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 [ edit ] Euler proved that the number e is represented as the infinite simple continued fraction [ 1] (sequence A003417 in the OEIS ):
e = [ 2 ; 1 , 2 , 1 , 1 , 4 , 1 , 1 , 6 , 1 , 1 , 8 , 1 , … , 1 , 2 n , 1 , … ] . {\displaystyle e=[2;1,2,1,1,4,1,1,6,1,1,8,1,\ldots ,1,2n,1,\ldots ].} Its convergence can be tripled[clarification needed ] [citation needed ] by allowing just one fractional number:
e = [ 1 ; 1 / 2 , 12 , 5 , 28 , 9 , 44 , 13 , 60 , 17 , … , 4 ( 4 n − 1 ) , 4 n + 1 , … ] . {\displaystyle e=[1;1/2,12,5,28,9,44,13,60,17,\ldots ,4(4n-1),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 + 1 1 + 1 2 + 2 3 + 3 4 + 4 5 + ⋱ = 2 + 2 2 + 3 3 + 4 4 + 5 5 + 6 6 + ⋱ {\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 \,}}}}}}}}}}} e = 2 + 1 1 + 2 5 + 1 10 + 1 14 + 1 18 + ⋱ = 1 + 2 1 + 1 6 + 1 10 + 1 14 + 1 18 + ⋱ {\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 \,}}}}}}}}}}} 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 + 2 x 2 y − x + x 2 6 y + x 2 10 y + x 2 14 y + x 2 18 y + ⋱ {\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 }}}}}}}}}}} As an infinite series [ edit ] The number e can be expressed as the sum of the following infinite series :
e x = ∑ k = 0 ∞ x k k ! {\displaystyle 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 = ∑ k = 0 ∞ 1 k ! {\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}} ,[ 2] and e − 1 = ∑ k = 0 ∞ ( − 1 ) k k ! . {\displaystyle e^{-1}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}.} Other series include the following:
e = [ ∑ k = 0 ∞ 1 − 2 k ( 2 k ) ! ] − 1 {\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {1-2k}{(2k)!}}\right]^{-1}} [ 3] e = 1 2 ∑ k = 0 ∞ k + 1 k ! {\displaystyle e={\frac {1}{2}}\sum _{k=0}^{\infty }{\frac {k+1}{k!}}} e = 2 ∑ k = 0 ∞ k + 1 ( 2 k + 1 ) ! {\displaystyle e=2\sum _{k=0}^{\infty }{\frac {k+1}{(2k+1)!}}} e = ∑ k = 0 ∞ 3 − 4 k 2 ( 2 k + 1 ) ! {\displaystyle e=\sum _{k=0}^{\infty }{\frac {3-4k^{2}}{(2k+1)!}}} e = ∑ k = 0 ∞ ( 3 k ) 2 + 1 ( 3 k ) ! = ∑ k = 0 ∞ ( 3 k + 1 ) 2 + 1 ( 3 k + 1 ) ! = ∑ k = 0 ∞ ( 3 k + 2 ) 2 + 1 ( 3 k + 2 ) ! {\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)!}}} e = [ ∑ k = 0 ∞ 4 k + 3 2 2 k + 1 ( 2 k + 1 ) ! ] 2 {\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {4k+3}{2^{2k+1}\,(2k+1)!}}\right]^{2}} e = ∑ k = 0 ∞ k n B n ( k ! ) {\displaystyle e=\sum _{k=0}^{\infty }{\frac {k^{n}}{B_{n}(k!)}}} where B n {\displaystyle B_{n}} is the n th Bell number . e = ∑ k = 0 ∞ 2 k + 3 ( k + 2 ) ! {\displaystyle e=\sum _{k=0}^{\infty }{\frac {2k+3}{(k+2)!}}} [ 4] Consideration of how to put upper bounds on e leads to this descending series:
e = 3 − ∑ k = 2 ∞ 1 k ! ( k − 1 ) k = 3 − 1 4 − 1 36 − 1 288 − 1 2400 − 1 21600 − 1 211680 − 1 2257920 − ⋯ {\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 } which gives at least one correct (or rounded up) digit per term. That is, if 1 ≤ n , then
