Question

What does partial-fraction decomposition mean?

Answer 1

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`