What is an Engel Expansion?

It turns out that each positive real number $x\in\mathbb{R}^+$ can be written in the form

$$x=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\cdots,$$

where ${a_1,a_2,a_3,\ldots}$ is a unique non-decreasing sequence, finite or infinite, of postive integers. This is called the Engel expansion of $x$, after the German mathematician Friedrich Engel.

Example

One striking sequence is that for the number $e$. It’s ${1,1,2,3,\cdots}$, so

$$e=\frac{1}{1}+\frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+\cdots.$$

This is simply the well-know infinite sum

$$e=\sum_{n=0}^\infty\,\frac{1}{n!},$$

but it’s funny that either $\pi$ or $e$ often show up in these kinds of things.

Algorithm

The find the values of $a_i$, use the following iterative algorithm. Set

$$u_1=x,$$

$$a_i=\left\lceil{\frac{1}{u_i}}\right\rceil,$$

where $\lceil y\rceil$ is the ceiling, the smallest integer greater than or equal to $y$, and

$$u_{i+1}=u_ia_i-1,$$

and continue this process until you hit a value of $u_i$ that is zero. And that’s it. Finding a value of $u_i$ that is zero may or may not happen, since an Engel expansion can be infinite.

As a numerical example, let’s apply this algorithm to $x=1.225$. We have $u_1=1.225$ and $a_1=\lceil{1/1.225}\rceil=1$. Then $u_2=1.225\cdot 1-1=0.225$ and $a_2=\lceil{1/0.225}\rceil=5$. And then $u_3=0.225\cdot 5-1=0.125$ and $a_3=\lceil{1/0.125}\rceil=8$. And then it stops, because $u_4=0.125\cdot 8-1=0$. Hence,

$$1.225=\frac{1}{1}+\frac{1}{1\cdot 5}+\frac{1}{1\cdot 5\cdot 8}.$$

Where Does the Algorithm Come From?

A way to understand where the algorithm comes from, is to start from the original Engel expansion again,

$$x=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\cdots.$$

The obvious value to set $a_1$ to would be $1/x$, because then we would be done. However, we want the $a_i$s to be integers, so we set $a_1=\lceil 1/x\rceil$. We know that $x$ is now not (necessarily) equal to $1/a_1$, but we do know that we want to pick the other $a_i$s so that the rest of the (possibly infinite) sum is equal to their difference, so

$$x-\frac{1}{a_1}=\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\cdots.$$

First we get rid of the $a_1$s in all the remaining terms by multiplying both sides of the equation by $a_1$, so

$$xa_1-1=\frac{1}{a_2}+\frac{1}{a_2a_3}+\frac{1}{a_2a_3a_4}\cdots.$$

It turns out that we then immediately end up in a situation that is very similar to what we started off with, so that we can repeat the whole process by setting $u_2=xa_1-1$, which is the next step in the algorithm.

Note that this algorithm leads to $a_2$ (and all further $a_i$s) being positive, since $x-1/a_1\geq 0$, because $\lceil 1/x\rceil\geq 1/x$.

Also note that this doesn’t prove that the algorithm converges to the correct solution (although it actually does). It simply shows where the algorithm comes from.