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)1x+1+2x+3=x+3(x+1)(x+3)+2(x+1)(x+1)(x+3)=3x+5x2+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)3x+5x2+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)3x+5x2+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)x2+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 AA, BB so that the following equation holds:

(3x + 5)/((x+3)(x+1)) = A/(x+3) + B / (x+1)3x+5(x+3)(x+1)=Ax+3+Bx+1

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

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

Now, in order to solve this equation, we need to "group" the color(red)(x)x terms and the terms color(blue)("without "x)without x (more precisely: x^0x0 terms):

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

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