Why is this true with partial fractions?

Question

Suppose you have a fraction like $\frac{x^2+2x}{x(x-2)^2}$. You can rewrite that as

$$\frac{x^2+2x}{x(x-2)^2}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}.$$

Why is it that you must put the linear version and then the quadratic too? Why isn't it just $\frac{A}{x}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^2}$?

You don't need to. Instead you can write $\dfrac{A}{x} + \dfrac{Bx+C}{(x-2)^2}$. But this reduces to $\dfrac{A}{x} + \dfrac{B}{x-2}+ \dfrac{D}{(x-2)^2}$ where $D=C+2B$. — David P, Apr 25 '14 at 23:46
Is there a typo in the last expression in the question? The last two terms seem to be the same... — DJohnM, Apr 25 '14 at 23:51
There are numerous ways to decompose this function. The one you use depends on what you can solve. — Shahar, Apr 25 '14 at 23:52
When you look at your numerator, after clearing denominators in your partial fractions decomposition, you will have $3$ linear equations in $2$ unknowns, which in general will be inconsistent. (Note that $$\frac{A}{x} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^2} = \frac{A}{x} + \frac{D}{(x-2)^2}$$ so you don't have to consider $B$ and $C$ separately, instead only considering $D$, which leads to it being $3$ equations in $2$ unknowns.) — Nicholas Stull, Apr 26 '14 at 00:18

score 2 · Answer 1 · answered Apr 26 '14 at 00:55

One possible reason as to "why" you need what you need is "dimension".

Clearing the denominators you have $x^2+2x=$ stuff. We want to be able to figure out what kind of "stuff" we need.

Let's go with your proposal of repeating $(x-2)^2$ twice.

In this case we end up with $x^2+2x=A(x-2)^2+Bx+Cx$. We then try to solve for $A,B,C$ and run into problems.

The reason for this is that the $B$ and $C$ terms aren't really different. The collection: $(x-2)^2$, $x$, $x$ is linearly dependent (in fact the last two things are equal).

On the other hand, if we use the proper partial fraction forms, after clearing denominators, we'll get: $x^2+2x=A(x-2)^2+Bx(x-2)+Cx$. This time our collection of polynomials: $(x-2)^2$, $x(x-2)$, $x$ are linearly independent (we cannot multiply any pair by some real numbers, add them together, and get the third).

Our 3 linearly independent polynomials guarantee that no matter what the left hand side is: $???x^2+???x+???$ we can successfully solve $???x^2+???x+???=A(x-2)^2+Bx(x-2)+Cx$ for $A,B,C$.

The general case for partial fractions works just the same. The forms for partial fractions are chosen so that you'll end up with the same number of equations as you have unknowns. Moreover, these equations will "independent" in some sense (you don't have redundancies). So "linear algebra" will tell us that there is a solution (and in fact there is only 1 such solution).

I wrote a paper with a student trying to provide a complete fairly elementary explanation of this. It's posted here: http://www.m-hikari.com/imf/imf-2012/29-32-2012/cookIMF29-32-2012.pdf — Bill Cook, Apr 26 '14 at 00:58
So what you are saying is it is acceptable to decompose the fraction the way I had proposed but it just doesn't get us anywhere? — King Squirrel, Apr 26 '14 at 02:44
Kind of. You can propose any sort of decomposition you want. But the sort of choice you proposed won't always exist - you'll end up with equations that have no solution. — Bill Cook, Apr 26 '14 at 23:40

score 0 · Answer 2 · answered Apr 26 '14 at 01:16

Because if we had a general rational function $f(z) = \dfrac{P(z)}{z}$ where $P(z)$ is a polynomial function such that $P(0) \neq 0$, the Laurent series expansion about $z = 0$ of $f(z)$ would be $f(z) = \dfrac{A_{-1}}{z} + A_0 + A_1z + \cdots A_nz^n.$ It is clear that the residue of $f(z)$ is $A_{-1}$ and the residue of $\dfrac{f(z)}{z}$ is $A_0$ but if we take $$\dfrac{f(z)}{z} = A_1 + A_2z + A_nz^{n-1} + \dfrac{A_0z + A_{-1}}{z^2} =A_1 + A_2z + A_nz^{n-1} + \dfrac{C_1}{z^2} + \dfrac{C_2}{z^2}$$ the residue of $\dfrac{f(z)}{z}$ is zero which is a contradiction.

Thus we cannot say that $\dfrac{A_0z + A_{-1}}{z^2} = \dfrac{C_1}{z^2} + \dfrac{C_2}{z^2}$.

Why is this true with partial fractions?

2 Answers2