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Thread: Direct Analytical Continuations

  1. #1
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    Direct Analytical Continuations

    Hi any help with this would be greatly appreciated:

    Direct Analytical Continuations-q3a.jpg

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  2. #2
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    Quote Originally Posted by moolimanj View Post
    Hi any help with this would be greatly appreciated:

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    Many thanks and "Mathshelpforum" rocks
    $\displaystyle
    f(z)=\sum_{n=0}^{\infty}(2+z)^n, \ \ (|z+2|<1)
    $

    $\displaystyle
    g(z)=-2 \sum_{n=1}^{\infty}\left( \frac{1}{3+2x}\right)^n, \ \ (|z+3/2|>1/2)
    $

    The series defining function $\displaystyle f(z)$ is a geometric series which sums to:

    $\displaystyle
    f(z)=\frac{-1}{1+z}, \ \ (|z+2|<1)
    $

    Similarly the series defining $\displaystyle g(z)$ may also be summed with the
    aid of the geometric series formula to:


    $\displaystyle
    g(z)=2-2 \sum_{n=0}^{\infty}\left( \frac{1}{3+2x}\right)^n
    $$\displaystyle =2-\frac{2}{1-\frac{1}{3+2z}}=\frac{-1}{1+z} , \ \ (|z+3/2|>1/2)
    $

    Thus $\displaystyle f(z)$ and $\displaystyle g(z)$ are identical on the overlap of their regions of convergence, and so are analytic continuations of one another.

    RonL
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