2的自然对数 2的自然对数 種類 無理數 符號 ln 2 {\displaystyle \ln {2}} 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] 值 ln 2 ≈ {\displaystyle \ln {2}\approx } 0.693147180...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 在1972年證明,不尋常數 的自然密度 等於 ln 2 {\displaystyle \ln 2} 。換言之,若 u ( n ) {\displaystyle u(n)} 表示不大於 n {\displaystyle n} 的自然數之中,有多少個數 a {\displaystyle a} 具有大於 a {\displaystyle {\sqrt {a}}} 的質因數,則有:
lim n → ∞ u ( n ) n = ln ( 2 ) = 0.693147 … . {\displaystyle \lim _{n\rightarrow \infty }{\frac {u(n)}{n}}=\ln(2)=0.693147\dots \,.} ∑ 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 = 0 ∞ ( − 1 ) n ( n + 1 ) ( n + 2 ) = 2 ln 2 − 1. {\displaystyle \sum _{n=0}^{\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 } 是欧拉-马歇罗尼常数 , ζ {\displaystyle \zeta } 是黎曼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}}}.} (基於反雙曲函數 ,可參見計算自然對數的級數 。) ∫ 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 } 是欧拉-马歇罗尼常数 。
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 )} . 此章节
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