What does partial-fraction decomposition mean?

1 Answer
Nov 11, 2015

Transforming a rational polynomial into a sum of simpler rational polynomials due to the factorization of the denominator.

Explanation:

Let me try and explain this to you. :)

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What is a partial-fraction decomposition?

We are all familiar with the process of adding and subtracting fractions, e.g.

#1/(x+1) + 2/(x+3) = (x+3)/((x+1)(x+3)) + (2(x+1))/((x+1)(x+3)) = (3x + 5)/(x^2 + 4x + 3)#

Now, partial-fraction decomposition does exactly the reverse thing.

It takes a rational polynomial, so a fraction like

#(3x + 5)/(x^2 + 4x + 3)#

from my example, and tries to decompose it into a sum of "simpler" fractions (to be more precise: into a sum of fractions which denominators are factors of the original fractions's denominator.)

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How to compute a partial-fraction decomposition?

1) Linear and unique factors

Let's stick with my example:

#(3x + 5)/(x^2 + 4x + 3)#

The first thing to do is a always to find a complete factorization of the denominator:

#x^2 + 4x + 3 = (x + 3)(x + 1)#

Here, all the factors are linear and unique, this is the simple case.

The goal is to to find #A#, #B# so that the following equation holds:

#(3x + 5)/((x+3)(x+1)) = A/(x+3) + B / (x+1)#

To do so, first we should multiply both sides with the denominator:

#<=> 3x+ 5 = A ( x+1) + B ( x+3)#

Now, in order to solve this equation, we need to "group" the #color(red)(x)# terms and the terms #color(blue)("without "x)# (more precisely: #x^0# terms):

#color(red)(3x)+ color(blue)(5) = color(red)(A * x) + color(blue)(A) + color(red)(B * x) + color(blue)(3 B)#

Thus, we can form an equation based on the "red" terms and an equation based on the "blue" terms:

#{ (3 = A + B), (5 = A + 3B):}#

The solution of this linear equation system is inded #A = 2# and #B = 1# which leads us to the following partial-fraction decomposition:

#(3x + 5)/((x+3)(x+1)) = A/(x+3) + B / (x+1) = 2/(x+3) + 1/(x+1)#

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2) Non-unique factors

The computation is more complicated in case that the complete factorization of the original denominator doesn't consist just of linear and unique factors.

Just a short example:

Let's say we could factorize our rational polynomial like follows:

#x/(x^2(x+1)^2(x-1)#

As the factors #x# and #x+1# occur twice in the denominator, we need to account for that by building fractions with increasing powers of those factors.
In this case, we would need to find #A#, #B#, #C#, #D# and #E# so that

#x/(x^2(x+1)^2(x-1)) = A/x + B/x^2 + C/(x+1) + D/(x+1)^2 + E/(x-1)#

holds.

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3) Non-linear factors

Last but not least, there might be non-linear factors in your factorized denominator.

Example:

#1/(x^3+x) = 1/(x(x^2+1))#

Unfortunately, you can't factorize #x^2+1# further (at least, for real numbers).

In this case, your partial-fraction decomposition needs to look like follows:
Find #A#, #B#, #C# so that the following equation holds:

#1/(x(x^2+1)) = A/x + (Bx+C)/(x^2+1)#

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When can a partial-fraction decomposition be useful?

Finally, I'd like to note that a very common application of this method is integration of rational polynomials:

If you need to compute

#int (3x + 5)/(x^2 + 4x + 3) "d"x#

and know that #1/(x+1) + 2/(x+3) = (3x + 5)/(x^2 + 4x + 3)# holds,

it's easy to integrate by "splitting" the original fraction into a sum of simpler fractions and integrating each one of them:

#int (3x + 5)/(x^2 + 4x + 3) "d"x = int 1/(x+1) "d"x + int 2/(x+3) "d"x#