Representations of e

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The mathematical constant, e

Natural logarithm

Applications in: compound interest · Euler's identity & Euler's formula · half-lives & exponential growth/decay

Defining e: proof that e is irrational · representations of e · Lindemann–Weierstrass theorem

People John Napier · Leonhard Euler

Schanuel's conjecture

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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 a fraction, but it can be represented as a continued fraction. Using calculus, e may also be represented as an infinite series, infinite product, or other sort of limit of a sequence.

1 As a continued fraction
2 As an infinite series
3 As an infinite product
4 As the limit of a sequence
5 Notes

[edit] As a continued fraction

The number e can be represented as an infinite simple continued fraction (sequence A003417 in OEIS):

$e = [2; 1, \textbf{2}, 1, 1, \textbf{4}, 1, 1, \textbf{6}, 1, 1, \textbf{8}, 1, \ldots,1, \textbf{2n}, 1,\ldots] \,$

Here are some infinite generalized continued fraction expansions of e. The second of these can be generated from the first by a simple equivalence transformation. The third one – with ... 6, 10, 14, ... in it – converges very quickly.

$e= 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{\ddots}}}}} \qquad e= 2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cfrac{5}{5+\cfrac{6}{\ddots\,}}}}}$

$e = 1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{\ddots\,}}}}}$

$e^{2m/n} = 1+\cfrac{2m}{(n-m)+\cfrac{m^2}{3n+\cfrac{m^2}{5n+\cfrac{m^2}{7n+\cfrac{m^2}{\ddots\,}}}}}$

Setting m=x and n=2 yields

$e^x = 1+\cfrac{2x}{(2-x)+\cfrac{x^2}{6+\cfrac{x^2}{10+\cfrac{x^2}{14+\cfrac{x^2}{\ddots\,}}}}}$

[edit] As an infinite series

The number e is also equal to the sum of the following infinite series:

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

$e = \left [ \sum_{k=0}^\infty \frac{(-1)^k}{k!} \right ]^{-1}$

$e = \left [ \sum_{k=0}^\infty \frac{1-2k}{(2k)!} \right ]^{-1}$ ^[1]

$e = \frac{1}{2} \sum_{k=0}^\infty \frac{k+1}{k!}$

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

$e = \sum_{k=0}^\infty \frac{3-4k^2}{(2k+1)!}$

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

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

$e = -\frac{12}{\pi^2} \left [ \sum_{k=1}^\infty \frac{1}{k^2} \ \cos \left ( \frac{9}{k\pi+\sqrt{k^2\pi^2-9}} \right ) \right ]^{-1/3}$

$e = \sum_{k=1}^\infty \frac{k^2}{2(k!)}$

$e = \sum_{k=1}^\infty \frac{k}{2(k-1)!}$

$e = \sum_{k=1}^\infty \frac{k^3}{5(k!)}$

$e = \sum_{k=1}^\infty \frac{k^4}{15(k!)}$

$e = \sum_{k=1}^\infty \frac{k^n}{B_n(k!)}$ where

B n

is the

n t h

Bell number.

[edit] 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 ^[2]

$e = \left ( \frac{2}{1} \right )^{1/1} \left (\frac{2^2}{1 \cdot 3} \right )^{1/2} \left (\frac{2^3 \cdot 4}{1 \cdot 3^3} \right )^{1/3} \left (\frac{2^4 \cdot 4^4}{1 \cdot 3^6 \cdot 5} \right )^{1/4} \cdots ,$

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

$e = \frac{2\cdot 2^{(\ln(2)-1)^2} \cdots}{2^{\ln(2)-1}\cdot 2^{(\ln(2)-1)^3}\cdots }.$

[edit] 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,

$e=\lim_{n \to \infty} \left [ \frac{(n+1)^{n+1}}{n^n}- \frac{n^n}{(n-1)^{n-1}} \right ]$ ^[3]

may be obtained by manipulation of the basic limit definition of e. Another limit is

$e= \lim_{n \to \infty}(p_n \#)^{1/p_n}$ ^[4]

where $p n$ is the nth prime and $p_n \#$ is the primorial of the nth prime.

Also:

$e^x= \lim_{n \to \infty}\left (1+ \frac{x}{n} \right )^n$

And when $x = 1$ the result is the famous statement:

$e= \lim_{n \to \infty}\left (1+ \frac{1}{n} \right )^n$

[edit] Notes

^ 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; pages 34-39.
^ J. Sondow, A faster product for pi and a new integral for ln pi/2, Amer. Math. Monthly 112 (2005) 729-734.
^ H. J. Brothers and J. A. Knox, New closed-form approximations to the Logarithmic Constant e. The Mathematical Intelligencer, Vol. 20, No. 4, 1998; pages 25-29.
^ S. M. Ruiz 1997