## Notes on the partial fraction decomposition: why it always works

If you’ve taken any intro to Calculus class, you’re probably familiar with partial fraction decomposition.

In case you’re not, the idea is that you’re given some rational function with an awful denominator that you want to integrate, like:

$\frac{4x-2}{(x-2)(x+4)}$

And you break it up into smaller, simpler fractions:

$\frac{1}{x-2} +\frac{3}{x+4}$

This is the idea. If we get into the details, it gets fairly ugly — in a typical calculus textbook, you’ll find a plethora of rules regarding what to do in all sorts of cases: what to do when there are repeated linear factors, quadratic factors, repeated quadratic factors, and so on.

Since the textbooks generously cover this for us, we’ll assume that we know what to do with a rational polynomial with some polynomial as the numerator, and some number of linear or quadratic factors in the denominator. We can do partial fraction decomposition on this. If we like, we could integrate it too. I’m talking about anything of this form:

$\frac{P(x)}{((ax+b)(cx+d) \cdots)((ex^2+fx+g)(hx^2+ix+j) \cdots)}$

Although we won’t prove this, this seems fairly believable. We’ll assume that once we get a fraction into this form, we’re done and we can let existing partial fraction methods take care of the rest.

### Can Partial Fractions Fail?

What if we have a polynomial greater than a quadratic in the denominator? So let’s say:

$\frac{1}{x^3+1}$

Fortunately, here the denominator can be factored, giving us a form we can deal with:

$\frac{1}{(x+1)(x^2-x+1)}$

But we were lucky that time. After all, not all polynomials can be factored, right? What if we have this:

$\frac{1}{x^3+5}$

We can’t factor this. What can we do?

It turns out that this isn’t a huge problem. We never required the coefficients of the factors to be integers! Although the factorization is awkward, it can still be factored:

$\frac{1}{(x + 5^{1/3})(x^2-5^{1/3}x+5^{2/3})}$

Other than making the next step somewhat algebraically tedious, this decomposition is perfectly valid. The coefficients need not be integers, or even be expressed with radicals. As long as every coefficient is real, partial fraction decomposition will work fine.

### Universality of Partial Fractions

The logical next question would be, can all radical functions be written in the previous partial fraction decomposition-suitable form? Looking through my calculus textbooks, none seemed to provide a proof of this — and failing to find a proof on the internet, I’ll give the proof here.

We need to prove that any polynomial that might appear in the denominator of a rational function, say $Q(x)$, can be broken down into linear or quadratic factors with real coefficients.

In order to prove this, we’ll need the following two theorems:

• Fundamental Theorem of Algebra — any polynomial of degree n can be written as a product of n linear complex factors: $Q(x) = (x-z_1) (x-z_2) \cdots (x-z_n)$
• Complex Conjugate Root Theorem — if some complex number $a + bi$ is a root of some polynomial with real coefficients, then its conjugate $a-bi$ is also a root.

Starting with the denominator polynomial $Q(x)$, we break it down using the Fundamental Theorem of Algebra into complex factors. Of these factors, some will be real, while others will be complex.

Consider the complex factors of $Q(x)$. By the complex conjugate root theorem, for every complex factor we have, its conjugate is also a factor. Hence we can take all of the complex factors and pair them up with their conjugates. Why? If we multiply a complex root by its complex conjugate root: $(x-z)(x-\bar{z})$ — we always end up with a quadratic with real coefficients. (you can check this for yourself if you want)

Before, we were left with real linear factors and pairs of complex factors. The pairs of complex factors multiply to form quadratic polynomials with real coefficients, so we are done.

At least in theory — partial fraction decomposition always works. The problem is just that we relied on the Fundamental Theorem of Algebra to hand us the roots of our polynomial. Often, these roots aren’t simple integers or radicals — often they can’t really be expressed exactly at all. So we should say — partial fraction decomposition always works, if you’re fine with having infinitely long decimals in the decomposed product.

### One Response to Notes on the partial fraction decomposition: why it always works

1. Patrick says:

it is one of the technique to solve calculus problem by using partial decomposition. You should use it in a practical way. If it is too difficult to decompose, it is then NOT practical to use the method. You should use other method.