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Easy partial fractions explanation

  1. Aug 8, 2007 #1
    1. The problem statement, all variables and given/known data

    I just want to know how to proceed to get
    1/s - s/(s^2+1)

    using partial fractions on the term

    1/(s(s^2 − 1))

    I know this is probably straight forward but I just don't get it.

  2. jcsd
  3. Aug 8, 2007 #2


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    [tex] \dfrac{1}{s(s^2-1)} = \dfrac{A}{s} + \dfrac{B}{s-1} + \dfrac{C}{s+1} [/tex]

    Since s^2 - 1 = (s+1)(s-1)

    now get LHS = RHS (left hand side, right hand side)

    [tex] \dfrac{1}{s(s^2-1)} = \dfrac{A(s-1)(s+1)}{s(s-1)(s+1)} + \dfrac{Bs(s-1)}{s(s-1)(s+1)} } + \dfrac{Cs(s+1)}{s(s-1)(s+1)} } = \dfrac{A(s-1)(s+1) + Bs(s-1) + Cs(s+1)}{s(s^2-1)}[/tex]

    Now do all the multiplication on the RHS, and solve the linear equation system with respect to A,B and C.
  4. Aug 8, 2007 #3


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    You DON'T. (And so it is certainly not "straight forward"!)

    1/s- s/(s^2+ 1) would come from something with denominator s(s^2+1)= s^3+ 1.

    1/(s(s^2-1)= 1/(s(s-1)(s+1)) gives partial fractions of the form
    A/s+ B/(s-1)+ C/(s+1)

    Malawi_glenn showed one way to do that. I would have done it slightly differently.
    Write 1/(s(s-1)(s+1))= A/s+ B/(s-1)+ C/(s+1) and, instead of adding the fractions on the right, get rid of the fractions by multiplying both sides by s(s-1)(s+1). That gives 1= A(s-1)(s+1)+ Bs(s+1)+ Cs(s-1).
    Letting s= 0 in that gives 1= A(-1)(1)= -A so A= -1.
    Letting s= 1 in that gives 1= B(1)(2)= 2B so B= 1/2.
    Letting s= -1 in that gives 1= C(-1)(-2)= 2C so C= 1/2
    1/(s(s^2-1))= -1/s+ (1/2)/(s-1)+ (1/2)/(s+1).

    Now, if you meant 1/s(s^2+1)) orginally, then you know that
    1/s(s^2+1))= A/s+ (Bs+ C)/(s^2+1)

    We can, again, eliminate the fractions by multiplying both sides by s(s^2+ 1) to get 1= A(s^2+1)+ (Bs+ C)s

    Taking s= 0 gives 1= A so A= 1

    Unfortunately, there is no real s that makes s^2+ 1= 0 so just take s= 1 and -1 to get 1= (1)(2)+ (B(1)+C)(1) and 1= (1)(2)+ (B(-1)+ C)(-1) or
    B+C= -1 and B- C= -1. Adding those two equations 2B= -2 so B= -1 and then -1+ C= -1 so C= 0.

    1/(s(s^2+ 1))= 1/s- s/(s^2+ 1)

    Was that what you meant?
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