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}} 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 integers, 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

Euler proved that the number {{math|e}} is represented as the infinite simple continued fraction{{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}} {{OEIS|id=A003417}}:

:

\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}

Here are some infinite generalized continued fraction expansions of {{math|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+{{}\atop\ddots}}}}}} = 2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cfrac{5}{5+\cfrac{6}{6+{{}\atop\ddots}\,}}}}}

: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}\,}}}}}

This last non-simple continued fraction {{OEIS|A110185}}, equivalent to e = [1; 0.5, 12, 5, 28, 9, ...], 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:

: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}}}}}}

As an infinite series

The number {{math|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!},{{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}} 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} 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.

: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)!} = \sum_{k=0}^\infty \frac{(3k+1)^2+1}{(3k+1)!} = \sum_{k=0}^\infty \frac{(3k+2)^2+1}{(3k+2)!}

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

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

:e = \sum_{k=0}^\infty \frac{2k+3}{(k+2)!}Formula 8: A. G. Llorente, [https://osf.io/3yzbj A Novel Simple Representation Series for Euler's Number e], preprint, 2023.

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)} \,.

As a recursive function

The series representation of e, given as e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdotscan also be expressed using a form of recursion. When \textstyle \frac{1}{n} is iteratively factored from the original series the result is the nested series{{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

}}. e = 1 + \frac{1}{1}\left(1 + \frac{1}{2}\left(1 + \frac{1}{3}\left(1 + \cdots \right)\right)\right)which equates to e = 1 + \cfrac{1 + \cfrac{1 + \cfrac{1 + \cdots }{3}}{2}}{1} This fraction is of the form \textstyle f(n) = 1 + \frac{f(n + 1)}{n}, where f(1) computes the sum of the terms from 1 to \infty.

As an infinite product

The number {{math|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 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.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.

: 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 }.

Also

: e = \lim\limits_{n\rightarrow\infty}\prod_{k=0}^n{n \choose k}^{2/{((n

+\alpha)(n+\beta))}}\ \forall\alpha,\beta\in\Bbb R

As the limit of a sequence

The number {{math|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!}} (both by Stirling's formula).

The symmetric limit,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.

: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 {{math|e}}.

The next two definitions are direct corollaries of the prime number theorem{{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|url-access=subscription }}

:

\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}

where p_n is the nth prime, p_n \# is the primorial of the nth prime, and \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.

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

:e= \lim_{n \to \infty} \frac{n!}{!n}.

As a limiting probability

If we consider an event which has a probability of \frac{1}{n} of occurring in any one trial, then the probability of the event not occurring in n trials tends to 1 / e .

That is, \lim_{n \to \infty}\left (1 - \frac{1}{n} \right )^n = \frac{1}{e}

As a [[binomial series]]

Consider the sequence:

:

e_n = \left(1 + \frac{1}{n}\right)^n

By the binomial theorem:{{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

}}

: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}

which converges to e as n increases. The term n^{\underline{k}} is the kth falling factorial power of n, which behaves like n^k when n is large. For fixed k and as \textstyle n \to \infty:

:

\frac{n^{\underline{k}}}{n^k} \approx 1 - \frac{k(k - 1)}{2n}

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 coefficients, 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:

:

\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}

:\exp x \sim e

The details of this relationship and its proof are outlined in the discussion on the properties of the rows of Pascal's triangle.{{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}}{{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}}

In trigonometry

Trigonometrically, {{math|e}} can be written in terms of the sum of two hyperbolic functions,

: e^x = \sinh(x) + \cosh(x) ,

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

See also

Notes