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BerandaWikiLogaritma alami dari 2
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Logaritma alami dari 2

Nilai desimal dari logaritma natural dari 2 (urutan kira-kira.

Wikipedia article
Diperbarui 31 Januari 2023

Sumber: Lihat artikel asli di Wikipedia

Nilai desimal dari logaritma natural dari 2 (urutan (barisan A002162 pada OEIS) kira-kira

ln ⁡ 2 ≈ 0.693 147 180 559 945 309 417 232 121 458 {\displaystyle \ln 2\approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458} {\displaystyle \ln 2\approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458}.

Logaritma dari 2 dalam basis lainnya diperoleh dengan rumus

log b ⁡ 2 = ln ⁡ 2 ln ⁡ b {\displaystyle \log _{b}2={\frac {\ln 2}{\ln b}}} {\displaystyle \log _{b}2={\frac {\ln 2}{\ln b}}}

Logaritma umum secara khusus adalah OEIS A007524

log 10 ⁡ 2 ≈ 0.301 029 995 663 981 195 {\displaystyle \log _{10}2\approx 0.301\,029\,995\,663\,981\,195} {\displaystyle \log _{10}2\approx 0.301\,029\,995\,663\,981\,195}.

Invers dari bilangannya ini adalah logaritma biner dari 10:

log 2 ⁡ 10 = 1 log 10 ⁡ 2 ≈ 3.321 928 095 {\displaystyle \log _{2}10={\frac {1}{\log _{10}2}}\approx 3.321\,928\,095} {\displaystyle \log _{2}10={\frac {1}{\log _{10}2}}\approx 3.321\,928\,095} OEIS A020862

Dengan menggunakan teorema Lindemann–Weierstrass, logaritma natural dari setiap bilangan asli selain 0 dan 1 (lebih umumnya, dari setiap positif bilangan aljabar selain 1) adalah sebuah bilangan transenden.

Wakilan deret

Faktorial bolak-balik menaik

  • ln ⁡ 2 = ∑ n = 1 ∞ ( − 1 ) n + 1 n = 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + ⋯ {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots } {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots }. Ini dikenal "deret harmonik bolak-balik".
  • ln ⁡ 2 = 1 2 + 1 2 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) {\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}} {\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}}.
  • ln ⁡ 2 = 5 8 + 1 2 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) {\displaystyle \ln 2={\frac {5}{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}} {\displaystyle \ln 2={\frac {5}{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}}.
  • ln ⁡ 2 = 2 3 + 3 4 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) {\displaystyle \ln 2={\frac {2}{3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}} {\displaystyle \ln 2={\frac {2}{3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}}.
  • ln ⁡ 2 = 131 192 + 3 2 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) {\displaystyle \ln 2={\frac {131}{192}}+{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}} {\displaystyle \ln 2={\frac {131}{192}}+{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}}.
  • ln ⁡ 2 = 661 960 + 15 4 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) {\displaystyle \ln 2={\frac {661}{960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}} {\displaystyle \ln 2={\frac {661}{960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}}.

Faktorial konstanta menaik biner

  • ln ⁡ 2 = ∑ n = 1 ∞ 1 2 n n {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}} {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}}.
  • ln ⁡ 2 = 1 − ∑ n = 1 ∞ 1 2 n n ( n + 1 ) {\displaystyle \ln 2=1-\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)}}} {\displaystyle \ln 2=1-\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)}}}.
  • ln ⁡ 2 = 1 2 + 2 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) {\displaystyle \ln 2={\frac {1}{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}} {\displaystyle \ln 2={\frac {1}{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}}.
  • ln ⁡ 2 = 5 6 − 6 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) {\displaystyle \ln 2={\frac {5}{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}} {\displaystyle \ln 2={\frac {5}{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}}
  • ln ⁡ 2 = 7 12 + 24 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) {\displaystyle \ln 2={\frac {7}{12}}+24\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}} {\displaystyle \ln 2={\frac {7}{12}}+24\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}}.
  • ln ⁡ 2 = 47 60 − 120 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) {\displaystyle \ln 2={\frac {47}{60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}} {\displaystyle \ln 2={\frac {47}{60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}}.

Wakilan deret lainnya

  • ∑ n = 0 ∞ 1 ( 2 n + 1 ) ( 2 n + 2 ) = ln ⁡ 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2} {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2}.
  • ∑ 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} {\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} {\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}}} {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}}.
  • ∑ n = 1 ∞ 1 4 n 2 − 2 n = ln ⁡ 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-2n}}=\ln 2} {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-2n}}=\ln 2}.
  • ∑ n = 1 ∞ 2 ( − 1 ) n + 1 ( 2 n − 1 ) + 1 8 n 2 − 4 n = ln ⁡ 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln 2} {\displaystyle \sum _{n=1}^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln 2}.
  • ∑ n = 0 ∞ ( − 1 ) n 3 n + 1 = ln ⁡ 2 3 + π 3 3 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}} {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}}.
  • ∑ n = 0 ∞ ( − 1 ) n 3 n + 2 = − ln ⁡ 2 3 + π 3 3 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}} {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}}.
  • ∑ n = 0 ∞ ( − 1 ) n ( 3 n + 1 ) ( 3 n + 2 ) = 2 ln ⁡ 2 3 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}} {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}}.
  • ∑ n = 1 ∞ 1 ∑ k = 1 n k 2 = 18 − 24 ln ⁡ 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} menggunakan lim N → ∞ ∑ n = N 2 N 1 n = ln ⁡ 2 {\displaystyle \lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} {\displaystyle \lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2}.
  • ∑ n = 1 ∞ 1 4 k 2 − 3 k = ln ⁡ 2 + π 6 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4k^{2}-3k}}=\ln 2+{\frac {\pi }{6}}} {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4k^{2}-3k}}=\ln 2+{\frac {\pi }{6}}} (jumlah timbal-balik dari bilangan dekagonal).

Melibatkan fungsi zeta Riemann

  • ∑ 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}}} {\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}}.
  • ∑ n = 2 ∞ 1 2 n + 1 [ ζ ( n ) − 1 ] = 1 − γ − ln ⁡ 2 2 {\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {\ln 2}{2}}} {\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {\ln 2}{2}}}.
  • ∑ n = 1 ∞ 1 2 2 n − 1 ( 2 n + 1 ) ζ ( 2 n ) = 1 − ln ⁡ 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{2n-1}(2n+1)}}\zeta (2n)=1-\ln 2} {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{2n-1}(2n+1)}}\zeta (2n)=1-\ln 2}.

