Natural logarithm of 2 – Wikipedia

The decimal value of the natural logarithm of 2 (sequence A002162 in the OEIS)
is approximately

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

The logarithm of 2 in other bases is obtained with the formula

${displaystyle log _{b}2={frac {ln 2}{ln b}}.}$

The common logarithm in particular is (OEIS: A007524)

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

The inverse of this number is the binary logarithm of 10:

${displaystyle log _{2}10={frac {1}{log _{10}2}}approx 3.321,928,095}$

(OEIS: A020862).

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.

Series representations[edit]

Rising alternate factorial[edit]

${displaystyle ln 2=sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n}}=1-{frac {1}{2}}+{frac {1}{3}}-{frac {1}{4}}+{frac {1}{5}}-{frac {1}{6}}+cdots .}$

This is the well-known “alternating harmonic series”.

${displaystyle ln 2={frac {1}{2}}+{frac {1}{2}}sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n(n+1)}}.}$

${displaystyle ln 2={frac {5}{8}}+{frac {1}{2}}sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n(n+1)(n+2)}}.}$

${displaystyle ln 2={frac {2}{3}}+{frac {3}{4}}sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}.}$

${displaystyle ln 2={frac {131}{192}}+{frac {3}{2}}sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}.}$

${displaystyle ln 2={frac {661}{960}}+{frac {15}{4}}sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}.}$

${displaystyle ln 2={frac {2}{3}}.(1+{frac {2}{4^{3}-4}}+{frac {2}{8^{3}-8}}+{frac {2}{12^{3}-12}}+………).}$

Binary rising constant factorial[edit]

${displaystyle ln 2=sum _{n=1}^{infty }{frac {1}{2^{n}n}}.}$

${displaystyle ln 2=1-sum _{n=1}^{infty }{frac {1}{2^{n}n(n+1)}}.}$

${displaystyle ln 2={frac {1}{2}}+2sum _{n=1}^{infty }{frac {1}{2^{n}n(n+1)(n+2)}}.}$

${displaystyle ln 2={frac {5}{6}}-6sum _{n=1}^{infty }{frac {1}{2^{n}n(n+1)(n+2)(n+3)}}.}$

${displaystyle ln 2={frac {7}{12}}+24sum _{n=1}^{infty }{frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}.}$

${displaystyle ln 2={frac {47}{60}}-120sum _{n=1}^{infty }{frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}.}$

Other series representations[edit]

${displaystyle sum _{n=0}^{infty }{frac {1}{(2n+1)(2n+2)}}=ln 2.}$

${displaystyle sum _{n=1}^{infty }{frac {1}{n(4n^{2}-1)}}=2ln 2-1.}$

${displaystyle sum _{n=1}^{infty }{frac {(-1)^{n}}{n(4n^{2}-1)}}=ln 2-1.}$

${displaystyle sum _{n=1}^{infty }{frac {(-1)^{n}}{n(9n^{2}-1)}}=2ln 2-{frac {3}{2}}.}$

${displaystyle sum _{n=1}^{infty }{frac {1}{4n^{2}-2n}}=ln 2.}$

${displaystyle sum _{n=1}^{infty }{frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=ln 2.}$

${displaystyle sum _{n=0}^{infty }{frac {(-1)^{n}}{3n+1}}={frac {ln 2}{3}}+{frac {pi }{3{sqrt {3}}}}.}$

${displaystyle sum _{n=0}^{infty }{frac {(-1)^{n}}{3n+2}}=-{frac {ln 2}{3}}+{frac {pi }{3{sqrt {3}}}}.}$

${displaystyle sum _{n=0}^{infty }{frac {(-1)^{n}}{(3n+1)(3n+2)}}={frac {2ln 2}{3}}.}$

${displaystyle sum _{n=1}^{infty }{frac {1}{sum _{k=1}^{n}k^{2}}}=18-24ln 2}$

using

${displaystyle lim _{Nrightarrow infty }sum _{n=N}^{2N}{frac {1}{n}}=ln 2}$

${displaystyle sum _{n=1}^{infty }{frac {1}{4n^{2}-3n}}=ln 2+{frac {pi }{6}}}$

(sums of the reciprocals of decagonal numbers)

Involving the Riemann Zeta function[edit]

${displaystyle sum _{n=1}^{infty }{frac {1}{n}}[zeta (2n)-1]=ln 2.}$

${displaystyle sum _{n=2}^{infty }{frac {1}{2^{n}}}[zeta (n)-1]=ln 2-{frac {1}{2}}.}$

