Is there a simple explanation for calculus classes of why partial fractions work?

Question

I'd be happy even with an explanation in the simplest case: an explanation of why expressions of the form $\frac{ax + b}{(x - c)(x - d)}$ with $c \neq d$ can always be rewritten in the form $\frac{A}{x - c} + \frac{B}{x - d}$. I'm following Stewart's book, and his explanation, in its entirety, is "A theorem in algebra guarantees that it is always possible to do this".

In some sense, that is enough, because the thing to do is to try to rewrite it as $\frac{A}{x - c} + \frac{B}{x - d}$, solve for $A$ and $B$, and the "theorem in algebra" says that you will always be able to. But can one be more convincing?

My students will not have seen linear algebra, so that explicit discussion of linear independence will be unhelpful, even though this is the proper language in which to formulate a proof. They will also not have seen number theory, so that comparisons with modular arithmetic and the Chinese Remainder Theorem are also likely to be unhelpful.

Answer 1

Here is what I might do, if I were teaching a class which taught some abstract proof skill (but not a true proof class) or if I was happy reaching only the top half of a standard Calculus class here at Michigan.

Suppose we want to expand $$\frac{f(x)}{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}$$ where all the $r_i$ are distinct, and $\deg f \leq 3$. I claim that $$\begin{array}{rcl} \frac{f(x)}{(x-r_1)(x-r_2) \cdots (x-r_4)} &=&\phantom{+} \frac{f(r_1)}{(r_1-r_2)(r_1-r_3)(r_1-r_4)} \frac{1}{x-r_1}\\ & &+ \frac{f(r_2)}{(r_2-r_1)(r_2-r_3)(r_2-r_4)} \frac{1}{x-r_2}\\ & &+ \frac{f(r_3)}{(r_3-r_1)(r_3-r_2)(r_3-r_4)} \frac{1}{x-r_3}\\ & &+ \frac{f(r_4)}{(r_4-r_1)(r_4-r_2)(r_4-r_3)} \frac{1}{x-r_4}\\ \end{array}.$$

Of course, you might wonder how on earth I found that formula, but let's first take it as handed down by magic and show that it works.

Clearing denominators, the claim is that $$\begin{array}{rcl} f(x) &=& \phantom{+}\frac{f(r_1)}{(r_1-r_2)(r_1-r_3)(r_1-r_4)} (x-r_2)(x-r_3)(x-r_4)\\ & & +\frac{f(r_2)}{(r_2-r_1)(r_2-r_3)(r_2-r_4)} (x-r_1)(x-r_3)(x-r_4)\\ & & +\frac{f(r_3)}{(r_3-r_1)(r_3-r_2)(r_3-r_4)} (x-r_1)(x-r_2)(x-r_4)\\ & & +\frac{f(r_4)}{(r_4-r_1)(r_4-r_2)(r_4-r_3)}(x-r_1)(x-r_2)(x-r_3)\\ \end{array}.$$

Now remember: $f$ and the $r$'s are all constants that we know. In the last problem we did, they were $8x-13$, $1$ and $2$ (or whatever). The supposed claim is that this is a true equation for all $x$.

Notice that both sides of this equation are polynomials in $x$. What is the degree of that polynomial? Right, $3$, or possibly less if the coefficients of $x^3$ cancel on the right hand side. A degree $3$ polynomial is determined by $4$ values, just like a linear polynomial is determined by $2$ values. Which $4$ values might we plug in for $x$? Let's try the $r_i$.

Pause for at least $10$ seconds. Do the computation. Pause for at least $10$ seconds again.

Okay, how did I find this formula? This approach is much more worth memorizing than the formula is. Suppose that there were a formula like $$\frac{f(x)}{(x-r_1)(x-r_2)(x-r_3)(x-r_4)} = \frac{c_1}{x-r_1} + \frac{c_2}{x-r_2} + \frac{c_3}{x-r_3} + \frac{c_4}{x-r_4}.$$ So the $c_i$ are the constants we want to find: in the last problem, they turned out to be $3$ and $5$ (or whatever). Here is the trick:

Multiply both sides by $x-r_i$ and take the limit as $x \to r_i$.

Pause for at least $10$ seconds. Do the limit. Pause for at least $10$ seconds again.

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As a general note, I've found that sum and product notation, and using variables for discrete quantities like the degree of a polynomial, are huge stumbling blocks. I pretty much always use specific values of $n$, like $4$, whenever working with non-math majors.

Answer 2

It is a matter of taste, but I find the following approach easier to understand for beginners (and thus more convincing) than trying to rewrite the fraction in a good form and solving for coefficients.

It might be good to start with stating that if $ax+b$ can be written in the form $A(x-d)+B(x-c)$, then $\frac{ax + b}{(x - c)(x - d)}$ can be written in the form $\frac{A}{x - c} + \frac{B}{x - d}$; don't expect that all students will think of this unless told. It is quite simple to explain that this is possible if for all $a,b$ there are $A,B$ so that $A+B=a$ and $Ad+Bc=-b$. This last statement can be tricky to prove (if one does not have more advanced tools and there is not much time to spend on it), but little experimentation will certainly make it plausible. Proving the last claim by hand can be given as an extra exercise for those that are interested in order not to disturb the flow of the course.

