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Multivariable factor decomposition

  1. Aug 18, 2008 #1
    (Keep in mind, I made this off the top of my head, so if something cancels easy, ignore it)

    Let's say I had this expression:


    I want to decompose this to:

    [tex]\frac{A}{x+y} + \frac{B}{x-y}[/tex]

    So i begin the process:
    [tex]y^2-xy+1=A(x-y) + B(x+y)[/tex]
    [tex]y^2-xy+1=x(A+B) + y(B-A)[/tex]

    At this point, I can't just plug in random numbers, I need to plug in points that correspond to this point:
    [tex](\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})[/tex]

    So I would pick a point, say (1,1), plug it into [tex]\frac{\partial f}{\partial x}[/tex] to get the x coordinate of a point, [tex]P_1[/tex], then I would plug in the same point, (1,1), into [tex]\frac{\partial f}{\partial y}[/tex] to get the y coordinate of the [tex]P_1[/tex]. After I get two points,[tex]P_1[/tex] and [tex]P_2[/tex], I can plug them into the equation and solve the system to find A and B.

    My question is, is this the right process to decompose the fraction, or am I way off?
  2. jcsd
  3. Aug 18, 2008 #2
    At this point you see a problem. There does not exist such numbers A and B that the last equation would be true for all x and y. It follows that your decomposition attempt doesn't work.

    What points are you talking about?

    I did not understand your process, but since the task you are trying to do is impossible, I might guess your process was not right.
  4. Aug 18, 2008 #3
    Why are you messing with partials? Let x = y in the first expression and you get

    [tex] x^2 - x^2 + 1 = 2Bx \Rightarrow B = \frac{1}{2x} [/tex]

    Similarly let x = -y and you get

    [tex] x^2 + x^2 + 1 = 2Ax \Rightarrow A = x^{2} + \frac{1}{2} [/tex]
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