2的自然对数 2的自然对数 識別 種類 無理數 符號
ln
2
{\displaystyle \ln {2}}
性質 連分數 [0; 1, 2, 3, 1, 6, 3, 1, 1, 2, 1, 1, 1, 1, 3, 10] （OEIS 數列A016730 ）
0
+
1
1
+
1
2
+
1
3
+
1
1
+
1
6
+
⋱
{\displaystyle 0+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {1}{1+{\cfrac {1}{6+\ddots }}}}}}}}}}}
以此為根 的多項式或函數
e
x
−
2
=
0
{\displaystyle e^{x}-2=0}
[1] 表示方式 值
ln
2
≈
{\displaystyle \ln {2}\approx }
二进制 0.10110001 0111 0010 0001 0111 … 十进制 0.69314718 0559 9453 0941 7232 … 十六进制 0.B17217F7 D1CF 79AB C9E3 B398 …
ln 2 （OEIS 數列A002162 ）约为：
ln
2
≈
0.693147
{\displaystyle \ln 2\approx 0.693147}
使用对数公式
log
b
2
=
ln
2
ln
b
.
{\displaystyle \log _{b}2={\frac {\ln 2}{\ln b}}.}
可以求出log2，它约为：（OEIS 數列A007524 ）
log
10
2
≈
0.301029995663981195
{\displaystyle \log _{10}2\approx 0.301029995663981195}
數學家理查德·施羅培爾 不尋常數 的自然密度 等於
ln
2
{\displaystyle \ln 2}
u
(
n
)
{\displaystyle u(n)}
n
{\displaystyle n}
a
{\displaystyle a}
a
{\displaystyle {\sqrt {a}}}
lim
n
→
∞
u
(
n
)
n
=
ln
(
2
)
=
0.693147
…
.
{\displaystyle \lim _{n\rightarrow \infty }{\frac {u(n)}{n}}=\ln(2)=0.693147\dots \,.}
公式
∑
n
=
1
∞
(
−
1
)
n
+
1
n
=
∑
n
=
0
∞
1
(
2
n
+
1
)
(
2
n
+
2
)
=
ln
2.
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2.}
∑
n
=
1
∞
(
−
1
)
n
(
n
+
1
)
(
n
+
2
)
=
2
ln
2
−
1.
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{(n+1)(n+2)}}=2\ln 2-1.}
∑
n
=
1
∞
1
n
(
4
n
2
−
1
)
=
2
ln
2
−
1.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1.}
∑
n
=
1
∞
(
−
1
)
n
n
(
4
n
2
−
1
)
=
ln
2
−
1.
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1.}
∑
n
=
1
∞
(
−
1
)
n
n
(
9
n
2
−
1
)
=
2
ln
2
−
3
2
.
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}.}
∑
n
=
2
∞
1
2
n
[
ζ
(
n
)
−
1
]
=
ln
2
−
1
2
.
{\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.}
∑
n
=
1
∞
1
2
n
+
1
[
ζ
(
n
)
−
1
]
=
1
−
γ
−
1
2
ln
2.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {1}{2}}\ln 2.}
∑
n
=
1
∞
1
2
2
n
(
2
n
+
1
)
ζ
(
2
n
)
=
1
2
(
1
−
ln
2
)
.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{2n}(2n+1)}}\zeta (2n)={\frac {1}{2}}(1-\ln 2).}
γ
{\displaystyle \gamma }
欧拉-马歇罗尼常数 ，
ζ
{\displaystyle \zeta }
黎曼ζ函數 。
ln
2
=
∑
k
≥
1
1
k
2
k
.
{\displaystyle \ln 2=\sum _{k\geq 1}{\frac {1}{k2^{k}}}.}
[2] :31
ln
2
=
∑
k
≥
1
(
1
3
k
+
1
4
k
)
1
k
.
{\displaystyle \ln 2=\sum _{k\geq 1}\left({\frac {1}{3^{k}}}+{\frac {1}{4^{k}}}\right){\frac {1}{k}}.}
ln
2
=
2
3
+
∑
k
≥
1
(
1
2
k
+
1
4
k
+
1
+
1
8
k
+
4
+
1
16
k
+
12
)
1
16
k
.
{\displaystyle \ln 2={\frac {2}{3}}+\sum _{k\geq 1}\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}.}
贝利－波尔温－普劳夫公式 ）
ln
2
=
2
3
∑
k
≥
0
1
(
2
k
+
1
)
9
k
.