( γ {\displaystyle \gamma } {\displaystyle \gamma } adalah konstanta Euler−Mascheroni dan ζ {\displaystyle \zeta } {\displaystyle \zeta } adalah fungsi zeta Riemann.)

Wakilan tipe-BBP

ln ⁡ 2 = 2 3 + 1 2 ∑ 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}}+{\frac {1}{2}}\sum _{k=1}^{\infty }\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}} {\displaystyle \ln 2={\frac {2}{3}}+{\frac {1}{2}}\sum _{k=1}^{\infty }\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}}

(Lihat lebih banyak mengenai wakilan tipe Bailey−Borwein−Plouffe (BBP).)

Menerapkan ketiga deret umum untuk logaritma natural ke 2 secara langsung memberikan:

  • ln ⁡ 2 = ∑ n = 1 ∞ ( − 1 ) n − 1 n {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}} {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}}.
  • ln ⁡ 2 = ∑ n = 1 ∞ 1 2 n n {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}} {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}}.
  • ln ⁡ 2 = 2 3 ∑ k = 0 ∞ 1 9 k ( 2 k + 1 ) {\displaystyle \ln 2={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}} {\displaystyle \ln 2={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}}.

Menerapkannya untuk 2 = 3 2 ⋅ 4 3 {\displaystyle \textstyle {2={\frac {3}{2}}\cdot {\frac {4}{3}}}} {\displaystyle \textstyle {2={\frac {3}{2}}\cdot {\frac {4}{3}}}} memberikan:

  • ln ⁡ 2 = ∑ n = 1 ∞ ( − 1 ) n − 1 2 n n + ∑ n = 1 ∞ ( − 1 ) n − 1 3 n n {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{2^{n}n}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{3^{n}n}}} {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{2^{n}n}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{3^{n}n}}}.
  • ln ⁡ 2 = ∑ n = 1 ∞ 1 3 n n + ∑ n = 1 ∞ 1 4 n n {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{3^{n}n}}+\sum _{n=1}^{\infty }{\frac {1}{4^{n}n}}} {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{3^{n}n}}+\sum _{n=1}^{\infty }{\frac {1}{4^{n}n}}}.
  • ln ⁡ 2 = 2 5 ∑ k = 0 ∞ 1 25 k ( 2 k + 1 ) + 2 7 ∑ k = 0 ∞ 1 49 k ( 2 k + 1 ) {\displaystyle \ln 2={\frac {2}{5}}\sum _{k=0}^{\infty }{\frac {1}{25^{k}(2k+1)}}+{\frac {2}{7}}\sum _{k=0}^{\infty }{\frac {1}{49^{k}(2k+1)}}} {\displaystyle \ln 2={\frac {2}{5}}\sum _{k=0}^{\infty }{\frac {1}{25^{k}(2k+1)}}+{\frac {2}{7}}\sum _{k=0}^{\infty }{\frac {1}{49^{k}(2k+1)}}}.

Menerapkannya untuk 2 = ( 2 ) 2 {\displaystyle 2=\left({\sqrt {2}}\right)^{2}} {\displaystyle 2=\left({\sqrt {2}}\right)^{2}} memberikan:

  • ln ⁡ 2 = 2 ∑ n = 1 ∞ ( − 1 ) n − 1 ( 2 + 1 ) n n {\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{({\sqrt {2}}+1)^{n}n}}} {\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{({\sqrt {2}}+1)^{n}n}}}.
  • ln ⁡ 2 = 2 ∑ n = 1 ∞ 1 ( 2 + 2 ) n n {\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {1}{(2+{\sqrt {2}})^{n}n}}} {\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {1}{(2+{\sqrt {2}})^{n}n}}}.
  • ln ⁡ 2 = 4 3 + 2 2 ∑ k = 0 ∞ 1 ( 17 + 12 2 ) k ( 2 k + 1 ) {\displaystyle \ln 2={\frac {4}{3+2{\sqrt {2}}}}\sum _{k=0}^{\infty }{\frac {1}{(17+12{\sqrt {2}})^{k}(2k+1)}}} {\displaystyle \ln 2={\frac {4}{3+2{\sqrt {2}}}}\sum _{k=0}^{\infty }{\frac {1}{(17+12{\sqrt {2}})^{k}(2k+1)}}}.

Menerapkannya untuk 2 = ( 16 15 ) 7 ⋅ ( 81 80 ) 3 ⋅ ( 25 24 ) 5 {\displaystyle \textstyle 2={\left({\frac {16}{15}}\right)}^{7}\cdot {\left({\frac {81}{80}}\right)}^{3}\cdot {\left({\frac {25}{24}}\right)}^{5}} {\displaystyle \textstyle 2={\left({\frac {16}{15}}\right)}^{7}\cdot {\left({\frac {81}{80}}\right)}^{3}\cdot {\left({\frac {25}{24}}\right)}^{5}} memberikan:

  • ln ⁡ 2 = 7 ∑ n = 1 ∞ ( − 1 ) n − 1 15 n n + 3 ∑ n = 1 ∞ ( − 1 ) n − 1 80 n n + 5 ∑ n = 1 ∞ ( − 1 ) n − 1 24 n n {\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{15^{n}n}}+3\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{80^{n}n}}+5\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{24^{n}n}}} {\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{15^{n}n}}+3\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{80^{n}n}}+5\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{24^{n}n}}}.
  • ln ⁡ 2 = 7 ∑ n = 1 ∞ 1 16 n n + 3 ∑ n = 1 ∞ 1 81 n n + 5 ∑ n = 1 ∞ 1 25 n n {\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {1}{16^{n}n}}+3\sum _{n=1}^{\infty }{\frac {1}{81^{n}n}}+5\sum _{n=1}^{\infty }{\frac {1}{25^{n}n}}} {\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {1}{16^{n}n}}+3\sum _{n=1}^{\infty }{\frac {1}{81^{n}n}}+5\sum _{n=1}^{\infty }{\frac {1}{25^{n}n}}}.
  • ln ⁡ 2 = 14 31 ∑ k = 0 ∞ 1 961 k ( 2 k + 1 ) + 6 161 ∑ k = 0 ∞ 1 25921 k ( 2 k + 1 ) + 10 49 ∑ k = 0 ∞ 1 2401 k ( 2 k + 1 ) {\displaystyle \ln 2={\frac {14}{31}}\sum _{k=0}^{\infty }{\frac {1}{961^{k}(2k+1)}}+{\frac {6}{161}}\sum _{k=0}^{\infty }{\frac {1}{25921^{k}(2k+1)}}+{\frac {10}{49}}\sum _{k=0}^{\infty }{\frac {1}{2401^{k}(2k+1)}}} {\displaystyle \ln 2={\frac {14}{31}}\sum _{k=0}^{\infty }{\frac {1}{961^{k}(2k+1)}}+{\frac {6}{161}}\sum _{k=0}^{\infty }{\frac {1}{25921^{k}(2k+1)}}+{\frac {10}{49}}\sum _{k=0}^{\infty }{\frac {1}{2401^{k}(2k+1)}}}.

