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Math Help - Infinite Product

  1. #1
    Junior Member
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    Infinite Product

    i am having some difficult to expand e^z - 1 in an infinite product. I was using the identity e^z-1 = 2ie^(z/2)sin(z/2i) but i couldn't really figure it out.

    also, similar question im trying to prove an identity
    1/(e^z-1) = 1/z - 1/2 + 2z
    (n=1 to infinite) 1/(z^2+4π^2n^2) by taking logarithmic derivatives in e^z - 1

    I do not know what to do after... plz help me
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  2. #2
    MHF Contributor chisigma's Avatar
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    Your approach is quite correct and You can obtain the desired result applying the following theorem...

    Weierstrass factorization theorem - Wikipedia, the free encyclopedia

    Starting from the identity...

    e^{z}-1 = 2i\cdot e^{\frac{z}{2}}\cdot \sin \frac{z}{2i} (1)

    ... and remembering that...

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

    ... setting in (2) p=\frac{z}{2i} and inserting it in (1) You obtain...

    e^{z}-1 = e^{\frac{z}{2}}\cdot z\cdot \prod_{n=1}^{\infty} \{1+\frac{z^{2}}{(2n\pi)^{2}}\} (3)

    Using (3) You should have no difficulties to resolve the second part of the problem...

    Kind regards

    \chi \sigma
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  3. #3
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    thank you so much
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