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Math Help - series

  1. #1
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    series

    (a)Expand each of the following functions as a series in ascending power of x up to and including the term in x^3 .

    (1) \frac{x-1}{x+1}


    (2) \frac{1}{(x-2)(1+2x)}
    Last edited by thereddevils; February 17th 2009 at 05:07 AM.
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  2. #2
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    Hello, thereddevils!

    (a) Expand each of the following functions as a series
    in ascending power of x up to and including the term in x^3

    (1)\;\;\frac{x-1}{x+1}
    Have you tried Long Division?


    We have: . \frac{x-1}{x+1} \;=\;\frac{-1+x}{1+x}


    . . \begin{array}{ccccccccc} & & & -1 & +2x &- 2x^2 & +2x^3 & + \hdots \\ & & -- & -- & -- & -- & -- & -- \\ 1+x & ) & -1 & +x \\ & & -1 & -x \\ & & -- & -- \\ & & & 2x \\ & & & 2x & +2x^2 \\ & & & -- & -- \\ & & & & -2x^2 \\ & & & & -2x^2 & -2x^3 \\ \end{array}

    . . . . . . . . . . . . . . . \begin{array}{ccccc}<br />
 -- & \;\;-- \\ & \;\; 2x^3 \\ &\;\; 2x^3 &\;\; + 2x^4 \\ &\;\; -- & \;\;-- \\<br />
& & \;\;\hdots \end{array}


    Therefore: . \frac{x-1}{x+1} \;=\;-1 + 2x - 2x^2 + 2x^3 - \hdots



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  3. #3
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    Thanks Soroban . So if we want to swap positions , we hv to swap the top and bottom as well right ? Is this the commutative law ?

    Wow , i have never realised we can do it by long division but can we do it using binomial theorem ?
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