Wakilan sebagai integral

Logaritma natural dari 2 sering terjadi sebagai hasil integrasi. Beberapa rumus eksplisit untuknya termasuk

  • ∫ 0 1 d x 1 + x = ∫ 1 2 d x x = ln ⁡ 2 {\displaystyle \int _{0}^{1}{\frac {dx}{1+x}}=\int _{1}^{2}{\frac {dx}{x}}=\ln 2} {\displaystyle \int _{0}^{1}{\frac {dx}{1+x}}=\int _{1}^{2}{\frac {dx}{x}}=\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} {\displaystyle \int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}\,dx=\ln 2}.
  • ∫ 0 π 3 tan ⁡ x d x = 2 ∫ 0 π 4 tan ⁡ x d x = ln ⁡ 2 {\displaystyle \int _{0}^{\frac {\pi }{3}}\tan x\,dx=2\int _{0}^{\frac {\pi }{4}}\tan x\,dx=\ln 2} {\displaystyle \int _{0}^{\frac {\pi }{3}}\tan x\,dx=2\int _{0}^{\frac {\pi }{4}}\tan x\,dx=\ln 2}.

Wakilan lainnya

Pengembangan Piercenya adalah OEIS A091846

ln ⁡ 2 = 1 − 1 1 ⋅ 3 + 1 1 ⋅ 3 ⋅ 12 − ⋯ {\displaystyle \ln 2=1-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\cdots } {\displaystyle \ln 2=1-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\cdots }

Pengembangan Engelnya adalah OEIS A059180

ln ⁡ 2 = 1 2 + 1 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 7 + 1 2 ⋅ 3 ⋅ 7 ⋅ 9 + ⋯ {\displaystyle \ln 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}}+\cdots } {\displaystyle \ln 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}}+\cdots }

Pengembangan kotangennya adalah OEIS A081785

ln ⁡ 2 = cot ⁡ ( arccot ⁡ ( 0 ) − arccot ⁡ ( 1 ) + arccot ⁡ ( 5 ) − arccot ⁡ ( 55 ) + arccot ⁡ ( 14187 ) − ⋯ ) {\displaystyle \ln 2=\cot({\operatorname {arccot}(0)-\operatorname {arccot}(1)+\operatorname {arccot}(5)-\operatorname {arccot}(55)+\operatorname {arccot}(14187)-\cdots })} {\displaystyle \ln 2=\cot({\operatorname {arccot} (0)-\operatorname {arccot} (1)+\operatorname {arccot} (5)-\operatorname {arccot} (55)+\operatorname {arccot} (14187)-\cdots })}

Pengembangan pecahan berlanjutnya adalah OEIS A016730

ln ⁡ 2 = [ 0 ; 1 , 2 , 3 , 1 , 6 , 3 , 1 , 1 , 2 , 1 , 1 , 1 , 1 , 3 , 10 , 1 , 1 , 1 , 2 , 1 , 1 , 1 , 1 , 3 , 2 , 3 , 1 , . . . ] {\displaystyle \ln 2=\left[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,...\right]} {\displaystyle \ln 2=\left[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,...\right]}

yang menghasilkan aproksimasi rasional, beberapa yang pertama adalah 0 {\displaystyle 0} {\displaystyle 0}, 1 {\displaystyle 1} {\displaystyle 1}, 2 3 {\displaystyle \textstyle {\frac {2}{3}}} {\displaystyle \textstyle {\frac {2}{3}}}, 7 10 {\displaystyle \textstyle {\frac {7}{10}}} {\displaystyle \textstyle {\frac {7}{10}}}, 9 13 {\displaystyle \textstyle {\frac {9}{13}}} {\displaystyle \textstyle {\frac {9}{13}}}, dan 61 88 {\displaystyle \textstyle {\frac {61}{88}}} {\displaystyle \textstyle {\frac {61}{88}}}.

Pecahan berlanjut yang digeneralisasi ini:

ln ⁡ 2 = [ 0 ; 1 , 2 , 3 , 1 , 5 , 2 3 , 7 , 1 2 , 9 , 2 5 , . . . , 2 k − 1 , 2 k , . . . ] {\displaystyle \ln 2=\left[0;1,2,3,1,5,{\tfrac {2}{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]} {\displaystyle \ln 2=\left[0;1,2,3,1,5,{\tfrac {2}{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]},[1] dapat diekspresikan sebagai
ln ⁡ 2 = 1 1 + 1 2 + 1 3 + 2 2 + 2 5 + 3 2 + 3 7 + 4 2 + ⋱ = 2 3 − 1 2 9 − 2 2 15 − 3 2 21 − ⋱ {\displaystyle \ln 2={\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {2}{5+{\cfrac {3}{2+{\cfrac {3}{7+{\cfrac {4}{2+\ddots }}}}}}}}}}}}}}}}={\cfrac {2}{3-{\cfrac {1^{2}}{9-{\cfrac {2^{2}}{15-{\cfrac {3^{2}}{21-\ddots }}}}}}}}} {\displaystyle \ln 2={\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {2}{5+{\cfrac {3}{2+{\cfrac {3}{7+{\cfrac {4}{2+\ddots }}}}}}}}}}}}}}}}={\cfrac {2}{3-{\cfrac {1^{2}}{9-{\cfrac {2^{2}}{15-{\cfrac {3^{2}}{21-\ddots }}}}}}}}}

