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Math Help - MacLaurin Series

  1. #1
    Junior Member SirOJ's Avatar
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    MacLaurin Series

    I know how to do the easy maclaurin series questions like;

    f(x) = 1/(2+x)

    but im a bit clueless on the more complicated ones like;

    (i) sin^x2

    (ii) coshx = (e^x + e^-x)/2

    (iii) integral from 0 to X of e^t2dt.

    Any help would be greatly appreciated
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by SirOJ View Post
    I know how to do the easy maclaurin series questions like;

    f(x) = 1/(2+x)

    but im a bit clueless on the more complicated ones like;

    (i) sin^x2

    (ii) coshx = (e^x + e^-x)/2

    (iii) integral from 0 to X of e^t2dt.

    Any help would be greatly appreciated
    (i) Use \sin^2 x = \frac{1}{2} \left( 1 - \cos(2x) \right).

    (ii) There is a simple pattern to the derivatives of cosh(x) evaluated at x = 0.

    (iii) Substitute x = t^2 into the Maclaurin series for e^x and integrate the result.
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  3. #3
    MHF Contributor

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    (ii) Or, if you already know that e^x= \sum_{n=0}^\infty \frac{x^n}{n!} = 1+ x+ \frac{1}{2}x^2+ \frac{1}{6}x^3+ \cdot\cdot\cdot, then you also know that e^{-x}= \sum_{n=0}^\infty \frac{(-x)^n}{n!} = 1- x+ \frac{1}{2}x^2- \frac{1}{6}x^3+ \cdot\cdot\cdot so that cosh(x)= \frac{e^x+ e^{-x}}{2}= 1+ \frac{1}{2}x^2+ \frac{1}{4!}x^4+ \cdot\cdot\cdot = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}
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  4. #4
    Junior Member SirOJ's Avatar
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    thanks alot guys, it's all clear to me now
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