View source for List of representations of e

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{{DISPLAYTITLE:List of representations of {{mvar|e}}}}
{{more citations needed|date=December 2007}}
{{E (mathematical constant)}}
The [[mathematical constant]] {{math|[[E (mathematical constant)|''e'']]}} can be represented in a variety of ways as a [[real number]].  Since {{math|''e''}} is an [[irrational number]] (see [[proof that e is irrational]]), it cannot be represented as the [[quotient]] of two [[integer]]s, but it can be represented as a [[continued fraction]].  Using [[calculus]], {{math|''e''}} may also be represented as an [[infinite series]], [[infinite product]], or other types of [[limit of a sequence]].

==As a continued fraction==

[[Leonhard Euler|Euler]] proved that the number {{math|''e''}} is represented as the infinite [[simple continued fraction]]<ref>{{cite web|url=http://eulerarchive.maa.org/hedi/HEDI-2006-02.pdf|title=How Euler Did It: Who proved ''e'' is Irrational?|last=Sandifer|first=Ed|date=Feb 2006|publisher=MAA Online|accessdate=2017-04-23}}</ref> {{OEIS|id=A003417}}:

:<math>
\begin{align}
e & = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, \ldots, 1, 2n, 1, \ldots] \\[8pt]
& = 2 + \cfrac1{1 + \cfrac1{2 + \cfrac1{1 + \cfrac1{1 + \cfrac1{4 + \cfrac1{1 + \cfrac1{1 + \cfrac1{6 + \cfrac1{1 + \cfrac1{1 + \cfrac1{8 + {{} \atop \ddots}}}}}}} }}}}}
\end{align}
</math>

Here are some infinite [[generalized continued fraction]] expansions of {{math|''e''}}. The second is generated from the first by a simple [[generalized continued fraction#The equivalence transformation|equivalence transformation]].

:<math>
e= 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{5+{{}\atop\ddots}}}}}} = 2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cfrac{5}{5+\cfrac{6}{6+{{}\atop\ddots}\,}}}}}
</math>

:<math>e = 2+\cfrac{1}{1+\cfrac{2}{5+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+{{}\atop\ddots}\,}}}}} = 1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+{{}\atop\ddots}\,}}}}}</math>

This last non-simple continued fraction {{OEIS|A110185}}, equivalent to <math>e = [1; 0.5, 12, 5, 28, 9, ...]</math>, has a quicker convergence rate compared to [[Euler's continued fraction formula]]{{clarify|reason=By what measure?|date = April 2017}} and is a special case of a general formula for the [[exponential function]]:

:<math>e^{x/y} = 1+\cfrac{2x} {2y-x+\cfrac{x^2} {6y+\cfrac{x^2} {10y+\cfrac{x^2} {14y+\cfrac{x^2} {18y+{{}\atop\ddots}}}}}}</math>

==As an infinite series==
The number {{math|''e''}} can be expressed as the sum of the following [[infinite series]]:

:<math>e^x = \sum_{k=0}^\infty \frac{x^k}{k!} </math> for any real number ''x''.

In the [[special case]] where ''x''&nbsp;=&nbsp;1 or &minus;1, we have:

:<math>e = \sum_{k=0}^\infty \frac{1}{k!}</math>,<ref>{{cite web|url=http://oakroadsystems.com/math/loglaws.htm|title=It's the Law Too — the Laws of Logarithms|last=Brown|first=Stan|date=2006-08-27|publisher=Oak Road Systems|accessdate=2008-08-14|url-status=usurped|archiveurl=https://web.archive.org/web/20080813175402/http://oakroadsystems.com/math/loglaws.htm|archivedate=2008-08-13}}</ref> and

:<math>e^{-1} = \sum_{k=0}^\infty \frac{(-1)^k}{k!}.</math>

Other series include the following:

:<math>e = \left [ \sum_{k=0}^\infty \frac{1-2k}{(2k)!} \right ]^{-1}</math> <ref>Formulas 2–7: H. J. Brothers,  [http://www.brotherstechnology.com/docs/Improving_Convergence_(CMJ-2004-01).pdf Improving the convergence of Newton's series approximation for ''e''],  ''The College Mathematics Journal'', Vol. 35, No. 1, (2004),  pp. 34–39.</ref>

:<math>e =  \frac{1}{2} \sum_{k=0}^\infty \frac{k+1}{k!}</math>

:<math>e =  2 \sum_{k=0}^\infty \frac{k+1}{(2k+1)!}</math>

:<math>e =   \sum_{k=0}^\infty \frac{3-4k^2}{(2k+1)!}</math>

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

:<math>e =  \sum_{k=0}^\infty \frac{k^n}{B_n(k!)}</math> where <math>B_n</math> is the {{mvar|n}}th [[Bell number]].

