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Math Help - Maclaurin Series

  1. #1
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    Maclaurin Series

    How would I find the maclaurin series for the following function: f(x) = \frac{3}{(1-x)^2}.

    The first, second, and third derivatives are: \frac{6}{(1-x)^3}, \frac{18}{(1-x)^4}, \frac{72}{(1-x)^5}.

    f(x)=\Sigma \frac{f^{(n)}(0)}{n!}x^n = 3 + 6x +\frac{18}{2!}x^2 + \frac{72}{3!}x^3 + ....

    Here's where I'm stuck.

    Any help would be greatly appreciated.
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  2. #2
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    Re: Maclaurin Series

    f(x) = \frac{3}{(1-x)^2}

    let F(x) be the antiderivative of f(x) ...

    F(x) = \frac{3}{1-x} = 3(1 + x + x^2 + x^3 + ... )

    F'(x) = f(x) = \frac{3}{(1-x)^2} = 3(1 + 2x + 3x^2 + ... )
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  3. #3
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    Re: Maclaurin Series

    Quote Originally Posted by statmajor View Post
    How would I find the maclaurin series for the following function: f(x) = \frac{3}{(1-x)^2}.

    The first, second, and third derivatives are: \frac{6}{(1-x)^3}, \frac{18}{(1-x)^4}, \frac{72}{(1-x)^5}.

    f(x)=\Sigma \frac{f^{(n)}(0)}{n!}x^n = 3 + 6x +\frac{18}{2!}x^2 + \frac{72}{3!}x^3 + ....

    Here's where I'm stuck.

    Any help would be greatly appreciated.
    The easiest way would be to notice that \displaystyle 3\,\frac{d}{dx}\left(\frac{1}{1 - x}\right) = \frac{3}{(1 - x)^2} and that \displaystyle \frac{1}{1 - x} is the closed form of the geometric series \displaystyle \sum_{n = 0}^{\infty}x^n for \displaystyle |x| < 1.
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