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Math Help - Maclaurin Polynomial

  1. #1
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    Maclaurin Polynomial

    How do we find the maclaurin polynomial for f(x)=\left\{\begin{array}{cc}\frac{cos(x)-1}{x},&\mbox{ if }x\not= 0\\0, & \mbox{ if } x=0\end{array}\right,
    Its not the maclaurin polynomial part that i dont understand its how i differentiate the pairwise function and find f(0),f'(0),f''(0)....
    Thanks for any help
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  2. #2
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    Quote Originally Posted by hmmmm View Post
    How do we find the maclaurin polynomial for f(x)=\left\{\begin{array}{cc}\frac{cos(x)-1}{x},&\mbox{ if }x\not= 0\\0, & \mbox{ if } x=0\end{array}\right,
    Its not the maclaurin polynomial part that i dont understand its how i differentiate the pairwise function and find f(0),f'(0),f''(0)....
    Thanks for any help
    Remember that

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

    This gives that

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

    and finally that

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

    Now what is the value of this series at 0? How does it compare with the limit

    \displaystyle \lim_{x \to 0}\frac{\cos(x)-1}{x}?
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  3. #3
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    sorry im still a bit confused with the pairwise part when x=0, I understand what you have do to get the maclaurin expansion when x is not 0 but what happens to the x=0 bit?
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  4. #4
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    Quote Originally Posted by hmmmm View Post
    sorry im still a bit confused with the pairwise part when x=0, I understand what you have do to get the maclaurin expansion when x is not 0 but what happens to the x=0 bit?
    What I was trying to point out is that

    \displaystyle \lim_{x \to 0}f(x)=\lim_{x \to 0}\frac{\cos(x)-1}{x}=0

    So the function f is actually continuous moreover it actually infinity differentiable there. They key idea is that the function can be extended to be defined at 0.

    The 2nd point is the series should sum to 0 at 0 right. So when we evaluate the series at zero we get 0. So the series does not care that the function was defined piecewise because of the continuous(analytic) extension. Maybe a plot of the function would help.Maclaurin Polynomial-plot.png
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  5. #5
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    ok Im pretty sure i understand what your getting at thanks very much sorry about that
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