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| {{E (mathematical constant)}}
| | The author's title is Christy Brookins. To perform lacross is one of the things she enjoys most. Her family members lives in Ohio. Since I was 18 I've been operating as a bookkeeper but soon my wife and I will start our own business.<br><br>Stop by my homepage ... tarot readings - [http://appin.co.kr/board_Zqtv22/688025 Read the Full Guide], |
| {{refimprove|date=December 2007}}
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| The [[mathematical constant]] [[E (mathematical constant)|{{math|''e''}}]] can be represented in a variety of ways as a [[real number]]. Since {{math|''e''}} is an [[irrational number]] (see [[proof that e is irrational]]), it cannot be represented as a [[fraction (mathematics)|fraction]], but it can be represented as a [[continued fraction]]. Using [[calculus]], {{math|''e''}} may also be represented as an [[infinite series]], [[infinite product]], or other sort of [[limit of a sequence]]. | |
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| ==As a continued fraction==
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| [[Leonhard Euler|Euler]] proved that the number {{math|''e''}} is represented as the infinite [[simple continued fraction]]<ref>{{cite web|url=http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2028%20e%20is%20irrational.pdf|title=How Euler Did It: Who proved ''e'' is Irrational?|last=Sandifer|first=Ed|date=Feb. 2006|publisher=MAA Online|accessdate=2010-06-18}}</ref> {{OEIS|id=A003417}}:
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| :<math>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]. \,</math>
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| Its convergence can be tripled by allowing just one fractional number:
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| :<math> e = [ 1 ; \textbf{0.5} , 12 , 5 , 28 , 9 , 44 , 13 , 60 , 17 , \ldots , \textbf{4(4n-1)} , \textbf{4n+1} , \ldots]. \,</math>
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| Here are some infinite [[generalized continued fraction]] expansions of {{math|''e''}}. The second is generated from the first by a simple [[generalized continued fraction#The equivalence transformation|equivalence transformation]].
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| :<math>
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| 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\,}}}}}
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| </math>
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| :<math>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\,}}}}}</math>
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| This last, equivalent to [1; 0.5, 12, 5, 28, 9, ...], is a special case of a general formula for the [[exponential function]]:
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| :<math>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}}}}}</math>
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| ==As an infinite series==
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| The number {{math|''e''}} can be expressed as the sum of the following [[infinite series]]:
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| :<math>e^x = \sum_{k=0}^\infty \frac{x^k}{k!} </math> for any real number ''x''.
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| In the special case where ''x'' = 1, or −1, we have:
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| :<math>e = \sum_{k=0}^\infty \frac{1}{k!}</math>,<ref>{{cite web|url=http://oakroadsystems.com/math/loglaws.htm|title=It’s the Law Too — the Laws of Logarithms|last=Brown|first=Stan|date=2006-08-27|publisher=Oak Road Systems|accessdate=2008-08-14}}</ref> and
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| :<math>e^{-1} = \sum_{k=0}^\infty \frac{(-1)^k}{k!}</math>
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| Other series include the following:
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| :<math>e = \left [ \sum_{k=0}^\infty \frac{1-2k}{(2k)!} \right ]^{-1}</math> <ref>Formulas 2–7: [[Harlan J. Brothers|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.</ref>
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| :<math>e = \frac{1}{2} \sum_{k=0}^\infty \frac{k+1}{k!}</math>
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| :<math>e = 2 \sum_{k=0}^\infty \frac{k+1}{(2k+1)!}</math>
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| :<math>e = \sum_{k=0}^\infty \frac{3-4k^2}{(2k+1)!}</math>
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| :<math>e = \sum_{k=0}^\infty \frac{(3k)^2+1}{(3k)!}</math>
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| :<math>e = \left [ \sum_{k=0}^\infty \frac{4k+3}{2^{2k+1}\,(2k+1)!} \right ]^2</math>
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| :<math>e = \left [ -\frac{12}{\pi^2} \sum_{k=1}^\infty \frac{1}{k^2} \ \cos \left ( \frac{9}{k\pi+\sqrt{k^2\pi^2-9}} \right ) \right ]^{-1/3} </math>
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| :<math>e = \sum_{k=1}^\infty \frac{k^n}{B_n(k!)}</math> where <math>B_n</math> is the <math>n^{th}</math> [[Bell number]]. Some few examples: (for ''n''=1,2,3)
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| :<math>e = \sum_{k=1}^\infty \frac{k}{k!} = \sum_{k=1}^\infty \frac{1}{(k-1)!} = \sum_{k=0}^\infty \frac{1}{k!}</math>
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| :<math>e = \sum_{k=1}^\infty \frac{k^2}{2(k!)}</math>
