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Weierstrass Product convergence

  1. Apr 24, 2012 #1
    Show that the infinite product [itex]f(z) = \prod\limits_{n = 0}^{\infty}(1 + z^{2^n})[/itex] converges on the open disc [itex]D(0,1)[/itex] to the function [itex]1/(1 - z)[/itex]. Is this convergence uniform on compact subsets of the disc?



    This should actually be done by the comparison test.

    For [itex]|z| < 1[/itex], we have that
    $$
    \frac{1}{1-z}=\sum_{n=0}^{\infty}|z|^{2^n}\leq \sum_{n=0}^{\infty}|z|^n
    $$

    So now I need to show by partial products that there are no zeros in [itex]K\subset D(0,1)[/itex]. So I need to find an N such that forall n > N this holds. I am looking for some guidance on this piece.

    $$
    f(z) = \prod_{n=0}^{N}\left(1+z^{2^n}\right) \prod_{n=N+1}^{\infty}\left(1+z^{2^n}\right)
    $$

    We need to fix [itex]R\in\mathbb{R}^+[/itex]. Let [itex]N\in\mathbb{N}[/itex] such that [itex]|z_N|\leq 2R < |z_{N+1}[/itex] (is this correct-the inequalities?).

    The first partial product is finite on [itex]D(0,1)[/itex] and the second partial product behaves well on [itex]D(0,1)[/itex].

    What would be my choice of [itex]k_n[/itex] for this product?
     
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 26, 2012 #2
    So continuing.

    Then for [itex]|z|\leq R[/itex] and [itex]n>N[/itex] we have
    $$
    \left|\frac{z}{z_n}\right| <\frac{1}{2}, \quad\forall n>N
    $$
    so by Lemma: If [itex]|z|\leq 1/2[/itex], then [itex]\log\left[\prod\limits_{n=1}^{\infty}\left(1-\frac{z}{z_n}\right)\right]\leq 2|z|^n[/itex],
    $$
    \log\left|\left[\prod\limits_{n=1}^{\infty}\left(1-z^{2^n}\right)\right]\right| = \sum_{n=1}^{\infty}\left|\log(1+z^{2^n})\right| \leq 2\left(\frac{R}{z_n}\right)^{k_n}.
    $$

    What choices of $k_n$ will allow convergence(uniform/absolute?).
     
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