:<math>e =   \sum_{k=0}^\infty \frac{2k+3}{(k+2)!}</math><ref>Formula 8: A. G. Llorente, [https://osf.io/3yzbj A Novel Simple Representation Series for Euler's Number ''e''], preprint, 2023.</ref>
Consideration of how to put upper bounds on ''e'' leads to this descending series:
:<math>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>e < 3 - \sum_{k=2}^n \frac{1}{k! (k-1) k} < e + 0.6 \cdot 10^{1-n} \,.</math>
More generally, if ''x'' is not in {2, 3, 4, 5, ...}, then
:<math>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 [[#As_an_infinite_series|series representation]] of <math>e</math>, given as <math display="block">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>\textstyle \frac{1}{n}</math> is iteratively factored from the original series the result is [[Engel expansion|the nested series]]<ref>{{citation
 | journal = Wolfram MathWorld
 | title = e
 | pages = ex. 17, 18, and 19
 | url = https://mathworld.wolfram.com/e.html
 | archive-url = https://web.archive.org/web/20230315161656/https://mathworld.wolfram.com/e.html
 | archive-date = 2023-03-15
 }}.</ref> <math display="block">e = 1 + \frac{1}{1}\left(1 + \frac{1}{2}\left(1 + \frac{1}{3}\left(1 + \cdots \right)\right)\right)</math>which equates to <math display="block">e = 1 + \cfrac{1 + \cfrac{1 + \cfrac{1 + \cdots }{3}}{2}}{1}</math> This fraction is of the form <math>\textstyle f(n) = 1 + \frac{f(n + 1)}{n}</math>, where <math>f(1)</math> computes the sum of the terms from <math>1</math> to <math>\infty</math>.

==As an infinite product==
The number {{math|''e''}} is also given by several [[infinite product]] forms including [[Nick Pippenger|Pippenger]]'s product

:<math> 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 <ref>J. Sondow, [https://arxiv.org/abs/math/0401406 A faster product for pi and a new integral for ln pi/2], ''Amer. Math. Monthly'' 112 (2005) 729–734.</ref><ref>J. Guillera and J. Sondow, [https://arxiv.org/abs/math.NT/0506319 Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent], ''Ramanujan Journal'' 16 (2008), 247–270.</ref>
:<math> 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 ''n''th factor is the ''n''th root of the product
:<math>\prod_{k=0}^n (k+1)^{(-1)^{k+1}{n \choose k}},</math>

as well as the infinite product

:<math> 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'' < ''e''<sup>2</sup> (which includes ''B'' = 2, 3, 4, 5, 6, or 7), then

:<math> e = \frac{B\cdot B^{(\ln(B)-1)^2} \cdots}{B^{\ln(B)-1}\cdot B^{(\ln(B)-1)^3}\cdots }.</math>

Also

:<math> 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 {{math|''e''}} is equal to the [[limit of a sequence|limit]] of several [[infinite sequences]]:

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

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

The symmetric limit,<ref>H. J. Brothers and J. A. Knox,  [http://www.brotherstechnology.com/docs/Closed-Form_Approximations_(MI-1998-12).pdf New closed-form approximations to the Logarithmic Constant ''e''], ''The Mathematical Intelligencer'', Vol. 20, No. 4, (1998), pp. 25–29.</ref>

:<math>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 {{math|''e''}}.

The next two definitions are direct corollaries of the [[prime number theorem]]<ref>{{Cite journal |last=Ruiz |first=Sebastian Martin |title=81.27 A result on prime numbers |journal=The Mathematical Gazette |year=1997 |volume=81 |issue=491 |page=269|publisher=Cambridge University Press |url=https://doi.org/10.2307/3619207 |doi=10.2307/3619207}}</ref>

:<math>
\begin{align}
e &= \lim_{n \to \infty}(p_n \#)^{1/p_n} \\
e &= \lim_{n \to \infty} n^{\pi(n)/n} \\
  &= \lim_{n \to \infty} n^{n/p_n}
\end{align}
</math>
where <math> p_n </math> is the ''n''th [[prime number|prime]], <math> p_n \# </math> is the [[primorial]] of the ''n''th prime, and <math> \pi(n) </math> is the [[prime-counting function]].

