3 ms·

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 =

by 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).

by panic 8y ago

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).