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Arctan(t) power series.

  1. Mar 28, 2013 #1

    Zondrina

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

    An exercise from advanced calculus by taylor :

    attachment.php?attachmentid=57226&stc=1&d=1364515036.png

    2. Relevant equations



    3. The attempt at a solution

    (a)

    ##\int_{0}^{x} tan^{-1}(t) dt = \int_{0}^{x} \sum_{n=0}^{∞} (-1)^n \frac{t^{2n+1}}{2n+1} dt = \sum_{n=0}^{∞} \frac{(-1)^n}{2n+1} \int_{0}^{x} t^{2n+1} dt = \sum_{n=0}^{∞} \frac{(-1)^n}{2n+1} \frac{x^{2n+2}}{2n+2}##

    Now I wont plaster everything here, but this series appears to converge for ##|x| < 1## by the ratio test. The question asks me to pay special attention to the endpoints of the interval of convergence, namely ##±1##.

    By some quick inspection, I see that at x = 1 the series will converge absolutely by p-comparison so that the series will be uniformly convergent on [0,1] by Abel's theorem.

    At x = -1 the series appears to be absolutely convergent by p-comparison so that the series converges uniformly on [-1,0] by Abel's theorem once again.

    This implies the series is uniformly convergent over the whole interval [-1,1].


    (b)

    I have no clue how to do part (b) for some reason though. I know that as ##x→1^-##, ##arctan(x)→π/4##

    Not quite sure how to use this though.
     

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    Last edited by a moderator: Mar 28, 2013
  2. jcsd
  3. Mar 28, 2013 #2

    SammyS

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    Here's an image of the problem:


    attachment.php?attachmentid=57220&stc=1&d=1364497551.png
     

    Attached Files:

    Last edited by a moderator: May 6, 2017
  4. Mar 28, 2013 #3

    Zondrina

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    Thanks Sammy.

    I was thinking a moment ago for part (b).

    Consider : ##lim_{x→1^-} \space arctan(x) - ln( \sqrt{x+1}) = lim_{x→1^-} \space arctan(x) - \frac{1}{2}ln(x+1)##

    Perhaps that could be of some usage? I could convert each function into its equivalent Taylor series and I'll take a stab in the dark that things will start cancelling out.
     
  5. Mar 28, 2013 #4
    ^Just a hint, it's not a matter so much of cancelling as it is evaluating arctan(x) and 1/2 ln(x+1) and noticing a pattern. Whatever series you use, be mindful of where they are centered, by which I mean be mindful of your choice of [tex] z_0 [/tex] in [tex] \sum_{n=0}^\infty a_n(z-z_0)^n[/tex]. If the series you're using are not given by your teacher or haven't been derived in class, you might want to state or show the radius of convergence and check endpoints.
     
  6. Mar 28, 2013 #5

    Zondrina

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    That seems like more of a first year thing learning what ln(x+1) is in terms of its series.

    ##ln(x+1) = \sum_{n=1}^{∞} (-1)^{n+1} \frac{x^n}{n}## which i know will converge for x inside the interval (-1,1].

    EDIT : Ohh so as x goes to 1 from the left, the desired result is achieved since the series for ln(x+1) converges for x in (-1,1].
     
    Last edited: Mar 28, 2013
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