TITLE: Which of the following sums is a closer approximate to $e^{-5}$?
QUESTION [7 upvotes]: Here are two sums given:

$$e^{-5}\approx \sum_{i=0}^9 \frac{(-1)^i 5^i}{i!}$$
$$e^{-5}=1/e^5\approx \frac{1}{\sum_{i=0}^9 \frac{5^i}{i!}}$$

Using Matlab ;) I've got (2) to be closer. But computer-aside, is there anyway to say which one is closer than the other, say by thinking about the errors?

REPLY [5 votes]: The approximation
$$
e^x\approx\sum_{i=0}^{n-1}\frac{x^i}{i!}
$$
will have error term $\approx x^n/n!$ when $n\gg x$ since at this stage the terms are getting closer to zero. Thus, for $e^x$ and $e^{-x}$, the error term is the same order of magnitude. However, for $x>0$, the error term relative to the value is larger for $e^{-x}$, which is what matters when we estimate $e^{-x}$ using $1/e^x$.
Thus, for $x>0$, it is better to approximate $e^{-x}$ as $1/e^x$.