What is the value of $e^\frac{-1}{e}$?

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.

• Would you prefer $e^{-1/e}$? Dec 4 '20 at 23:12
• Yes, I'll edit that Dec 4 '20 at 23:25
• 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. Dec 4 '20 at 23:29
• $e^{-1/e}$ 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.
– user239203
Dec 4 '20 at 23:29
• I don't see any point of writing anything other than $e^{-1/e}$ Dec 4 '20 at 23:29

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.