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Partial Fraction Decomposition

  1. Jul 26, 2009 #1
    1. The problem statement, all variables and given/known data
    1/[s*(s^2 + 4)]

    2. Relevant equations



    3. The attempt at a solution
    1/[s*(s^2 + 4)] = A/(s) + (Bs + C)/(s^2 + 4)
    => 1 = A(s^2 + 4) + (Bs + C)s

    s = 0

    1 = A(0 + 4) + (B*0 + C)*0
    A = 1/4


    s = i

    1 = A(i^2 + 4) + (Bi + C)i
    1 = A(i^2 + 4) + Bi^2 + Ci
    1 = (1/4)(-1 + 4) - B + Ci
    1 = 3/4 - B - Ci
    1/4 = -B - Ci

    No clue how to do the rest,

    Thanks!
     
  2. jcsd
  3. Jul 26, 2009 #2
    Now, I have NO idea how to do this, but try this:

    On this equation, [tex]\frac {1} {4} = -B - Ci[/tex] what values of i would eliminate C, apart from 0?
    After that, you solve for B and then you can insert that in [tex]1 = \frac {1} {4}\cdot(s^2 +4) + (Bs +C)\cdot s[/tex]
     
    Last edited: Jul 26, 2009
  4. Jul 26, 2009 #3
    I dont know. i = 1/C ?
     
  5. Jul 26, 2009 #4
    I think the substitution s=0 was a good idea, but I don't really see a need to introduce complex numbers here. It would clearly be hard to eliminate either B or C with one substitution so we seek two substitutions that will give us two equations which can easily be used to find B and C. Pretty much any two substitutions will work, but if we want to get rid of fractions and get some symmetry a good choice would be s=2 and s=-2 (symmetry comes from them being negations of each other, elimination of fraction comes from the fact that s^2 + 4 is divisible by 4 for these choices). This should give you two equations that you can combine to find B and C.
     
  6. Jul 26, 2009 #5
    Im sorry but I dont follow
     
  7. Jul 26, 2009 #6
    I solved it with coefficients!! XD I compared the right and left side to the coefficient s

    I got that 0 = C

    now if i set s = 1

    I get 1 = 5/4 + B
    B = -1/4

    Woot :)
     
  8. Jul 26, 2009 #7
    Yeah that's correct.

    In case anyone was curious what I suggested was to make the two separate substitutions s=-2 and s=2 to get:
    1 = 1/4(2^2 + 4) + (2B + C)2
    1 = 1/4((-2)^2 + 4) + ((-2)B + C)(-2)
    which can be written as:
    -1 = 4B + 2C
    -1 = 4B - 2C
    Adding these we get:
    B = -1/4
    and then we get C = (4B + 1)/2 = 0.
     
  9. Jul 26, 2009 #8

    HallsofIvy

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    Yes, 1=A(s^2 + 4) + (Bs + C)s and taking s= 0 gives 1= 4A so A= 1/4.

    I am not sure why to take s= i. You can, instead, take s= 2i so that [itex]s^2+ 4= (2i)^2+ 4= -4+ 4= 0[/itex]. That gives 1= A(0)+ (B(2i)+ C)(2i)= -4B+ 2Ci and then take s= -2i so that you have 1= -4B- 2Ci.

    But, in fact, after you know that A= 1/4, you have [itex]1= (1/4)(s^2+ 4)+ (Bs+ C)s[/itex] and you really just need two equations to solve for B and C.

    Yes, you certainly can multiply out the right side to get [itex]1= (B+ 1/4)s^2+ Cs+ 1[/itex] and then argue that, because this is true for all s, the corresponding coefficients must be equal: B+ 1/4= 0 and C= 0. That was the way you found- excellent!

    But again, you just need to get two equations. (You got the two equations B+ 1/4= 0 and C= 0.) And putting any two values for s will give two equations. Since s= 0 (my favorite value to put in!) has already been used, let's try s= 1 and s= -1 (almost as easy as 0). Setting s= 1 gives 1= (1/4)(1+ 4)+ (B(1)+ C)(1)= 5/4+ B+ C or B+ C= -1/4. Setting s= -1 gives 1= (1/4)(1+ 4)+ (B(-1)+ C)(-1)= 5/4+ B- C or B- C= -1/4. Adding the two equations eliminates C leaving 2B= -2/4 so B= -1/4. Of course, then, B+ C= -1/4+ C= -1/4 so C= 0.

    Just think "I want to find two numbers. How can I get two equations?" Any method that gives you two equations is good.
     
  10. Jul 31, 2009 #9
    you have A/s+(Bs+C)/(s^2+4). Multiplying each numerator by the other's denominator gives:
    1=A(s^2+4)+(Bs+C)s=(A+B)s^2+Cs+4A so now what are your equations for the coefficients?
     
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