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Math Help - Taylor series arctan(x) without computations

  1. #1
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    Taylor series arctan(x) without computations

    Hello forum! I was wondering if somebody knows of a fancy way of determining the Taylor series expansion of \arctan (x) about x = 1 without computing all the derivatives.
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    MHF Contributor ebaines's Avatar
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    Re: Taylor series arctan(x) without computations

    Determining the series does indeed require computing derivatives, which can appear fairly daunting at first. But after determining the first, second and third derivatives at x=1 you'll see that a simple pattern emerges: arctan(1)= 1/3 + 1/5 - 1/7 +1/9 -....
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    Junior Member Barioth's Avatar
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    Re: Taylor series arctan(x) without computations

    none that I can think of sorry!

    You should never be afraid of derivating altough! Don't be lazy and get to it!
    Last edited by Barioth; January 16th 2013 at 08:07 AM.
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    Re: Taylor series arctan(x) without computations

    Okay thanks!
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    Re: Taylor series arctan(x) without computations

    It can be derived from geometric series, if that's okay.


    \begin{aligned}\arctan{x} & = \int_{0}^{x}\frac{1}{1+t^2}\ dt  = \int_{0}^{x}\sum_{n=0}^{\infty}(-1)^n t^{2n} \ dt \\& = \sum_{n=0}^{\infty}(-1)^n\int_{0}^{x} t^{2n}\ dt   = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}. \end{aligned}
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    Re: Taylor series arctan(x) without computations

    Just noticed that you wanted the expansion at about x = 1. Shouldn't be hard to tweak the above.
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