I tried to organize the above argument so that the problem is reduced to a more tractable one in a couple of steps and the first step is (hopefully) the simplest one to grasp. I assume your goal is not to give a full proof, but to give a justification for why it works so that all students can understand at least a little. If the problem is linked to some other problems (formally equivalent does not mean equivalent in all students' minds), it suffices that the students believe that at least one of them can be solved; students will probably disagree with each other about the simplest formulation.

Trying to give a fuller proof before appropriate machinery is available might — depending on the exact situation — be a waste of time.

Answer 3

I have always found it instructive and fun to address this by thinking graphically. Students in Calculus II will know the graphs of the rational functions $f(x)=\frac{ax+b}{(x-c)(x-d)}$ and $g(x)=\frac{A}{x-c}+\frac{B}{x-d}$ have the same vertical asymptotes and same horizontal asymptote. Varying $a$ and $b$ moves the $x$ and $y$ intercepts of $y=f(x)$. It is also clear that varying $A$ and $B$ moves the $x$ and $y$ intercepts of $y=g(x)$.

Fixing $c$ and $d$, we have a two-parameter family of curves $y=f(x)$ and similarly for $y=g(x)$. The key point here is that two families are identical--we can find $A$ and $B$ so that the $g$-curve matches any given $f$-curve. It is straightforward to illustrate this in class with animations generated by software--Geometers Sketchpad or Maple for example. As you vary the two parameters of one family, the curves fill out the curves from the other family.

This is not a proof, but rather a conceptual understanding that calculus students can appreciate. The proof that Stewart alludes to really is straightforward and mundane, and is covered in a precalculus course--it reduces to the existence of a solution to two linear equations in two unknowns.

Answer 4

If you see this in calculus, there is a rather natural approach: to use the methods of calculus to study it!

Significant features of $f(x) = \frac{ax + b}{(x - c)(x - d)}$ happen around $x=c$ and $x=d$. Suppose we study the $x \approx c$ case.

One basic thing is to observe that $x-c$ is the only part of the expression that behaves interestingly; consequently, we should expect

$$ x \approx c \implies f(x) \approx \frac{ac+b}{(x-c)(c-d)} $$

Following the pattern of subtracting off features, you write

$$ f(x) = \frac{ac+b}{(x-c)(c-d)} + g(x) $$

and have some expectation that $g(x)$ doesn't have a 'feature' at $c$, or at least, has a simpler feature. Then, either you do a few examples to see that this really does eliminate the pole at $c$, or maybe you do something with Taylor/Laurent series, or you estimate it with asymptotic analysis.

A more generic lemma is probably simpler to use, though:

$$ \frac{r(x)}{x-c} = \frac{r(c)}{x-c} + \frac{r(x)-r(c)}{x-c} $$

and we recognize the second term as a difference quotient, which should remain well-behaved at $x=c$, as it converges to $r'(c)$.

The remaining clever idea to convey (and it is a clever idea until you're used to such things) is that once you know the end form of the manipulation, you can skip straight there and solve for the unknowns by hook or by crook. (or even by the formula indicated by the lemma above)

Answer 5

As in @user52817's graphically-oriented answer, I think it is fair to secretly borrow some ideas nowadays labelled "complex analysis" to help explain why this should work: that is, while thinking in terms of the residue at a simple pole $x_1$, one can explain it (perhaps quasi-geometrically...) as giving the "asymptotics" of the rational function as $x\to x_1$. That is, for low-degree numerator, $f(x)/(x-x_1)\ldots(x-x_n)\sim {1\over x-x_1}\cdot {f(x_1)\over (x_1-x_2)\ldots(x_1-x_n)}$ near $x_1$, and so on. Subtracting all these "polar parts" (or whatever one chooses to call them for calculus students) leaves a rational function that has no poles, and does goes to $0$ for large $x$, so has to be $0$ (a tiny prototype for Liouville's theorem, perhaps).