{\displaystyle \ln 2={\frac {2}{3}}\sum _{k\geq 0}{\frac {1}{(2k+1)9^{k}}}.}
反雙曲函數 ，可參見計算自然對數的級數 。）积分公式
∫
0
1
d
x
1
+
x
=
ln
2
{\displaystyle \int _{0}^{1}{\frac {dx}{1+x}}=\ln 2}
∫
1
∞
d
x
(
1
+
x
2
)
(
1
+
x
)
2
=
1
4
(
1
−
ln
2
)
{\displaystyle \int _{1}^{\infty }{\frac {dx}{(1+x^{2})(1+x)^{2}}}={\frac {1}{4}}(1-\ln 2)}
∫
0
∞
d
x
1
+
e
n
x
=
1
n
ln
2
;
∫
0
∞
d
x
3
+
e
n
x
=
2
3
n
ln
2
{\displaystyle \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}
∫
0
∞
(
1
e
x
−
1
−
2
e
2
x
−
1
)
=
ln
2
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {2}{e^{2x}-1}}\right)=\ln 2}
∫
0
∞
e
−
x
1
−
e
−
x
x
d
x
=
ln
2
{\displaystyle \int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}dx=\ln 2}
∫
0
1
ln
x
2
−
1
x
ln
x
d
x
=
−
1
+
ln
2
+
γ
{\displaystyle \int _{0}^{1}\ln {\frac {x^{2}-1}{x\ln x}}dx=-1+\ln 2+\gamma }
∫
0
π
3
tan
x
d
x
=
2
∫
0
π
4
tan
x
d
x
=
ln
2
{\displaystyle \int _{0}^{\frac {\pi }{3}}\tan xdx=2\int _{0}^{\frac {\pi }{4}}\tan xdx=\ln 2}
∫
−
π
4
π
4
ln
(
sin
x
+
cos
x
)
d
x
=
−
π
4
ln
2
{\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\ln(\sin x+\cos x)dx=-{\frac {\pi }{4}}\ln 2}
∫
0
1
x
2
ln
(
1
+
x
)
d
x
=
2
3
ln
2
−
5
18
{\displaystyle \int _{0}^{1}x^{2}\ln(1+x)dx={\frac {2}{3}}\ln 2-{\frac {5}{18}}}
∫
0
1
x
ln
(
1
+
x
)
ln
(
1
−
x
)
d
x
=
1
4
−
ln
2
{\displaystyle \int _{0}^{1}x\ln(1+x)\ln(1-x)dx={\frac {1}{4}}-\ln 2}
∫
0
1
x
3
ln
(
1
+
x
)
ln
(
1
−
x
)
d
x
=
13
96
−
2
3
ln
2
{\displaystyle \int _{0}^{1}x^{3}\ln(1+x)\ln(1-x)dx={\frac {13}{96}}-{\frac {2}{3}}\ln 2}
∫
0
1
ln
x
(
1
+
x
)
2
d
x
=
−
ln
2
{\displaystyle \int _{0}^{1}{\frac {\ln x}{(1+x)^{2}}}dx=-\ln 2}
∫
0
1
ln
(
1
+
x
)
−
x
x
2
d
x
=
1
−
2
ln
2
{\displaystyle \int _{0}^{1}{\frac {\ln(1+x)-x}{x^{2}}}dx=1-2\ln 2}
∫
0
1
d
x
x
(
1
−
ln
x
)
(
1
−
2
ln
x
)
=
ln
2
{\displaystyle \int _{0}^{1}{\frac {dx}{x(1-\ln x)(1-2\ln x)}}=\ln 2}
∫
1
∞
ln
ln
x
x
3
d
x
=
−
1
2
(
γ
+
ln
2
)
{\displaystyle \int _{1}^{\infty }{\frac {\ln \ln x}{x^{3}}}dx=-{\frac {1}{2}}(\gamma +\ln 2)}
γ
{\displaystyle \gamma }
欧拉-马歇罗尼常数 。
其他公式 用皮尔斯展开式（A091846 ）表达ln2：
log
2
=
1
1
−
1
1
⋅
3
+
1
1
⋅
3
⋅
12
−
…
{\displaystyle \log 2={\frac {1}{1}}-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\ldots }
用恩格尔展开式 A059180 表达ln2：
log
2
=
1
2
+
1
2
⋅
3
+
1
2
⋅
3
⋅
7
+
1
2
⋅
3
⋅
7
⋅
9
+
…
{\displaystyle \log 2={\frac {1}{2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3\cdot 7}}+{\frac {1}{2\cdot 3\cdot 7\cdot 9}}+\ldots }
用余切展开式A081785 表达ln2：
log
2
=
cot
(
arccot
0
−
arccot
1
+
arccot
5
−
arccot
55
+
arccot
14187
−
…
)
{\displaystyle \log 2=\cot(\operatorname {arccot} 0-\operatorname {arccot} 1+\operatorname {arccot} 5-\operatorname {arccot} 55+\operatorname {arccot} 14187-\ldots )}
其他對數 範例 此章节尚無任何内容，需要扩充 。 (2020年4月30日 )
10的自然對數 此章节尚無任何内容，需要扩充 。 (2020年4月30日 )
參考文獻
Brent, Richard P. Fast multiple-precision evaluation of elementary functions. J. ACM. 1976, 23 (2): 242–251. doi:10.1145/321941.321944 MR 0395314 . Uhler, Horace S. Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17. Proc. Nat. Acac. Sci. U. S. A. 1940, 26 : 205–212. MR 0001523 . Sweeney, Dura W. On the computation of Euler's constant. Mathematics of Computation. 1963, 17 . MR 0160308 . Chamberland, Marc. Binary BBP-formulae for logarithms and generalized Gaussian-Mersenne primes (PDF) . Journal of Integer Sequences. 2003, 6 : 03.3.7 [2011-01-08 ] . MR 2046407 . （原始内容 (PDF) 存档于2011-06-06）. Gourévitch, Boris; Guillera Goyanes, Jesus. Construction of binomial sums for π and polylogarithmic constants inspired by BBP formulas (PDF) . Applied Math. E-Notes. 2007, 7 : 237–246 [2011-01-08 ] . MR 2346048 . （原始内容存档 (PDF) 于2020-02-06）. Wu, Qiang. On the linear independence measure of logarithms of rational numbers . Mathematics of Computation. 2003, 72 (242): 901–911. doi:10.1090/S0025-5718-02-01442-4
外部連結 參見 
![{\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b61c76e20b8727253043baa534c03f6a85808e72)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {1}{2}}\ln 2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52102e7a19d5707bdb52e85feebca116766023d3)
![[icon]](/wiki/File:Wiki_letter_w_cropped.svg){kind=small}
