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Partial fraction decomposition

  1. Nov 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Go here: http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1999s.pdf

    You do not need to read the problem. In solution two for 1999 A3, I can do the partial fraction decomposition but I get something different from that... I am wondering whether that comes from the partial fraction decomposition or something else...

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 6, 2007 #2


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    We don't need to read the problem and you won't tell us what you did- but you want someone to tell you if what you did was right? How, exactly?
  4. Nov 6, 2007 #3
    It would take me like 6 lines of latex to show what I did, so I am asking whether it was the right choice of technique not whether I performed it right. I want to know how they arrived at the equality after "Note that".
  5. Nov 6, 2007 #4


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    1- 2x- x2= -(x2+ 2x+ 1-1)+1= 2- (x-1)2 so 1- 2x- x2= 0 has roots [itex]x= 1\pm \sqrt{2}[/itex] and so can 1-2x-x2 can be factored as [itex](x-1-\sqrt{2})(x-1+\sqrt{2})[/itex]. Using "partial fractions" then
    [tex]\frac{1}{1-2x-x^2}= \frac{A}{x-1-\sqrt{2}}+ \frac{B}{x-1+\sqrt{2}}[/tex]
    gives A= [itex]-1/2\sqrt{2}[/itex] and B= [itex]1/2\sqrt{2}[/itex]. The final form is got by dividing both numerator and denominator of the first fraction by [itex]1+\sqrt{2}[/itex] and the numerator and denominator of the second fraction by [itex]1-\sqrt{2}[/itex].
  6. Nov 6, 2007 #5
    I tried that and then x is divided by what you want it to be multiplied by!
  7. Nov 7, 2007 #6
    Can you explain how that gives the final form? I cannot figure out how to manipulate that any more.

    EDIT: never mind I see
    Last edited: Nov 7, 2007
  8. Nov 7, 2007 #7
    You're factorization is wrong. The roots of it should be x = -1 +/- sqrt(2).
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