Partial Fraction Decomposition

[jowrose]

Ok, I remember learning about partial fraction decomposition a while ago, and how it involved separating a fraction into the 2 fractions that existed before they were multiplied by a fraction of 1 to obtain identical denominators. However, I have forgotten how to do this, and I was wondering if someone could refresh my memory on how it's done.

 

Thanks,

 

john

[Dave]

Here's a simple example. Let's split up the fraction [math]\frac{1}{x(x-1)}[/math] by partial fractions.

 

First, since both of our factors are linear, let's just suppose that:

 

[math]\frac{1}{x(x-1)} \equiv \frac{A}{x} + \frac{B}{x-1}[/math]

 

where A and B are some co-efficients. Then we re-combine the two fractions on the right hand side to get:

 

[math]\frac{1}{x(x-1)} \equiv \frac{A(x-1) + Bx}{x(x-1)}[/math]

 

Since both of the denominators are equal we can say:

 

[math]1 \equiv A(x-1) + Bx[/math]

 

Now, since this holds for any x we care to choose, all we have to do is let x = 1 to find that B = 1, and x = 0 to see that A = -1.

 

So [math]\frac{1}{x(x-1)} \equiv \frac{1}{x-1} - \frac{1}{x}[/math].

 

This is a really simple example; it gets trickier when you have non-linear factors, for example. But this is the general idea.

[jowrose]

thanks.