List of representations of e

As a continued fraction

Euler proved that the number e is represented as the infinite simple continued fraction :

: \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 + }}}}} }}}}} \end{align}

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

:e = 2+\cfrac{1}{1+\cfrac{2}{5+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+}}} = 1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+}}}

This last non-simple continued fraction , equivalent to e = [1; 0.5, 12, 5, 28, 9, ...], has a quicker convergence rate compared to Euler's continued fraction formula 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+}}}}

As an infinite series

The number 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 or -1, we have:

:e = \sum_{k=0}^\infty \frac{1}{k!}, 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}

: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 nth Bell number.

:e^{-1} = \sum_{k=0}^\infty \frac{(-1)^k B_{2k}(2\pi)^{2k}}{(2k+1)!} where B_{n} is the nth Bell number.

:e = \sum_{k=0}^\infty \frac{2k+3}{(k+2)!} 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 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 | archive-url = https://web.archive.org/web/20230315161656/https://mathworld.wolfram.com/e.html | archive-date = 2023-03-15

As an infinite product

The number 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 : 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 2 (which includes , 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 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,

: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

: \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 , 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

: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 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 n increases:

: \begin{align} P_b(n) &= \sum_{k = 0}^{n} \log_b \binom{n}{k} \ A &= P_b(n - 1), B = P_b(n), C = P_b(n + 1) \ x &= (A - B) + (C - B) \sim 1 \ \end{align} For b = e, \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.

In trigonometry

Trigonometrically, e can be written in terms of the sum of two hyperbolic functions,

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

at .

Notes

References

1. Sandifer, Ed. (Feb 2006). "How Euler Did It: Who proved ''e'' is Irrational?". MAA Online.
2. Brown, Stan. (2006-08-27). "It's the Law Too — the Laws of Logarithms". Oak Road Systems.
3. 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.
4. Formula 8: A. G. Llorente, [https://osf.io/3yzbj A Novel Simple Representation Series for Euler's Number ''e''], preprint, 2023.
5. 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.
6. 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.
7. 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.
8. Ruiz, Sebastian Martin. (1997). "81.27 A result on prime numbers". Cambridge University Press.
9. Brothers, Harlan. (2012). "Pascal's Triangle: The Hidden Stor-e". The Mathematical Gazette.
10. Brothers, Harlan. (2012). "Math Bite: Finding e in Pascal's Triangle". Mathematics Magazine.