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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 w

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).
learnstats2 8y ago
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.
- siproprio 8y ago
  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.
- edflsafoiewq 8y ago
  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.
  - JadeNB 8y ago
    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.)
siproprio 8y ago
> 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#Residue_method https://en.wikipedia.org/wiki/Partial_fraction_decomposition...
- monochromatic 8y ago
  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.
contravariant 8y ago
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.
- edflsafoiewq 8y ago
  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.
  - JadeNB 8y ago
    > 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`."
    - edflsafoiewq 8y ago
      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.
      - JadeNB 8y ago
        (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.

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).

learnstats2 8y ago

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.

siproprio 8y ago
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.
edflsafoiewq 8y ago
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.
- JadeNB 8y ago
  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.)

siproprio 8y ago

> 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#Residue_method https://en.wikipedia.org/wiki/Partial_fraction_decomposition...

monochromatic 8y ago
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.

contravariant 8y ago

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.

edflsafoiewq 8y ago
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.
- JadeNB 8y ago
  > 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`."
  - edflsafoiewq 8y ago
    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.
    - JadeNB 8y ago
      (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.