e < 3 − ∑ k = 2 n 1 k ! ( k − 1 ) k < e + 0.6 ⋅ 10 1 − n . {\displaystyle 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 = 2 + x 2 − x + ∑ k = 2 ∞ − x k + 1 k ! ( k − x ) ( k + 1 − x ) . {\displaystyle 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 [ edit ] The series representation of e {\displaystyle e} , given as e = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! ⋯ {\displaystyle e={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}\cdots } can also be expressed using a form of recursion. When 1 n {\displaystyle {\frac {1}{n}}} is iteratively factored from the original series the result is the nested series[ 5] e = 1 + 1 1 ( 1 + 1 2 ( 1 + 1 3 ( 1 + ⋯ ) ) ) {\displaystyle e=1+{\frac {1}{1}}(1+{\frac {1}{2}}(1+{\frac {1}{3}}(1+\cdots )))} which equates to e = 1 + 1 + 1 + 1 + ⋯ 3 2 1 {\displaystyle e=1+{\cfrac {1+{\cfrac {1+{\cfrac {1+\cdots }{3}}}{2}}}{1}}} This fraction is of the form f ( n ) = 1 + f ( n + 1 ) n {\displaystyle f(n)=1+{\frac {f(n+1)}{n}}} , where f ( 1 ) {\displaystyle f(1)} computes the sum of the terms from 1 {\displaystyle 1} to ∞ {\displaystyle \infty } .
As an infinite product [ edit ] The number e is also given by several infinite product forms including Pippenger 's product
e = 2 ( 2 1 ) 1 / 2 ( 2 3 4 3 ) 1 / 4 ( 4 5 6 5 6 7 8 7 ) 1 / 8 ⋯ {\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 } and Guillera's product [ 6] [ 7]
e = ( 2 1 ) 1 / 1 ( 2 2 1 ⋅ 3 ) 1 / 2 ( 2 3 ⋅ 4 1 ⋅ 3 3 ) 1 / 3 ( 2 4 ⋅ 4 4 1 ⋅ 3 6 ⋅ 5 ) 1 / 4 ⋯ , {\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 ,} where the n th factor is the n th root of the product
∏ k = 0 n ( k + 1 ) ( − 1 ) k + 1 ( n k ) , {\displaystyle \prod _{k=0}^{n}(k+1)^{(-1)^{k+1}{n \choose k}},} as well as the infinite product
e = 2 ⋅ 2 ( ln ( 2 ) − 1 ) 2 ⋯ 2 ln ( 2 ) − 1 ⋅ 2 ( ln ( 2 ) − 1 ) 3 ⋯ . {\displaystyle 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 < e 2 (which includes B = 2, 3, 4, 5, 6, or 7), then
e = B ⋅ B ( ln ( B ) − 1 ) 2 ⋯ B ln ( B ) − 1 ⋅ B ( ln ( B ) − 1 ) 3 ⋯ . {\displaystyle e={\frac {B\cdot B^{(\ln(B)-1)^{2}}\cdots }{B^{\ln(B)-1}\cdot B^{(\ln(B)-1)^{3}}\cdots }}.} Also
e = lim n → ∞ ∏ k = 0 n ( n k ) 2 / ( ( n + α ) ( n + β ) ) ∀ α , β ∈ R {\displaystyle e=\lim \limits _{n\rightarrow \infty }\prod _{k=0}^{n}{n \choose k}^{2/{((n+\alpha )(n+\beta ))}}\ \forall \alpha ,\beta \in {\mathbb {R}}} As the limit of a sequence [ edit ] The number e is equal to the limit of several infinite sequences :
e = lim n → ∞ n ⋅ ( 2 π n n ! ) 1 / n {\displaystyle e=\lim _{n\to \infty }n\cdot \left({\frac {\sqrt {2\pi n}}{n!}}\right)^{1/n}} and e = lim n → ∞ n n ! n {\displaystyle e=\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}} (both by Stirling's formula ). The symmetric limit,[ 8]
e = lim n → ∞ [ ( n + 1 ) n + 1 n n − n n ( n − 1 ) n − 1 ] {\displaystyle 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 [ 9]
e = lim n → ∞ ( p n # ) 1 / p n {\displaystyle e=\lim _{n\to \infty }(p_{n}\#)^{1/p_{n}}} where p n {\displaystyle p_{n}} is the n th prime and p n # {\displaystyle p_{n}\#} is the primorial of the n th prime.
