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Math Help - Euler's Reflection Formula...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Euler's Reflection Formula...

    What is the first formula in "Proof" and how can I prove it?

    Euler's Reflection Formula - ProofWiki


    Thank you...
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    MHF Contributor chisigma's Avatar
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    Re: Euler's Reflection Formula...

    Quote Originally Posted by Also sprach Zarathustra View Post
    What is the first formula in "Proof" and how can I prove it?...
    If as 'first formula' You mean that...

    \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n} (1)

    ... then that's a perfect rubbish and I suggest You don't waste time trying to 'demonstrate' it...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Euler's Reflection Formula...

    Quote Originally Posted by chisigma View Post
    If as 'first formula' You mean that...

    \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n} (1)

    ... then that's a perfect rubbish and I suggest You don't waste time trying to 'demonstrate' it...

    Kind regards

    \chi \sigma


    I came up with idea, tell me if that is wrong....

    We need to show that:

    \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n}

    If we know that \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2}), then we can re-write it in the form:


     \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2})=\pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})(1+\frac{z}{n})


    And the question now is:

    Is it true that,


     \prod_{n \ne 0} (1+\frac{z}{n})= \prod_{n \ne 0} e^{\frac{z}{n}}
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    Re: Euler's Reflection Formula...

    Quote Originally Posted by Also sprach Zarathustra View Post
    What is the first formula in "Proof" and how can I prove it?

    Euler's Reflection Formula - ProofWiki


    Thank you...
    Weierstrass factorization theorem - Wikipedia, the free encyclopedia
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: Euler's Reflection Formula...

    Quote Originally Posted by Also sprach Zarathustra View Post
    I came up with idea, tell me if that is wrong....

    We need to show that:

    \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n}

    If we know that \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2}), then we can re-write it in the form:


     \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2})=\pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})(1+\frac{z}{n})


    And the question now is:

    Is it true that,


     \prod_{n \ne 0} (1+\frac{z}{n})= \prod_{n \ne 0} e^{\frac{z}{n}}
    I said the the formula was 'a perfect rubbish' because it is written in ambigous form... the usual and fully clear formula is...

    \sin \pi\ z = \pi\ z\ \prod_{n=1}^{\infty} (1-\frac{z^{2}}{n^{2}}) (1)

    The 'demonstration' of (1) is done using the Weierstrass factorisation theorem...

    Kind regards

    \chi \sigma
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