TheEngel expansion of apositive real numberx is the uniquenon-decreasingsequence of positiveintegers${\displaystyle (a_1,a_2,a_3,\dots )}$ such that

${\displaystyle x=\frac{1}{a_1}+\frac{1}{a_1{a}_2}+\frac{1}{a_1{a}_2{a}_3}+\cdots =\frac{1}{a_1}\left(1+\frac{1}{a_2}\left(1+\frac{1}{a_3}\left(1+\cdots \right)\right)\right)}$

For instance,Euler's numbere has the Engel expansion^[1]

1, 1, 2, 3, 4, 5, 6, 7, 8, ...

corresponding to theinfinite series

${\displaystyle e=\frac{1}{1}+\frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{1\cdot 2\cdot 3\cdot 4}+\cdots }$

Rational numbers have a finite Engel expansion, whileirrational numbers have an infinite Engel expansion. Ifx is rational, its Engel expansion provides a representation ofx as anEgyptian fraction. Engel expansions are named afterFriedrich Engel, who studied them in 1913.

An expansion analogous to anEngel expansion, in which alternating terms are negative, is called a Pierce expansion.

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

${\displaystyle x=\frac{1+\frac{1+\frac{1+\cdots }{a_3}}{a_2}}{a_1}.}$

They claim that ascending continued fractions such as this have been studied as early asFibonacci'sLiber 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:

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

If such a notation has all numerators 0 or 1, as occurs in several instances inLiber 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 ofx, let

${\displaystyle u_1=x,}$

${\displaystyle a_k=\lceil \frac{1}{u_k}\rceil ,}$

and

${\displaystyle u_{k+1}=u_k{a}_k-1}$

where${\displaystyle \lceil r\rceil }$ is theceiling function (the smallest integer not less thanr).

If${\displaystyle u_i=0}$ for anyi, halt the algorithm.

Another equivalent method is to consider themap^[2]

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

and set

${\displaystyle u_k=1+\lfloor \frac{1}{g^{(n-1)}(x)}\rfloor }$

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

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

and

The Frobenius–Perrontransfer operator of the Engel map${\displaystyle g(x)}$ acts on functions${\displaystyle f(x)}$ with

${\displaystyle [{\mathcal{L}}_{g}f](x)=\sum _{y:g(y)=x}\frac{f(y)}{{\left|\frac{d}{dz}g(z)\right|}_{z=y}}=\sum _{n=1}^{\mathrm{\infty }}\frac{f\left(\frac{x+1}{n+1}\right)}{n+1}}$

since

${\displaystyle \frac{d}{dx}[x(n+1)-1]=n+1}$ and the inverse of then-th component is${\displaystyle \frac{x+1}{n+1}}$ which is found by solving${\displaystyle x(n+1)-1=y}$ for${\displaystyle x}$.

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

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

${\displaystyle u_1=1.175,a_1=\lceil \frac{1}{1.175}\rceil =1;}$

${\displaystyle u_2=u_1{a}_1-1=1.175\cdot 1-1=0.175,a_2=\lceil \frac{1}{0.175}\rceil =6}$

${\displaystyle u_3=u_2{a}_2-1=0.175\cdot 6-1=0.05,a_3=\lceil \frac{1}{0.05}\rceil =20}$

${\displaystyle u_4=u_3{a}_3-1=0.05\cdot 20-1=0}$

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, ifu_i is a rational numberx/y, thenu_i + 1 = (−y modx)/y. Therefore, at each step, the numerator in the remaining fractionu_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

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

the final digitn 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 finitedecimal representation also has an infinite decimal representation (see0.999...).An infinite Engel expansion in which all terms are equal is ageometric series.

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

Consider this sum:

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{\prod _{i=0}^{k-1}(\alpha +i\beta )}=\frac{1}{\alpha }+\frac{1}{\alpha (\alpha +\beta )}+\frac{1}{\alpha (\alpha +\beta )(\alpha +2\beta )}+\cdots ,}$

where${\displaystyle \alpha ,\beta \in \mathbb{N}}$ and. Thus, in general

${\displaystyle {\left(\frac{1}{\beta }\right)}^{1-\frac{\alpha }{\beta }}{e}^{\frac{1}{\beta }}\gamma \left(\frac{\alpha }{\beta },\frac{1}{\beta }\right)=\{\alpha ,\alpha (\alpha +\beta ),\alpha (\alpha +\beta )(\alpha +2\beta ),\dots \}}$,

where${\displaystyle \gamma }$ represents the lowerIncomplete gamma function.

Specifically, if${\displaystyle \alpha =\beta }$,

${\displaystyle e^{1/\beta }-1=\{1\beta ,2\beta ,3\beta ,4\beta ,5\beta ,6\beta ,\dots \}}$.

The Gauss identity of theq-analog can be written as:

${\displaystyle \prod _{n=1}^{\mathrm{\infty }}\frac{1-\frac{1}{q^{2n}}}{1-\frac{1}{q^{2n-1}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{q^{\frac{n(n+1)}{2}}},q\in \mathbb{N}.}$

Using this identity, we can express the Engel expansion for powers of${\displaystyle q}$ as follows:

${\displaystyle \prod _{n=1}^{\mathrm{\infty }}{\left(1-\frac{1}{q^n}\right)}^{(-1{)}^n}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\prod _{i=1}^n{q}^i}.}$

Furthermore, this expression can be written in closed form as:

${\displaystyle \frac{q^{1/8}{\vartheta }_2\left(\frac{1}{\sqrt{q}}\right)}{2}=\{1,q,q^3,q^6,q^{10},\dots \}}$

where${\displaystyle {\vartheta }_2}$ is the secondTheta function.

${\displaystyle \pi }$ = (1, 1, 1, 8, 8, 17, 19, 300, 1991, 2492, ...) (sequenceA006784 in theOEIS)

${\displaystyle \sqrt{2}}$ = (1, 3, 5, 5, 16, 18, 78, 102, 120, 144, ...) (sequenceA028254 in theOEIS)

${\displaystyle e}$ = (1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, ...) (sequenceA028310 in theOEIS)

More Engel expansions for constants can be foundhere.

Thecoefficientsa_i of the Engel expansion typically exhibitexponential growth; more precisely, foralmost all numbers in theinterval (0,1], thelimit${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n^{1/n}}$ exists and is equal toe. However, thesubset of the interval for which this is not the case is still large enough that itsHausdorff dimension is one.^[4]

The same typical growth rate applies to the terms in expansion generated by thegreedy 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.

• Mathematical analysis
• Continued fractions
• Egyptian fractions