As the title suggests, what is the value of $e^\frac{-1}{e}$?? I never understood how things to the power of irrational numbers are calculated. The more broad question would be what is the minimum value for the function $y=x^x$, such that $x \geq 0$? I have tried taking the derivative to get $x^x(\ln(x)+1) = 0$, then simplifying to get $\ln(x)+1=0$, and $\ln(x)=-1$, so $x = \frac{1}{e}$. This simplifies to $e^\frac{-1}{e}$. What do I put for my final answer? Any help would be appreciated.
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Would you prefer $e^{-1/e}$? – Hagen von Eitzen Dec 04 '20 at 23:12
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Yes, I'll edit that – John Liu Dec 04 '20 at 23:25
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Also, I would like to note that this isn't a contest question - there is not necessarily a different answer. If that is how I am supposed to write my answer, then that is very possible too. – John Liu Dec 04 '20 at 23:29
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$e^{-1/e}$ is just $e^{-1/e}$. You could write it in a couple of ways, for instance $e^{-e^{-1}}$, but there is no reason to prefer one over the other. – Dec 04 '20 at 23:29
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2I don't see any point of writing anything other than $e^{-1/e}$ – morrowmh Dec 04 '20 at 23:29
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The only other possibility would be to use $e^{-x} = \dfrac1{e^x}$, so you could write $\dfrac1{e^{1/e}}$, but I wouldn't bother. – Ted Shifrin Dec 04 '20 at 23:53
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But how would a calculator determine the value of e^/-1/e? – John Liu Dec 05 '20 at 00:34
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As the replies indicate, you should put $e^{\frac{-1}{e}}$ since there is no simpler form.
Nevertheless, you seem to seek an intuition into irrational exponents, namely $\frac{-1}{e}$.
The basic idea is that every real number is the limit of a sequence of rational numbers, in this case with the help of the Taylor series of $e^x$ $$\frac{-1}{e}=-e^{-1}=-\sum_{n=0}^\infty \frac{(-1)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n!} =-1+\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\cdots$$
Which follows $$ \exp\left(\frac{-1}{e}\right) = \exp\left(\sum\limits_{n=0}^\infty \frac{(-1)^{n+1}}{n!} \right) = \prod\limits_{n=0}^\infty \exp\left( \frac{(-1)^{n+1}}{n!} \right) = e^{-1}\cdot e^{\frac{1}{1!}}\cdot e^{-\frac{1}{2!}}\cdot e^{\frac{1}{3!}}\cdot e^{-\frac{1}{4!}}\cdot\dots $$ Which translates into a product of roots of $e$ and $\frac{1}{e}$.
$$e^{\frac{-1}{e}} = e^{-1}\cdot e\cdot \sqrt{e^{-1}}\cdot \sqrt[3!]{e}\cdot \sqrt[4!]{e^{-1}}\cdot \ldots$$
While the obtained expression might make sense, it is not mathematically significant.
However, one might use this expression to write a small program to compute $e^\frac{-1}{e}$ in the desired precision.
Check out Real exponents on Wikipedia.
rand
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