List of representations of e

Short description: none

The mathematical constant e can be represented in a variety of ways as a real number. Since e is an irrational number (see proof that e is irrational), it cannot be represented as the quotient of two integers, but it can be represented as a continued fraction. Using calculus, e may also be represented as an infinite series, infinite product, or other types of limit of a sequence.

As a continued fraction

Euler proved that the number e is represented as the infinite simple continued fraction (sequence A003417 in the OEIS):

[math]\displaystyle{ e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, \ldots, 1, 2n, 1, \ldots]. }[/math]

Its convergence can be tripled by allowing just one fractional number:

[math]\displaystyle{ e = [1; 1/2, 12, 5, 28, 9, 44, 13, 60, 17, \ldots, 4(4n-1), 4n+1, \ldots]. }[/math]

Here are some infinite generalized continued fraction expansions of e. The second is generated from the first by a simple equivalence transformation.

[math]\displaystyle{ e= 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{5+\ddots}}}}} = 2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cfrac{5}{5+\cfrac{6}{6+\ddots\,}}}}} }[/math]

[math]\displaystyle{ e = 2+\cfrac{1}{1+\cfrac{2}{5+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+\ddots\,}}}}} = 1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+\ddots\,}}}}} }[/math]

This last, equivalent to [1; 0.5, 12, 5, 28, 9, ...], is a special case of a general formula for the exponential function:

[math]\displaystyle{ e^{x/y} = 1+\cfrac{2x} {2y-x+\cfrac{x^2} {6y+\cfrac{x^2} {10y+\cfrac{x^2} {14y+\cfrac{x^2} {18y+\ddots}}}}} }[/math]

As an infinite series

The number e can be expressed as the sum of the following infinite series:

[math]\displaystyle{ e^x = \sum_{k=0}^\infty \frac{x^k}{k!} }[/math] for any real number x.

In the special case where x = 1 or −1, we have:

[math]\displaystyle{ e = \sum_{k=0}^\infty \frac{1}{k!} }[/math], and

[math]\displaystyle{ e^{-1} = \sum_{k=0}^\infty \frac{(-1)^k}{k!}. }[/math]

Other series include the following:

[math]\displaystyle{ e = \left [ \sum_{k=0}^\infty \frac{1-2k}{(2k)!} \right ]^{-1} }[/math]

[math]\displaystyle{ e = \frac{1}{2} \sum_{k=0}^\infty \frac{k+1}{k!} }[/math]

[math]\displaystyle{ e = 2 \sum_{k=0}^\infty \frac{k+1}{(2k+1)!} }[/math]

[math]\displaystyle{ e = \sum_{k=0}^\infty \frac{3-4k^2}{(2k+1)!} }[/math]

[math]\displaystyle{ e = \sum_{k=0}^\infty \frac{(3k)^2+1}{(3k)!} = \sum_{k=0}^\infty \frac{(3k+1)^2+1}{(3k+1)!} = \sum_{k=0}^\infty \frac{(3k+2)^2+1}{(3k+2)!} }[/math]

[math]\displaystyle{ e = \left [ \sum_{k=0}^\infty \frac{4k+3}{2^{2k+1}\,(2k+1)!} \right ]^2 }[/math]

[math]\displaystyle{ e = \sum_{k=0}^\infty \frac{k^n}{B_n(k!)} }[/math] where [math]\displaystyle{ B_n }[/math] is the nth Bell number.

[math]\displaystyle{ e = \sum_{k=0}^\infty \frac{2k+3}{(k+2)!} }[/math]

Consideration of how to put upper bounds on e leads to this descending series:

[math]\displaystyle{ e = 3 - \sum_{k=2}^\infty \frac{1}{k! (k-1) k} = 3 - \frac{1}{4} - \frac{1}{36} - \frac{1}{288} - \frac{1}{2400} - \frac{1}{21600} - \frac{1}{211680} - \frac{1}{2257920} - \cdots }[/math]

which gives at least one correct (or rounded up) digit per term. That is, if 1 ≤ n, then

[math]\displaystyle{ e \lt 3 - \sum_{k=2}^n \frac{1}{k! (k-1) k} \lt e + 0.6 \cdot 10^{1-n} \,. }[/math]

More generally, if x is not in {2, 3, 4, 5, ...}, then

[math]\displaystyle{ e^x = \frac{2+x}{2-x} + \sum_{k=2}^\infty \frac{- x^{k+1}}{k! (k-x) (k+1-x)} \,. }[/math]

As a recursive function

The series representation of [math]\displaystyle{ e }[/math], given as [math]\displaystyle{ e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} \cdots }[/math]can also be expressed using a form of recursion. When [math]\displaystyle{ \frac{1}{n} }[/math] is iteratively factored from the original series the result is the nested series [math]\displaystyle{ e = 1 + \frac{1}{1}(1 + \frac{1}{2}(1 + \frac{1}{3}(1 + \cdots ))) }[/math]which equates to [math]\displaystyle{ e = 1 + \cfrac{1 + \cfrac{1 + \cfrac{1 + \cdots }{3}}{2}}{1} }[/math] This fraction is of the form [math]\displaystyle{ f(n) = 1 + \frac{f(n + 1)}{n} }[/math], where [math]\displaystyle{ f(1) }[/math] computes the sum of the terms from [math]\displaystyle{ 1 }[/math] to [math]\displaystyle{ \infty }[/math].

