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}
.
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}}}
Logaritma umum secara khusus adalah 
A007524
-
log
10
2
≈
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}
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 }
. 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)}}}
.
-
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)}}}
.
-
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)}}}
.
-
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)}}}
.
-
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)}}}
.
Faktorial konstanta menaik biner
-
ln
2
=
∑
n
=
1
∞
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)}}}
.
-
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)}}}
.
-
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)}}}
-
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)}}}
.
-
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)}}}
.
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}
.
-
∑
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
=
1
∞
1
4
n
2
−
2
n
=
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}
.
-
∑
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}}}}}
.
-
∑
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}}}}}
.
-
∑
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}}}
.
-
∑
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}
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}
.
-
∑
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}}}
(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}}}
.
-
∑
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}}}
.
-
∑
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}
.
(
γ
{\displaystyle \gamma }
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}}}}
(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}}}
.
-
ln
2
=
∑
n
=
1
∞
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)}}}
.
Menerapkannya untuk
2
=
3
2
⋅
4
3
{\displaystyle \textstyle {2={\frac {3}{2}}\cdot {\frac {4}{3}}}}
-
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}}}
.
-
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}}}
.
-
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)}}}
.
Menerapkannya untuk
2
=
(
2
)
2
{\displaystyle 2=\left({\sqrt {2}}\right)^{2}}
-
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}}}
.
-
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}}}
.
-
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)}}}
.
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}}
-
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}}}
.
-
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}}}
.
-
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)}}}
.
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}
.
-
∫
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
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}
.
Wakilan lainnya
Pengembangan Piercenya adalah 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 }
Pengembangan Engelnya adalah 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 }
Pengembangan kotangennya adalah 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 })}
Pengembangan pecahan berlanjutnya adalah 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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f8173c7c5f057cbbf1d0292b9cf10e9eee6e208)
yang menghasilkan aproksimasi rasional, beberapa yang pertama adalah
0
{\displaystyle 0}
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]}
,[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 }}}}}}}}}
Bootstrap logaritma lainnya
Diberikan sebuah nilai dari
ln
2
{\displaystyle \ln 2}
-
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 }
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
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}
 |
A016720 |
DI lapisan ketiga, logaritma bilangan rasional
r
=
a
b
{\displaystyle r={\frac {a}{b}}}
![{\displaystyle \ln {\sqrt[{n}]{c}}={\frac {1}{n}}\ln c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c430198b7b8550dc70e62c23077a8c14eaa80389)
Logaritma dari 2 berguna dalam arti bahwa pangkat dari 2 tersebar agak padat, mencari
2
i
{\displaystyle 2^{i}}
Contoh
Jika
p
s
=
q
t
+
d
{\displaystyle p^{s}=q^{t}+d}
-
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}}
Memilih
q
=
2
{\displaystyle q=2}
-
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}}
:
Ini sebenarnya baris ketiga dalam tabel ekspansi tipe ini:
s
{\displaystyle s}
 |
p
{\displaystyle p}
 |
t
{\displaystyle t}
 |
q
{\displaystyle q}
 |
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 }
 |
| 1 | 3 | 2 | 2 |
−
1
4
=
−
0.250
000
00
…
{\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 }
 |
| 5 | 3 | 8 | 2 |
−
13
256
=
−
0.050
781
25
…
{\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 }
 |
| 1 | 5 | 2 | 2 |
1
4
=
0.250
000
00
…
{\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 }
 |
| 1 | 7 | 2 | 2 |
3
4
=
0.750
000
00
…
{\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 }
 |
| 5 | 7 | 14 | 2 |
423
16384
=
0.025
817
87
…
{\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 }
 |
| 2 | 11 | 7 | 2 |
−
7
128
=
−
0.054
687
50
…
{\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 }
 |
| 1 | 13 | 3 | 2 |
5
8
=
0.625
000
00
…
{\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 }
 |
| 3 | 13 | 11 | 2 |
149
2048
=
0.072
753
91
…
{\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 }
 |
| 10 | 13 | 37 | 2 |
419538377
137438953472
=
0.003
052
54
…
{\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 }
 |
| 1 | 19 | 4 | 2 |
3
16
=
0.187
500
00
…
{\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 }
 |
| 1 | 23 | 4 | 2 |
7
16
=
0.437
500
00
…
{\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 }
 |
| 2 | 23 | 9 | 2 |
17
512
=
0.033
203
12
…
{\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 }
 |
| 1 | 29 | 5 | 2 |
−
3
32
=
−
0.093
750
00
…
{\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 }
 |
| 1 | 31 | 5 | 2 |
−
1
32
=
−
0.031
250
00
…
{\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 }
 |
| 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 }
 |
| 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 }
 |
| 1 | 41 | 5 | 2 |
9
32
=
0.281
250
00
…
{\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 }
 |
| 3 | 41 | 16 | 2 |
3385
65
536
=
0.051
651
00
…
{\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 }
 |
| 2 | 43 | 11 | 2 |
−
199
2048
=
−
0.097
167
97
…
{\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 }
 |
| 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 }
 |
Dimulai dari logaritma natural dari
q
=
10
{\displaystyle q=10}
|
s
{\displaystyle s}
|
p
{\displaystyle p}
|
t
{\displaystyle t}
|
q
{\displaystyle q}
|
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 }
 |
| 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 }
 |
| 3 | 5 | 2 | 10 |
1
4
=
0.250
000
…
{\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 }
|
| 6 | 7 | 5 | 10 |
17
649
100
000
=
0.176
490
00
…
{\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 }
 |
| 1 | 11 | 1 | 10 |
1
10
=
0.100
000
00
…
{\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 }
 |
| 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 }
 |
| 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 }
 |
| 1 | 17 | 1 | 10 |
7
10
=
0.700
000
00
…
{\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 }
 |
| 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 }
 |
| 3 | 19 | 4 | 10 |
−
3141
10
000
=
−
0.314
100
00
…
{\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 }
 |
| 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 }
 |
| 2 | 23 | 3 | 10 |
−
471
1000
=
−
0.471
000
00
…
{\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 }
 |
| 2 | 29 | 3 | 10 |
−
159
1000
=
−
0.159
000
00
…
{\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 }
 |
Digit yang diketahui
Ini adalah sebuah tabel catatan terbaru dalam menghitung digit
ln
2
{\displaystyle \ln 2}
| 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}
- Waktu-paruh#Rumus untuk waktu-paruh dalam peluruhan eksponensial, di mana
ln
2
{\displaystyle \ln 2}
- Persamanan Erdős–Moser, semua penyelesaian harus datang dari sebuah konvergen dari
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.
- ↑ 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.
- ↑ "y-cruncher". numberworld.org. Diakses tanggal 10 Desember 2018.
{{cite web}}: Pemeliharaan CS1: Tanggal diterjemahkan otomatis (link) - ↑ "Natural log of 2". numberworld.org. Diakses tanggal 10 Desember 2018.
{{cite web}}: Pemeliharaan CS1: Tanggal diterjemahkan otomatis (link) - ↑ "Records set by y-cruncher". Diarsipkan dari asli tanggal 2020-09-15. Diakses tanggal September 15, 2020.
{{cite web}}: Pemeliharaan CS1: Tanggal diterjemahkan otomatis (link) - ↑ "Natural logarithm of 2 (Log(2)) world record by Seungmin Kim". Diakses tanggal September 15, 2020.
{{cite web}}: Pemeliharaan CS1: Tanggal diterjemahkan otomatis (link)
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".