The mathematical constant e real number . Since e 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 infinite series , infinite product , or other types of limit of a sequence .
As a continued fraction [ edit ] Euler proved that the number e simple continued fraction [ 1] A003417 OEIS ):
e
=
[
2
;
1
,
2
,
1
,
1
,
4
,
1
,
1
,
6
,
1
,
1
,
8
,
1
,
…
,
1
,
2
n
,
1
,
…
]
=
2
+
1
1
+
1
2
+
1
1
+
1
1
+
1
4
+
1
1
+
1
1
+
1
6
+
1
1
+
1
1
+
1
8
+
⋱
{\displaystyle {\begin{aligned}e&=[2;1,2,1,1,4,1,1,6,1,1,8,1,\ldots ,1,2n,1,\ldots ]\\[8pt]&=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{6+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{8+{{} \atop \ddots }}}}}}}}}}}}}}}}}}}}}}}\end{aligned}}}
Here are some infinite generalized continued fraction expansions of e 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+{{} \atop \ddots }}}}}}}}}}}=2+{\cfrac {2}{2+{\cfrac {3}{3+{\cfrac {4}{4+{\cfrac {5}{5+{\cfrac {6}{6+{{} \atop \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+{{} \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 (sequence A110185 OEIS ), equivalent to
e
=
[
1
;
0.5
,
12
,
5
,
28
,
9
,
.
.
.
]
{\displaystyle e=[1;0.5,12,5,28,9,...]}
Euler's continued fraction formula [clarification needed and 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+{{} \atop \ddots }}}}}}}}}}}}
As an infinite series [ edit ] The number e infinite series :
e
x
=
∑
k
=
0
∞
x
k
k
!
{\displaystyle e^{x}=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}}
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]
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!)}}}
B
n
{\displaystyle B_{n}}
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}
e
=
1
0
!
+
1
1
!
+
1
2
!
+
1
3
!
+
⋯
{\displaystyle e={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+\cdots }
1
n
{\displaystyle {\frac {1}{n}}}
the nested series [ 5]
e
=
1
+
1
1
(
1
+
1
2
(
1
+
1
3
(
1
+
⋯
)
)
)
{\displaystyle e=1+{\frac {1}{1}}\left(1+{\frac {1}{2}}\left(1+{\frac {1}{3}}\left(1+\cdots \right)\right)\right)}
e
=
1
+
1
+
1
+
1
+
⋯
3
2
1
{\displaystyle e=1+{\cfrac {1+{\cfrac {1+{\cfrac {1+\cdots }{3}}}{2}}}{1}}}
f
(
n
)
=
1
+
f
(
n
+
1
)
n
{\displaystyle f(n)=1+{\frac {f(n+1)}{n}}}
f
(
1
)
{\displaystyle f(1)}
1
{\displaystyle 1}
∞
{\displaystyle \infty }
As an infinite product [ edit ] The number e 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 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}}
e
=
lim
n
→
∞
n
n
!
n
{\displaystyle e=\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}}
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}}
n th prime and
p
n
#
{\displaystyle p_{n}\#}
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)}
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}
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!}
permutations of an ordered set S with cardinality
n
{\displaystyle n}
derangement function)
!
n
{\displaystyle !n}
e
{\displaystyle e}
n
{\displaystyle n}
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 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 e rows of Pascal's Triangle .[ 10] [ 11]
Trigonometrically, e 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 .
![{\displaystyle {\begin{aligned}e&=[2;1,2,1,1,4,1,1,6,1,1,8,1,\ldots ,1,2n,1,\ldots ]\\[8pt]&=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{6+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{8+{{} \atop \ddots }}}}}}}}}}}}}}}}}}}}}}}\end{aligned}}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/495fc2c8e684ddab9700423d2bf2578b67b6822d)
![{\displaystyle e=[1;0.5,12,5,28,9,...]}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/2c0bd16556922b64f610acfbfed8ec9e682175ae)
![{\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {1-2k}{(2k)!}}\right]^{-1}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/a3fed598ba8fe6acd4dae74a3348d7780f1b1e07)
![{\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {4k+3}{2^{2k+1}\,(2k+1)!}}\right]^{2}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d26c23d9900e39fbd0c5a8062a393db51ebc445f)
![{\displaystyle e=\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/310afedad5e855a28eb0d1b232a29031bd682548)
![{\displaystyle e=\lim _{n\to \infty }\left[{\frac {(n+1)^{n+1}}{n^{n}}}-{\frac {n^{n}}{(n-1)^{n-1}}}\right]}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/9cffa4b98fac483870fb5da1d93940ef414d8319)
