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Partial Fraction Decomposition #3

  1. Jul 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the partial fraction decomposition

    1.
    [tex]\frac{x^4}{(x-1)^3}[/tex]

    2.
    [tex]\frac{1}{a^2-x^2}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    1.
    [tex]\frac{x^4}{(x-1)^3}[/tex]
    [tex]\frac{x^4}{(x-1)^3}=\frac{a}{(x-1)}+\frac{b}{(x-1)^2}+\frac{c}{(x-1)^3}[/tex]
    [tex]x^4=(x-1)^2+(x-1)b+c[/tex]
    [tex]x^4=ax^2+(-2a+b)x+(a-b+c)[/tex]
    I can determine that c is 1 by subbing in 1 for x to get rid of the [tex](x-1)[/tex] terms.

    If I plug in a number that dosen't eliminate any terms I get a two variable equation (plugging in my value for c) and I can't solve that to get a numerical answer.

    System of equations?
    [tex]x^4=ax^2+(-2a+b)x+(a-b+c)[/tex]
    all the right sides of the system would be 0 because none of the terms on the right side correspond to the terms on the left side and that would yield a=0 b=0 c=0 which is wrong.

    2.
    [tex]\frac{1}{a^2-x^2}[/tex]
    [tex]\frac{1}{(a+x)(a-x)}[/tex]
    [tex]\frac{1}{a^2-x^2}=\frac{a}{(a+x)}+\frac{b}{(a-x)}[/tex]
    [tex]1=(a-x)a+(a+x)b[/tex]
    From here I'm not quite sure what to do, if you sub in x=a you get

    [tex]1=2ab[/tex]
    [tex]\frac{1}{2a}=b[/tex]
    could you also have
    [tex]\frac{1}{2b}=a[/tex]

    is x=-a you get [tex]\frac{1}{\sqrt{2}}=a[/tex]

    and if you foil and get it into polynomial form...
    [tex]1=(-a+b)x+(ba+a^2)[/tex]

    If you created a matrix from this you would have a non linear system of equations, I don't know how to solve those and I don't think you need to know how to solve this problem.
     
  2. jcsd
  3. Jul 4, 2010 #2
    1. It is not a proper fraction. Convert it first before you decompose it.

    2. It might help to use "A and B" instead of "a and b" as your unknown constants...so you wouldn't mix up your 'a's?
     
  4. Jul 4, 2010 #3
    oh. Thanks! Those hints helped me solve my problems.
     
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