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Does anyone know a simple common function which has undefined values for many domain values?

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closed as too broad by Famous Blue Raincoat, Przemysław Scherwentke, dustin, user7530, Jonas Meyer Feb 12 at 3:39

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How about $$\arcsin x?$$ –  lab bhattacharjee Jan 30 '14 at 18:45
Like what? Like $\sqrt{x}$ is undefined for $x<0$? –  GEdgar Jan 30 '14 at 18:46
Or $$\frac 1{\sin x}$$ –  Ragnar Jan 30 '14 at 18:56
The trigonometric functions $\tan x$ and such. –  vonbrand Jan 30 '14 at 18:57
How about $f(x) = \sqrt{x} + \sqrt{-x}$? –  Dave L. Renfro Jan 30 '14 at 19:02

2 Answers 2

up vote 1 down vote accepted

Consider the indicator function of the rationals, ${\bf 1}_{\Bbb Q}$. There is quite a big collection of numbers whose (ir)rationality is unknown, so there are many inputs $x$ whose output is unknown.

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In this context, is "unknown" the same as "undefined"? –  Newb Jan 30 '14 at 19:25
@Newb I have interpreted the question in a different way you did, it seems. –  Pedro Tamaroff Jan 30 '14 at 19:29

The problem with your question is that the domain of a function is the largest set on which the function is defined. The domain of a function $f$ is the set of values $x$ for which $f(x)$ is defined. So by definition, a function does not have domain values for which the function is not defined. If it does, then you say that the function is ill-defined, and then the function is not a function.

For example, you might say what about $\tan(x)$? It is undefined at $x=\frac{\pi}{2}$, etc. As I learned when I asked the same thing, we formally regard the domain of $\tan(x)$ to be $$ X=\bigcup_{k\in\Bbb{Z}} \left(-\frac{\pi}{2}+k\pi,\frac{\pi}{2}+k\pi\right) $$ Instead of the entire real line, as you might otherwise have guessed.

So, to answer your question: by definition of a function, there are no functions which are not defined on any values of their domain.

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+1 Cool if one can help others here with things one learned here :) –  Hagen von Eitzen Jan 30 '14 at 19:02

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