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Math Help - find maclaurin series

  1. #1
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    find maclaurin series

    integral of 1/(2+x^3)dx bounds:0 to .5
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  2. #2
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    Quote Originally Posted by cm3pyro View Post
    integral of 1/(2+x^3)dx bounds:0 to .5
    \frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)

    integrate ...

    \frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}

    using the first three terms should get you within 0.00001 of the actual value
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)

    integrate ...

    \frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}

    using the first three terms should get you within 0.00001 of the actual value
    i need a maclaurin series
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  4. #4
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    Quote Originally Posted by cm3pyro View Post
    i need a maclaurin series
    what do you think I posted?
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