Also:
:<math>e^x= \lim_{n \to \infty}\left (1+ \frac{x}{n} \right )^n.</math>

In the special case that <math>x = 1</math>, the result is the famous statement:

:<math>e= \lim_{n \to \infty}\left (1+ \frac{1}{n} \right )^n.</math>

The ratio of the [[factorial]] <math>n!</math>, that counts all [[permutation]]s of an ordered set S with [[cardinality]] <math>n</math>, and the subfactorial (a.k.a. the [[derangement]] function) <math>!n</math>, which counts the amount of permutations where no element appears in its original position, tends to <math>e</math> as <math>n</math> grows.

:<math>e= \lim_{n \to \infty} \frac{n!}{!n}.</math>

== As a [[binomial series]] ==
Consider the sequence:

:<math>
e_n = \left(1 + \frac{1}{n}\right)^n
</math>

By the [[Binomial_theorem#Series_for_e|binomial theorem]]:<ref>{{cite book
 | last=Stewart
 | first=James
 | title=Calculus: Early Transcendentals
 | edition=6th
 | publisher=Brooks/Cole Cengage Learning
 | year=2008
 | page=742
 |url=https://archive.org/details/calculusearlytra0000jame/page/742/mode/2up
}}</ref>

:<math>e_n = \sum_{k=0}^n {n \choose k} \frac{1}{n^k} = \sum_{k=0}^{n} \frac{n^{\underline{k}}}{k!} \frac{1}{n^k}</math>

which converges to <math>e</math> as <math>n</math> increases. The term <math>n^{\underline{k}}</math> is the <math>k</math>th [[Falling_and_rising_factorials|falling factorial power]] of <math>n</math>, which [[Asymptotic_analysis|behaves]] like <math>n^k</math> when <math>n</math> is [[Extended_real_number_line#Arithmetic_operations|large]]. For fixed <math>k</math> and as <math>\textstyle n \to \infty</math>:

:<math>
\frac{n^{\underline{k}}}{n^k} \approx 1 - \frac{k(k - 1)}{2n}
</math>

==As a ratio of ratios==

A unique representation of {{math|''e''}} can be found within the structure of [[Pascal's Triangle]], as discovered by [[Harlan Brothers]]. Pascal's Triangle is composed of [[binomial coefficient]]s, which are traditionally summed to derive polynomial expansions. However, Brothers identified a product-based relationship between these coefficients that links to {{math|''e''}}. Specifically, the ratio of the products of binomial coefficients in adjacent rows of Pascal's Triangle tends to {{math|''e''}} as the row number {{math|''n''}} increases:

:<math>
\begin{align}
P(n) &= \sum_{k = 0}^{n} \ln \binom{n}{k} \\
A &= P(n - 1), B = P(n), C = P(n + 1) \\
x &= (A - B) + (C - B) \sim 1 \\
\end{align}
</math>
:<math>\exp x \sim e</math>

The details of this relationship and its proof are outlined in the discussion on the properties of the [[Pascal's Triangle#Rows|rows of Pascal's Triangle]].<ref>{{Cite journal |last=Brothers |first=Harlan |title=Pascal's Triangle: The Hidden Stor-e |journal=The Mathematical Gazette |volume=96 |pages=145–148 |year=2012 |doi=10.1017/S0025557200004204}}</ref><ref>{{Cite journal |last=Brothers |first=Harlan |title=Math Bite: Finding e in Pascal's Triangle |journal=Mathematics Magazine |volume=85 |issue=1 |pages=51 |year=2012 |doi=10.4169/math.mag.85.1.51}}</ref>

==In trigonometry==
Trigonometrically, {{math|''e''}} can be written in terms of the sum of two [[hyperbolic function]]s,

: <math>e^x = \sinh(x) + \cosh(x) ,</math>

at {{math|1=''x'' = 1}}.

==See also==
* [[List of formulae involving π]]

==Notes==
{{reflist}}

{{DEFAULTSORT:e, Representations}}
[[Category:Exponentials]]
[[Category:Logarithms]]
[[Category:E (mathematical constant)]]