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I've just learned about removable discontinuities. So, when we have such a function we redefine it, making a new function that is defined at the point the first isn't. What is the point of this? What advantages do we get? Wouldn't making it continuous cause problems when we use it instead of the old one?

LearningMath
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  • Because we don't want to deal with a point where the function is undefined when it could be without changing anything else. An analogy with real numbers is the function f(x) = x / x. It's technically not defined at x=0 and is equal to 1 everywhere else. We'd rather just work with the constant function g(x) = 1. – Not a grad student Dec 19 '14 at 23:27
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    It is the opposite, you make it continuous so you do not have anymore problems – Mosk Dec 19 '14 at 23:27
  • Well, you never know when a discontinuity will break down, so if it is removable, it is easy to replace. – copper.hat Dec 20 '14 at 00:20

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Many times we know from context that the solution of a problem will necessarily have certain continuity properties (e.g. physics laws may imply that). However, "apparent" singularities may be introduced as artifacts of the solution methods employed. In such case we can safely remove these apparent singularities.

This can be done quite spectacularly in algebra by universally cancelling apparent singularites, e.g. follow the links I gave in this answer, which includes a folklore slick purely algebraic proof of Sylvester's determinant identity (avoiding any topology/analysis such as density arguments).

Community
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Bill Dubuque
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  • I must say I don't understand you. I was asking about discontinuity like in this function: $f(x) = \frac{x-1}{x-1}$, where 1 is removable discontinuity. Thanks. – LearningMath Dec 20 '14 at 15:22
  • @notorious The linked examples are simply multivariate generalizations of that, i.e. instead of cancelling $,x-1,$ they cancel more complicated multivariate polynomials to effectively remove apparent singularities. – Bill Dubuque Dec 20 '14 at 16:17
  • Could you please tell me why we remove the discontinuity? What benefits do we get from that? Is it even necessary to do that? – LearningMath Dec 20 '14 at 16:55
  • @notorious Let's consider a very simple example. Suppose that we are give some algebraic equations that specify a function $, f(x),,$ and suppose that we know additionally that $,f,$ is a polynomial function of $,x,$ (or is continuous). After some algebraic manipulations we deduce that the given equations are equivalent to $, x f(x) = x^2.,$ Therefore the solution is $,f(x) = x,,$ because this is the unique solution that is a polynomial (or continuous). – Bill Dubuque Dec 20 '14 at 17:09