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Thread: power series of arctan'x

  1. #1
    Junior Member
    Aug 2007

    power series of arctan'x

    how could i expand something such as arctan'x (derivate of arctanx ... i.e. d/dx arctanx) into a power series. also how would you be able to find the radius of convergence for it?

    so far i have managed to work out that:

    arctan'x =  \frac{1}{1 + x^2}

    \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +...+ (- 1)^n x^{2n}

    how do you work out the "radius of convergence" though: i know it is : |x|< 1.. but how do you work it out please?

    i tried it on (-1)^n x^{2n}
    i ended up with

    a_{n+1} / a_{n} = \frac{|x|^{2n + 2}}{|x|^{2n}} = |x|^2/1 as n tends to infinity... ...

    so radius of convergence is |x|< 1...

    is this working out correct?
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  2. #2
    Global Moderator

    Nov 2005
    New York City
    If |x|<1 then,
    1-x^2+x^4-x^6 + ... = \frac{1}{1+x^2}

    Integrate both side from 0\mbox{ to }t where |t|<1 to get,
    \tan^{-1}t = t - \frac{t^3}{3}+...

    Now this hold if -1 < t < 1.
    But it also hold when t=1.
    That is a little more difficult to show.
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