Multivariable factor decomposition

[epkid08]

(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:

f(x,y)=\frac{y^2-xy+1}{(x+y)(x-y)}

I want to decompose this to:

\frac{A}{x+y} + \frac{B}{x-y}

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


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

So I would pick a point, say (1,1), plug it into \frac{\partial f}{\partial x} to get the x coordinate of a point, P_1, then I would plug in the same point, (1,1), into \frac{\partial f}{\partial y} to get the y coordinate of the P_1. After I get two points,P_1 and P_2, 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?

[jostpuur]

Let's say I had this expression:

f(x,y)=\frac{y^2-xy+1}{(x+y)(x-y)}

I want to decompose this to:

\frac{A}{x+y} + \frac{B}{x-y}

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


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.


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


What points are you talking about?


My question is, is this the right process to decompose the fraction, or am I way off?

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.

[NoMoreExams]

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


Why are you messing with partials? Let x = y in the first expression and you get

x^2 - x^2 + 1 = 2Bx \Rightarrow B = \frac{1}{2x}

Similarly let x = -y and you get

x^2 + x^2 + 1 = 2Ax \Rightarrow A = x^{2} + \frac{1}{2}