Bootstrap logaritma lainnya

Diberikan sebuah nilai dari ln ⁡ 2 {\displaystyle \ln 2} {\displaystyle \ln 2}, sebuah skema menghitung logaritma dari bilangan bulat lainnya adalah untuk mentabulasi logaritma dari bilangan prima dan di lapisan berikutnya, logaritma dari bilangan komposit c {\displaystyle c} {\displaystyle c} berdasarkan faktorisasinya

c = 2 i 3 j 5 k 7 l ⋯ → ln ⁡ ( c ) = i ln ⁡ ( 2 ) + j ln ⁡ ( 3 ) + k ln ⁡ ( 5 ) + l ln ⁡ ( 7 ) + ⋯ {\displaystyle c=2^{i}3^{j}5^{k}7^{l}\cdots \rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots } {\displaystyle c=2^{i}3^{j}5^{k}7^{l}\cdots \rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots }

Ini memakai

Bilangan prima Memperkirakan logaritma natural OEIS
2 0.693 147 180 559 945 309 417 232 121 458 {\displaystyle 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458} {\displaystyle 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458} A002162
3 1.098 612 288 668 109 691 395 245 236 92 {\displaystyle 1.098\,612\,288\,668\,109\,691\,395\,245\,236\,92} {\displaystyle 1.098\,612\,288\,668\,109\,691\,395\,245\,236\,92} A002391
5 1.609 437 912 434 100 374 600 759 333 23 {\displaystyle 1.609\,437\,912\,434\,100\,374\,600\,759\,333\,23} {\displaystyle 1.609\,437\,912\,434\,100\,374\,600\,759\,333\,23} A016628
7 1.945 910 149 055 313 305 105 352 743 44 {\displaystyle 1.945\,910\,149\,055\,313\,305\,105\,352\,743\,44} {\displaystyle 1.945\,910\,149\,055\,313\,305\,105\,352\,743\,44} A016630
11 2.397 895 272 798 370 544 061 943 577 97 {\displaystyle 2.397\,895\,272\,798\,370\,544\,061\,943\,577\,97} {\displaystyle 2.397\,895\,272\,798\,370\,544\,061\,943\,577\,97} A016634
13 2.564 949 357 461 536 736 053 487 441 57 {\displaystyle 2.564\,949\,357\,461\,536\,736\,053\,487\,441\,57} {\displaystyle 2.564\,949\,357\,461\,536\,736\,053\,487\,441\,57} A016636
17 2.833 213 344 056 216 080 249 534 617 87 {\displaystyle 2.833\,213\,344\,056\,216\,080\,249\,534\,617\,87} {\displaystyle 2.833\,213\,344\,056\,216\,080\,249\,534\,617\,87} A016640
19 2.944 438 979 166 440 460 009 027 431 89 {\displaystyle 2.944\,438\,979\,166\,440\,460\,009\,027\,431\,89} {\displaystyle 2.944\,438\,979\,166\,440\,460\,009\,027\,431\,89} A016642
23 3.135 494 215 929 149 690 806 752 831 81 {\displaystyle 3.135\,494\,215\,929\,149\,690\,806\,752\,831\,81} {\displaystyle 3.135\,494\,215\,929\,149\,690\,806\,752\,831\,81} A016646
29 3.367 295 829 986 474 027 183 272 032 36 {\displaystyle 3.367\,295\,829\,986\,474\,027\,183\,272\,032\,36} {\displaystyle 3.367\,295\,829\,986\,474\,027\,183\,272\,032\,36} A016652
31 3.433 987 204 485 146 245 929 164 324 54 {\displaystyle 3.433\,987\,204\,485\,146\,245\,929\,164\,324\,54} {\displaystyle 3.433\,987\,204\,485\,146\,245\,929\,164\,324\,54} A016654
37 3.610 917 912 644 224 444 368 095 671 03 {\displaystyle 3.610\,917\,912\,644\,224\,444\,368\,095\,671\,03} {\displaystyle 3.610\,917\,912\,644\,224\,444\,368\,095\,671\,03} A016660
41 3.713 572 066 704 307 803 866 763 373 04 {\displaystyle 3.713\,572\,066\,704\,307\,803\,866\,763\,373\,04} {\displaystyle 3.713\,572\,066\,704\,307\,803\,866\,763\,373\,04} A016664
43 3.761 200 115 693 562 423 472 842 513 35 {\displaystyle 3.761\,200\,115\,693\,562\,423\,472\,842\,513\,35} {\displaystyle 3.761\,200\,115\,693\,562\,423\,472\,842\,513\,35} A016666
47 3.850 147 601 710 058 586 820 950 669 77 {\displaystyle 3.850\,147\,601\,710\,058\,586\,820\,950\,669\,77} {\displaystyle 3.850\,147\,601\,710\,058\,586\,820\,950\,669\,77} A016670
53 3.970 291 913 552 121 834 144 469 139 03 {\displaystyle 3.970\,291\,913\,552\,121\,834\,144\,469\,139\,03} {\displaystyle 3.970\,291\,913\,552\,121\,834\,144\,469\,139\,03} A016676
59 4.077 537 443 905 719 450 616 050 373 72 {\displaystyle 4.077\,537\,443\,905\,719\,450\,616\,050\,373\,72} {\displaystyle 4.077\,537\,443\,905\,719\,450\,616\,050\,373\,72} A016682
61 4.110 873 864 173 311 248 751 389 103 43 {\displaystyle 4.110\,873\,864\,173\,311\,248\,751\,389\,103\,43} {\displaystyle 4.110\,873\,864\,173\,311\,248\,751\,389\,103\,43} A016684
67 4.204 692 619 390 966 059 670 071 996 36 {\displaystyle 4.204\,692\,619\,390\,966\,059\,670\,071\,996\,36} {\displaystyle 4.204\,692\,619\,390\,966\,059\,670\,071\,996\,36} A016690
71 4.262 679 877 041 315 421 329 454 532 51 {\displaystyle 4.262\,679\,877\,041\,315\,421\,329\,454\,532\,51} {\displaystyle 4.262\,679\,877\,041\,315\,421\,329\,454\,532\,51} A016694
73 4.290 459 441 148 391 129 092 108 857 44 {\displaystyle 4.290\,459\,441\,148\,391\,129\,092\,108\,857\,44} {\displaystyle 4.290\,459\,441\,148\,391\,129\,092\,108\,857\,44} A016696
79 4.369 447 852 467 021 494 172 945 541 48 {\displaystyle 4.369\,447\,852\,467\,021\,494\,172\,945\,541\,48} {\displaystyle 4.369\,447\,852\,467\,021\,494\,172\,945\,541\,48} A016702
83 4.418 840 607 796 597 923 475 472 223 29 {\displaystyle 4.418\,840\,607\,796\,597\,923\,475\,472\,223\,29} {\displaystyle 4.418\,840\,607\,796\,597\,923\,475\,472\,223\,29} A016706
89 4.488 636 369 732 139 838 317 815 540 67 {\displaystyle 4.488\,636\,369\,732\,139\,838\,317\,815\,540\,67} {\displaystyle 4.488\,636\,369\,732\,139\,838\,317\,815\,540\,67} A016712
97 4.574 710 978 503 382 822 116 721 621 70 {\displaystyle 4.574\,710\,978\,503\,382\,822\,116\,721\,621\,70} {\displaystyle 4.574\,710\,978\,503\,382\,822\,116\,721\,621\,70} A016720

