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Math Help - Show fnfinite products are equal

  1. #1
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    TheEmptySet's Avatar
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    Show fnfinite products are equal

    I have been asked to show that for |z|<1 that

    \displaystyle \left[ \prod_{j=0}^{\infty}(1-z^{2j+1})\right]^{-1}=\prod_{j=0}^{\infty}(1+z^{j+1})

    I know since \sum_{j=0}^{\infty}|z^{j+1}| and \sum_{j=0}^{\infty}|z^{2j+1}|
    Converge uniformly on compact subsets of the unit disk that both of the infinite products converge to a holomorphic function of the open disk.

    So here is my idea let

    f= \prod_{j=0}^{\infty}(1-z^{2j+1}) and

    g=\prod_{j=0}^{\infty}(1+z^{j+1})

    I would be sufficient to show that fg=1

    My idea was to take the deriative and show that it is equal to 0

    let h=fg \implies \ln(h)=\ln(fg)

    This gives

    \displaystyle \frac{h'(z)}{h(z)}=\frac{d}{dz} \left( \sum_{j=0}^{\infty}\ln(1+z^{j+1})+\ln(1-z^{2j+1})\right)=\sum_{j=0}^{\infty}\frac{(j+1)z^j  }{1+z^{j+1}}-\frac{(2j+1)z^{2j}}{1-z^{2j+1}}

    I cannot sum this series or show that it is equal to 0. Also if you see a different approach please let me know.

    Thanks
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I have been asked to show that for |z|<1 that

    \displaystyle \left[ \prod_{j=0}^{\infty}(1-z^{2j+1})\right]^{-1}=\prod_{j=0}^{\infty}(1+z^{j+1})

    I know since \sum_{j=0}^{\infty}|z^{j+1}| and \sum_{j=0}^{\infty}|z^{2j+1}|
    Converge uniformly on compact subsets of the unit disk that both of the infinite products converge to a holomorphic function of the open disk.

    So here is my idea let

    f= \prod_{j=0}^{\infty}(1-z^{2j+1}) and

    g=\prod_{j=0}^{\infty}(1+z^{j+1})

    It would be sufficient to show that fg=1.
    With that notation f(z)= \prod_{j=0}^{\infty}(1-z^{2j+1}) and g(z)=\prod_{j=0}^{\infty}(1+z^{j+1}), both infinite products converge uniformly to analytic (therefore continuous) functions on the open unit disk. Also, any rearrangement of the product converges to the same limit. It follows that

    \begin{aligned}f(z)g(z) &=  \prod_{j=0}^{\infty}(1-z^{2j+1})\prod_{j=0}^{\infty}(1+z^{2j+1})\prod_{j=  0}^{\infty}(1+z^{2j+2}) \\ &= \prod_{j=0}^{\infty}(1-z^{4j+2})\prod_{j=0}^{\infty}(1+z^{2j+2}) = f(z^2)g(z^2).\end{aligned}

    By induction, f(z)g(z) = f(z^{2^n})g(z^{2^n}) for all n. Then by continuity, as n\to\infty we get f(z)g(z) = f(0)g(0) = 1.
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