查看“︁2的自然对数”︁的源代码

{{Infobox number
| range = <table style= "width:100%; margin:0;"><tr style="align:center;white-space: nowrap; ">
<td>[[數表]]—{{無理數}}</td>
</tr></table>
| name=2的自然对数
| nav=no
| number=<math>\ln{2}</math>
| value=<math>\ln{2}\approx</math>0.693147180...
| root of=<math>e^x-2 = 0</math><ref>{{Cite WolframAlpha| equation= e^x-2=0}}</ref>
| symbol=<math>\ln{2}</math>
| other name=
| type=[[無理數]]
| 連分數=[0; 1, 2, 3, 1, 6, 3, 1, 1, 2, 1, 1, 1, 1, 3, 10] {{OEIS|A016730}} <br/> <math>0 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{6 + \ddots}}}}}</math>
| basedata = {{Infobox number/base
| 二進制 = {{FractionalGaps|{{進制|2|0.69314718055994530941723212145817|precision=24}}|4|…}}
| 十進位 = {{FractionalGaps|0.693147180559945309417232|4|…}}
| 十六進位 = {{FractionalGaps|{{進制|16|0.69314718055994530941723212145817|precision=24}}|4|…}}}}
}}
[[自然對數|ln]][[2]]{{OEIS|A002162}}约为：
:<math>\ln 2 \approx 0.693147</math>
使用对数公式
:<math>\log_b 2 = \frac{\ln 2}{\ln b}.</math>
可以求出log2，它约为：{{OEIS|A007524}}
:<math>\log_{10} 2 \approx 0.301029995663981195</math>。

數學家{{le|理查德·施羅培爾|Richard Schroeppel}}在1972年證明，[[不尋常數]]的[[自然密度]]等於 <math>\ln 2</math>。換言之，若 <math>u(n)</math> 表示不大於 <math>n</math> 的自然數之中，有多少個數 <math>a</math> 具有大於 <math>\sqrt a</math> 的質因數，則有：

:<math>\lim_{n \rightarrow \infty} \frac{u(n)}{n} = \ln(2) = 0.693147 \dots\, .</math>

==公式==

:<math>\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} = \ln 2.</math>

:<math>\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)(n+2)} = 2\ln 2 -1.</math>

:<math>\sum_{n=1}^\infty \frac{1}{n(4n^2-1)} = 2\ln 2 -1.</math>

:<math>\sum_{n=1}^\infty \frac{(-1)^n}{n(4n^2-1)} = \ln 2 -1.</math>

:<math>\sum_{n=1}^\infty \frac{(-1)^n}{n(9n^2-1)} = 2\ln 2 -\frac{3}{2}.</math>

:<math>\sum_{n=2}^\infty \frac{1}{2^n}[\zeta(n)-1] = \ln 2 -\frac{1}{2}.</math>

:<math>\sum_{n=1}^\infty \frac{1}{2n+1}[\zeta(n)-1] = 1-\gamma-\frac{1}{2}\ln 2.</math>

:<math>\sum_{n=1}^\infty \frac{1}{2^{2n}(2n+1)}\zeta(2n) = \frac{1}{2}(1-\ln 2).</math>
<math>\gamma</math>是[[欧拉-马歇罗尼常数]]，<math>\zeta</math>是[[黎曼ζ函數]]。

:<math>\ln 2 = \sum_{k\ge 1} \frac{1}{k2^k}.</math><ref>{{cite book | author=Bailey, D. H.; Borwein, J. M.; Calkin, N. J.; Girgensohn, R.; Luke, D. R.; and Moll, V. H. |chapter=§2.2 Integer Relation Detection | title=Experimental Mathematics in Action | publisher=  A K Peters/CRC Press | isbn=978-1568812717 | page=pp. 29-31 | year=2007}}</ref>{{rp|31}}

:<math>\ln 2 = \sum_{k\ge 1}\left(\frac{1}{3^k}+\frac{1}{4^k}\right)\frac{1}{k}.</math>

:<math>\ln 2 = \frac{2}{3}+\sum_{k\ge 1}\left(\frac{1}{2k}+\frac{1}{4k+1}+\frac{1}{8k+4}+\frac{1}{16k+12}\right)\frac{1}{16^k}.</math>（[[贝利－波尔温－普劳夫公式]]）

