Why is the function $f(x) = e^x$ not a rational function?

Question

This is probably a misunderstanding of how e works on my part, but I figured that since $e^x$ is a polynomial and $f(x)$ can be rewritten as $(e^x)/1$ that it was rational, but that turned out to be incorrect. Can someone explain where my error came from?

Answer 1

$e^x$ is not a polynomial. The power series defining it involves infinitely many terms.

However, a priori this doesn't rule out the possibility that $e^x$ could be written as a rational function in some more complicated way. So here is a proof of that. Any non-constant rational function $f(z) = \frac{p(z)}{q(z)}$ takes on the value either $0$ or $\infty$ for some $z \in \mathbb{C}$, given by a root of either the numerator or denominator. But $e^z \neq 0, \infty$ for any $z \in \mathbb{C}$. (Many other proofs could also be given here.)

Answer 2

to see that $e^{x}$ is not polynomial,use the fact that polynomials are characterized by the fact that their $n$-derivative is identically zero, for some n,and this is not the case for $e^{x}$ .to see that $e^{x}$ is not a rational function, one can use the fact that the limits of any rational function at $\pm \infty$ are both either zero, a constant or $\pm \infty $ but for $e^{x}$ one limit is zero, while the other is $+\infty$.