# The logic behind partial fraction decomposition

In the general case of any function would be interesting but my question is concerning the general case of polynomials with integer powers. I can use the method of partial fractions in the simple case required for an introductory course on integration, but I'm not sure I really understand it or could describe WHY it works.

For an arbitrary example $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are some string of polynomials, I know to check that this is indeed a proper fraction before continuing. Next I know that the number of constants I need to find is determined by the degree of $Q$. But why is this? I don't really get how this is known in advance. Further, in the case of repeated roots, what is the mechanism at play behind how a fraction like this is expanded? [i.e. $(x+1)^3$ in the denominator would be expanded with $\frac{A}{(x+1)} + \frac{B}{(x+1)^2} + \frac{C}{(x+1)^3}]$. I'm not really that advanced in mathematics but I know that in the case of a repeated root it would be necessary to include all the powers of the root to "account for" all the possible cases of what you may have started with. From the theoretical aspect though, why does this work? Why do we know that the degree of $Q$ determines the number of constants and that we can find all the constants by incrementing every power of the root in the case of a repeated root? I want lift the rug up and see what mechanisms are at play in order to understand this better.

• You may get something from the answers to math.stackexchange.com/questions/48051/… Commented Aug 22, 2012 at 1:39
• Thanks for your comment. I looked at that question before posting but it didn't help me in considering how we know that the degree of Q in my example will be equal to the number of unknown constants that must be found. Perhaps this is trivially knowable but I lack that knowledge and would be interested in acquiring it.
– R R
Commented Aug 22, 2012 at 1:42

Great question! First, let's tackle just the case where the roots of $Q(x)$ are all distinct. One way to conceptualize what's going on is the following: if $r$ is a root of $Q(x)$, then as $x \to r$, the function $f(x) = \frac{P(x)}{Q(x)}$ (we always assume $P, Q$ have no common roots) goes to infinity. How quickly does it go to infinity? Well, write $Q(x) = (x - r) R(x)$. Then

$$\frac{P(x)}{Q(x)} = \frac{1}{x - r} \left( \frac{P(x)}{R(x)} \right)$$

and $R(r) \neq 0$. So we see that as $x \to r$, this expression goes to infinity like $\frac{1}{x - r}$; more precisely, it goes to infinity like $\frac{c_r}{x - r}$ where $c_r = \frac{P(r)}{R(r)}$. This number is referred to in complex analysis as the residue of the pole at $x = r$. So the upshot of all of this is that we can subtract this pole away and write

$$f(x) - \frac{c_r}{x - r} = \frac{1}{x - r} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right).$$

The expression in parentheses approaches $0$ as $x \to r$, and in fact it is a rational function whose numerator is divisible by $x - r$, so we can actually divide by $x - r$. The result is a new rational function which no longer has a pole at $r$.

We can repeat this process for every root of $Q$ until we get a rational function with no poles whatsoever. But this must be a polynomial. So now we've written $f$ as a sum of fractions of the form $\frac{c_r}{x - r}$ plus a polynomial. (Note that in general we need to consider the complex roots of $Q$.)

Okay, so what if $Q$ has repeated roots? Then $f$ might go to infinity more quickly as $x \to r$. If $r$ is a root with multiplicity $n$, then writing $Q(x) = (x - r)^n R(x)$ we now have

$$\frac{P(x)}{Q(x)} = \frac{1}{(x - r)^n} \left( \frac{P(x)}{R(x)} \right)$$

where $R(r) \neq 0$. So we see that as $x \to r$, this expression goes to infinity like $\frac{1}{(x - r)^n}$; more precisely, it goes to infinity like $\frac{c_{r,n}}{(x - r)^n}$ where $c_{r, n} = \frac{P(r)}{R(r)}$. So we can do the same thing as before and just subtract this off, getting

$$f(x) - \frac{c_{r,n}}{(x - r)^n} = \frac{1}{(x - r)^n} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right).$$

The expression in parentheses approaches $0$ as $x \to r$, so again it is divisible by $x - r$, but this time we're not done! We still have to subtract off terms that look like $\frac{1}{(x - r)^k}$ where $k < n$ until the resulting rational function no longer goes to infinity as $x \to r$.

The above is nice as far as it goes, but let me mention that algebraically partial fraction decomposition is more general than rational functions over $\mathbb{C}$. It also generalizes to, for example, rational numbers! Like rational functions, rational numbers also have partial fraction decompositions, like

$$\frac{5}{12} = \frac{2}{3} - \frac{1}{4}.$$

Explaining all of this in a unified framework requires the language of abstract algebra, in particular the notion of a group, of a field, and of a principal ideal domain. Partial fraction decomposition in this setting describes the additive group of the field of fractions of a principal ideal domain using essentially the Chinese remainder theorem, but that's a story for another day...

(It might not seem like uniqueness of partial fraction decomposition holds for rational numbers because we can also write $\frac{5}{12} = \frac{3}{4} - \frac{1}{3}$, but the correct notion of uniqueness here is subtle; it is uniqueness "mod $1$.")

