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I have plotted function $(1+\frac{1}{x})^{x}$ using Maple and got following graph:

enter image description here

So it seems that function isn't defined on $(-1,0)$ interval , but if I take that $x=\frac{-1}{3}$ I can write:


which is real number.So my question is :Is function $(1+\frac{1}{x})^{x}$ defined on $(-1,0)$ interval or not ?

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up vote 2 down vote accepted

Another way to solve this is to allow complex values, and choose the principal value in the usual way. Then your function IS defined and continuous on $(-1,0)$, but has non-real values everywhere there!

Here it is, real part in green, imaginary part in red:


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so,I can conclude that function is continuous only if codomain is $\mathbb{C}$ ? – pedja Oct 30 '11 at 12:51
That function in my picture is continuous. Don't know about your "only if" though. – GEdgar Oct 30 '11 at 13:08

I guess here there is kind o a philosophical problem: for instance, if you take $x=-\frac{1}{2}$, you face the problem $\sqrt{-1}$. So one can say that there are some points in $(-1,0)$ where the function is defined and some other where is not. In general, it's quite difficult to say a priori which are the right points and so, in order to avoid contradiction, one prefers to exclude $(-1,0)$ from the domain.

This i by the way also what happens for the function $x^x$. One says that its domain is $(0,\infty)$, even if it is clear that it is well defined also for particular negative real numbers.

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Let's suppose that I have some number of infected persons among healthy ones..does it mean that I should quarantined all of them because I don't know which of them are infected ? – pedja Oct 30 '11 at 11:40
@pedja: Yes. At least When the non-infected people form a set of measure zero. Also when allowing them to escape the quarantine poses a hazard to public safety! – Jyrki Lahtonen Oct 30 '11 at 12:10
@pedja: Yes, Jyrki's reason above shows that it would be impossible for a computer to write down a picture. Also because I don't think there is a closed formula to find the non-infected persons. – Valerio Capraro Oct 30 '11 at 12:17
I will stop thinking about math as an exact science – pedja Oct 30 '11 at 12:37
He, he... – J. M. Oct 30 '11 at 12:50

The expression $x^y$ is defined for $x>0$ if $y\not \in Z $ in the case of real numbers. Otherwise we could write, for example, $-\sqrt[3]{2}=\sqrt[3]{-2}=(-2)^{1/3}=(-2)^{2/6}=\sqrt[6]{(-2)^2}=\sqrt[6]{2^2}=\sqrt[3]{2}$. Contradiction.



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