Answer 6

Given they have no algebra background, the simplest thing is just to do the algebra. $$ \frac{ax + b}{(x - c)(x - d)} = \frac{A}{x-c}+\frac{B}{x-d} $$ Multiplying by $(x-c)(x-d)$ yields (for $x \neq c,d$) $$ ax+b= A(x-d)+B(x-c) = (A+B)x-(Ad+Bc) $$ From which we read $a=A+B$ and $b = -Ad-Bc$. Multiplying by $d$ we obtain $ad=Ad+Bd$ and adding equations yields $b+ad=Bd-Bc=B(d-c)$ and as $d \neq c$ we find $B = \frac{b+ad}{d-c}$. It follows that $A=a-B = a-\frac{b+ad}{d-c} = \frac{a(d-c)-b-ad}{d-c} = \frac{-ac-b}{d-c} = -\frac{ac+b}{d-c}$. Now, I will reverse engineer the algebra: \begin{align} \frac{-\frac{ac+b}{d-c}}{x-c}+\frac{\frac{b+ad}{d-c}}{x-d} &= \frac{1}{d-c} \left[\frac{-(ac+b)}{x-c}\cdot\frac{x-d}{x-d}+\frac{b+ad}{x-d}\cdot\frac{x-c}{x-c} \right] \\ &= \frac{1}{d-c} \left[\frac{-(ac+b)(x-d)+(b+ad)(x-c)}{(x-c)(x-d)}\right] \\ &= \frac{1}{d-c} \left[\frac{(d-c)ax-bx+bx-acd+acd+b(d-c)}{(x-c)(x-d)}\right] \\ &= \frac{ax + b}{(x - c)(x - d)}. \end{align} Ok, so, you can start with the last expression, multiply by $1$ and add $0$ until you obtain the partial fractions decomposition $\displaystyle \frac{-\frac{ac+b}{d-c}}{x-c}+\frac{\frac{b+ad}{d-c}}{x-d} $. If the student thirsts for formulas, here they are: $$ A = -\frac{ac+b}{d-c} \qquad \& \qquad B=\frac{b+ad}{d-c}. $$ Of course, for specific examples, it is far easier to simply work out the algebra. That said, I think what I have written above can be formalized into a proof. You just have to add zero and multiply by $1$ creatively.

Answer 7

As long as your goal is not proof, but just helping people get a general feel for it, there are several ways you can go.

The first is to say that, even if there were no proof, it is a worthwhile thing to try. That is, we can imagine not having the proof that it will work, but still attempting to do it anyway, just like we do with a lot of other methods. We can see that adding some numerator over $(x - c)$ and some numerator over $(x - d)$ will give us a sum that has a denominator of $(x - c)(x - d)$. We also know that it would be a heck of a lot easier to do the integral if we could find some isolated number to put in the numerator. So, why not see if there is one there.

The procedure of partial fractions is fairly straightforward if we just see if something will turn up. If we proceed just on the hunch that maybe we'll find something appropriate, then we can move forward. If we succeed we have found something we can integrate, and if we fail then, just like any other attempt, we'll have to find some other way.

In fact (I don't know if this is the case here), sometimes in math this is the way it works. You have a method that you know can work sometimes, and then, you just kind of notice that it works always. You don't know if it is "always always" or just "always for my problems", so then you search to see if you can prove it one way or the other. In any case, seeing that it could work is the first step to seeing how it works.

Now, given that it could work, it isn't too hard to see why it should always work. To see this, imagine what would happen if $(x - c)$ or $(x - d)$ had an $x$ in the numerator. When cross-multiplying, that would bring the degree up to $x^2$! Since we don't have any $x^2$ to work with, we can say for certain that either neither of them or both of them have $x$ (they could both have $x$ if one were a negative of the other to cancel them out). However, if they both had an $x$ in the numerator, then they could both be simplified through polynomial long division, which would get rid of the $x$ and leave a $1$ and a $-1$ by themselves, and a fraction without the $x$ in the numerator. When the $1$ and the $-1$ cancel out, that leaves just the fractions without the $x$ in the numerator!

So, we identified two possible cases - either both have an $x$ or neither do, and realized that the only way that both of them had an $x$ is if it were in a way that they canceled out and reduced to the case where neither one had an $x$. Thus, we can always assume that neither one has an $x$.

It's kind of convoluted, and not entirely a proof, but I think it goes a long way to showing why this method can work, and why they should trust that a proof exists.

Answer 8

1. I suspect the issue is more one for the teacher than the students. Having a proof (or at a lighter level, a derivation) is nice motivation, but the key thing to getting confidence is just drilling the method itself. I suspect teacher is implicitly unhappy with teaching non-proved topics and that the student questions come from some implicit presentation (alluding to the proof gap by teacher).

2. If I wanted to show a little derivation for motivation, I would do something like this (showing the backwards derivation):

a. Show them a specific quadratic reciprocal integral (with numbers, not general). Make a joke about how scary and evil it looks--we don't know what the heck to do with it.

b. hey, y'all...remember in grade school, when you combined fractions? Well let's do one for practice.

[work through $\frac23 + \frac45 = \frac{22}{15}$]

c. OK, then. Well now we aren't in grade school any more, we need to do stuff with variables and all! Remember in Algebra 2, when we learned about partial fractions. "Oh....don't you groan now...because it is finally going to have a use!"

[Work through combination of $\frac{A}{x-c} + \frac{B}{x-d}$. In other words, GO BACKWARDS.]

d. Do the algebraic derivation forwards. (from the quadratic on the bottom to the two fractions.)

e. Solve the specific "scary looking" integral using partial fractions. Make a joke about how we kicked the butt of that scary looking integral using grade school fractions and high school algebra. Not so scary any more!

f. Have the kids do one in class (simple, simple version).

g. "Ok, guys, this is just a bunch of algebra. It's not as hard as that darned integration by parts or those evil related rate word problems. But you gotta practice it. If you don't, you will mess up the grade school fractions and high school algebra. So, I want each of you to do 10 of these problems tonight. And we will have a one question quiz tomorrow. And next week on the test, I guarantee you at least one problem will be a partial fraction integral. And remember, you need to kick the integral's butt. Not let it kick yours!"