As an infinite product

The number e is also given by several infinite product forms including Pippenger's product

[math]\displaystyle{ e= 2 \left ( \frac{2}{1} \right )^{1/2} \left ( \frac{2}{3}\; \frac{4}{3} \right )^{1/4} \left ( \frac{4}{5}\; \frac{6}{5}\; \frac{6}{7}\; \frac{8}{7} \right )^{1/8} \cdots }[/math]

and Guillera's product

[math]\displaystyle{ e = \left ( \frac{2}{1} \right )^{1/1} \left (\frac{2^2}{1 \cdot 3} \right )^{1/2} \left (\frac{2^3 \cdot 4}{1 \cdot 3^3} \right )^{1/3} \left (\frac{2^4 \cdot 4^4}{1 \cdot 3^6 \cdot 5} \right )^{1/4} \cdots , }[/math]

where the nth factor is the nth root of the product

[math]\displaystyle{ \prod_{k=0}^n (k+1)^{(-1)^{k+1}{n \choose k}}, }[/math]

as well as the infinite product

[math]\displaystyle{ e = \frac{2\cdot 2^{(\ln(2)-1)^2} \cdots}{2^{\ln(2)-1}\cdot 2^{(\ln(2)-1)^3}\cdots }. }[/math]

More generally, if 1 < B < e² (which includes B = 2, 3, 4, 5, 6, or 7), then

[math]\displaystyle{ e = \frac{B\cdot B^{(\ln(B)-1)^2} \cdots}{B^{\ln(B)-1}\cdot B^{(\ln(B)-1)^3}\cdots }. }[/math]

Also

[math]\displaystyle{ e = \lim\limits_{n\rightarrow\infty}\prod_{k=0}^n{n \choose k}^{2/{((n +\alpha)(n+\beta))}}\ \forall\alpha,\beta\in\Bbb R }[/math]

As the limit of a sequence

The number e is equal to the limit of several infinite sequences:

[math]\displaystyle{ e= \lim_{n \to \infty} n\cdot\left ( \frac{\sqrt{2 \pi n}}{n!} \right )^{1/n} }[/math] and

[math]\displaystyle{ e=\lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}} }[/math] (both by Stirling's formula).

The symmetric limit,

[math]\displaystyle{ e=\lim_{n \to \infty} \left [ \frac{(n+1)^{n+1}}{n^n}- \frac{n^n}{(n-1)^{n-1}} \right ] }[/math]

may be obtained by manipulation of the basic limit definition of e.

The next two definitions are direct corollaries of the prime number theorem

[math]\displaystyle{ e= \lim_{n \to \infty}(p_n \#)^{1/p_n} }[/math]

where [math]\displaystyle{ p_n }[/math] is the nth prime and [math]\displaystyle{ p_n \# }[/math] is the primorial of the nth prime.

[math]\displaystyle{ e= \lim_{n \to \infty}n^{\pi(n)/n} }[/math]

where [math]\displaystyle{ \pi(n) }[/math] is the prime-counting function.

Also:

[math]\displaystyle{ e^x= \lim_{n \to \infty}\left (1+ \frac{x}{n} \right )^n. }[/math]

In the special case that [math]\displaystyle{ x = 1 }[/math], the result is the famous statement:

[math]\displaystyle{ e= \lim_{n \to \infty}\left (1+ \frac{1}{n} \right )^n. }[/math]

The ratio of the factorial [math]\displaystyle{ n! }[/math], that counts all permutations of an ordered set S with cardinality [math]\displaystyle{ n }[/math], and the derangement function [math]\displaystyle{ !n }[/math], which counts the amount of permutations where no element appears in its original position, tends to [math]\displaystyle{ e }[/math] as [math]\displaystyle{ n }[/math] grows.

[math]\displaystyle{ e= \lim_{n \to \infty} \frac{n!}{!n}. }[/math]

In trigonometry

Trigonometrically, e can be written in terms of the sum of two hyperbolic functions,

[math]\displaystyle{ e^x = \sinh(x) + \cosh(x) , }[/math]

at x = 1.

See also

• List of formulae involving π