jowrose Posted February 2, 2006 Share Posted February 2, 2006 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 Link to comment Share on other sites More sharing options...

Dave Posted February 2, 2006 Share Posted February 2, 2006 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. Link to comment Share on other sites More sharing options...

jowrose Posted February 3, 2006 Author Share Posted February 3, 2006 thanks. Link to comment Share on other sites More sharing options...

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