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Math Help - Sequence of converging polynomials

  1. #1
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    Sequence of converging polynomials

    I need help proving the following true or false: there is a sequence of polynomials {p_n} which converge uniformly to f(z)=1/z on the unit circle |z|=1
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  2. #2
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    Use the Stone-Wierstrass theorem which states that the set of polynomials viewed as a subset of C(X), form a dense subset. (where X is a compact set).

    Note that on the unit circle f(x)=\frac{1}{x} is continuous.
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  3. #3
    Super Member Gamma's Avatar
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    polynomials have residue 0, while 1/z has residue 1.

    It is false.
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  4. #4
    Super Member Gamma's Avatar
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    Oh another nice proof is just to use cauchy's formula. polynomials are entire, so take the usual path \gamma(t)=e^{i t} for t\in [0,2\pi ] so their integral is 0, while \int_{\gamma}z^{-1}=2\pi i, but since p(z) \rightarrow z^{-1} uniformly,

    o=\int_{\gamma}p_i(x)\rightarrow \int_{\gamma}z^{-1}=2\pi i a contradiction.
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  5. #5
    MHF Contributor chisigma's Avatar
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    On the unit circle is z=e^{i\cdot \omega} so that for f(z)=\frac{1}{z} is...

    f(e^{i\cdot \omega})= e^{-i\cdot \omega} = \sum_{k=0}^{\infty} (-1)^{k} \cdot \frac{(i\cdot \omega)^{k}}{k!} (1)

    From (1) it derives that the polynomial sequence...

    p_{n} (z) = \sum_{k=0}^{n} (-1)^{k} \cdot \frac{z^{k}}{k!} (2)

    ... converges uniformly to \frac{1}{z} for z=e^{i\cdot \omega} ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; June 7th 2009 at 11:18 PM.
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