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Math Help - representation of a arccos as taylor expansio

  1. #1
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    representation of a arccos as taylor expansio

    I'm having trouble expressing arccos(x) as a Maclaurin series. I realized that once you differentiate it, the function will just expand rapidly, so is there a easy way approaching this problem.
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    MHF Contributor Mathstud28's Avatar
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    Maybe this is what you are talking about

    Quote Originally Posted by lllll View Post
    I'm having trouble expressing arccos(x) as a Maclaurin series. I realized that once you differentiate it, the function will just expand rapidly, so is there a easy way approaching this problem.
    but did you consider using a similar method to your last one by saying arcos(x)=\int{\frac{-1}{\sqrt{1-x^2}}}?
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    Quote Originally Posted by lllll View Post
    so is there a easy way approaching this problem.
    Yes.

    Have you considered the Binomial series - Wikipedia, the free encyclopedia?
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    MHF Contributor Mathstud28's Avatar
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    Just wondering

    Quote Originally Posted by Krizalid View Post
    This is what I was implying in my post...do waht I said then say (1-x^2)^{\alpha} with \alpha=\frac{-1}{2} in this case...so does that work or is the binomial expansion limited to x not x^{y},y>1?
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  5. #5
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    Quote Originally Posted by lllll View Post
    I'm having trouble expressing arccos(x) as a Maclaurin series. I realized that once you differentiate it, the function will just expand rapidly, so is there a easy way approaching this problem.
    Yes

    f(x)=\cos^{-1}(x) then

    f'(x)=-\frac{1}{\sqrt{1-x^2}}

    expand this in a power series using the binomial series then integrate

    \binom{k}{n}=\frac{k(k-1)(k-2) \cdot \cdot \cdot (k-n+1)}{n!}

    (1-x^2)^{-1/2}=1-\frac{1}{2}x^2+\frac{\frac{-1}{2}\frac{-3}{2}}{2!}x^4- \cdot \cdot \cdot =

     -f'(x)=\sum_{n=0}^{\infty} \binom{k}{n}(-1)^nx^{2n} \iff f'(x)=\sum_{n=0}^{\infty} \binom{k}{n}(-1)^{n+1}x^{2n}

    Integrating both sides we get

    \cos^{-1}(x)=\int f'(x)dx=\int \sum_{n=0}^{\infty} \binom{k}{n}(-1)^{n+1}x^{2n}

    Finally

    f(x)=\sum_{n=0}^{\infty} \binom{k}{n} \frac{(-1)^{n+1}x^{2n+1}}{2n+1}
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    Quote Originally Posted by TheEmptySet View Post
    Yes

    (1-x^2)^{-1/2}=1-\frac{1}{2}x^2+\frac{\frac{-1}{2}\frac{-3}{2}}{2!}x^4- \cdot \cdot \cdot =
    how did you get this?
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by lllll View Post
    how did you get this?
    This is newtons extention of the binomial theorem to fractional and negative values

    <br />
\binom{k}{n}=\frac{k(k-1)(k-2) \cdot \cdot \cdot (k-n+1)}{n!}

    lets expand a few

    \binom{-\frac{1}{2}}{0} =1 by definition

    \binom{-\frac{1}{2}}{1} =\frac{\frac{-1}{2}}{1!}

    \binom{-\frac{1}{2}}{2} =\frac{\frac{-1}{2}\frac{-3}{2}}{2!}

    \binom{-\frac{1}{2}}{3} =\frac{\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}}{3!}

    and so on
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  8. #8
    Behold, the power of SARDINES!
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    Sorry I think this is what you need

    (1+x)^k=\sum_{n=0}^{\infty}\binom{k}{n}x^n

    so

    (1+(-x^2))^k=\sum_{n=0}^{\infty}\binom{k}{n}(-x^2)^n=\sum_{n=0}^{\infty}\binom{k}{n}(-1)^nx^{2n}
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