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

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