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Exhibit a function f

  1. Apr 24, 2012 #1
    Use the Weierstrass Product Theorem to exhibit a function [itex]f[/itex] such that each positive integer [itex]n[/itex], [itex]f[/itex] has a pole of order [itex]n[/itex], and [itex]f[/itex] is analytic and nonzero at every other complex number

    For [itex]f[/itex] to have a pole of order [itex]n[/itex], we have that [itex]f = \prod\limits_{n = 1}^{\infty}\left(1 - \frac{z}{z_n}\right)^{-1}e^{-P_n(z/z_n)}[/itex].
    Let [itex]z_n[/itex] be the [itex]\text{n}^{\text{th}}[/itex] term in the sequence, i.e. [itex]1, 2, 2, 3,\ldots[/itex].
    So taking [itex]k_n[/itex] to be 3, we have that (why is it 3?)
    $$
    \sum_{n = 1}^{\infty}\frac{1}{\left|z_n\right|^3} = \sum_{n = 1}^{\infty}\frac{1}{n^2} < \infty
    $$
    which converges since we have a p-series of degree two.
    Now [itex]P_n\left(\dfrac{z}{z_n}\right) = \dfrac{z}{z_n} + \dfrac{\left(\frac{z}{z_n}\right)^2}{2} + \cdots + \dfrac{\left(\frac{z}{z_n}\right)^{k - 1}}{k - 1}[/itex] so the Weierstrass Product for [itex]k_n = 3[/itex] is

    $$
    \prod_{n = 1}^{\infty}\left(1 - \frac{z}{z_n}\right)^{-1}e^{-\left[\frac{z}{z_n} + \left(\frac{z}{z_n}\right)^2/2\right]}
    $$

    I was told that the above product can be simplified down. How?
     
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 24, 2012 #2
    So looking at the product, we have
    $$
    \left[\left(1-\frac{z}{1}\right)^{-1}\left(1-\frac{z}{2}\right)^{-2}\cdots\left(1-\frac{z}{n}\right)^{-n}\cdots\right]\exp\left[-z-\frac{z}{2}-\cdots -\frac{z}{n} - \cdots + \frac{z^2}{2^2}+ \frac{z^2}{6} + \frac{z^2}{8}\cdots\right]
    $$

    So can this be written as a different infinite product then from the final one I obtained?

    All I see is
    $$
    \prod_{n=1}^{\infty}\left(1-\frac{z}{n}\right)^{-n}\exp\left[-\frac{z}{n}+\frac{z^2}{2^n}\right]
    $$
    but is this even the right observation?
     
    Last edited: Apr 24, 2012
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