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Complex Functions (Power Series)

  1. Oct 9, 2006 #1
    Hi, Power Series' were not covered in my cal II class, so I dont know how to solve these. Is there a certain way to solve these?

    Find the Radiusof convergence and open disk of convergence of the power series:

    [tex]\frac{n^2}{2n+1}(z+6+2i)^n[/tex]

    I dont know how to latex the summation but it is there, n=0 - inf.

    THanks
     
  2. jcsd
  3. Oct 9, 2006 #2

    quasar987

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    See the formula for r here:

    http://en.wikipedia.org/wiki/Radius_of_convergence

    Note that is you have a function f(z), and you find its power series expansion about a point a, then its radius of convergence is just the distance from a to the nearest singularity of f. You can use this fact to find the radius of convergence of the power series of a real-valued function f(x) by finding the nearest singularity of the corresponding complex valued function f(z).
     
  4. Oct 9, 2006 #3

    quasar987

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    I should have written "Note that if you have an holomorphic function f(z),..."

    I need to refresh my memory about all this... :grumpy:
     
    Last edited: Oct 9, 2006
  5. Oct 9, 2006 #4
    So you just take the lim as n->infinity of [tex]|\frac{C_n}{C_n+1}|[/tex]?

    or [tex]\frac{\frac{n^2}{2n+1}}{\frac{n^2}{2n+1}+1}[/tex]?
     
  6. Oct 9, 2006 #5

    quasar987

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    That's probably the easiest way.
     
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