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Math Help - Help... Mclaurin Series,

  1. #1
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    Post Help... Mclaurin Series,

    The formula is F(x) = (1-x)^-1
    and x is given at X = 1

    Can anyone solve this for me cause I don't have a clue where to start
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  2. #2
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    Re: Help... Mclaurin Series,

    The McLaurin series is just the Taylor expansion of $F(x)$ about $x=0$

    If you're clever you'll notice $F(x)$ is the sum of a well known series.

    Note that F(1) is undefined and indeed the series will not converge for x=1.
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  3. #3
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    Re: Help... Mclaurin Series,

    Well I haven't seen Taylor series before only learnt about Mclaurin series, ive got little knowledge as I havnt learnt this before, would help if you could explain it, this is just for reference as the question ive got to solve is x=0.01
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  4. #4
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    Re: Help... Mclaurin Series,

    Quote Originally Posted by Baitbook View Post
    Well I haven't seen Taylor series before only learnt about Mclaurin series, ive got little knowledge as I havnt learnt this before, would help if you could explain it, this is just for reference as the question ive got to solve is x=0.01
    I don't understand how you can be given a problem without having been taught the reference material.

    The Mclaurin series way of doing this problem is to recognize that

    $\displaystyle{\sum_{k=0}^\infty}x^n = \dfrac 1 {1-x}~~\mbox{ for } |x| < 1$

    This is the geometric series. Have you seen this before?

    So given that your $F(x)=\dfrac 1 {1-x}$ it's obvious that your Mclaurin series is

    $F(x) = \displaystyle{\sum_{k=0}^\infty}x^n$

    and this converges for $|x| < 1$

    It's not possible to try to evaluate this at $x=0.01$ via the series as there an infinite number of terms. So just plug $0.01$ into the formula for the sum, i.e. $F(x)$.

    I'll let you compute $\dfrac 1 {1-0.01}$
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