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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 Irelan

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.)
  - gizmo686 8y ago
    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.
    - JadeNB 8y ago
      > 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.
  - edflsafoiewq 8y ago
    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?
    - JadeNB 8y ago
      > 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).
contravariant 8y ago
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.

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.)
- gizmo686 8y ago
  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.
  - JadeNB 8y ago
    > 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.
- edflsafoiewq 8y ago
  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?
  - JadeNB 8y ago
    > 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).

contravariant 8y ago

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.