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Partial fractions

  1. Dec 15, 2009 #1
    Hi, had to learn partial fractions last year for laplace transforms, but have forgotten the general rules, and now i cant work out how to turn this into partial fractions:

    [tex]\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}[/tex]

    [tex]\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}=\frac{A}{s^{2}+4}+\frac{B}{s^{2}+9}[/tex]
    so

    [tex]s=A\left(s^{2}+9\right)+B\left(s^{2}+4\right)[/tex]

    i understand that i am supposed to strategically choose s to make A or B zero, but sith the s^2 i'm not sure what to do

    thanks
     
  2. jcsd
  3. Dec 15, 2009 #2
    make the coefficient 0
     
  4. Dec 15, 2009 #3
    by setting s to what?
     
  5. Dec 15, 2009 #4

    Mark44

    Staff: Mentor

    This is not the way to decompose irreducible quadratic factors. Here's what you should have:
    [tex]\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}=\frac{As + B}{s^{2}+4}+\frac{Cs + D}{s^{2}+9}[/tex]

     
  6. Dec 15, 2009 #5
    ok, yeah i wondered, but mupad's soln looked so simple i didnt think there could be 4 unkowns. So can i now multiply everything by the denomiator of s on the LHS.

    What i dont understand is usually you would choose values for s to make most unkowns 0, then calculate for each unkown. but here, there is no s that can make
    S^2 + 4 =0 without imaginary numbers?
     
  7. Dec 15, 2009 #6
    You should then write the partial fraction expansion as follows:

    s/[(s^2+4)(s^2+9)] = A/(s+2i) + A*/(s-2i) + B/(s+3i) + B*(s-3i)

    Where the star denotes complex conjugation. The "strategic value" method now becomes trivial, you don't have to actually write down any equations. If you know about complex analysis, you'll recognize the coefficients as the residues of the function at the poles and compute them directly. If you don't know this, you can easily see that e.g. multiplying boith sides by s + 2i and taking the limit s to -2i will yield A.
     
  8. Dec 15, 2009 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    An obvious way to simplify
    [tex]\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)} =\frac{As+B}{s^{2}+4}+\frac{Cs+D}{s^{2}+9}[/tex]
    is to multiply both sides by [itex](s^2+ 4)(s^2+ 9)[/itex] to get
    s= (As+B)(s^2+9)+ (Cs+D)(s^2+4).

    Because those denominators were "irreducible" you cannot "make a coefficient 0" but you still need only 4 equations to solve for the 4 constants.

    One way would be to take s to be 4 simple numbers, say s= 0, s= 1, s= -1, and s= 2.

    If n= 0, your equation is 0= 9B+ 4D. If s= 1, it is 1= 10A+ 10B +5s+ 5D, etc.

    Another way to get 4 equations is to multiply out that right side:

    s= (As+B)(s^2+9)+ (Cs+D)(s^2+9)= As^3+ Bs^2+ As+ 9B+ Cs^3+ Ds^2+ Cs+ D.
    0s^3+ 0s^2+ s+ 0= (A+ C)s^3+ (B+D)s^2+ (A+C)+ B+D.

    In order for those to be equal for all s, the "corresponding coefficients" must be the same: we must have A+ C= 0, B+ D= 0, A+ C= 1, and B+ D= 0.
     
  9. Dec 15, 2009 #8
    Why can't this be done:
    As^2 +9A +Bs^2 +4B=0
    A+B=0
    9A+4B=s
    A=s/5, B=-s/5
     
  10. Dec 16, 2009 #9

    Mark44

    Staff: Mentor

    Mainly because it isn't correct. It took me some time to figure out what you had done, but going back to the OP, I was able to follow your reasoning. In the first post in this thread, a.mlw.walker thought that (incorrectly) that this would be the way to go:
    [tex]\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)} =\frac{A}{s^{2}+4}+\frac{B}{s^{2}+9}[/tex]
    And this leads to:
    [tex]s=A\left(s^{2}+9\right)+B\left(s^{2}+4\right)[/tex]

    Your first equation is As^2 +9A +Bs^2 +4B=0, but where you have zero on the right side, you should have s, but I think that was a typo on your part. You are reasoning that since there is no s^2 term on the right side, A + B (the coefficient of s^2) has to be zero, and that 9A + 4B has to be s.

    Your values for A and B satisfy the equation A + B = 0, but they don't satisfy 9A + 4B = s.
     
  11. Dec 19, 2009 #10
    Why not? What is 9s/5 -4s/5 ?
     
  12. Dec 19, 2009 #11
    Solved it by doing what Mark44 said.
     
  13. Dec 19, 2009 #12

    Mark44

    Staff: Mentor

    9s/5 - 4s/5 = s

    The main problem with your technique is the difference between equations that are identically true (true for all values of the variable) and those that are conditionally true (true only for certain values of the variable). In partial fractions decomposition, the idea is to rewrite the product on the left side below as a sum in such a way that the equation is identically true. A and B on the right side have to be constants.
    [tex]\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)} =\frac{A}{s^{2}+4}+\frac{B}{s^{2}+9}[/tex]

    By multiplying both sides by (s2 + 4)(s2 + 9), you get the next equation, which also has to be identically true.
    [tex]s=A\left(s^{2}+9\right)+B\left(s^{2}+4\right)[/tex]

    If we group like terms together on the right side, we get
    s = (A + B)s2 + (9A + 4B)

    This is the same as 0s2 + 1s + 0 = (A + B)s2 + (9A + 4B)

    For this equation to be identically true, we have to have A + B = 0 and 9A + 4B = 0 so that the coefficients of s2, s, and the constant term are equal on both sides of the equation. The only way two polynomials of the same degree can be identically equal is for their corresponding coefficients to be equal.

    Using you technique, we have A + B = 0 and 9A + 4B = 0, which gives A = B = 0, and we also have an s term on the left side that is not matched on the right side.

    As it turns out in this problem, your values of A = s/5 and B = -s/5 happen to work out, but I'm reasonably sure this is only an accident.
     
  14. Dec 19, 2009 #13
    you were talking to vin300 not me right?
     
  15. Dec 20, 2009 #14

    HallsofIvy

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    The whole point of "partial fractions" is that A and B must be constants, not functions of s.
     
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