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Natural logarithm of 2

Mathematical constant

Mathematical constant

FieldValue
imageNatural Logarithm of 2.png
image_captionThe natural logarithm of 2 as an area under the curve 1/x.
rationalityIrrational
decimal

In mathematics, the natural logarithm of 2 is the unique real number argument such that the exponential function equals two. It appears frequently in various formulas and is also given by the alternating harmonic series. The decimal value of the natural logarithm of 2 truncated at 30 decimal places is given by:

:\ln 2 \approx 0.693,147,180,559,945,309,417,232,121,458.

The logarithm of 2 in other bases is obtained with the formula :\log_b 2 = \frac{\ln 2}{\ln b}. The common logarithm in particular is () :\log_{10} 2 \approx 0.301,029,995,663,981,195. The inverse of this number is the binary logarithm of 10: : \log_2 10 =\frac{1}{\log_{10} 2} \approx 3.321,928,095 ().

By the Lindemann–Weierstrass theorem, the natural logarithm of any natural number other than 0 and 1 (more generally, of any positive algebraic number other than 1) is a transcendental number. It is also contained in the ring of algebraic periods.

Series representations

Rising alternate factorial

:\ln 2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots. This is the well-known "alternating harmonic series". :\ln 2 = \frac{1}{2} +\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}. :\ln 2 = \frac{5}{8} +\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)}. :\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 = \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 = \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)}. :\ln 2 = \frac{2}{3}\left(1+\frac{2}{4^3-4}+\frac{2}{8^3-8}+\frac{2}{12^3-12}+\dots\right) .

Binary rising constant factorial

:\ln 2 = \sum_{n=1}^\infty \frac{1}{2^{n}n}. :\ln 2 = 1 -\sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)}. :\ln 2 = \frac{1}{2} + 2 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)} . :\ln 2 = \frac{5}{6} - 6 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)(n+3)} . :\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 = \frac{47}{60} - 120 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)} .

Other series representations

:\sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} = \ln 2. :\sum_{n=1}^\infty \frac{1}{n(4n^2-1)} = 2\ln 2 -1. :\sum_{n=1}^\infty \frac{(-1)^n}{n(4n^2-1)} = \ln 2 -1. :\sum_{n=1}^\infty \frac{(-1)^n}{n(9n^2-1)} = 2\ln 2 -\frac{3}{2}. :\sum_{n=1}^\infty \frac{1}{4n^2-2n} = \ln 2. :\sum_{n=1}^\infty \frac{2(-1)^{n+1}(2n-1)+1}{8n^2-4n} = \ln 2. :\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+1} = \frac{\ln 2}{3}+\frac{\pi}{3\sqrt{3}}. :\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+2} = -\frac{\ln 2}{3}+\frac{\pi}{3\sqrt{3}}. :\sum_{n=0}^\infty \frac{(-1)^{n}}{(3n+1)(3n+2)} = \frac{2\ln 2}{3}. :\sum_{n=1}^\infty \frac{1}{\sum_{k=1}^n k^2} = 18 - 24 \ln 2 using \lim_{N\rightarrow \infty} \sum_{n=N}^{2N} \frac{1}{n} = \ln 2 :\sum_{n=1}^\infty \frac{1}{4n^2-3n} = \ln 2 + \frac{\pi}{6} (sums of the reciprocals of decagonal numbers)

Involving the Riemann Zeta function

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

(γ is the Euler–Mascheroni constant and ζ Riemann's zeta function.)

BBP-type representations

:\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} . (See more about Bailey–Borwein–Plouffe (BBP)-type representations.)

Applying the three general series for natural logarithm to 2 directly gives: :\ln 2 = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n}. :\ln 2 = \sum_{n = 1}^\infty \frac{1}{2^{n}n}. :\ln 2 = \frac{2}{3} \sum_{k = 0}^\infty \frac{1}{9^{k}(2k+1)}.

Applying them to \textstyle 2 = \frac{3}{2} \cdot \frac{4}{3} gives: :\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 = \sum_{n = 1}^\infty \frac{1}{3^n n} + \sum_{n = 1}^\infty \frac{1}{4^n n} . :\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)} .

Applying them to \textstyle 2 = (\sqrt{2})^2 gives: :\ln 2 = 2 \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(\sqrt{2} + 1)^n n} . :\ln 2 = 2 \sum_{n = 1}^\infty \frac{1}{(2 + \sqrt{2})^n n} . :\ln 2 = \frac{4}{3 + 2 \sqrt{2}} \sum_{k = 0}^\infty \frac{1}{(17 + 12 \sqrt{2})^{k}(2k+1)} .

Applying them to \textstyle 2 = { \left( \frac{16}{15} \right) }^{7} \cdot { \left( \frac{81}{80} \right) }^{3} \cdot { \left( \frac{25}{24} \right) }^{5} gives: :\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 \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 = \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)} .

Representation as integrals

The natural logarithm of 2 occurs frequently as the result of integration. Some explicit formulas for it include:

:\int_0^1 \frac{dx}{1+x} = \int_1^2 \frac{dx}{x} = \ln 2

:\int_0^\infty e^{-x}\frac{1-e^{-x}}{x} , dx= \ln 2

:\int_0^\infty 2^{-x} dx= \frac{1}{\ln 2}

:\int_0^\frac{\pi}{3} \tan x , dx=2\int_0^\frac{\pi}{4} \tan x , dx = \ln 2

:-\frac{1}{\pi i}\int_{0}^{\infty} \frac{\ln x \ln\ln x}{(x+1)^2} , dx= \ln 2

Other representations

The Pierce expansion is : \ln 2 = 1 -\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 12} -\cdots. The Engel expansion is : \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. The cotangent expansion is : \ln 2 = \cot({\arccot(0) -\arccot(1) + \arccot(5) - \arccot(55) + \arccot(14187) -\cdots}). The simple continued fraction expansion is : \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], which yields rational approximations, the first few of which are 0, 1, 2/3, 7/10, 9/13 and 61/88.

