I was messing around on desmos and all of these functions seem to be only defined for $x=0$ and $x>0$. Is this an error on desmos' part or are they really only defined for $x=0$ and $x>0$ and if so, why? Does it have something to do with the exponents being irrational and what about the exponents being irrational makes these functions undefined for negative values?
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1Desmos is likely ignoring the case for $x<0$ because these functions are complex-valued in this domain. – Aaron Hendrickson Mar 04 '22 at 15:43
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Do you happen to know why they are imaginary in this domain? – finnbratfisch Mar 04 '22 at 15:45
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1@finnbratfisch You can use the fact that $e^{i\pi} = -1$ as a starting point. – soupless Mar 04 '22 at 15:50
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@finnbratfisch Well, with a little hand waving we could write $x^\pi=(\sqrt x)^{2\pi}$. What do we know about the square roots of negative real numbers? – Aaron Hendrickson Mar 04 '22 at 15:58
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Yes that makes sense, but could you tell me why $x^\pi=(\sqrt{x})^{2\pi}$? – finnbratfisch Mar 04 '22 at 16:20
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1@AaronHendrickson And moreover multivalued , so we would have to choose a specific branch. Therefore, it makes sense to declare $a^b$ undefined for $a<0$ and irrational $b$. – Peter Mar 04 '22 at 16:30
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@Peter It is multivalued but I thought this was TMI and would confuse the OP. – Aaron Hendrickson Mar 04 '22 at 17:09
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@finnbratfisch This is a rule of exponents (although I am ignoring some technicalities such as multivaluedness). Simply write $x^\pi=x^{2\pi/2}=(x^{1/2})^{2\pi}=(\sqrt x)^{2\pi}$. – Aaron Hendrickson Mar 04 '22 at 17:15
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@AaronHendrickson This exponention rule is not correct if $x$ is not a nonnegative real number. This incorrectness is used in many fake-proofs, so it is no good idea to write $x^{\pi}$ this way , if $x$ is negative. – Peter Mar 05 '22 at 10:56
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@Peter Yes Peter. I know. But the OP does not have sufficient background for the correct explanation you gave. What I wrote gives a heuristic approach to understanding why it's complex valued at an appropriate level for the OP. – Aaron Hendrickson Mar 05 '22 at 12:11
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Recall that $x^y = e^{y \ln x}$. This raises the question of how to take the logarithm of a negative number. It can be done by using Euler's formula $e^{i\theta}=\cos \theta + i \sin \theta$. So, if you can express a complex number in polar form as $x = r(\cos \theta + i \sin \theta) = re^{i\theta}$, then $\ln x = \ln r + i\theta$.
If $x$ is a negative real number, then we must have $\cos \theta = -1$ and $\sin \theta = 0$, which means that $\theta$ must be an odd multiple of $\pi$, i.e., $\theta = (2n + 1)\pi$ for some integer $n$. So $\ln x = \ln r + i(2n + 1)\pi$.
So, for the case of $f(x) = x^e$ (with $x$ negative), that works out to $f(x) = e^{e(\ln r + i(2n + 1)\pi)} = e^{e \ln r}e^{ie(2n + 1)\pi} = r^e(\cos (e(2n + 1)\pi) + i \sin (e(2n + 1)\pi))$. This is a real number only if $e(2n + 1)$ is an integer. But because $e$ is an irrational number, that never happens. Therefore, the function must be complex-valued.
Dan
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I wish I could understand this because it is most likely right but I don't. – finnbratfisch Mar 04 '22 at 16:44