An easier approach to partial fractions decomposition | Hacker News Reader

An easier approach to partial fractions decomposition

jaydaigle.net

32 pointsluu29 comments

I learned this in undergrad calc 1 as the Heaviside cover-up method [1], named after Oliver Heaviside.

Heaviside was a remarkably productive scientist [2], contributing to a range of topics from vector calculus to electromagnetic theory.

[1] https://en.wikipedia.org/wiki/Heaviside_cover-up_method

[2] https://en.wikipedia.org/wiki/Heaviside

Didn't Heaviside condense Maxwell's nine (or more?) original equations down to the four we know and use today?

In the wiki linked, that is in fact his first-listed innovation ... he condensed 20 equations into 4 using vector calculus.

And those equations can be condensed into a single manifestly covariant equation:

□A = J

where □ is the d'Alembertian, A is the 4-potential, and J is the 4-current.

"His Wormship, Professor Oliver Heaviside, (W.O.R.M.), presents his compliments to ..."

Well.

As the author presents it, it seems to me difficult to discover the idea of "partially clearing denomninators"; but notice that we can still clear all denominators—we just have to be lazy about what we do next. (This is one of the things that I have trouble convincing my calculus students to do; I'll frequently see them encounter a product of two binomials, reflexively multiply them together, and then spend time factoring them to find the roots of the product.)

Anyway, if we clear denominators in the author's example, then we get:

    7x + 2 = A(x - 1)(x + 2) + B(x - 1) + C(x + 2)^2.

As long as we don't multiply and re-group as instinct may suggest, we can plug in `x = 1` just as easily as the author describes:

    9 = 0 + 0 + 9C.

As he suggests, we can also plug in `x = -2` (and now it's maybe less mysterious why this is still OK) to get

    -12 = -3B

; and then we can proceed as the author does, subtracting off the B term and factoring out `(x + 2)` from what's left, or (which is substantially the same procedure) we can differentiate:

    7 = A(x - 1) + B + 2C(x + 2) => 7 = -3A + B.

(The differentiation would look more complicated if we had more terms, but notice that we don't actually need to know that the derivative of `C(x + 2)^2` is `2C(x + 2)`, only that the derivative of any polynomial divisible by `(x + 2)^2` is divisible by `(x + 2)`, and so vanishes at `x = -2`.)

Another way to find A (which is only another form of differentiation) is not to evaluate, that is reduce mod (x-a), like for B and C, but to reduce mod the higher-order term, (x+2)^2. The C term drops out immediately and (x-1)(x+2) = (x+2 - 3)(x+2) = -3(x+2) so we get

    7x + 2 = -3A(x+2) + B(x-1) => (evaluate at x=1) => 9 = -9A

This method may work, but the explanation is nonsense.

We start out with an equation that’s undefined at x = 1. First we multiply both sides by (x - 1), then we evaluate at x = 1. None of that is valid.

It does seem sketchy, but I don't think it's nonsense. If A = B for all x, then A / (x - 1) = B / (x - 1) for x not equal to 1. You start by evaluating at x = 1 to find where A and B are equal; then, after making the two sides equal, you divide. The reason it's written in the other order is to motivate the particular choice of A and B.

> If A = B for all x, then A / (x - 1) = B / (x - 1) for x not equal to 1.

This is true, but it is the converse of what we want: we know (or guess) that we can arrange that `A/(x - 1) = B/(x - 1)` for all `x \ne 1`, and want to deduce that `A = B` for all `x`. Here we need to know something about `A` and `B` for the deduction to be valid; it suffices that they are continuous at `x = 1`, which is the case here.

We don't know that A/(x - 1) = B/(x - 1) -- that's what we're trying to make true! We start by making A equal to B, then using that to show that A/(x - 1) = B/(x - 1). The process goes backwards from how it's written on the page.

For example, say we have functions f_a(t, x) which are equal to x for t > a and 0 for t <= a. To make f_1(x, 10) / (x - 1) = f_1(x, y) / (x - 1), we can solve for f_1(x, 10) = f(x, y) — y = 10 — then divide on both sides, despite the function being discontinuous at x = 1.

