6 ms·

An easier approach to partial fractions decomposition

32 points by luu 8y ago

kwantam 8y ago
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 https://en.wikipedia.org/wiki/Heaviside_cover-up_method [2] https://en.wikipedia.org/wiki/Heaviside https://en.wikipedia.org/wiki/Heaviside
- tigerlily 8y ago
  Didn't Heaviside condense Maxwell's nine (or more?) original equations down to the four we know and use today?
  - ekun 8y ago
    In the wiki linked, that is in fact his first-listed innovation ... he condensed 20 equations into 4 using vector calculus.
  - antidesitter 8y ago
    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.
- blattimwind 8y ago
  "His Wormship, Professor Oliver Heaviside, (W.O.R.M.), presents his compliments to ..." Well.
JadeNB 8y ago
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`.)
- edflsafoiewq 8y ago
  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
monochromatic 8y ago
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.
- panic 8y ago
  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.
  - JadeNB 8y ago
    > 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.
    - panic 8y ago
      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.
      - JadeNB 8y ago
        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).

32 points by luu 8y ago

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

tigerlily 8y ago
Didn't Heaviside condense Maxwell's nine (or more?) original equations down to the four we know and use today?
- ekun 8y ago
  In the wiki linked, that is in fact his first-listed innovation ... he condensed 20 equations into 4 using vector calculus.
- antidesitter 8y ago
  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.
blattimwind 8y ago
"His Wormship, Professor Oliver Heaviside, (W.O.R.M.), presents his compliments to ..." Well.

JadeNB 8y ago

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`.)

edflsafoiewq 8y ago
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

monochromatic 8y ago

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.

panic 8y ago
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.
- JadeNB 8y ago
  > 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.
  - panic 8y ago
    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.
    - JadeNB 8y ago
      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).