:<math>\ln 2 = \frac{2}{3} \sum_{k\ge 0} \frac{1}{(2k+1)9^k}.</math>（基於[[反雙曲函數]]，可參見[[對數#计算自然对数的级数|計算自然對數的級數]]。）

==积分公式==

:<math>\int_0^1 \frac{dx}{1+x} = \ln 2</math>

:<math>\int_1^\infty \frac{dx}{(1+x^2)(1+x)^2} = \frac{1}{4}(1-\ln 2)</math>

:<math>\int_0^\infty \frac{dx}{1+e^{nx}} = \frac{1}{n}\ln 2;
\int_0^\infty \frac{dx}{3+e^{nx}} = \frac{2}{3n}\ln 2</math>

:<math>\int_0^\infty \left(\frac{1}{e^x-1}-\frac{2}{e^{2x}-1}\right)=\ln 2</math>

:<math>\int_0^\infty e^{-x}\frac{1-e^{-x}}{x} dx= \ln 2</math>

:<math>\int_0^1 \ln\frac{x^2-1}{x\ln x}dx=-1+\ln 2+\gamma</math>

:<math>\int_0^{\frac{\pi}{3}} \tan x dx=2\int_0^{\frac{\pi}{4}} \tan x dx=\ln 2</math>

:<math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x+\cos x)dx=-\frac{\pi}{4}\ln 2</math>

:<math>\int_0^1 x^2\ln(1+x)dx=\frac{2}{3}\ln 2-\frac{5}{18}</math>

:<math>\int_0^1 x\ln(1+x)\ln(1-x)dx=\frac{1}{4}-\ln 2</math>

:<math>\int_0^1 x^3\ln(1+x)\ln(1-x)dx=\frac{13}{96}-\frac{2}{3}\ln 2</math>

:<math>\int_0^1 \frac{\ln x}{(1+x)^2}dx = -\ln 2</math>

:<math>\int_0^1 \frac{\ln(1+x)-x}{x^2}dx=1-2\ln2</math>

:<math>\int_0^1 \frac{dx}{x(1-\ln x)(1-2\ln x)} = \ln 2</math>

:<math>\int_1^\infty \frac{\ln\ln x}{x^3}dx = -\frac{1}{2}(\gamma+\ln 2)</math>

<math>\gamma</math>是[[欧拉-马歇罗尼常数]]。

==其他公式==
用皮尔斯展开式（{{OEIS2C|A091846}}）表达ln2： 
:<math> \log 2 = \frac{1}{1} -\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 12} -\ldots</math>.
用[[恩格尔展开式]]{{OEIS2C|A059180}}表达ln2： 
:<math>\log 2 = \frac{1}{2}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot3\cdot 7}+\frac{1}{2\cdot 3\cdot 7\cdot 9}+\ldots </math>.
用余切展开式{{OEIS2C|A081785}}表达ln2：
:<math>\log 2 = \cot(\arccot 0 -\arccot 1 +\arccot 5 -\arccot 55+\arccot 14187-\ldots)</math>.

==其他對數==

===範例===
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==10的自然對數==
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==參考文獻==
{{Refbegin}}
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* {{cite journal
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* {{cite journal
|first1=Marc
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|archive-date=2011-06-06
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* {{cite journal
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* {{cite journal
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{{Refend}}

==外部連結==
* {{MathWorld|urlname=NaturalLogarithmof2|title=Natural logarithm of 2}}
* {{planetmath reference|id=10699|title=table of natural logarithms|urlname=tableofnaturallogarithms}}
* {{cite web|url=http://numbers.computation.free.fr/Constants/Log2/log2.html|title=The logarithm constant:log 2|last1=Gourdon|first1=Xavier|last2=Sebah|first2=Pascal|accessdate=2011-01-08|archive-date=2020-02-23|archive-url=https://web.archive.org/web/20200223074007/http://numbers.computation.free.fr/Constants/Log2/log2.html|dead-url=no}}
==參見==
*[[對數]]
*[[自然對數]]
*[[常用對數]]
*[[超越数]]

{{無理數導航}}

[[Category:对数]]
[[Category:數學常數]]
[[Category:实数超越数]]