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Math Help - Maclaurin series

  1. #1
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    Maclaurin series

    use a maclaurin series to obtain the maclaurin series for the given function f(x)= (sinx)^2 hint use (sinx)^2 = 1/2(1-cos2x)
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by twilightstr View Post
    use a maclaurin series to obtain the maclaurin series for the given function f(x)= (sinx)^2 hint use (sinx)^2 = 1/2(1-cos2x)
    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}

    \cos(2x)=\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n}(x)^{2n}}{(2n)!}

    Just use this with the hint and you are done
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  3. #3
    Super Member Gamma's Avatar
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    f(x)=sin^2(x)=\frac{1}{2}-\frac{1}{2}cos(2x)

    Basically this amounts to finding the maclaurin series for cos(u) since then you can just put in a 2x for u (then multiply by your 1/2 scalar and subtract from 1/2). Note this trick works because the radius of convergence for this series turns out to be infinite, more care should in general be used.

    let
    f(u)=cos(u) f(0)=1
    f'(u)=-sin(u) f'(0)=0
    f"(u)=-cos(u) f"(0)=-1
    f"'(u)=sin(u) f"'(0)=0
    and repeat.

    so:
    cos(u)=1 + 0u -\frac{1}{2!}u^2 + 0u^3 + \frac{1}{4!}u^4... =\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n}}{2n!}

    so finally you see:
    sin^2(x)=\frac{1}{2}(1-cos(2x))=\frac{1}{2}\left(1-\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n}}{2n!}\right)

    Think you can take her from here?
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