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Math Help - Partial fraction decomposition

  1. #1
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    Partial fraction decomposition

    \frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{d  x+e}{x^2}\\
    ax^4+bx^3+cx^2+dx^4+ex^3=x^4(a+d)+x^3(b+e)+cx^2=1\  \
    c=0\\
    b+e=0\\
    a+d=0\\
    i got 4 variables with 2 equations
    what to do??
    Last edited by mr fantastic; August 17th 2009 at 01:21 PM.
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  2. #2
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    Hi transgalactic

    Your partial fraction is wrong

    <br />
\frac{1}{x^3(x^2+1)}\neq\frac{ax^2+bx+c}{x^3}+\fra  c{dx+e}{x^2}\\<br />
because if we put the denominator of RHS together, it won't become x^3(x^2+1) but x^3 (different from LHS)

    It should be :
    <br />
\frac{1}{x^3(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\fr  ac{C}{x^3}+\frac{Dx+E}{x^2+1}
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    \frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{d  x+e}{x^2}\\
    ax^4+bx^3+cx^2+dx^4+ex^3=x^4(a+d)+x^3(b+e)+cx^2=1\  \
    c=0\\
    b+e=0\\
    a+d=0\\
    i got 4 variables with 2 equations
    what to do??
    \frac{1}{x^3(x^2+1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{dx+e}{x^2+1}

    1 = ax^2(x^2+1) + bx(x^2+1) + c(x^2+1) + (dx+e)x^3<br />

    1 = (a+d)x^4 + (b+e)x^3 +(a+c)x^2 + bx + c

    equating coefficients ...

    a+d = 0

    b+e = 0

    a+c = 0

    b = 0

    c = 1


    from the above equations ...

    a = -1 , b = 0 , c = 1 , d = 1 , e = 0

    \frac{1}{x^3(x^2+1)} = -\frac{1}{x} + \frac{1}{x^3} + \frac{x}{x^2+1}
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  4. #4
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    can you give me a link to the manual of this stuff
    in order to know thurely all the laws
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  5. #5
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    Quote Originally Posted by transgalactic View Post
    can you gove me a link to this stuff
    in order to know thurely all the laws
    partial fraction decomposition - Google Search
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  6. #6
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    thanks
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