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Representations of e

 
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The mathematical constant e
Euler's formula.svg

Natural logarithm · Exponential function

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.

Contents

As a continued fraction

The number e is 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, 1, \ldots, \textbf{2n}, 1, 1, \ldots], \,

whose convergence can be tripled by allowing just one decimal number:

e = [ 1 , \textbf{0.5} , 12 , 5 , 28 , 9 , 44 , 13 , 60 , 17 , \ldots , \textbf{4(4n-1)} , \textbf{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. The third is equivalent to [1, 0.5, 12, 5, 28, 9, ...].

e= 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{5+\ddots}}}}} \qquad e= 2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cfrac{5}{5+\cfrac{6}{6+\ddots\,}}}}}
e = 1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+\ddots\,}}}}}

This last is a special case of a general formula for the exponential function:

e^\frac{2x}{y} = 1+\cfrac{2x}{y-x+\cfrac{x^2}{3y+\cfrac{x^2}{5y+\cfrac{x^2}{7y+\cfrac{x^2}{9y+\ddots\,}}}}}

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!}[1]
e = \left [ \sum_{k=0}^\infty \frac{(-1)^k}{k!} \right ]^{-1}
e = \left [ \sum_{k=0}^\infty \frac{1-2k}{(2k)!} \right ]^{-1} [2]
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(k-1)!}
e = \sum_{k=1}^\infty \frac{k^2}{2(k!)}
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 Bn is the nth Bell number.

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 [3][4]

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 }.

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 ] [5][6]

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} [7]

where pn 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

In the special case that x = 1, the result is the famous statement:

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

Notes

  1. ^ Brown, Stan (2006-08-27). "It’s the Law Too — the Laws of Logarithms". Oak Road Systems. http://oakroadsystems.com/math/loglaws.htm. Retrieved 2008-08-14. 
  2. ^ 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.
  3. ^ J. Sondow, A faster product for pi and a new integral for ln pi/2, Amer. Math. Monthly 112 (2005) 729-734.
  4. ^ 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.
  5. ^ 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.
  6. ^ Khattri, Sanjay. "From Lobatto Quadrature to the Euler constant e". http://ans.hsh.no/home/skk/Publications/Lobatto/PRIMUS_KHATTRI.pdf. 
  7. ^ S. M. Ruiz 1997

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