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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

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.

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.