List of representations of e

mathematical constant $e$
Part of a series of articles on the

Properties
Natural logarithm Exponential function
Applications
compound interest Euler's identity Euler's formula half-lives exponential growth and decay
Defining $e$
proof that $e$ is irrational representations of $e$ Lindemann–Weierstrass theorem
People
John Napier Leonhard Euler
Related topics
Schanuel's conjecture

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 a fraction, but it can be represented as a continued fraction. Using calculus, $e$ may also be represented as an infinite series, infinite product, or other sort of limit of a sequence.

As a continued fraction

Euler proved that the number $e$ is represented as the infinite simple continued fraction^[1] (sequence A003417 in the OEIS):

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

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

e=[1;{\textbf {0.5}},12,5,28,9,44,13,60,17,\ldots ,{\textbf {4(4n-1)}},{\textbf {4n+1}},\ldots ].\,

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

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 \,}}}}}}}}}}

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 \,}}}}}}}}}}

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

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 }}}}}}}}}}

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

e^{x}=\sum _{{k=0}}^{\infty }{\frac {x^{k}}{k!}}

for any real number x.

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

e=\sum _{{k=0}}^{\infty }{\frac {1}{k!}}

,^[2] and

e^{{-1}}=\sum _{{k=0}}^{\infty }{\frac {(-1)^{k}}{k!}}

Other series include the following:

e=\left[\sum _{{k=0}}^{\infty }{\frac {1-2k}{(2k)!}}\right]^{{-1}}

^[3]

e={\frac {1}{2}}\sum _{{k=0}}^{\infty }{\frac {k+1}{k!}}

e=2\sum _{{k=0}}^{\infty }{\frac {k+1}{(2k+1)!}}

e=\sum _{{k=0}}^{\infty }{\frac {3-4k^{2}}{(2k+1)!}}

e=\sum _{{k=0}}^{\infty }{\frac {(3k)^{2}+1}{(3k)!}}

e=\left[\sum _{{k=0}}^{\infty }{\frac {4k+3}{2^{{2k+1}}\,(2k+1)!}}\right]^{2}

e=\sum _{{k=1}}^{\infty }{\frac {k^{n}}{B_{n}(k!)}}

where

B_{n}

is the

n^{th}

Bell number. Some few examples: (for n = 2..7)

e=\sum _{{k=1}}^{\infty }{\frac {k^{2}}{2(k!)}}

e=\sum _{{k=1}}^{\infty }{\frac {k^{3}}{5(k!)}}

e=\sum _{{k=1}}^{\infty }{\frac {k^{4}}{15(k!)}}

e=\sum _{{k=1}}^{\infty }{\frac {k^{5}}{52(k!)}}

e=\sum _{{k=1}}^{\infty }{\frac {k^{6}}{203(k!)}}

e=\sum _{{k=1}}^{\infty }{\frac {k^{7}}{877(k!)}}

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

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

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

e<3+\sum _{k=2}^{n}{\frac {-1}{k!(k-1)k}}<e+0.6\cdot 10^{1-n}\,.

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

e^{x}={\frac {2+x}{2-x}}+\sum _{k=2}^{\infty }{\frac {-x^{k+1}}{k!(k-x)(k+1-x)}}\,.

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

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

and Guillera's product ^[4]^[5]

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 ,

where the nth factor is the nth root of the product

\prod _{{k=0}}^{n}(k+1)^{{(-1)^{{k+1}}{n \choose k}}},

as well as the infinite product

e={\frac {2\cdot 2^{{(\ln(2)-1)^{2}}}\cdots }{2^{{\ln(2)-1}}\cdot 2^{{(\ln(2)-1)^{3}}}\cdots }}.

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

e={\frac {B\cdot B^{(\ln(B)-1)^{2}}\cdots }{B^{\ln(B)-1}\cdot B^{(\ln(B)-1)^{3}}\cdots }}.

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

e=\lim _{{n\to \infty }}n\cdot \left({\frac {{\sqrt {2\pi n}}}{n!}}\right)^{{1/n}}

and

e=\lim _{{n\to \infty }}{\frac {n}{{\sqrt[ {n}]{n!}}}}

The symmetric limit,^[6]^[7]

e=\lim _{{n\to \infty }}\left[{\frac {(n+1)^{{n+1}}}{n^{n}}}-{\frac {n^{n}}{(n-1)^{{n-1}}}}\right]

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

The next two definitions are direct corollaries of the prime number theorem^[8]

e=\lim _{{n\to \infty }}(p_{n}\#)^{{1/p_{n}}}

where $p_{n}$ is the nth prime and $p_{n}\#$ is the primorial of the nth prime.

e=\lim _{{n\to \infty }}n^{{\pi (n)/n}}

where $\pi (n)$ is the prime counting function.

Also:

e^{x}=\lim _{{n\to \infty }}\left(1+{\frac {x}{n}}\right)^{n}.

In the special case that $x=1$ , the result is the famous statement:

e=\lim _{{n\to \infty }}\left(1+{\frac {1}{n}}\right)^{n}.

Trigonometrically, $e$ can be written as the sum of two hyperbolic functions:

e^{x}=\sinh(x)+\cosh(x)\,

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