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  1. #1
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    sum

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

    Obtain an expression of f(x) in ascending power of 1/x up to the term in 1/x^3 . Determine the set of values of x which this expansion is valid .

    My work .

    $\displaystyle f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

    $\displaystyle =(x+2)^{-1}-(x-3)^{-1}$

    $\displaystyle =\frac{1}{x}(1-\frac{3}{x})^{-1}-\frac{1}{x}(1+\frac{2}{x})^{-1}$

    When i continue simplifying , i got $\displaystyle \frac{5}{x^2}+\frac{5}{x^3}$

    Am i correct ?

    The other thing is to find the set of values of x , is it
    $\displaystyle

    |-\frac{3}{x}|<1
    $

    and $\displaystyle |\frac{2}{x}|<1 $ ??
    Last edited by thereddevils; Sep 8th 2009 at 05:39 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by thereddevils View Post
    f(x)=\frac{x^2-x-1}{(x+2)(x-3)}

    Obtain an expression of f(x) in ascending power of 1/x up to the term in 1/x^3 . Determine the set of values of x which this expansion is valid .

    My work .

    $\displaystyle f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

    $\displaystyle =(x+2)^{-1}-(x-3)^{-1}$

    $\displaystyle =\frac{1}{x}(1-\frac{3}{x})^{-1}-\frac{1}{x}(1+\frac{2}{x})^{-1}$

    When i continue simplifying , i got $\displaystyle \frac{5}{x^2}+\frac{5}{x^3}$

    Am i correct ?

    The other thing is to find the set of values of x , is it
    $\displaystyle

    |-\frac{3}{x}|<1
    $

    and $\displaystyle |\frac{2}{x}|<1 $ ??

    your function

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

    or

    $\displaystyle f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

    if your question at the first form I think the solution will be

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

    $\displaystyle f(x)=\frac{1-\dfrac{1}{x}-\dfrac{1}{x^2}}{\left(1+\dfrac{2}{x}\right)\left(1-\dfrac{3}{x}\right)}$

    the denominator should not equal zero in each fraction

    so $\displaystyle x\ne 0 $ and

    $\displaystyle 1+\frac{2}{x}\ne 0 \Rightarrow x\ne -2 $ and

    $\displaystyle 1-\frac{3}{x} \ne 0 \Rightarrow x\ne 3 $

    so the values of x all real numbers expect {-2,0,3}


    if

    $\displaystyle f(x)=\frac{1}{x-3} - \frac{1}{x+2} $

    $\displaystyle f(x)=\frac{5}{(x-3)(x+2)}$

    $\displaystyle f(x)=\dfrac{\dfrac{5}{x^2}}{\left(1-\dfrac{3}{x}\right)\left(1+\dfrac{2}{x}\right)}$

    and the values of x is the same as the first form
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  3. #3
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    Quote Originally Posted by Amer View Post
    your function

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

    or

    $\displaystyle f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

    if your question at the first form I think the solution will be

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

    $\displaystyle f(x)=\frac{1-\dfrac{1}{x}-\dfrac{1}{x^2}}{\left(1+\dfrac{2}{x}\right)\left(1-\dfrac{3}{x}\right)}$

    the denominator should not equal zero in each fraction

    so $\displaystyle x\ne 0 $ and

    $\displaystyle 1+\frac{2}{x}\ne 0 \Rightarrow x\ne -2 $ and

    $\displaystyle 1-\frac{3}{x} \ne 0 \Rightarrow x\ne 3 $

    so the values of x all real numbers expect {-2,0,3}


    if

    $\displaystyle f(x)=\frac{1}{x-3} - \frac{1}{x+2} $

    $\displaystyle f(x)=\frac{5}{(x-3)(x+2)}$

    $\displaystyle f(x)=\dfrac{\dfrac{5}{x^2}}{\left(1-\dfrac{3}{x}\right)\left(1+\dfrac{2}{x}\right)}$

    and the values of x is the same as the first form

    Well , that doesn't answer my question . However , the answer to this question is $\displaystyle 1+\frac{5}{x^2}+\frac{5}{x^3}$ , i just cant figure out where is my '1' .

    And also the ranges of x which makes this expansion valid ??
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