1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial fraction decomposition

  1. Mar 21, 2006 #1

    dnt

    User Avatar

    can someone help me set up this problem. it asks for the partial fraction decomposition of:

    (7x^3 - 2)/[(x^2)(x+1)^3]

    i thought you put A/x^2 + B/(x+1)^3 and solve but it doesnt work that way.
     
  2. jcsd
  3. Mar 21, 2006 #2

    TD

    User Avatar
    Homework Helper

    It's not that simple: when your factors in the denominator are raised to a power n, you need n instances of that factor in your decomposition proposal (one for each exponent, 1 -> n). In your case:

    [tex]
    \frac{{7x^3 - 2}}{{x^2 \left( {x + 1} \right)^3 }} = \frac{A}{{x^2 }} + \frac{B}{x} + \frac{C}{{\left( {x + 1} \right)^3 }} + \frac{D}{{\left( {x + 1} \right)^2 }} + \frac{E}{{x + 1}}
    [/tex]

    Do you get the idea?
     
  4. Mar 21, 2006 #3

    dnt

    User Avatar

    yeah that makes sense. is there another way to do it where you put Ax+B over the x^2 and Cx^2 + Dx + E over the (x+1)^3 and do it with just two fractions?
     
  5. Mar 21, 2006 #4

    TD

    User Avatar
    Homework Helper

    Sure, since (my A and B are the other way arround though)

    [tex]\frac{A}{{x^2 }} + \frac{B}{x} = \frac{A}{{x^2 }} + \frac{{Bx}}{{x^2 }} = \frac{{A + Bx}}{{x^2 }}[/tex]

    I don't see how this would make any significant difference though...
     
  6. Mar 21, 2006 #5

    dnt

    User Avatar

    ok cool - im sure it wouldnt but i just wanted to be sure.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Partial fraction decomposition
Loading...