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Partial Fraction Decomposition Problem

  1. Dec 10, 2007 #1

    Can this be decomposed, and how?
  2. jcsd
  3. Dec 10, 2007 #2


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    I'd say, yes; it can be decomposed; without my first relearning the method and trying to decompose to partial fractions. Your denominators might be [tex] \[
    [/tex] and [tex] \[
    3s + 1
  4. Dec 10, 2007 #3
    Yep. I understood that much, but I'm confused about what I can do from there because I end up with:

    [tex]\frac{A}{3s^{2}}[/tex] + [tex]\frac{B}{3s+1}[/tex]

    which would equal out to

    [tex]\frac{A}{3s^{2}}[/tex] + [tex]\frac{B}{3s+1}[/tex] = [tex]\frac{7}{3s^{2}(3s+1)}[/tex]


    7 = [tex]A(3s+1)[/tex] + [tex]3Bs^{2}[/tex]

    --which is where I'm stuck, because eliminating "s" leaves me with A = 7, which doesn't make sense if I want to separate the fraction.
  5. Dec 10, 2007 #4
    It is important to remember that if one of the factors are squared, then you need to fractions for that expression, i.e. for (7/3)//(s^2(3s+1)), you need to decompose it into a/s AND b/s^2, in addition to c/(3s+1).
  6. Dec 10, 2007 #5
    Last edited: Dec 10, 2007
  7. Dec 10, 2007 #6


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    You may need to fill in some missing steps, but this seemed to work:
    [tex] \[
    \frac{a}{{3s^2 }} + \frac{b}{{3s + 1}} = performSteps = \frac{{(3s + 1)a + 3s^2 b}}{{3s^2 (3s + 1)}} \\
    fromOriginalDeno\min ator,\;0s + 7 = 3s^2 b + 3sa + a \\
    Answer:\;\;a = 7\quad b = \frac{{ - 7}}{s} \\
  8. Dec 10, 2007 #7
    Hmm, I'll have to try to this in a second.

    --wow the method gao xiong did works but I don't recall seeing this in my initial searches through the textbook or online.

    Nice method, thanks.

    Big-T, I don't understand where the a/s would come from if the highest polynomial is 9s^3 (or the highest term is cubic) when you combine the denominator.
    Last edited: Dec 10, 2007
  9. Dec 10, 2007 #8


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    gao_xiong's form makes the best sense, and he obtained constants as answers. He decided that [tex] \[
    \] [/tex]
    is an irreducible quadratic expression and gave a suitable linear binomial expression with it.
  10. Dec 11, 2007 #9
    another problem that's confusing me

    I have one more problem.

    Would this:[tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

    simplify to something like

    [tex]\frac{Ax+B}{x^{2}+13x+38} + \frac{C}{x}[/tex] = [tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]
  11. Dec 11, 2007 #10
    well first look if [tex]\{x^{2}+13x+38}[/tex] has any roots over reals so the problem could get more simplified, since you could express that to some form of (x+-a)(x+-b) where a,b are the real roots.
    then it would be of the form
    [tex]\frac{A}{x+-a} + \frac{B}{x+-b}+\frac{C}{x}[/tex]
    Last edited: Dec 11, 2007
  12. Dec 11, 2007 #11


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    With that "s2", you are going to need both 1/s and 1/s2.
    [tex]\frac{7}{3s^2(3s+1)}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{3s+1}[/tex]
    Multiplying through by the common denominator, [itex]7= As(3s+1)+ B(3s+1)+ Cs^2[/itex]. Taking s= 0, 7= B. Taking s= -1/3, 7= C/9 so C= 63. Finally, taking s= 1, 7= 4A+ 4B+ C= 4A+ 28+ 63. 4A= 7- 91= -84, A= -21.
  13. Dec 11, 2007 #12
    I don't doubt what you're saying, but how do [you] rationalize

    a 1/s and 1/s^2 from one 1/s^2. I've never seen that before.
    Last edited: Dec 11, 2007
  14. Dec 11, 2007 #13
    You lost the 7 in the original numerator. The correct values are A = -3, B = 1, C = 9.
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