Engel expansion

Kraaikamp & Wu (2004) observe that an Engel expansion can also be written as an ascending variant of a continued fraction:

${\displaystyle x={\cfrac {1+{\cfrac {1+{\cfrac {1+\cdots }{a_{3}}}}{a_{2}}}}{a_{1}}}.}$

They claim that ascending continued fractions such as this have been studied as early as Fibonacci's Liber Abaci (1202). This claim appears to refer to Fibonacci's compound fraction notation in which a sequence of numerators and denominators sharing the same fraction bar represents an ascending continued fraction:

${\displaystyle {\frac {a\ b\ c\ d}{e\ f\ g\ h}}={\dfrac {d+{\cfrac {c+{\cfrac {b+{\cfrac {a}{e}}}{f}}}{g}}}{h}}.}$

If such a notation has all numerators 0 or 1, as occurs in several instances in Liber Abaci, the result is an Engel expansion. However, Engel expansion as a general technique does not seem to be described by Fibonacci.

To find the Engel expansion of x, let

${\displaystyle u_{1}=x,}$

${\displaystyle a_{k}=\left\lceil {\frac {1}{u_{k}}}\right\rceil \!,}$

and

${\displaystyle u_{k+1}=u_{k}a_{k}-1}$

where ${\displaystyle \left\lceil r\right\rceil }$ is the ceiling function (the smallest integer not less than r).

If ${\displaystyle u_{i}=0}$ for any i, halt the algorithm.

Another equivalent method is to consider the map[2]

${\displaystyle g(x)=x\!\left(1+\left\lfloor x^{-1}\right\rfloor \right)-1}$

and set

${\displaystyle u_{k}=1+\left\lfloor {\frac {1}{g^{(n-1)}(x)}}\right\rfloor }$

where

${\displaystyle g^{(n)}(x)=g(g^{(n-1)}(x))}$ and ${\displaystyle g^{(0)}(x)=x.}$

Yet another equivalent method, called the modified Engel expansion calculated by

${\displaystyle h(x)=\left\lfloor {\frac {1}{x}}\right\rfloor g(x)=\left\lfloor {\frac {1}{x}}\right\rfloor \!\left(x\left\lfloor {\frac {1}{x}}\right\rfloor +x-1\right)}$

and

${\displaystyle u_{k}={\begin{cases}1+\left\lfloor {\frac {1}{x}}\right\rfloor &n=1\\\left\lfloor {\frac {1}{h^{(k-2)}(x)}}\right\rfloor \!\left(1+\left\lfloor {\frac {1}{h^{(k-1)}(x)}}\right\rfloor \right)&n\geq 2\end{cases}}}$

The Mellin transform of the map ${\displaystyle g(x)}$ is related to the Riemann zeta function by the formula

${\displaystyle {\begin{aligned}\int _{0}^{1}g(x)x^{s-1}\,dx&=\sum _{n=1}^{\infty }\int _{\frac {1}{n+1}}^{\frac {1}{n}}(x(n+1)-1)x^{s-1}\,dx\\[5pt]&=\sum _{n=1}^{\infty }{\frac {n^{-s}(s-1)+(n+1)^{-s-1}(n^{2}+2n+1)+n^{-s-1}s-n^{1-s}}{(s+1)s(n+1)}}\\[5pt]&={\frac {\zeta (s+1)}{s+1}}-{\frac {1}{s(s+1)}}\end{aligned}}.}$

To find the Engel expansion of 1.175, we perform the following steps.

${\displaystyle u_{1}=1.175,a_{1}=\left\lceil {\frac {1}{1.175}}\right\rceil =1;}$

${\displaystyle u_{2}=u_{1}a_{1}-1=1.175\cdot 1-1=0.175,a_{2}=\left\lceil {\frac {1}{0.175}}\right\rceil =6}$

${\displaystyle u_{3}=u_{2}a_{2}-1=0.175\cdot 6-1=0.05,a_{3}=\left\lceil {\frac {1}{0.05}}\right\rceil =20}$

${\displaystyle u_{4}=u_{3}a_{3}-1=0.05\cdot 20-1=0}$

The series ends here. Thus,

${\displaystyle 1.175={\frac {1}{1}}+{\frac {1}{1\cdot 6}}+{\frac {1}{1\cdot 6\cdot 20}}}$

and the Engel expansion of 1.175 is (1, 6, 20).

Every positive rational number has a unique finite Engel expansion. In the algorithm for Engel expansion, if u_i is a rational number x/y, then u_i + 1 = (−y mod x)/y. Therefore, at each step, the numerator in the remaining fraction u_i decreases and the process of constructing the Engel expansion must terminate in a finite number of steps. Every rational number also has a unique infinite Engel expansion: using the identity

${\displaystyle {\frac {1}{n}}=\sum _{r=1}^{\infty }{\frac {1}{(n+1)^{r}}}.}$

the final digit n in a finite Engel expansion can be replaced by an infinite sequence of (n + 1)s without changing its value. For example,

${\displaystyle 1.175=(1,6,20)=(1,6,21,21,21,\dots ).}$

This is analogous to the fact that any rational number with a finite decimal representation also has an infinite decimal representation (see 0.999...). An infinite Engel expansion in which all terms are equal is a geometric series.

Erdős, Rényi, and Szüsz asked for nontrivial bounds on the length of the finite Engel expansion of a rational number x/y ; this question was answered by Erdős and Shallit, who proved that the number of terms in the expansion is O(y^1/3 + ε) for any ε > 0.[3]

${\displaystyle \pi }$ = (1, 1, 1, 8, 8, 17, 19, 300, 1991, 2492, ...) (sequence A006784 in the OEIS)

${\displaystyle {\sqrt {2}}}$ = (1, 3, 5, 5, 16, 18, 78, 102, 120, 144, ...) (sequence A028254 in the OEIS)

${\displaystyle e}$ = (1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, ...) (sequence A028310 in the OEIS)

And in general,

${\displaystyle e^{1/r}-1=(r,2r,3r,4r,5r,6r,\dots )}$

More Engel expansions for constants can be found here.

The coefficients a_i of the Engel expansion typically exhibit exponential growth; more precisely, for almost all numbers in the interval (0,1], the limit ${\displaystyle \lim _{n\to \infty }a_{n}^{1/n}}$ exists and is equal to e. However, the subset of the interval for which this is not the case is still large enough that its Hausdorff dimension is one.[4]

The same typical growth rate applies to the terms in expansion generated by the greedy algorithm for Egyptian fractions. However, the set of real numbers in the interval (0,1] whose Engel expansions coincide with their greedy expansions has measure zero, and Hausdorff dimension 1/2.[5]

• Euler's continued fraction formula

• Weisstein, Eric W. "Engel Expansion". MathWorld–A Wolfram Web Resource.