Why does partial fraction decomposition always work?

Say you have a function $p(x)/q(x)$ for some polynomials $p(x)$ and $q(x)$ and $p$ has a lower degree than $q$.

Say $q$ has degree three and $p$ has degree two. If you partially decompose it, you'll get three variables $A$,$B$,$C$ and three equations with these variables. You can solve these variables and find the values of the numerators.

But not all systems of three equations have solutions, so how do we know that $A$, $B$ and $C$ always exists for all rational functions that meet the criteria for decomposition?

• I believe the answer lies in abstract algebra, not linear algebra. The number-theoretic analogue of partial fraction decomposition is the Prufer decomposition of $\Bbb Q/\Bbb Z$. – anon Feb 27 '14 at 2:54
• @anon Haha, that's way above my head. Is there any chance I'll be able to understand this only knowing calculus and linear algebra? – dfg Feb 27 '14 at 2:57
• The answers here are helpful, but it still seems that partial fraction decomposition doesn't work in cases like $x \over (x - 1)(1 - x)$. If I assume that constants $A_1$ and $A_2$ exist, and try a couple values for $x$, it leads me to the conclusion that $2 = 3$. Does partial fraction decomposition simply not apply in this case? – Alex D May 11 '17 at 8:49
• – Watson Nov 26 '18 at 10:09

One nice way to get the particular constants (and that gives a hint of why it works) is to take: $$q(x) = (x - a_1) (x - a_2) \ldots (x - a_n)$$ For simplicity, take all $a_i$ different. If you can write: $$\frac{p(x)}{q(x)} = \frac{A_1}{x - a_1} + \cdots + \frac{A_n}{x - a_n}$$ Now compute: $$\lim_{x \to a_k} \frac{(x - a_k) p(x)}{q(x)} = A_k$$ The limit on the left hand side is easy to compute, and it is clear that this is consistent. It is even: $$A_k = \lim_{x \to a_k} \frac{x - a_k}{q(x)} \cdot p(x) = \frac{p(a_k)}{q'(a_k)}$$

A similar trick works for repeat roots of $q(x)$. And the case of quadratic factors (irreducible over the reals) can be deduced factoring over the complex numbers.

• Isn't this assuming that you already have a decomposition? This trick is useful though in actually computing the values of the $A_i$, so it's worth mentioning. – Sergio Da Silva Feb 27 '14 at 20:06
• I'm not claiming it is a proof, just a heuristic of why it works (and give a formula for the constants in an important special case). – vonbrand Feb 27 '14 at 21:27

In your specific case, where the degree of $p$ and $q$ are 2 and 3 respectively, we can prove such a decomposition directly. Assuming we are working over the reals, we can decompose $q(x)$ as the product of three linear terms, or the product of an irreducible quadratic and a linear term. Knowing this alone, and writing $q(x) = ax^3+bx^2+cx+d$ and keeping track of all the variables will let you decompose your fraction (albeit a rather messy process).

Of course this brute force method isn't what we would use to prove that a decomposition always exists. If you look in Spivak's Calculus book, in chapter 19, he discusses the partial fraction decomposition theorem, and states "The integration of an arbitrary rational function depends on two facts; the first follows from the Fundamental Theorem of Algebra, but the second will not be proven in this book". The Fundamental Theorem of Algebra portion allows you to factor out $q(x)$ into linear and irreducible quadratic terms. The book is quite rigorous in all it's proofs, so if the proof was left out, it is probably because it is too difficult at an elementary "calculus and linear algebra" level.

I think a step towards the proof would be the following. Let's look at $\frac{f(x)}{g(x)}$ where $f,g\in \mathbb{R}[x]$, and factor $g(x) = a(x)b(x)$ where $a$ and $b$ have no common factors. Since $\mathbb{R}[x]$ is a Euclidean domain (we have a division algorithm), we can find $c(x)$ and $d(x)$ such that $c(x)a(x)+d(x)b(x)=1$ (this is Bezout's identity). Now multiply by $p(x)$ and divide by $q(x)$ to get $$\frac{c(x)a(x)p(x)}{q(x)}+ \frac{d(x)b(x)p(x)}{q(x)}=\frac{p(x)}{q(x)} \Rightarrow \frac{c(x)p(x)}{b(x)}+ \frac{d(x)p(x)}{a(x)}=\frac{p(x)}{q(x)}$$

Now proceeding by induction on the number of factors might get you somewhere. There still is the issue of forcing the numerators to be linear or quadratic, which may require more work. Hopefully this sheds some light onto a possible direction for a proof and on the potential problems that arise.

• Allright, fair enough. Is there a way I can understand why it always works intuitively instead of a formal proof? – dfg Feb 27 '14 at 19:18
• dfg, I added an addition to my answer which might help you understand where a possible decomposition is coming from. – Sergio Da Silva Mar 1 '14 at 22:07

$$\text{Why partial fraction decomposition works?}$$

Assume that $\frac{p(x)}{q(x)}=\frac{p(x)}{(x-a)^nR(x)}$, where $R(a)\neq0$. Since $R(x)$ and $(x-a)^n$ don't have a root in common, there are polynomials $s(x)$ and $r(x)$ with $\text{deg}(s)<n$ and $\text{deg}(r)<\text{deg}(R)$ such that

$$s(x)R(x)+r(x)(x-a)^n=1$$

Recall that this is Bezout's identity which follows from applying Euclid's division algorithm to $R(x)$ and $(x-a)^n$. Therefore (multiplying by $p(x)$)

$$s(x)p(x)R(x)+r(x)p(x)(x-a)^n=p(x)$$

We can divide $s(x)p(x)=(x-a)^nm(x)+S(x)$ with $deg(S)<n$ and write

$$S(x)R(x)+(x-a)^n[r(x)p(x)+m(x)]=p(x).$$

Notice that $\text{deg}(S(x)R(x))<n+\text{deg}(R)$. Since $\text{deg}(p)<\text{deg}(q)=n+\text{deg}(R)$, we must have $$\color{red}{\text{deg}(r(x)p(x)+m)<\text{deg}(R)}$$

Therefore $$\frac{p(x)}{q(x)}=\frac{S(x)}{(x-a)^n}+\frac{r(x)p(x)+m(x)}{R(x)}$$

Since $\text{deg}(S)<n$ we can write $$S(x)=A_1(x-a)^{n-1}+A_2(x-a)^{n-2}+...+A_{n-1}(x-a)+A_n$$

From where we get

$$\frac{p(x)}{q(x)}=\frac{A_1}{(x-a)}+...+\frac{A_n}{(x-a)^n}+\frac{r(x)p(x)+m(x)}{R(x)}$$

From the inequality in $\color{red}{\text{red}}$ we see that $\frac{r(x)p(x)+m(x)}{R(x)}$ is a proper fraction, so we can repeat the same procedure above with it to extract the principal parts of the partial fraction decomposition corresponding to the other factors of $R(x)$.