DI lapisan ketiga, logaritma bilangan rasional r = a b {\displaystyle r={\frac {a}{b}}} {\displaystyle r={\frac {a}{b}}} dihitung dengan menggunakan ln ⁡ ( r ) = ln ⁡ ( a ) − ln ⁡ ( b ) {\displaystyle \ln(r)=\ln(a)-\ln(b)} {\displaystyle \ln(r)=\ln(a)-\ln(b)}, dan logaritma akar melalui ln ⁡ c n = 1 n ln ⁡ c {\displaystyle \ln {\sqrt[{n}]{c}}={\frac {1}{n}}\ln c} {\displaystyle \ln {\sqrt[{n}]{c}}={\frac {1}{n}}\ln c}.

Logaritma dari 2 berguna dalam arti bahwa pangkat dari 2 tersebar agak padat, mencari 2 i {\displaystyle 2^{i}} {\displaystyle 2^{i}} yang mendekati dengan pangkat b j {\displaystyle b^{j}} {\displaystyle b^{j}} dari bilangan b {\displaystyle b} {\displaystyle b} lainnya relatif mudah, dan representasi deret ln ⁡ b {\displaystyle \ln b} {\displaystyle \ln b} dengan menggabungkan 2 {\displaystyle 2} {\displaystyle 2} ke b {\displaystyle b} {\displaystyle b} dengan perubahan logaritmik.

Contoh

Jika p s = q t + d {\displaystyle p^{s}=q^{t}+d} {\displaystyle p^{s}=q^{t}+d} dengan beberapa d {\displaystyle d} {\displaystyle d}, maka p s q t = 1 + d q t {\displaystyle {\frac {p^{s}}{q^{t}}}=1+{\frac {d}{q^{t}}}} {\displaystyle {\frac {p^{s}}{q^{t}}}=1+{\frac {d}{q^{t}}}} dan karena itu

s ln ⁡ ( p ) − t ln ⁡ ( q ) = ln ⁡ ( 1 + d q t ) = ∑ m = 1 ∞ ( − 1 ) m + 1 ( d q t ) m m = ∑ n = 0 ∞ 2 2 n + 1 ( d 2 q t + d ) 2 n + 1 {\displaystyle s\ln(p)-t\ln(q)=\ln \left(1+{\frac {d}{q^{t}}}\right)=\sum _{m=1}^{\infty }(-1)^{m+1}{\frac {({\frac {d}{q^{t}}})^{m}}{m}}=\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {d}{2q^{t}+d}}\right)}^{2n+1}} {\displaystyle s\ln(p)-t\ln(q)=\ln \left(1+{\frac {d}{q^{t}}}\right)=\sum _{m=1}^{\infty }(-1)^{m+1}{\frac {({\frac {d}{q^{t}}})^{m}}{m}}=\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {d}{2q^{t}+d}}\right)}^{2n+1}}

Memilih q = 2 {\displaystyle q=2} {\displaystyle q=2} mewakili ln ⁡ ( p ) {\displaystyle \ln(p)} {\displaystyle \ln(p)} oleh ln ⁡ 2 {\displaystyle \ln 2} {\displaystyle \ln 2} dan sebuah deret dari sebuah parameter d q t {\displaystyle {\frac {d}{q^{t}}}} {\displaystyle {\frac {d}{q^{t}}}} yang ingin tetap kecil untuk konvergen cepat. Mengambil 3 2 = 2 3 + 1 {\displaystyle 3^{2}=2^{3}+1} {\displaystyle 3^{2}=2^{3}+1}, sebagai contoh, menghasilkan

2 ln ⁡ ( 3 ) = 3 ln ⁡ 2 − ∑ k ≥ 1 ( − 1 ) k 8 k k = 3 ln ⁡ 2 + ∑ n = 0 ∞ 2 2 n + 1 ( 1 2 ⋅ 8 + 1 ) 2 n + 1 {\displaystyle 2\ln(3)=3\ln 2-\sum _{k\geq 1}{\frac {(-1)^{k}}{8^{k}k}}=3\ln 2+\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {1}{2\cdot 8+1}}\right)}^{2n+1}} {\displaystyle 2\ln(3)=3\ln 2-\sum _{k\geq 1}{\frac {(-1)^{k}}{8^{k}k}}=3\ln 2+\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {1}{2\cdot 8+1}}\right)}^{2n+1}}:

