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Math Help - Radius of convergence of an analytic function

  1. #1
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    Radius of convergence of an analytic function

    This is from a practice exam:

    Let f(z)=\sum_{k=0}^\infty a_kz^k have a radius of convergence 2, and let g(z)=\sum_{k=0}^\infty b_k(z-3)^k have a radius of convergence 2. Let I=\{z:f(z),g(z)\text{ both converge}\}.

    (a) If f(z)=g(z) for z\in I, prove there is an analytic function w for which w(z)=f(z) when |z|<2 and w(z)=g(z) when |z-3|<2.

    (b) For the function w of part (a), determine the radius of convergence R for the power series w(z)=\sum_{k=0}^\infty c_k\left(z-\frac{3}{2}\right)^k.

    (c) What relationship exists between the coefficient sequence \{a_k\} and the coefficient sequence \{b_k\}?
    Part (a) is fairly straightforward. Let A=\{z:|z|<2\} and B=\{z:|z-3|<2\}. Then A\cap B\subseteq I, which means we can define w by w=f'\cup g', where f' and g' are the restrictions of f and g to A and B, respectively. It follows immediately that w is analytic on A\cup B.

    However, I'm lost for parts (b) and (c). Any help would be much appreciated!
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  2. #2
    Super Member girdav's Avatar
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    The series is convergent for the z=-2+0.00000001 and z=5-0.0000000001 for example, hence we can expect that the radius of convergence is greater than \frac 72. Now you have to show it's indeed \frac 72.
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  3. #3
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    Hmm. I'm sorry to say I don't understand your reasoning here.

    By Taylor's theorem R\geq \sup\{r : D(3/2;r)\subseteq A\cup B\}=\frac{\sqrt{7}}{2}. (Here I use the notation D (z_0;r)=\{z:|z-z_0|<r\}). However, I don't see how we can get R\geq\frac{7}{2}, nor R\leq\frac{7}{2}.
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  4. #4
    Super Member girdav's Avatar
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    The series is convergent on set a set which a diameter \geq 7, because the series converges on points of the form -2+\delta end 5-\delta for 1>\delta>0
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  5. #5
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    Oh, of course... I had forgotten every power series converges precisely on a disc and a subset of its boundary. Sorry about that.
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