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Math Help - Help with Series Expansion

  1. #1
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    Help with Series Expansion

    Hi - I ask for a bit of help in trying to find the series expansion of
    Integral(arctan(x^3)) - I was having quite a bit of trouble, and eventually I was forced to look at Wolfram Alpha to get done with the homework, but I want to know where I messes up when I tried to do it.

    Basically, this was where I started.

    arctan x^3 = integral (1/(1-(-x^6)) dx
    = Sum [(-1)^n*x^6n,n,0,infinity]
    Then integrating both sides to get my answer, but this has been wrong.

    Is my first step the bad part?
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  2. #2
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    Quote Originally Posted by Lord Voldemort View Post
    Hi - I ask for a bit of help in trying to find the series expansion of
    Integral(arctan(x^3)) - I was having quite a bit of trouble, and eventually I was forced to look at Wolfram Alpha to get done with the homework, but I want to know where I messes up when I tried to do it.

    Basically, this was where I started.

    arctan x^3 = integral (1/(1-(-x^6)) dx
    = Sum [(-1)^n*x^6n,n,0,infinity]
    Then integrating both sides to get my answer, but this has been wrong.

    Is my first step the bad part?
    \frac{d}{dX}\left(\arctan{X}\right) = \frac{1}{1 + X^2}.


    Notice that you can rewrite \frac{1}{1 + X^2} as

    \frac{1}{1 - (-X^2)}.

    This is the closed form of the geometric series with a = 1, r = -x^2.

    So you can say

    \frac{1}{1 + X^2} = \sum_{n = 0}^{\infty} (-X^2)^n

     = \sum_{n = 0}^{\infty}(-1)^nX^{2n}.


    So \frac{d}{dX}(\arctan{X}) = \sum_{n = 0}^{\infty} (-1)^nX^{2n}

    which means

    \arctan{X} = \int{\sum_{n = 0}^{\infty}(-1)^nX^{2n}\,dX}

    \arctan{X} = \sum_{n = 0}^{\infty}\int{(-1)^n X^{2n}\,dX}

    \arctan{X} = \sum_{n = 0}^{\infty}\frac{(-1)^nX^{2n + 1}}{2n + 1}.


    This means \arctan{(x^3)} = \sum_{n = 0}^{\infty}\frac{(-1)^n(x^3)^{2n + 1}}{2n + 1}

    \arctan{(x^3)} = \sum_{n = 0}^{\infty}\frac{(-1)^n x^{6n + 3}}{2n + 1}.



    So that means:

    \int{\arctan{(x^3)}\,dx} = \int{\sum_{n = 0}^{\infty}\frac{(-1)^n x^{6n + 3}}{2n + 1}\,dx}

     = \sum_{n = 0}^{\infty}\int{\frac{(-1)^n x^{6n + 3}}{2n + 1}\,dx}

     = \sum_{n = 0}^{\infty}\frac{(-1)^n x^{6n + 4}}{(2n + 1)(6n + 4)}.
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