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Partial fractions

  1. Dec 11, 2008 #1
    [tex]\frac{s-1}{s(s-2)^2}[/tex]

    How can I expand this fraction?

    [tex]\frac{A}{s} + \frac{B}{(s-2)} + \frac{C}{(s-2)^2}[/tex]

    right?

    This gives me the equation

    [tex]As^3 - 6As^2 + 12As - 8A Bs^3 - 4Bs^2 + 4Bs + Cs^2 - 2Cs = s-1[/tex]

    so that

    (1) A + B =0
    (2)- 6A - 4B + C = 0
    (3) 12A + 4B - 2C = 1
    (4) -8A = -1

    (4) gives A = 1/8
    (1) gives B = -1/8
    (2) gives C = 1/4

    The correct answer is

    A = -1/4
    B = 1/4
    C = 2/4

    What's wrong?
     
  2. jcsd
  3. Dec 11, 2008 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    So far so good :smile:...

    Are you sure about that?:wink:
     
  4. Dec 11, 2008 #3
    I figured it out, thanks!
     
  5. Dec 11, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    leopard, for future use, you may find it easier (and less error prone) to do this:
    to find A, B, C so that
    [tex]\frac{A}{s}+ \frac{B}{s-2}+ \frac{C}{(x-2)^2}= \frac{s-1}{s(s-2)^2}[/tex]
    Multiply both sides by the denominator, s(s-2)2, leaving it as
    [tex]A(s-2)^2+ Bs(s-2)+ Cs= s- 1[/tex]

    Now, since this must be true for all s, select simple values of s:
    if s= 0, 4A= -1
    if s= 2, 2C= 1
    if s= 1, A- B+ C= 0
     
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