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Radius of Convergence

  1. Apr 5, 2008 #1
    [SOLVED] Radius of Convergence

    1. The problem statement, all variables and given/known data

    1/(1+x^2) = sum ( (-1)^k*x^(2k) ; 0 ; inf) - A


    arctan (x) = sum ((-1)^k * x^(2k+1) / (2k+1) ; 0; inf) B

    I know A has radius of converge of 1, and I calculated B to be 2.

    My assignment solution says "Similarly, the series for 1/(1+x^2) has R = 1 and integrating does not affect this. so R for atan (x) series is 1"

    Obviously, they are wrong :biggrin:. Right?

  2. jcsd
  3. Apr 5, 2008 #2


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    Your notation ;0;inf)-A and ;0;inf)B is confusing. Can you clarify?
  4. Apr 5, 2008 #3
    Sorry for the confusion.
    It says sum of "(-1)^k * x^(2k+1) / (2k+1) "
    k is from 0 to inf

    [they are series]
  5. Apr 5, 2008 #4


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    Well, there is one other possiblity!

    Unfortunately, since you don't say HOW you got 2 as the radius of convergence for B, there isn't a whole lot I can say.

    Using the ratio test,
    [tex]\frac{|x^{2k+3}|}{2k+3}\frac{2k+1}{|x^{2k+1}|}= \frac{2k+1}{2k+3}|x|^2< 1[/tex]
    gives |x|< 1. Radius of convergence 1.

    Did you forget the "2" on 2k+ 3?
  6. Apr 5, 2008 #5
    Thanks a lot,

    I saw 2 in there (2^2k+1), and made it
    abs(x^2)/4 <1
    without going through all the steps.
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