Ini sebenarnya baris ketiga dalam tabel ekspansi tipe ini:

s {\displaystyle s} {\displaystyle s} p {\displaystyle p} {\displaystyle p} t {\displaystyle t} {\displaystyle t} q {\displaystyle q} {\displaystyle q} d q t {\displaystyle {\frac {d}{q^{t}}}} {\displaystyle {\frac {d}{q^{t}}}}
1 3 1 2 1 2 = 0.500 000 00 … {\displaystyle {\frac {1}{2}}=0.500\,000\,00\dots } {\displaystyle {\frac {1}{2}}=0.500\,000\,00\dots }
1 3 2 2 − 1 4 = − 0.250 000 00 … {\displaystyle -{\frac {1}{4}}=-0.250\,000\,00\dots } {\displaystyle -{\frac {1}{4}}=-0.250\,000\,00\dots }
2 3 3 2 1 8 = 0.125 000 00 … {\displaystyle {\frac {1}{8}}=0.125\,000\,00\dots } {\displaystyle {\frac {1}{8}}=0.125\,000\,00\dots }
5 3 8 2 − 13 256 = − 0.050 781 25 … {\displaystyle -{\frac {13}{256}}=-0.050\,781\,25\dots } {\displaystyle -{\frac {13}{256}}=-0.050\,781\,25\dots }
12 3 19 2 7153 524288 = 0.013 643 26 … {\displaystyle {\frac {7153}{524288}}=0.013\,643\,26\dots } {\displaystyle {\frac {7153}{524288}}=0.013\,643\,26\dots }
1 5 2 2 1 4 = 0.250 000 00 … {\displaystyle {\frac {1}{4}}=0.250\,000\,00\dots } {\displaystyle {\frac {1}{4}}=0.250\,000\,00\dots }
3 5 7 2 − 3 128 = − 0.023 437 50 … {\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots } {\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots }
1 7 2 2 3 4 = 0.750 000 00 … {\displaystyle {\frac {3}{4}}=0.750\,000\,00\dots } {\displaystyle {\frac {3}{4}}=0.750\,000\,00\dots }
1 7 3 2 − 1 8 = − 0.125 000 00 … {\displaystyle -{\frac {1}{8}}=-0.125\,000\,00\dots } {\displaystyle -{\frac {1}{8}}=-0.125\,000\,00\dots }
5 7 14 2 423 16384 = 0.025 817 87 … {\displaystyle {\frac {423}{16384}}=0.025\,817\,87\dots } {\displaystyle {\frac {423}{16384}}=0.025\,817\,87\dots }
1 1 3 2 3 8 = 0.375 000 00 … {\displaystyle {\frac {3}{8}}=0.375\,000\,00\dots } {\displaystyle {\frac {3}{8}}=0.375\,000\,00\dots }
2 11 7 2 − 7 128 = − 0.054 687 50 … {\displaystyle -{\frac {7}{128}}=-0.054\,687\,50\dots } {\displaystyle -{\frac {7}{128}}=-0.054\,687\,50\dots }
11 11 38 2 10433763667 274877906944 = 0.037 957 81 … {\displaystyle {\frac {10433763667}{274877906944}}=0.037\,957\,81\dots } {\displaystyle {\frac {10433763667}{274877906944}}=0.037\,957\,81\dots }
1 13 3 2 5 8 = 0.625 000 00 … {\displaystyle {\frac {5}{8}}=0.625\,000\,00\dots } {\displaystyle {\frac {5}{8}}=0.625\,000\,00\dots }
1 13 4 2 − 3 16 = − 0.187 500 00 … {\displaystyle -{\frac {3}{16}}=-0.187\,500\,00\dots } {\displaystyle -{\frac {3}{16}}=-0.187\,500\,00\dots }
3 13 11 2 149 2048 = 0.072 753 91 … {\displaystyle {\frac {149}{2048}}=0.072\,753\,91\dots } {\displaystyle {\frac {149}{2048}}=0.072\,753\,91\dots }
7 13 26 2 − 4360347 67108864 = − 0.064 974 23 … {\displaystyle -{\frac {4360347}{67108864}}=-0.064\,974\,23\dots } {\displaystyle -{\frac {4360347}{67108864}}=-0.064\,974\,23\dots }
10 13 37 2 419538377 137438953472 = 0.003 052 54 … {\displaystyle {\frac {419538377}{137438953472}}=0.003\,052\,54\dots } {\displaystyle {\frac {419538377}{137438953472}}=0.003\,052\,54\dots }
1 17 4 2 1 16 = 0.062 500 00 … {\displaystyle {\frac {1}{16}}=0.062\,500\,00\dots } {\displaystyle {\frac {1}{16}}=0.062\,500\,00\dots }
1 19 4 2 3 16 = 0.187 500 00 … {\displaystyle {\frac {3}{16}}=0.187\,500\,00\dots } {\displaystyle {\frac {3}{16}}=0.187\,500\,00\dots }
4 19 17 2 − 751 131072 = − 0.005 729 68 … {\displaystyle -{\frac {751}{131072}}=-0.005\,729\,68\dots } {\displaystyle -{\frac {751}{131072}}=-0.005\,729\,68\dots }
1 23 4 2 7 16 = 0.437 500 00 … {\displaystyle {\frac {7}{16}}=0.437\,500\,00\dots } {\displaystyle {\frac {7}{16}}=0.437\,500\,00\dots }
1 23 5 2 − 9 32 = − 0.281 250 00 … {\displaystyle -{\frac {9}{32}}=-0.281\,250\,00\dots } {\displaystyle -{\frac {9}{32}}=-0.281\,250\,00\dots }
2 23 9 2 17 512 = 0.033 203 12 … {\displaystyle {\frac {17}{512}}=0.033\,203\,12\dots } {\displaystyle {\frac {17}{512}}=0.033\,203\,12\dots }
1 29 4 2 13 16 = 0.812 500 00 … {\displaystyle {\frac {13}{16}}=0.812\,500\,00\dots } {\displaystyle {\frac {13}{16}}=0.812\,500\,00\dots }
1 29 5 2 − 3 32 = − 0.093 750 00 … {\displaystyle -{\frac {3}{32}}=-0.093\,750\,00\dots } {\displaystyle -{\frac {3}{32}}=-0.093\,750\,00\dots }
7 29 34 2 70007125 17179869184 = 0.004 074 95 … {\displaystyle {\frac {70007125}{17179869184}}=0.004\,074\,95\dots } {\displaystyle {\frac {70007125}{17179869184}}=0.004\,074\,95\dots }
1 31 5 2 − 1 32 = − 0.031 250 00 … {\displaystyle -{\frac {1}{32}}=-0.031\,250\,00\dots } {\displaystyle -{\frac {1}{32}}=-0.031\,250\,00\dots }
1 37 5 2 5 32 = 0.156 250 00 … {\displaystyle {\frac {5}{32}}=0.156\,250\,00\dots } {\displaystyle {\frac {5}{32}}=0.156\,250\,00\dots }
4 37 21 2 − 222 991 2 097 152 = − 0.106 330 39 … {\displaystyle -{\frac {222\,991}{2\,097\,152}}=-0.106\,330\,39\dots } {\displaystyle -{\frac {222\,991}{2\,097\,152}}=-0.106\,330\,39\dots }
5 37 26 2 2 235 093 67 108 864 = 0.033 305 48 … {\displaystyle {\frac {2\,235\,093}{67\,108\,864}}=0.033\,305\,48\dots } {\displaystyle {\frac {2\,235\,093}{67\,108\,864}}=0.033\,305\,48\dots }
1 41 5 2 9 32 = 0.281 250 00 … {\displaystyle {\frac {9}{32}}=0.281\,250\,00\dots } {\displaystyle {\frac {9}{32}}=0.281\,250\,00\dots }
2 41 11 2 − 367 2048 = − 0.179 199 22 … {\displaystyle -{\frac {367}{2048}}=-0.179\,199\,22\dots } {\displaystyle -{\frac {367}{2048}}=-0.179\,199\,22\dots }
3 41 16 2 3385 65 536 = 0.051 651 00 … {\displaystyle {\frac {3385}{65\,536}}=0.051\,651\,00\dots } {\displaystyle {\frac {3385}{65\,536}}=0.051\,651\,00\dots }
1 43 5 2 11 32 = 0.343 750 00 … {\displaystyle {\frac {11}{32}}=0.343\,750\,00\dots } {\displaystyle {\frac {11}{32}}=0.343\,750\,00\dots }
2 43 11 2 − 199 2048 = − 0.097 167 97 … {\displaystyle -{\frac {199}{2048}}=-0.097\,167\,97\dots } {\displaystyle -{\frac {199}{2048}}=-0.097\,167\,97\dots }
5 43 27 2 12 790 715 134 217 728 = 0.095 298 25 … {\displaystyle {\frac {12\,790\,715}{134\,217\,728}}=0.095\,298\,25\dots } {\displaystyle {\frac {12\,790\,715}{134\,217\,728}}=0.095\,298\,25\dots }
7 43 38 2 3 059 295 837 274 877 906 944 = 0.011 129 65 … {\displaystyle {\frac {3\,059\,295\,837}{274\,877\,906\,944}}=0.011\,129\,65\dots } {\displaystyle {\frac {3\,059\,295\,837}{274\,877\,906\,944}}=0.011\,129\,65\dots }

