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Partial fraction decomposition

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the partial fraction decomposition of :
    [tex]\frac{x^2}{(1-x^4)^2}[/tex]



    3. The attempt at a solution

    [tex]\frac{x^2}{(1-x^4)^2}=\frac{A}{(1-x^4)}+\frac {B}{(1-x^4)^2}[/tex]

    [tex]=A(1-x^4)+B[/tex]

    when x=1

    [tex]1=A(1-1^4)+B[/tex]

    Hence B=1 and A=0

    [tex]\frac{x^2}{(1-x^4)^2}=\frac{0}{(1-x^4)}+ \frac{1}{(1-x^4)^2}[/tex]

    How come according to the answers at the back of the book l am wrong. Where have l done a mistake ?
     
  2. jcsd
  3. Mar 20, 2009 #2
    When x = 1, you have
    [itex]1 = A(1-1^{4})+B = A(0)+B = B[/itex]
    So B = 1 works, but what values of A satisfy that equation? What values don't? If there are multiple values A can have, then you need to determine which is the correct one.
     
  4. Mar 20, 2009 #3

    Dick

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    Well, 0/(1-x^4)^2+1/(1-x^4)^2 clearly does not equal x^2/(1-x^4)^2, now does it? You should probably factor the denominator completely and then look up what partial fractions look like if you have a quadratic in the denominator.
     
    Last edited: Mar 20, 2009
  5. Mar 20, 2009 #4

    Isn't the denominator already in a factored form ? Should l expand it ?
     
  6. Mar 20, 2009 #5

    gabbagabbahey

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    The denominator is a reducible quartic...I'll give you a hint to factoring it: [itex](1-x^a)(1+x^a)=[/itex]____?
     
  7. Mar 20, 2009 #6
    [tex](1-x^4)(1-x^4)=(1-x^2)^2 (1-x^2)^2[/tex]
     
  8. Mar 20, 2009 #7

    gabbagabbahey

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    No, I meant [itex]1-x^4[/itex] is a reducible quartic....factor that:smile:
     
  9. Mar 20, 2009 #8
    I am getting very confused. Show me an example which has nothing to do with this question maybe l will understand what you are trying to say
     
  10. Mar 20, 2009 #9

    gabbagabbahey

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    I'm not sure how much simpler I can make it....do you really not know how to factor [itex]1-x^4[/itex]? It is something you should have been taught in highschool....
     
  11. Mar 21, 2009 #10

    "Reducible quartic" send me a little bit off track:smile:


    [tex]1-x^4=(1-x^2)(1+x^2)=(1+x)(1-x)(1+x^2)[/tex]


    Will this mean l will be left with :

    [tex]\frac{x^2}{(1-x^4)^2}=\frac{A}{(1+x)}+\frac {B}{(1-x)}+\frac{Cx+D}{1+x^2}[/tex]
     
  12. Mar 21, 2009 #11

    gabbagabbahey

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    Yes!:smile:


    Careful:

    [tex]\frac{x^2}{(1-x^4)^2}=\frac{x^2}{(1+x)^2(1-x)^2(1+x^2)^2}[/tex]
     
  13. Mar 21, 2009 #12

    Uhhmmm, which means

    [tex]\frac{x^2}{(1-x^4)^2}=\frac {x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{1-x}+\frac{D}{1-x}+\frac{Ex+F}{1+x^2)}^2[/tex]
     
  14. Mar 21, 2009 #13

    gabbagabbahey

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    Your LaTeX is a little a sloppy; surely you mean:

    [tex]\frac{x^2}{(1-x^4)^2}=\frac {x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{A}{(1+x)}+\frac{B}{(1+x)^2}+\frac{C}{(1-x)}+\frac{D}{(1-x)^2}+\frac{Ex+F}{(1+x^2)}+\frac{Gx+H}{(1+x^2)^2}[/tex]

    right?
     
  15. Mar 21, 2009 #14

    Yes that what l mean, is it correct ?
     
  16. Mar 21, 2009 #15

    gabbagabbahey

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    Yes, now determine the constants....
     
  17. Mar 21, 2009 #16
    [tex]x^2=A(1-x)^2(1+x)+B(1-x)^2+C(1-x)(1+x)^2+D(1+x^2)^2+(Ex+F)(1-x)^2+(Gx+H)(1-x)^2[/tex]

    when x=1

    [tex]1= A(1-1)^2(1+1)+ B(1-1)^2+C(1-1)(1+1)^2+D(1+1)^2+(Ex+F)(1-1)^2+(Gx+H)(1-1)^2[/tex]

    [tex]4D=1[/tex]

    [tex]D=\frac{1}{4}[/tex]

    when x=-1

    [tex]1=A(1+1)^2(1-1)+B(1+1)^2+C(1+1)(1-1)^2+\frac{1}{4}(1+1)^2+(Ex+F)(1+1)^2+(Gx+H)(1--1)^2[/tex]


    when x=o

    0=A+B+C+\frac{1}{4}+(Ex+F)+ (Gx+H)

    It got so many unknowns and only three equations, how do l get around this last hurdle
     
    Last edited: Mar 21, 2009
  18. Mar 21, 2009 #17

    djeitnstine

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    Compare coefficients.
     
  19. Mar 21, 2009 #18

    gabbagabbahey

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    That doesn't look right at all, be careful with your multiplication!
     
  20. Mar 21, 2009 #19

    gabbagabbahey

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    There is an easier way than solving the system of equation this method results in.

    Try plugging in x=i and x=-i for starters
     
  21. Mar 21, 2009 #20
    Does this look correct:

    [tex]A(1+x)(1-x)^2(1+x^2)^2+B(1-x)^2(1+x^2)^2+C(1-x)(1+x^2)^2(1+x)^2+D(1+x^2)^2(1+x)^2+(Ex+F)(1+x)^2(1+x^2)(1-x)^2+(Gx+H)(1-x)^2(1+x)^2[/tex]
     
    Last edited: Mar 21, 2009
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