• @R R: what kind of statement do you want for polynomials with non-integer powers? If the powers are rational, then you can replace the variable $x$ with $x^{1/d}$ where $d$ is a common denominator and then you can just use ordinary partial fraction decomposition. Otherwise, functions like $x^{\pi}$ aren't even well-defined for all values of $x$ (what does $(-1)^{\pi}$ mean?), or else depending on your tastes are multi-valued, so there are already a lot of subtleties in just describing these functions, let alone showing that there is some kind of partial fraction decomposition for them... Commented Aug 22, 2012 at 2:16
• Thank you, this really helped! I have a question though. You said: "The expression in parentheses approaches $0$ as $x \to r$, so again it is divisible by $x - r$, but this time we're not done! We still have to subtract off terms that look like $\frac{1}{(x - r)^k}$ where $k < n$ until the resulting rational function no longer goes to infinity as $x \to r$." I see why it's divisible by $x-r$, but how do we know it's not divisible by $(x-r)^k$ for $1< k \le n$? Because if that's the case, we wouldn't need a fraction for every $1\le k \le n$ to get rid of the pole correct?
– dfg
Commented Aug 24, 2014 at 23:24
• @dfg: yes, sometimes we don't need all possible terms, but in general we do (you can check this explicitly). There's no harm in including all possible terms to be safe; it's just the case that sometimes their coefficients are $0$. Commented Aug 25, 2014 at 1:48
• @QiaochuYuan Sorry, one more question. Your answer states that $f$ has been decomposed into a sum of fractions, plus a polynomial. All the resources I've come upon though don't account for a potential polynomial and just assume the function can be decomposed into only fractions. Why is that?
– dfg
Commented Aug 25, 2014 at 2:18
• @dfg: you don't need to take the polynomial into account if you assume that the degree of $P$ is less than the degree of $Q$. I think it's sloppy to just omit this though. Commented Aug 25, 2014 at 3:30

It looks like you have two questions, and they are definitely good ones. It looks like you are asking:

1. Why do we know that the degree of $$Q$$ determines the number of constants?
2. Why do we use constants in the numerator when incrementing every power of the root in the case of a repeated root in the denominator of the original expression?

Let's tackle the first one. Say we want to perform partial fraction decomposition on $$\frac{P(x)}{Q(x}$$. The key here is that the degree of the polynomial $$P(x)$$ in the numerator must be less than the degree of the polynomial $$Q(x)$$ in the denominator. So if deg$$(Q(x))=n$$, then the largest degree that $$P(x)$$ can be is $$n-1$$. So we can express $$P(x)$$ as:

$$P(x)=a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_1x+a_0$$

Where $$a_0,\ a_1,\dots ,\ a_{n-1}$$ are real coefficients. Wait a minute, how many coefficients does $$P(x)$$ have? It has $$n$$ coefficients. So $$Q$$ determines the number of coefficients in $$P$$. In the case when $$Q(x)$$ factors into $$n$$-many distinct linear terms like this:

$$\frac{P(x)}{Q(x)}=\frac{A_1}{x-r_1}+\frac{A_2}{x-r_2}+\cdots +\frac{A_n}{x-r_n}$$

we have $$n$$-many unknowns ($$A_1,\dots ,\ A_n$$) and $$n$$-known coefficients ($$a_0, \dots ,\ a_{n-1}$$). So we have a system of $$n$$ equations with $$n$$ unknowns, which we can solve. I hope that makes some sense.

I'll take a stab at the second one. It's a tricky one. Let's use your example of having $$(x+1)^3$$ in the denominator of the original expression. If we use the rational that the degree of the numerator in a rational expression must be less than the denominator then shouldn't we write:

$$\frac{a_2x^2+a_1x+a_0}{(x+1)^3}=\frac{A_0}{x+1}+\frac{B_1x+B_0}{(x+1)^2}+\frac{C_2x^2+C_1x+C_0}{(x+1)^3}$$

This is no bueno. We have 6 unknown constants ($$A_0,\ B_0,\ B_1,\ C_0,\ C_1,\ C_2$$) and 3 known coefficients $$(a_0,\ a_1,\ a_2)$$. There is no way we will get a unique answer out of this.

Let's try this again but with a twist. Instead of expressing the numerator as a polynomial in terms of $$x$$, let's express the numerator as a polynomial in terms of $$x+1$$, like this:

$$\frac{a_2x^2+a_1x+a_0}{(x+1)^3}=\frac{A_0}{x+1}+\frac{B_1(x+1)+B_0}{(x+1)^2}+\frac{C_2(x+1)^2+C_1(x+1)+C_0}{(x+1)^3}$$

When we expand this we get:

$$\frac{a_2x^2+a_1x+a_0}{(x+1)^3}=\frac{A_0+B_1+C_2}{x+1}+\frac{B_0+C_1}{(x+1)^2}+\frac{C_0}{(x+1)^3}$$

We don't need 6 constants anymore, we just need 3. Let $$A=A_0+B_1+C_2$$ and let $$B=B_0+C_1$$, and might as well replace $$C_0$$ with $$C$$ to keep up the pattern. There we go! We have 3 known coefficients $$(a_0,\ a_1,\ a_2)$$ and 3 unknown constants $$(A,\ B,\ C)$$.

Now is this valid? Sure it is! We do it when we do Taylor polynomials. It's basically a change of coordinates. Here using $$x$$ ends up being a pain, so let's use something to make our lives easier. So really there is the one rule: the degree of the numerator is less than the degree of the denominator. I hope this makes a little more sense.