I think that we are in violent agreement. The logical process could go backwards from how it's written on the page (although it doesn't have to, if sufficient theoretical machinery is in place); but the pedagogical process goes in the direction that it's written on the page. Technically speaking, any class that uses partial-fractions decompositions should prove that they always exist (in which case the procedure that starts by assuming that they exist is justified, and no backwards run is needed), or should finish every such problem by checking that the putative decomposition is valid (the backwards process to which you refer).

This type of method for partial fraction decomposition is commonly taught at A-level Maths (typical end of high school exams in England, Wales & Northern Ireland)

Can anyone explain why it works to use x=1, even though the fraction is undefined at x=1?

Presumably an explanation would involve continuity and limits.

It works because you are evaluating the limit at that point.

Think of it as multiplying both sides by (x-1) and then applying the limit as x->1 to both sides.

When you do that, you are going to end with an equality, 1 = C.

You don't need limits. It works perfectly fine as an equation in the ring of rational fractions on R[x]. That is, an equation between rational functions shouldn't be read as saying "if you evaluate these for a specific value of x and the denominators do not vanish the results are equal" but as saying "these are equal as polynomials after formally clearing the denominators".

Concretely the equations he computes with

(7x+2) / (x+2)^2 (x-1) = A/(x+2) + B/(x+2)^2 + C/(x-1)

(7x+2) / (x+2)^2 = A(x-1)/(x+2) + B(x-1)/(x+2)^2 + C

(7x+2) / (x-1) = A(x+2) + B + C(x+2)^2/(x-1)

(7x+2) / (x+2) (x-1) = A + B/(x+2) + C(x+2)/(x-1)

are all equivalent by definition (because f/g = f'/g' means fg' = f'g) to

7x+2 = A(x+2)(x-1) + B(x-1) + C(x+2)^2

Evaluate at x=1 to get C=1. Evaluate at x=-2 to get B=4. IMO it's easier to get C by using the system of equations approach rather than the evaluation method at this point, since the coefficient of X^2 of the RHS is evidently A+C and must vanish, so C=-1.

The problem with viewing this as only a computation in the ring `R(x)` (I think that's what you mean, not "the ring of rational fractions on R[x]"—maybe you meant the fraction field of `R[x]`?) is that, while all the equalities of rational functions make perfectly good sense in `R(x)`, there is no "plug in `x = 1`" function on `R(x)`. One must 'promote' the equality in `R(x)` to one in `R[x]` by clearing denominators, and only then plug in `x = 1`.

(Of course this is what you do, and I agree that it's the right, or at least a good, way to explain it; but I think that it's important to emphasise explicitly how the algebraic point of view avoids any concern with taking limits.)

You could view R[x] as a subring of R(x). At that point, once you have tranformed the problem into one contained entirely within R[x], evaluation at a point becomes a well defined function.

I suppose you could view evaluation as a non total function on R(x)XR (or rather, a function on the annoying to write domain where it is defined), but I have never seen anyone do that explicitly.

> You could view R[x] as a subring of R(x). At that point, once you have tranformed the problem into one contained entirely within R[x], evaluation at a point becomes a well defined function.

Indeed, this is exactly the process that I meant to describe by:

> One must 'promote' the equality in `R(x)` to one in `R[x]` by clearing denominators, and only then plug in `x = 1`.

> I suppose you could view evaluation as a non total function on R(x)XR (or rather, a function on the annoying to write domain where it is defined), but I have never seen anyone do that explicitly.

I agree that it would be strange, for many reasons. Mathematicians in the main tend to avoid the language of partial functions. An individual evaluation function, say at `x = c`, is naturally viewed as a function on the localisation R[x]_{(x - c)}; but, if one wants to view the evaluation as (informally) a "two-place" function, then I would say that it lives naturally on R[x] \times R.

Ahhh, apparently the ring of rational fractions on R means the field of fractions on R[x]. I did mean R(x).

