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An Infinite Product

  1. May 28, 2006 #1

    benorin

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    Prof. Putinar,

    Guillera & Sondow gave

    [tex]e^{x}=\prod_{n=1}^{\infty}\left(\prod_{k=1}^{n} (1+kx)^{(-1)^{k+1}\left(\begin{array}{c}n\\k\end{array}\right)} \right) ^{\frac{1}{n}}[/tex]​

    for [tex]x\geq 0[/tex], to which I add

    [tex]\boxed{\frac{e^{u}\Gamma (u)}{\sqrt{2\pi e}}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n} (k+u)^{(-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right) (k+u)} \right) ^{\frac{1}{n+1}}}[/tex]​

    for [tex]\mbox{Re} \geq 0[/tex].

    -Ben Orin
     
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  3. May 28, 2006 #2

    Zurtex

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    Cool.



















    :uhh:
     
  4. May 29, 2006 #3

    benorin

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  5. Jun 27, 2011 #4

    benorin

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    An Infinite Product for e^x

    Edit: Fixing my post for TeX and updating link to paper.

    Prof. Putinar,

    Guillera & Sondow1 gave [itex] e^{x} = \prod_{n=1}^{\infty}\left( \prod_{k=1}^{n} (kx+1) ^{(-1)^{k+1} \left( \begin{array}{c}n\\k\end{array}\right) } \right) ^{\frac{1}{n}}\mbox{ for }x\geq 0,[/itex]
    to it's company I add [itex]\frac{e^{u}\Gamma (u)}{\sqrt{2\pi e}}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}(k+u)^{
    (-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right)(k+u)}\right)^{\frac{1}{n+1}}\mbox{ for }\mbox{Re} \geq 0.[/itex]



    -Ben Orin

    benorin@umail.ucsb.edu

    1 The infinite product for [itex]e^{x}[/itex] is (60) on pg. 20 of Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent (pdf.)
     
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