This generalized continued fraction: : \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] , :also expressible as : \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}}}} The following continued fraction representation (J.L.Lagrange) gives (asymptotically) 1.53 new correct decimal places per cycle: Mathematical Analysis Functions, Limits, Series, Continued Fractions. Edited by L.A. Lyusternik and A. R. Yanpol’Skii Translated by D. E. Brown, Translation edited by E. Spence, 1965, p.273. Pergamon Press, Oxford, London, Edinburgh, New York :\ln 2=\frac{\frac{1}{2}}{L_{0}}, L_{k}=2k+1+\frac{k+1}{2+\frac{k+1}{L_{k+1}}} or :\ln 2=\frac{2}{G_{0}}, G_{k}=6k+3-\frac{(k+1)^2}{G_{k+1}}

Bootstrapping other logarithms

Given a value of ln 2, a scheme of computing the logarithms of other integers is to tabulate the logarithms of the prime numbers and in the next layer the logarithms of the composite numbers c based on their factorizations :c=2^i3^j5^k7^l\cdots\rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots This employs

PrimeApproximate natural logarithmOEIS
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97

In a third layer, the logarithms of rational numbers are computed with , and logarithms of roots via .

The logarithm of 2 is useful in the sense that the powers of 2 are rather densely distributed; finding powers 2 close to powers b of other numbers b is comparatively easy, and series representations of ln(b) are found by coupling 2 to b with logarithmic conversions.

Example

If with some small d, then and therefore : s\ln p -t\ln q = \ln\left(1+\frac{d}{q^t}\right) = \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m}\left(\frac{d}{q^t}\right)^m = \sum_{n=0}^\infty \frac{2}{2n+1} {\left(\frac{d}{2 q^t + d}\right)}^{2n+1} . Selecting represents ln p by ln 2 and a series of a parameter that one wishes to keep small for quick convergence. Taking , for example, generates :2\ln 3 = 3\ln 2 -\sum_{k\ge 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} . This is actually the third line in the following table of expansions of this type: : 5\ln 3 = 8\ln 2 -\sum_{k\ge 1}\frac{1}{k}(\frac{13}{256})^k. : 6\ln 5 = 14\ln 2 -\sum_{k\ge 1}\frac{1}{k}(\frac{759}{16384})^k. : \ln 7 = 3\ln 2 -\sum_{k\ge 1}\frac{1}{8^{k}k}. : 5\ln 7 = 14\ln 2 -\sum_{k\ge 1}\frac{(-1)^k}{k}(\frac{423}{16384})^k. : \ln 17 = 4\ln 2 -\sum_{k\ge 1}\frac{(-1)^k}{16^{k}k}.

sptq
1312
1322
2332
5382
123192
1522
3572
1722
1732
57142
11132
21172
1111382
11332
11342
313112
713262
1013372
11742
11942
419172
12342
12352
22392
12942
12952
729342
13152
13752
437212
537262
14152
241112
341162
14352
243112
543272
743382

Starting from the natural logarithm of one might use these parameters:

sptq
102310
2131010
35210
105710
67510
1371110
111110
113110
813910
9131010
117110
417510
9171110
319410
419510
719910
223310
323410
229310
231310

Known digits

This is a table of recent records in calculating digits of ln 2. As of December 2018, it has been calculated to more digits than any other natural logarithm of a natural number, except that of 1.

DateNameNumber of digits
January 7, 2009A.Yee & R.Chan15,500,000,000
February 4, 2009A.Yee & R.Chan31,026,000,000
February 21, 2011Alexander Yee50,000,000,050
May 14, 2011Shigeru Kondo100,000,000,000
February 28, 2014Shigeru Kondo200,000,000,050
July 12, 2015Ron Watkins250,000,000,000
January 30, 2016Ron Watkins350,000,000,000
April 18, 2016Ron Watkins500,000,000,000
December 10, 2018Michael Kwok600,000,000,000
April 26, 2019Jacob Riffee1,000,000,000,000
August 19, 2020Seungmin Kim1,200,000,000,100
September 9, 2021William Echols1,500,000,000,000
February 12, 2024Jordan Ranous3,000,000,000,000
November 15, 2025Mamdouh Barakat3,100,000,000,000

References

  • {{cite journal |doi-access=free
  • {{cite journal |doi-access=free
  • {{cite journal |doi-access=free
  • {{cite journal |access-date=2010-04-29 |archive-url=https://web.archive.org/web/20110606014342/http://www.emis.de/journals/JIS/VOL6/Chamberland/chamberland60.pdf |archive-date=2011-06-06 |url-status=dead
  • {{cite journal | author2-link=Jesús Guillera
  • {{cite journal |doi-access=free

References

  1. (2004). "On the Ramanujan AGM Fraction, I: The Real-Parameter Case". Exper. Math..
  2. A. N. Khovanski, The applications of continued fractions and their Generalisation to problemes in approximation theory,1963, Noordhoff, Groningen, The Netherlands
  3. "y-cruncher". numberworld.org.
  4. "Natural log of 2". numberworld.org.
  5. "Records set by y-cruncher".
  6. (19 August 2020). "Natural logarithm of 2 (Log(2)) world record by Seungmin Kim".
  7. "William Echols".
  8. "Records set by y-cruncher".
  9. "Records set by y-cruncher".
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