e = lim n → ∞ n π ( n ) / n {\displaystyle e=\lim _{n\to \infty }n^{\pi (n)/n}} where π ( n ) {\displaystyle \pi (n)} is the prime-counting function .
Also:
e x = lim n → ∞ ( 1 + x n ) n . {\displaystyle e^{x}=\lim _{n\to \infty }\left(1+{\frac {x}{n}}\right)^{n}.} In the special case that x = 1 {\displaystyle x=1} , the result is the famous statement:
e = lim n → ∞ ( 1 + 1 n ) n . {\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.} The ratio of the factorial n ! {\displaystyle n!} , that counts all permutations of an ordered set S with cardinality n {\displaystyle n} , and the subfactorial (a.k.a. the derangement function) ! n {\displaystyle !n} , which counts the amount of permutations where no element appears in its original position, tends to e {\displaystyle e} as n {\displaystyle n} grows.
e = lim n → ∞ n ! ! n . {\displaystyle e=\lim _{n\to \infty }{\frac {n!}{!n}}.} As a ratio of ratios [ edit ] A unique representation of 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 e . Specifically, the ratio of the products of binomial coefficients in adjacent rows of Pascal's Triangle tends to e as the row number increases. This relationship and its proof are outlined in the discussion on the properties of the rows of Pascal's Triangle .[ 10] [ 11]
Trigonometrically, e can be written in terms of the sum of two hyperbolic functions ,
e x = sinh ( x ) + cosh ( x ) , {\displaystyle e^{x}=\sinh(x)+\cosh(x),} at x = 1 .
^ Sandifer, Ed (Feb 2006). "How Euler Did It: Who proved e is Irrational?" (PDF) . MAA Online. Retrieved 2017-04-23 . ^ Brown, Stan (2006-08-27). "It's the Law Too — the Laws of Logarithms" . Oak Road Systems. Archived from the original on 2008-08-13. Retrieved 2008-08-14 . ^ Formulas 2–7: H. J. Brothers, Improving the convergence of Newton's series approximation for e , The College Mathematics Journal , Vol. 35, No. 1, (2004), pp. 34–39. ^ Formula 8: A. G. Llorente, A Novel Simple Representation Series for Euler’s Number e , preprint, 2023. ^ "e" , Wolfram MathWorld : ex. 17, 18, and 19, archived from the original on 2023-03-15 . ^ J. Sondow, 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, Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent , Ramanujan Journal 16 (2008), 247–270. ^ H. J. Brothers and J. A. Knox, New closed-form approximations to the Logarithmic Constant e , The Mathematical Intelligencer , Vol. 20, No. 4, (1998), pp. 25–29. ^ S. M. Ruiz 1997 ^ Brothers, Harlan (2012). "Pascal's Triangle: The Hidden Stor-e". The Mathematical Gazette . 96 : 145–148. doi :10.1017/S0025557200004204 . ^ Brothers, Harlan (2012). "Math Bite: Finding e in Pascal's Triangle". Mathematics Magazine . 85 (1): 51. doi :10.4169/math.mag.85.1.51 .