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Partial fractions

  1. Jan 27, 2013 #1

    trollcast

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    1. The problem statement, all variables and given/known data

    Use the method of partial fractions to show that:

    $$\frac{2x^2}{(1-x(1+x)} $$

    , may be written as:

    $$-2+\frac{1}{1-x}+\frac{1}{1+x}$$

    , where $$\lvert x\rvert\neq1 $$.

    2. Relevant equations
    3. The attempt at a solution
    I obviously know how to do it but in the solution am I allowed to do, let x = 1, to cancel some of the terms to figure out the constants in the partial fractions? Or do I have to do simultaneous equations by comparing the coefficients?
     
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  3. Jan 27, 2013 #2

    SammyS

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    It depends upon the stage in the solution at which you would set x = 1, or x = -1 .
     
  4. Jan 27, 2013 #3

    I like Serena

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    Hi trollcast! :smile:

    I guess you could let x=1, but I don't see how that would help you...
    Neither expression is defined if you do that, and you won't be able to properly compare them.
     
  5. Jan 27, 2013 #4

    trollcast

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    Ok I'll type my solution out to check its right:

    $$Let \frac{2x^2}{(1-x)(1+x)} = A + \frac{B}{(1-x)}+\frac{C}{(1+x)} $$
    $$\frac{A(1+x)(1-x)+B(1+x)+C(1-x)}{(1-x)(1+x)}=2x^2$$
    $$\Rightarrow A(1+x)(1-x)+B(1+x)+C(1-x) = 2x^2$$
    Compare coefficients of x^2: $$-A=2\\A=-2$$
    Let x = 1: $$2B=2\\B=1$$
    Let x = -1: $$2c=2\\c=1$$

    Is that ok?
     
  6. Jan 27, 2013 #5

    I like Serena

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    Nice!

    Slightly unconventional, but it seems fine to me. :wink:
     
  7. Jan 27, 2013 #6

    SammyS

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    The above line should be:
    [itex]\displaystyle \frac{A(1+x)(1-x)+B(1+x)+C(1-x)}{(1-x)(1+x)}=\frac{2x^2}{(1-x)(1+x)}[/itex]
    It's a fairly well known method (trick).

    Another variation is to do those substitutions for x immediately after you have

    [itex]\displaystyle A(1+x)(1-x)+B(1+x)+C(1-x)=2x^2\ [/itex]

    to get B and C.

    Then plug those values in & let x be some convenient number, like x = 0, to find A .
     
    Last edited: Jan 27, 2013
  8. Jan 27, 2013 #7

    trollcast

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    Thanks, I just wasn't sure whether the |x|=/= 1 , bit meant I couldn't use let x = 1.
     
  9. Jan 27, 2013 #8

    I like Serena

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    Well spotted.
    In this case you can get away with it.

    When you have your final equation, the solution for A, B, and C is unique and valid for any x, not just for |x|≠1.
    After that the same solution applies if you restrict x to |x|≠1.
    Formally, you should write something like that down.

    Btw, the ultimate check is when you substitute your A, B, and C, and verify they form a proper solution.
    When you have that, the steps how you got there become irrelevant (although your teacher may want to see them ;)).
     
  10. Jan 27, 2013 #9

    trollcast

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    So the where |x|≠1 is only so it makes mathematical sense otherwise you would imply the denominator could be 0.
     
  11. Jan 27, 2013 #10

    I like Serena

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    Yes.
     
  12. Jan 27, 2013 #11

    SammyS

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    The way I like to think of this is as follows.

    If there are values of A, B, and C which make the following true for all x ,
    [itex]\displaystyle A(1+x)(1-x)+B(1+x)+C(1-x)=2x^2\,,[/itex]​
    then certainly the same values of A, B, and C will make the same equation true if we exclude x = ±1 .
     
  13. Jan 28, 2013 #12

    vela

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    This is called the Heaviside cover-up method. Many claim "cover-up" means you cover up the term in the denominator that vanishes, but I suspect it really refers to the fact you're doing something sketchy because you're multiplying both sides of the equation by 0. :wink:
     
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