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Math Help - Show that complex product does not converge

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation Show that complex product does not converge

    Dear MHF members,

    I have the following product of complex numbers
    \prod_{j=1}^{\infty}\bigg[\frac{1}{2^{j}}+\mathrm{i}\bigg(1-\frac{1}{2^{j}}\bigg)\bigg]
    and Mathematica shows me that the product accumulates to 4 different points.
    Each is located in different quadrant and they are outside of the circle with a radius of half.
    However, I cannot compute them explicitly.
    Please show me some directions.

    Thank you.
    bkarpuz
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    Re: Show that complex product does not converge

    Hey bkarpuz.

    With regards to your original thread title (i.e. convergence), you can show that if the sum of the logs converges then the original product converges.

    See if you can prove convergence first because if you can't then you won't have a unique answer.
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    Senior Member bkarpuz's Avatar
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    Arrow Re: Show that complex product does not converge

    Quote Originally Posted by chiro View Post
    Hey bkarpuz.

    With regards to your original thread title (i.e. convergence), you can show that if the sum of the logs converges then the original product converges.

    See if you can prove convergence first because if you can't then you won't have a unique answer.
    Thank you very much chiro.
    I can simply prove that non-convergence (actually accumulation) as follows.

    Proof. As we know for any complex number z=x+\mathrm{i}y,
    |x|+|y|\geq|z|\geq\max\{|x|,|y|\}.
    Let x_{n}:=\frac{1}{2^{n}} and z_{n}:=x_{n}+\mathrm{i}(1-x_{n}),
    then we have |z_{j}|\geq|1-x_{j}|, which yields
    \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|=\prod_{i=1}^  {\infty}|z_{i}|\geq\prod_{i=1}^{\infty}|1-x_{i}|\approx\ell:=0.2887880951.
    This shows that product cannot tend to 0 (i.e., 0+\mathrm{i}0).
    On the other hand, we have
    \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|\leq\prod_{i=  1}^{\infty}(|x_{i}|+|1-x_{i}|)=1
    showing that the product can not also diverge (in the sense of complex modulus).
    Then, the complex sequence w_{n}:=\prod_{i=1}^{n}z_{i} lies in the annulus \mathcal{A}:\ \ell\leq|z|\leq1.
    Due to Bolzano-Weierstrass theorem (considered separetely for the real and imaginary parts),
    there exists a subsequence w_{n_{k}} of w_{n} which converges to some w\in\mathcal{A}.
    Recall that \lim_{n\to\infty}z_{n}=\mathrm{i} and w_{n_{k}+p}=w_{n_{k}}\prod_{i=n_{k}+1}^{p}z_{i},
    which shows that \lim_{k\to\infty}w_{n_{k}+p}=\mathrm{i}^{p}w for p=0,1,\cdots.
    Therefore, the product accumulates to at least (also at most) four different points \pm w,\pm\mathrm{i}w since w\neq0. \rule{0.2cm}{0.2cm}

    But I just wanted to know if it is possible to find (by hand) what w is.
    Last edited by bkarpuz; June 4th 2013 at 03:31 AM.
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    Senior Member bkarpuz's Avatar
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    Lightbulb Add: Show that complex product does not converge

    Quote Originally Posted by bkarpuz View Post
    Proof. As we know for any complex number z=x+\mathrm{i}y,
    |x|+|y|\geq|z|\geq\max\{|x|,|y|\}.
    Let x_{n}:=\frac{1}{2^{n}} and z_{n}:=x_{n}+\mathrm{i}(1-x_{n}),
    then we have |z_{j}|\geq|1-x_{j}|, which yields
    \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|=\prod_{i=1}^  {\infty}|z_{i}|\geq\prod_{i=1}^{\infty}|1-x_{i}|\approx\ell:=0.2887880951.

    Addentum.
    Recall that
    \prod_{j=1}^{\infty}\bigg(1-\frac{\lambda^{2}}{(j\pi)^{2}}\bigg)=\frac{\sin( \lambda)}{\lambda}.
    (See Basel problem - Euler's Approach)
    If \lambda\geq\frac{3\pi}{2\sqrt{2}}, we have \lambda\geq\frac{j\pi}{\sqrt{2^{j}}} for all j=1,2,\cdots.
    The sequence on the right-hand side increases for j=1,2,
    reaches its maximum value \frac{3\pi}{2\sqrt{2}}\approx3.33216 at j=3,
    and then decreasses for j=4,5,\cdots.
    This yields
    1-x_{j}=1-\frac{1}{2^{j}}\geq1-\frac{\lambda^{2}}{(j\pi)^{2}} for all j=1,2,\cdots
    provided that \lambda\geq\frac{3\pi}{2\sqrt{2}}.
    Letting \lambda\geq\frac{3\pi}{2\sqrt{2}} and we have for the real product
    \prod_{j=1}^{\infty}(1-x_{j})=\prod_{j=1}^{\infty}\bigg(1-\frac{1}{2^{j}}\bigg)\geq\prod_{j=1}^{\infty}\bigg  (1-\frac{\lambda^{2}}{(j\pi)^{2}}\bigg)=\frac{\sin( \lambda)}{\lambda},
    which is strictly positive for \lambda=2\pi+\frac{1}{2}\pi>\frac{3\pi}{2\sqrt{2}}. \rule{0.2cm}{0.2cm}

    Quote Originally Posted by bkarpuz View Post
    This shows that product cannot tend to 0 (i.e., 0+\mathrm{i}0).
    On the other hand, we have
    \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|\leq\prod_{i=  1}^{\infty}(|x_{i}|+|1-x_{i}|)=1
    showing that the product can not also diverge (in the sense of complex modulus).
    Then, the complex sequence w_{n}:=\prod_{i=1}^{n}z_{i} lies in the annulus \mathcal{A}:\ \ell\leq|z|\leq1.
    Due to Bolzano-Weierstrass theorem (considered separetely for the real and imaginary parts),
    there exists a subsequence w_{n_{k}} of w_{n} which converges to some w\in\mathcal{A}.
    Recall that \lim_{n\to\infty}z_{n}=\mathrm{i} and w_{n_{k}+p}=w_{n_{k}}\prod_{i=n_{k}+1}^{n_{k}+p}z_  {i},
    which shows that \lim_{k\to\infty}w_{n_{k}+p}=\mathrm{i}^{p}w for p=0,1,\cdots.
    Therefore, the product accumulates to at least (also at most) four different points \pm w,\pm\mathrm{i}w since w\neq0. \rule{0.2cm}{0.2cm}

    But I just wanted to know if it is possible to find (by hand) what w is.
    Last edited by bkarpuz; June 6th 2013 at 05:05 AM. Reason: Typos in the previous post are corrected.
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