Dimulai dari logaritma natural dari q = 10 {\displaystyle q=10} {\displaystyle q=10}, salah satunya dapat menggunakan parameter-parameter ini:

s {\displaystyle s} {\displaystyle s} p {\displaystyle p} {\displaystyle p} t {\displaystyle t} {\displaystyle t} q {\displaystyle q} {\displaystyle q} d q t {\displaystyle {\frac {d}{q^{t}}}} {\displaystyle {\frac {d}{q^{t}}}}
10 2 3 10 3 125 = 0.024 000 00 … {\displaystyle {\frac {3}{125}}=0.024\,000\,00\dots } {\displaystyle {\frac {3}{125}}=0.024\,000\,00\dots }
21 3 10 10 460 353 203 10 000 000 000 = 0.046 035 32 … {\displaystyle {\frac {460\,353\,203}{10\,000\,000\,000}}=0.046\,035\,32\dots } {\displaystyle {\frac {460\,353\,203}{10\,000\,000\,000}}=0.046\,035\,32\dots }
3 5 2 10 1 4 = 0.250 000 … {\displaystyle {\frac {1}{4}}=0.250\,000\,\dots } {\displaystyle {\frac {1}{4}}=0.250\,000\,\dots }
10 5 7 10 − 3 128 = − 0.023 437 50 … {\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots } {\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots }
6 7 5 10 17 649 100 000 = 0.176 490 00 … {\displaystyle {\frac {17\,649}{100\,000}}=0.176\,490\,00\dots } {\displaystyle {\frac {17\,649}{100\,000}}=0.176\,490\,00\dots }
13 7 11 10 − 3 110 989 593 100 000 000 000 = − 0.031 109 90 … {\displaystyle -{\frac {3\,110\,989\,593}{100\,000\,000\,000}}=-0.031\,109\,90\dots } {\displaystyle -{\frac {3\,110\,989\,593}{100\,000\,000\,000}}=-0.031\,109\,90\dots }
1 11 1 10 1 10 = 0.100 000 00 … {\displaystyle {\frac {1}{10}}=0.100\,000\,00\dots } {\displaystyle {\frac {1}{10}}=0.100\,000\,00\dots }
1 13 1 10 3 10 = 0.300 000 00 … {\displaystyle {\frac {3}{10}}=0.300\,000\,00\dots } {\displaystyle {\frac {3}{10}}=0.300\,000\,00\dots }
8 13 9 10 − 184 269 279 1 000 000 000 = − 0.184 269 28 … {\displaystyle -{\frac {184\,269\,279}{1\,000\,000\,000}}=-0.184\,269\,28\dots } {\displaystyle -{\frac {184\,269\,279}{1\,000\,000\,000}}=-0.184\,269\,28\dots }
9 13 10 10 604 499 373 10 000 000 000 = 0.060 449 94 … {\displaystyle {\frac {604\,499\,373}{10\,000\,000\,000}}=0.060\,449\,94\dots } {\displaystyle {\frac {604\,499\,373}{10\,000\,000\,000}}=0.060\,449\,94\dots }
1 17 1 10 7 10 = 0.700 000 00 … {\displaystyle {\frac {7}{10}}=0.700\,000\,00\dots } {\displaystyle {\frac {7}{10}}=0.700\,000\,00\dots }
4 17 5 10 − 16 479 100 000 = − 0.164 790 00 … {\displaystyle -{\frac {16\,479}{100\,000}}=-0.164\,790\,00\dots } {\displaystyle -{\frac {16\,479}{100\,000}}=-0.164\,790\,00\dots }
9 17 11 10 18 587 876 497 100 000 000 000 = 0.185 878 76 … {\displaystyle {\frac {18\,587\,876\,497}{100\,000\,000\,000}}=0.185\,878\,76\dots } {\displaystyle {\frac {18\,587\,876\,497}{100\,000\,000\,000}}=0.185\,878\,76\dots }
3 19 4 10 − 3141 10 000 = − 0.314 100 00 … {\displaystyle -{\frac {3141}{10\,000}}=-0.314\,100\,00\dots } {\displaystyle -{\frac {3141}{10\,000}}=-0.314\,100\,00\dots }
4 19 5 10 30 321 100 000 = 0.303 210 00 … {\displaystyle {\frac {30\,321}{100\,000}}=0.303\,210\,00\dots } {\displaystyle {\frac {30\,321}{100\,000}}=0.303\,210\,00\dots }
7 19 9 10 − 106 128 261 1 000 000 000 = − 0.106 128 26 … {\displaystyle -{\frac {106\,128\,261}{1\,000\,000\,000}}=-0.106\,128\,26\dots } {\displaystyle -{\frac {106\,128\,261}{1\,000\,000\,000}}=-0.106\,128\,26\dots }
2 23 3 10 − 471 1000 = − 0.471 000 00 … {\displaystyle -{\frac {471}{1000}}=-0.471\,000\,00\dots } {\displaystyle -{\frac {471}{1000}}=-0.471\,000\,00\dots }
3 23 4 10 2167 10 000 = 0.216 700 00 … {\displaystyle {\frac {2167}{10\,000}}=0.216\,700\,00\dots } {\displaystyle {\frac {2167}{10\,000}}=0.216\,700\,00\dots }
2 29 3 10 − 159 1000 = − 0.159 000 00 … {\displaystyle -{\frac {159}{1000}}=-0.159\,000\,00\dots } {\displaystyle -{\frac {159}{1000}}=-0.159\,000\,00\dots }
2 31 3 10 39 1000 = − 0.039 000 00 … {\displaystyle {\frac {39}{1000}}=-0.039\,000\,00\dots } {\displaystyle {\frac {39}{1000}}=-0.039\,000\,00\dots }

