4 ms·

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:

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.
contravariant 8y ago
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.

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.

contravariant 8y ago

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.