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I was doing some exercises on PFD(Partial Fraction Decomposition) to fresh up my memory on how to do it as I'll have use of it on my next test. I came across the following piece of text:

 

 

We need values of A and B so that the numerator of the expression on the left is the same as the numerator of the term on the right.  Or,

                                                 

This needs to be true regardless of the x that we plug into this equation.  As noted above there are several ways to do this.  One way will always work, but can be messy and will often require knowledge that we don’t have yet.  The other way will not always work, but when it does it will greatly reduce the amount of work required.

 

In this set of examples the second (and easier) method will always work so we’ll be using that here.  Here we are going to make use of the fact that this equation must be true regardless of the x that we plug in. 

 

So let’s pick an x, plug it in and see what happens.  For no apparent reason let’s try plugging in .   Doing this gives,

 

                                                

 

 

This question isn't regarding that specific PDF but instead something that was mentioned here, this part: 

 

"One way will always work, but can be messy and will often require knowledge that we don’t have yet.  The other way will not always work, but when it does it will greatly reduce the amount of work required."

 

 

I am curious about two things here.

 

 

1. What is the first way to solve these that always works. Solving it like a system of equations or something else?

 

2. Why would the other way not work? I can kind of imagine it wouldn't always work, probably something to do with the fact that a solution is lost because of something to do with a denominator? Probably a trigonometric function of x? But that's my guess. 

 

 

Thank you in advance.

Quazars  Apr 17, 2017
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Perhaps the example below will help answer your questions:

 

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I should add that we could still use the substitution method to find B above.  Having found A, we just use another value for x (say, x = 1) to find B.  So this probably doesn't address the issue of a method that doesn't always work!

.

Alan  Apr 17, 2017
edited by Alan  Apr 17, 2017

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