TITLE: Which expansion of $e$ is more accurate?
QUESTION [10 upvotes]: We have two forms  of $e^x$ 

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....$$
  and
  $$e^x=\frac{1}{\displaystyle 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+....}$$
  The second form comes from $e^x=1/e^{-x}$

Which one is more accurate if I want to find any value of $e^x$

REPLY [2 votes]: Here's one quick observation.  Letting
$$f_{1,n}(x)=\sum_{k=0}^n \frac{x^n}{n!}$$
and 
$$f_{2,n}(x)=1/\left(\sum_{k=0}^n \frac{(-x)^n}{n!}\right),$$
we have
$$\lim_{x\rightarrow\pm\infty}f_{1,n}(x) = \pm\infty \: \text{ and } \: 
\lim_{x\rightarrow\pm\infty}f_{2,n}(x) = 0.$$
Thus, I suppose we'd expect $f_{1,n}(x)$ to be a better approximation for positive $x$ and $f_{2,n}(x)$ to be a better approximation for negative $x$.
In the image below, for example, the thick, blue graph shows $f(x)=e^x$, the green graph shows $f_{1,6}(x)$, and the red graph shows $f_{2,6}(x)$.  We can clearly see that $f_{2,6}$ veers away from $f$ for positive $x$, while $f_{1,6}$ is still a reasonable approximation - at least on this scale.  The reverse is true for negative $x$.