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Math Help - Non-removable singularity and radius of Taylor expansion

  1. #1
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    Non-removable singularity and radius of Taylor expansion

    Hello,
    I'm trying to prove the follwing:
    For a function f(z) analytic in C, the radius of convergence of its Taylor expantion around \beta \in C equals r, where r is the minimal distance between \beta and a non-removable singularity of f.

    I have attempted to do so by first showing that the radius cannot be greater than r, for in that case the Taylor expantion will still be valid at the singuality.
    However, I cannot find a way to show the radius cannot be smaller than r.
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  2. #2
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    Quote Originally Posted by dudyu View Post
    Hello,
    I'm trying to prove the follwing:
    For a function f(z) analytic in C, the radius of convergence of its Taylor expantion around \beta \in C equals r, where r is the minimal distance between \beta and a non-removable singularity of f.

    I have attempted to do so by first showing that the radius cannot be greater than r, for in that case the Taylor expantion will still be valid at the singuality.
    However, I cannot find a way to show the radius cannot be smaller than r.
    The basic idea is to use the Cauchy integral formula on any circle of radius less than r, centred at \beta.

    You will find a more detailed proof on this Wikipedia page.
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