Digit yang diketahui

Ini adalah sebuah tabel catatan terbaru dalam menghitung digit ln ⁡ 2 {\displaystyle \ln 2} {\displaystyle \ln 2}. Mulai Desember 2018, ini telah dihitung lebih banyak digit dari setiap logaritma natural [2][3] dari sebuah bilangan asli, kecuali 1.

Tanggal Nama Jumlah digit
7 Januari 2009 A Yee & R Chan 15,500,000,000
4 Februari 2009 A Yee & R Chan 31,026,000,000
21 Februari 2011 Alexander Yee 50,000,000,050
14 Maret, 2011 Shigeru Kondo 100,000,000,000
28 Februari 2014 Shigeru Kondo 200,000,000,050
12 Juli 2015 Ron Watkins 250,000,000,000
30 Januari 2016 Ron Watkins 350,000,000,000
18 April 2016 Ron Watkins 500,000,000,000
10 Desember 2018 Michael Kwok 600,000,000,000
26 April 2019 Jacob Riffee 1,000,000,000,000
19 Agustus 2020 Seungmin Kim[4][5] 1,2000,000,000,100

Lihat pula

  • Aturan 72#Penggabungan kontinu, di mana ln ⁡ 2 {\displaystyle \ln 2} {\displaystyle \ln 2} sangat menonjol
  • Waktu-paruh#Rumus untuk waktu-paruh dalam peluruhan eksponensial, di mana ln ⁡ 2 {\displaystyle \ln 2} {\displaystyle \ln 2} sangat menonjol
  • Persamanan Erdős–Moser, semua penyelesaian harus datang dari sebuah konvergen dari ln ⁡ 2 {\displaystyle \ln 2} {\displaystyle \ln 2}.

Referensi

  • Brent, Richard P. (1976). "Fast multiple-precision evaluation of elementary functions". J. ACM. 23 (2): 242–251. doi:10.1145/321941.321944. MR 0395314.
  • Uhler, Horace S. (1940). "Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17". Proc. Natl. Acad. Sci. U.S.A. 26 (3): 205–212. doi:10.1073/pnas.26.3.205. MR 0001523. PMC 1078033. PMID 16588339.
  • Sweeney, Dura W. (1963). "On the computation of Euler's constant". Mathematics of Computation. 17 (82): 170–178. doi:10.1090/S0025-5718-1963-0160308-X. MR 0160308.
  • Chamberland, Marc (2003). "Binary BBP-formulae for logarithms and generalized Gaussian–Mersenne primes" (PDF). Journal of Integer Sequences. 6: 03.3.7. MR 2046407. Diarsipkan dari asli (PDF) tanggal 2011-06-06. Diakses tanggal 2010-04-29.
  • Gourévitch, Boris; Guillera Goyanes, Jesús (2007). "Construction of binomial sums for π and polylogarithmic constants inspired by BBP formulas" (PDF). Applied Math. E-Notes. 7: 237–246. MR 2346048.
  • Wu, Qiang (2003). "On the linear independence measure of logarithms of rational numbers". Mathematics of Computation. 72 (242): 901–911. doi:10.1090/S0025-5718-02-01442-4.
  1. ↑ Borwein, J.; Crandall, R.; Free, G. (2004). "On the Ramanujan AGM Fraction , I: The Real-Parameter Case" (PDF). Exper. Math. 13 (3): 278–280. doi:10.1080/10586458.2004.10504540.
  2. ↑ "y-cruncher". numberworld.org. Diakses tanggal 10 December 2018.
  3. ↑ "Natural log of 2". numberworld.org. Diakses tanggal 10 December 2018.
  4. ↑ "Records set by y-cruncher". Diarsipkan dari asli tanggal 2020-09-15. Diakses tanggal September 15, 2020.
  5. ↑ "Natural logarithm of 2 (Log(2)) world record by Seungmin Kim". Diakses tanggal September 15, 2020.

Pranala luar

  • (Inggris) Weisstein, Eric W. "Natural logarithm of 2". MathWorld.
  • table of natural logarithms, PlanetMath.org.
  • Gourdon, Xavier; Sebah, Pascal. "The logarithm constant:log 2".

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Daftar Isi

  1. Wakilan deret
  2. Faktorial bolak-balik menaik
  3. Faktorial konstanta menaik biner
  4. Wakilan deret lainnya
  5. Melibatkan fungsi zeta Riemann
  6. Wakilan tipe-BBP
  7. Wakilan sebagai integral
  8. Wakilan lainnya
  9. Bootstrap logaritma lainnya
  10. Contoh
  11. Digit yang diketahui
  12. Lihat pula
  13. Referensi
  14. Pranala luar

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