Why don't I have a "plug in x=1" (partial) function R(x) -> R? And what do you mean by "promote"? An equation in R(x) "is" a (set of) equations in R[x] right?

> Why don't I have a "plug in x=1" (partial) function R(x) -> R?

Precisely for the reason that you say: mathematicians in the main tend to avoid the language of partial functions, so I would say not "there is a partial function R(x) \to R" but "there is a function R[x]_{(x - 1)} \to R" (where R[x]_{(x - 1)} is a localisation).

> And what do you mean by "promote"? An equation in R(x) "is" a (set of) equations in R[x] right?

My language was somewhat fanciful. I don't think that one can say that an equation in R(x) is an equation in R[x], for essentially syntactic reasons; but, just as you said, if the rational function a/b equals the rational function c/d (a/b = c/d in R(x)), then a * d = b * c in R[x]. That's what I meant by 'promoting' the equality (although it is not great, and certainly not technical, language).

If you want an explanation in terms of continuity and limits then it works because both sides of the equation agree everywhere except possibly x=1 and so must have the same limit at x=1, and both are continuous at x=1 so they must have the same value at x=1.

> the explanation is nonsense

No, it is not.

You are literally applying the residue theorem [0] in a non-fancy way. You use that a lot on a signals and systems course to find the inverse laplace transform.

[0]: https://en.wikipedia.org/wiki/Partial_fraction_decomposition...

You’re applying it in a non-rigorous way. Like I said, it might work, but it’s not a valid explanation of why it works.

It's justified because both sides are continuous and agree on a dense set (all x not equal to 0). You're right that this requires some clarification though.

Both sides agree everywhere on their domain of definition, which is all real numbers sans -2 and +1. I don't see how denseness is entering in.

The question is: how do you know they agree everywhere? In fact, how do you know they agree anywhere at all, since the value for eg. C was derived only under the assumption that x=+1 and x=+1 is not even in the domain.

> I don't see how denseness is entering in.

Density enters in in the implicit translation of the solecism "plug in `x = 1`" to the mathematically valid "plug in numbers near `x = 1` [to the denominator-cleared expression], and take the limit as `x \to 1`."

How does density enter into that? AFAICT, that gloss depends on two things:

1. that you already know that a partial fraction decomposition exists (or else this argument proves that exp(x)/x(x-1) = -1/x + e/(x-1))

2. that every numerator is continuous at any zero of a denominator

It seems like we only need a finite number of limits (enough to determine the constants in our decomposition) and as long as the function is continuous enough at those points that we can compute the limits, it can do whatever it likes elsewhere.

That is, our global control over the function comes from our knowledge that the partial fraction decomposition exists in the first place; actually computing it only requires local information at a small number of points.

(I guess your "How does density enter into that?" questions are rhetorical, in the literal sense that you are using them as tools of rhetoric. I took it to be a question from the point of view of someone who didn't understand how density could be relevant, which is why my answer was at the informal level that it was.)

> AFAICT, that gloss depends on two things:

> 1. that you already know that a partial fraction decomposition exists (or else this argument proves that exp(x)/x(x-1) = -1/x + e/(x-1))

> 2. that every numerator is continuous at any zero of a denominator

Agreed on both counts.

> It seems like we only need a finite number of limits (enough to determine the constants in our decomposition) and as long as the function is continuous enough at those points that we can compute the limits, it can do whatever it likes elsewhere.

Agreed on this precise statement. In particular, we need `x = 1` to be an accumulation point for the domains of definition of the original (rational) functions. One easy way to accomplish that is for the intersection of the domains to be dense in ℝ; but of course this is far from necessary.

They agree everywhere on their domain of definition by assumption. Multiplying by (x-1) doesn't change that. What this does change however is that the resulting functions can be continuously extended to x=1, and because they agree on a dense set (in this case everywhere else) the extended functions must agree as well.

In some sense agreeing on a dense set and being continuous are the minimum properties for this to hold, there are alternative ways to reach the same conclusion but those must be based on those or even stronger properties, if the functions are either not continuous or don't agree on a dense set around x=1 the proof fails.

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