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Math Help - Maclaurin series

  1. #1
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    Maclaurin series

    Given (sin^{-1}x)^2=x^2+\frac{1}{3}x^4+..., use the standard series expansions for ln(1+x) and e^x to show that when x is small enough for powers of x above fourth to be neglected, (sin^{-1}x)^2=\frac{1}{6}[ln(1+x^2)+5e^{x^2}-5]
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  2. #2
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    Re: Maclaurin series

    Quote Originally Posted by Punch View Post
    Given (sin^{-1}x)^2=x^2+\frac{1}{3}x^4+..., use the standard series expansions for ln(1+x) and e^x to show that when x is small enough for powers of x above fourth to be neglected, (sin^{-1}x)^2=\frac{1}{6}[ln(1+x^2)+5e^{x^2}-5]
    following the given directions, I was able to approximate ...

    \ln(1+x^2) \approx x^2

    e^{x^2} \approx 1 + x^2

    \frac{1}{6}[\ln(1+x^2)+5e^{x^2}-5] \approx \frac{1}{6}[x^2 + 5(1+x^2) - 5] = x^2

    finish it ...
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    Re: Maclaurin series

    Hence find an approximatevalue for \pi^2-30e^{\frac{1}4}} giving your answer in the form cln\frac{5}{4}+ d
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  4. #4
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    Re: Maclaurin series

    Quote Originally Posted by Punch View Post
    Hence find an approximatevalue for \pi^2-30e^{\frac{1}4}} giving your answer in the form cln\frac{5}{4}+ d
    why did you leave out the "hence" part in the original post?
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  5. #5
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    Re: Maclaurin series

    Quote Originally Posted by skeeter View Post
    why did you leave out the "hence" part in the original post?
    Sorry, I didn't thought I would have had a problem with this.
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