5 ms·

Not a bug. Alpha is right. There is no singularity. let a=1.5 for brevity and let f(x) = (x-a)(x-1)/(x-a). Relabel f(x)= N(x)/D(x) (N for numerator, D for d

Not a bug. Alpha is right. There is no singularity. let a=1.5 for brevity and let f(x) = (x-a)(x-1)/(x-a). Relabel f(x)= N(x)/D(x) (N for numerator, D for denominator) N(x) and D(x) are analytic and so tend to zero as x--> a, so f(x) tends to the indeterminate form 0/0. In that case, according to L'hopital's rule, f(a) = N'(a) / D'(a). N'(x) = (x-a) + (x-1) by the product rule. D'(x) = 1. Plugging in x=a: f(a) = (a-1) / 1 = .5 Which is what Alpha implies. See also: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule http://en.wikipedia.org/wiki/Removable_singularity http://en.wikipedia.org/wiki/Removable_singularity http://xkcd.com/123/ http://xkcd.com/123/

lmkg 17y ago
Well, kinda. But not really, sometimes. L'Hopital's rule says that the limit of the function exists, not the function itself. A removable singularity is still a singularity, just one that's more well-behaved than a pole (or gods forbid, an essential singularity). While the function can be analytically extended to be defined at the singularity, that's an extension of the function, rather than the function itself... ... sorta. Since extensions of analytic function are unique, we usually treat the extension and the function as the same thing. Whether this is revealing a deeper truth about the function, or sloppy notation, is up to the individual practitioner, and the context. Certainly, the definition of the function that has the removable singularity is sufficient to define the function that has the singularity removed. For analysis they're basically the same and we usually extend as far as we can. But for topology, a single point separates open and closed sets, which are worlds apart, and the extended function is a very different beast from the original one. This just goes to show how limits identify the points that are most important to a (structured) set. However, for the purposes of Wolfram|alpha, I don't fault them for making that mistake. Removable singularities can be safely extended over for the purposes of computational mathematics.
dfranke 17y ago
A removable singularity is still a singularity. If I'm asking for f(x) and it's giving me lim(i->x,f(i)), it should tell me so.

lmkg 17y ago

Well, kinda. But not really, sometimes. L'Hopital's rule says that the limit of the function exists, not the function itself. A removable singularity is still a singularity, just one that's more well-behaved than a pole (or gods forbid, an essential singularity). While the function can be analytically extended to be defined at the singularity, that's an extension of the function, rather than the function itself... ... sorta. Since extensions of analytic function are unique, we usually treat the extension and the function as the same thing. Whether this is revealing a deeper truth about the function, or sloppy notation, is up to the individual practitioner, and the context. Certainly, the definition of the function that has the removable singularity is sufficient to define the function that has the singularity removed. For analysis they're basically the same and we usually extend as far as we can. But for topology, a single point separates open and closed sets, which are worlds apart, and the extended function is a very different beast from the original one. This just goes to show how limits identify the points that are most important to a (structured) set. However, for the purposes of Wolfram|alpha, I don't fault them for making that mistake. Removable singularities can be safely extended over for the purposes of computational mathematics.

dfranke 17y ago

A removable singularity is still a singularity. If I'm asking for f(x) and it's giving me lim(i->x,f(i)), it should tell me so.