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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 = $\displaystyle \frac{1}{1 + x^2} $

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

    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 $\displaystyle (-1)^n x^{2n}$
    i ended up with

    $\displaystyle 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 $\displaystyle |x|<1$ then,
    $\displaystyle 1-x^2+x^4-x^6 + ... = \frac{1}{1+x^2}$

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

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