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  1. #1
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    Taylor theorem

    Please help me out with this problem.

    Prove the given equation using Taylor series theorem.

    sin^{-1}\frac{2x}{1+x^{2}}=2\left \{ x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+... \right \}
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    Last edited by rishilaish; October 1st 2011 at 05:25 AM.
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  2. #2
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    Re: Taylor theorem

    Quote Originally Posted by rishilaish View Post
    Please help me out with this problem.

    Prove the given equation using Taylor series theorem.

    sin^{-1}\frac{2x}{1+x^{2}}=2\left \{ x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+... \right \}
    What have you tried? Where are you stuck?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Taylor theorem

    Quote Originally Posted by rishilaish View Post
    Please help me out with this problem.

    Prove the given equation using Taylor series theorem.

    sin^{-1}\frac{2x}{1+x^{2}}=2\left \{ x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+... \right \}
    My 'proof' doesn't use the 'Taylor series theorem' but avoids tedious computation of derivatives...

    Let's strart with the hypothesis that it exists the McLaurin expansion...

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

    ...and we want to compute the a_{n}. Now, considering that is...

    \int \sin^{-1} t\ dt= t\ \sin^{-1} t + \sqrt{1-t^{2}}+c (2)

    ... setting in (2) t=\frac{2 x}{1+x^{2}} we derive using (1) and a little of patience...


    \int_{0}^{x} \sin^{-1} t\ dt =  \frac{2 x}{1+x^{2}}\ \sin^{-1} \frac{2x}{1+x^{2}} - \frac{2\ x^{2}}{1+x^{2}} =

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

    ... and the identity (3) permits the computation of the a_{n}...

    Kind regards

    \chi \sigma
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