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Math Help - Taylor Series Expansion Around 0.25

  1. #1
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    Taylor Series Expansion Around 0.25

    Hi, I am having to expand the taylor series 1/(1-x) around the point 0.25 to 3 degrees.

    I am okay expanding it around x=0 and x=.5 but for this question I am having difficulties due to the point being 0.25.

    Currently I have gotten 4/3 for the first one and (16/9x - 4/9) for the 1st order.

    The second order I got is (32/27)x^2 - (16/27)x + (2/27).

    Due to all these extra terms being generated, I am unable to cancel out terms to make it into a geometric series. Any help would be appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by todo View Post
    Hi, I am having to expand the taylor series 1/(1-x) around the point 0.25 to 3 degrees.

    I am okay expanding it around x=0 and x=.5 but for this question I am having difficulties due to the point being 0.25.

    Currently I have gotten 4/3 for the first one and (16/9x - 4/9) for the 1st order.

    The second order I got is (32/27)x^2 - (16/27)x + (2/27).

    Due to all these extra terms being generated, I am unable to cancel out terms to make it into a geometric series. Any help would be appreciated.
    Taylor expansion about x_0:

    \displaymstyle f(x)=\sum_{n=0}^{\infty} \frac{1}{n!}(x-x_0)^nf^{(n)}(x_0)

    \displaymstyle f(x)=\frac{1}{0.75}+(x-0.25)\frac{1}{0.75^2}+\frac{(x-0.25)^2}{2}\frac{2}{0.75^3}+ ...

    which simplifies to:

    f(x)=\frac{28}{27}+\frac{16}{27} x+\frac{64}{27} x^2+O(x-0.25)^3

    CB
    Last edited by CaptainBlack; October 25th 2010 at 12:15 AM.
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  3. #3
    MHF Contributor

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    If you can do it for x_0= 0.5, why would x_0= 0.25 be any different?

    Captain Black gives the formula for a general Taylor's series. Since you specifically mention "geometric series", you can also do it this way-

    Write \frac{1}{1- x} as \frac{1}{1-x+ .25- .25}= \frac{1}{1-.25- (x- .25)}= \frac{1}{.75- (x- .25)}.

    Now divide both numerator and denominator by .75: \frac{1}{1- x}= \frac{\frac{4}{3}}{1- \frac{x- .25}{.75}}

    Now, you should be able to recognise that as \frac{A}{1- r} with A= 4/3 and r= \frac{x- .25}{.75} and you know that that is the sum of the geometric series \sum_{n=0}^\infty Ar^n.
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