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.

  • $\begingroup$ Would you prefer $e^{-1/e}$? $\endgroup$ Commented Dec 4, 2020 at 23:12
  • $\begingroup$ Yes, I'll edit that $\endgroup$
    – John Liu
    Commented Dec 4, 2020 at 23:25
  • $\begingroup$ 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. $\endgroup$
    – John Liu
    Commented Dec 4, 2020 at 23:29
  • $\begingroup$ $e^{-1/e}$ 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. $\endgroup$
    – user239203
    Commented Dec 4, 2020 at 23:29
  • 2
    $\begingroup$ I don't see any point of writing anything other than $e^{-1/e}$ $\endgroup$
    – wormram
    Commented Dec 4, 2020 at 23:29

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .