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
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
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 an infinite product
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))}
In trigonometry
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