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    Any idea how to solve this?

  3. anonymous
    • 4 years ago
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    As you might know this is my problem which I posted here http://math.stackexchange.com/questions/105923/

  4. anonymous
    • 4 years ago
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    I have a feeling that the proving that only (2,2) and (2,3) are the only solutions is somewhat difficult.

  5. anonymous
    • 4 years ago
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    Try using the gamma function.

  6. anonymous
    • 4 years ago
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    So you have a solution Aron?

  7. AravindG
    • 4 years ago
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    nic question aron

  8. anonymous
    • 4 years ago
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    No,I don't have a solution but I was just wondering if I could solve it using the gamma function.

  9. anonymous
    • 4 years ago
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    Dig deep into that Wikipedia page.

  10. anonymous
    • 4 years ago
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    \[\Gamma (Z) = \int\limits\limits_{0}^{\infty}e^{-t}t^{z-1}dt\]

  11. anonymous
    • 4 years ago
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    $$\stackrel{\circ.\circ}{\huge o}\angle\huge\lbrace\normalsize\text{I am tired, time for a nap! } $$

  12. anonymous
    • 4 years ago
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    Is there a way to format text in the math formatter so it isn't italicized?

  13. anonymous
    • 4 years ago
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    \text{}

  14. anonymous
    • 4 years ago
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    The graph looks weird!

  15. anonymous
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=Solve%20a!%20%2B%20b!%20%3D%20a%5Eb&t=crmtb01

  16. anonymous
    • 4 years ago
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    It looks like an alien spaceship

  17. anonymous
    • 4 years ago
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    haha :D

  18. anonymous
    • 4 years ago
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    yeah

  19. anonymous
    • 4 years ago
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    Anyone able to solve it yet?

  20. precal
    • 4 years ago
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    Cool graph! You guys are so funny!

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