# Different definitions of e (Part 12): Numerical computation

In this series of posts, we have seen that the number $e$ can be thought about in three different ways.

1. $e$ defines a region of area 1 under the hyperbola $y = 1/x$. 2. We have the limits $e = \displaystyle \lim_{h \to 0} (1+h)^{1/h} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$.

These limits form the logical basis for the continuous compound interest formula.

3. We have also shown that $\frac{d}{dx} \left(e^x \right) = e^x$. From this derivative, the Taylor series expansion for $e^x$ about $x = 0$ can be computed: $e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$

Therefore, we can let $x = 1$ to find $e$: $e = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} = \displaystyle 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots$ In yesterday’s post, I showed that using the original definition (in terms of an area under a hyperbola) does not lend itself well to numerically approximating $e$. Let’s now look at the other two methods.

2. The limit $e = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ gives a somewhat more tractable way of approximating $e$, at least with a modern calculator. However, you can probably imagine the fun of trying to use this formula without a calculator. 3. The best way to compute $e$ (or, in general, $e^x$) is with Taylor series. The fractions $\frac{1}{n!}$ get very small very quickly, leading to rapid convergence. Indeed, with only terms up to $1/6!$, this approximation beats the above approximation with $n = 1000$. Adding just two extra terms comes close to matching the accuracy of the above limit when $n = 1,000,000$. More about approximating $e^x$ via Taylor series can be found in my previous post.