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| :<math>e = \sum_{k=1}^\infty \frac{k^3}{5(k!)}</math>
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| :<math>e = \sum_{k=1}^\infty \frac{k^4}{15(k!)}</math>
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| :<math>e = \sum_{k=1}^\infty \frac{k^5}{52(k!)}</math>
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| :<math>e = \sum_{k=1}^\infty \frac{k^6}{203(k!)}</math>
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| :<math>e = \sum_{k=1}^\infty \frac{k^7}{877(k!)}</math>
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| ==As an infinite product==
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| The number {{math|''e''}} is also given by several [[infinite product]] forms including [[Nick Pippenger|Pippenger]]'s product
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| :<math> 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 </math>
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| and Guillera's product <ref>J. Sondow, [http://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.</ref><ref>J. Guillera and J. Sondow, [http://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.</ref>
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| :<math> 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}
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| \left (\frac{2^4 \cdot 4^4}{1 \cdot 3^6 \cdot 5} \right )^{1/4} \cdots ,</math>
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| where the ''n''th factor is the ''n''th root of the product
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| :<math>\prod_{k=0}^n (k+1)^{(-1)^{k+1}{n \choose k}},</math>
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| as well as the infinite product
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| :<math> e = \frac{2\cdot 2^{(\ln(2)-1)^2} \cdots}{2^{\ln(2)-1}\cdot 2^{(\ln(2)-1)^3}\cdots }.</math>
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| ==As the limit of a sequence==
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| The number {{math|''e''}} is equal to the [[limit of a sequence|limit]] of several [[infinite sequences]]:
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| :<math> e= \lim_{n \to \infty} n\cdot\left ( \frac{\sqrt{2 \pi n}}{n!} \right )^{1/n} </math> and
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| :<math> e=\lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}} </math> (both by [[Stirling's formula]]).
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| The symmetric limit,<ref>[[Harlan J. Brothers|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.</ref><ref>{{cite web|url=http://ans.hsh.no/home/skk/Publications/Lobatto/PRIMUS_KHATTRI.pdf|title=From Lobatto Quadrature to the Euler constant e|last=Khattri|first=Sanjay}}</ref>
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| :<math>e=\lim_{n \to \infty} \left [ \frac{(n+1)^{n+1}}{n^n}- \frac{n^n}{(n-1)^{n-1}} \right ]</math>
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| may be obtained by manipulation of the basic limit definition of {{math|''e''}}. Another limit is<ref>S. M. Ruiz 1997</ref>
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| :<math>e= \lim_{n \to \infty}(p_n \#)^{1/p_n} </math>
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| where <math> p_n </math> is the ''n''th [[prime number|prime]] and <math> p_n \# </math> is the [[primorial]] of the ''n''th prime.
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| :<math>e= \lim_{n \to \infty}n^{\pi(n)/n} </math>
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| where <math> \pi(n) </math> is the prime counting function. This definition is a direct corollary of the [[prime number theorem]].
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| Also:
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| :<math>e^x= \lim_{n \to \infty}\left (1+ \frac{x}{n} \right )^n.</math>
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| In the special case that <math>x = 1</math>, the result is the famous statement:
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| :<math>e= \lim_{n \to \infty}\left (1+ \frac{1}{n} \right )^n.</math>
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| == In trigonometry ==
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| Trigonometrically, {{math|''e''}} can be written as the sum of two [[hyperbolic functions]]:
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| <math>e = \sinh(1) + \cosh(1)\,</math>
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| ==Notes==
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| {{reflist}}
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| [[Category:Transcendental numbers]]
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| [[Category:Mathematical constants]]
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| [[Category:Exponentials]]
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| [[Category:Logarithms]]
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| [[Category:E (mathematical constant)]]
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The author's title is Christy Brookins. To perform lacross is one of the things she enjoys most. Her family members lives in Ohio. Since I was 18 I've been operating as a bookkeeper but soon my wife and I will start our own business.
Stop by my homepage ... tarot readings - Read the Full Guide,