${displaystyle sum _{n=1}^{infty }{frac {1}{2n+1}}[zeta (2n+1)-1]=1-gamma -{frac {ln 2}{2}}.}$

${displaystyle sum _{n=1}^{infty }{frac {1}{2^{2n-1}(2n+1)}}zeta (2n)=1-ln 2.}$

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

BBP-type representations[edit]

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

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

Applying the three general series for natural logarithm to 2 directly gives:

${displaystyle ln 2=sum _{n=1}^{infty }{frac {(-1)^{n-1}}{n}}.}$

${displaystyle ln 2=sum _{n=1}^{infty }{frac {1}{2^{n}n}}.}$

${displaystyle ln 2={frac {2}{3}}sum _{k=0}^{infty }{frac {1}{9^{k}(2k+1)}}.}$

Applying them to

${displaystyle textstyle 2={frac {3}{2}}cdot {frac {4}{3}}}$

gives:

${displaystyle ln 2=sum _{n=1}^{infty }{frac {(-1)^{n-1}}{2^{n}n}}+sum _{n=1}^{infty }{frac {(-1)^{n-1}}{3^{n}n}}.}$

${displaystyle ln 2=sum _{n=1}^{infty }{frac {1}{3^{n}n}}+sum _{n=1}^{infty }{frac {1}{4^{n}n}}.}$

${displaystyle ln 2={frac {2}{5}}sum _{k=0}^{infty }{frac {1}{25^{k}(2k+1)}}+{frac {2}{7}}sum _{k=0}^{infty }{frac {1}{49^{k}(2k+1)}}.}$

Applying them to

${displaystyle textstyle 2=({sqrt {2}})^{2}}$

gives:

${displaystyle ln 2=2sum _{n=1}^{infty }{frac {(-1)^{n-1}}{({sqrt {2}}+1)^{n}n}}.}$

${displaystyle ln 2=2sum _{n=1}^{infty }{frac {1}{(2+{sqrt {2}})^{n}n}}.}$

${displaystyle ln 2={frac {4}{3+2{sqrt {2}}}}sum _{k=0}^{infty }{frac {1}{(17+12{sqrt {2}})^{k}(2k+1)}}.}$

Applying them to

${displaystyle textstyle 2={left({frac {16}{15}}right)}^{7}cdot {left({frac {81}{80}}right)}^{3}cdot {left({frac {25}{24}}right)}^{5}}$

gives:

${displaystyle ln 2=7sum _{n=1}^{infty }{frac {(-1)^{n-1}}{15^{n}n}}+3sum _{n=1}^{infty }{frac {(-1)^{n-1}}{80^{n}n}}+5sum _{n=1}^{infty }{frac {(-1)^{n-1}}{24^{n}n}}.}$

${displaystyle ln 2=7sum _{n=1}^{infty }{frac {1}{16^{n}n}}+3sum _{n=1}^{infty }{frac {1}{81^{n}n}}+5sum _{n=1}^{infty }{frac {1}{25^{n}n}}.}$

${displaystyle ln 2={frac {14}{31}}sum _{k=0}^{infty }{frac {1}{961^{k}(2k+1)}}+{frac {6}{161}}sum _{k=0}^{infty }{frac {1}{25921^{k}(2k+1)}}+{frac {10}{49}}sum _{k=0}^{infty }{frac {1}{2401^{k}(2k+1)}}.}$

Representation as integrals[edit]

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

${displaystyle int _{0}^{1}{frac {dx}{1+x}}=int _{1}^{2}{frac {dx}{x}}=ln 2}$

${displaystyle int _{0}^{infty }e^{-x}{frac {1-e^{-x}}{x}},dx=ln 2}$

${displaystyle int _{0}^{infty }2^{-x}dx={frac {1}{ln 2}}}$

${displaystyle int _{0}^{frac {pi }{3}}tan x,dx=2int _{0}^{frac {pi }{4}}tan x,dx=ln 2}$

${displaystyle -{frac {1}{pi i}}int _{0}^{infty }{frac {ln xln ln x}{(x+1)^{2}}},dx=ln 2}$

Other representations[edit]

The Pierce expansion is OEIS: A091846

${displaystyle ln 2=1-{frac {1}{1cdot 3}}+{frac {1}{1cdot 3cdot 12}}-cdots .}$

The Engel expansion is OEIS: A059180

${displaystyle ln 2={frac {1}{2}}+{frac {1}{2cdot 3}}+{frac {1}{2cdot 3cdot 7}}+{frac {1}{2cdot 3cdot 7cdot 9}}+cdots .}$

The cotangent expansion is OEIS: A081785

${displaystyle ln 2=cot({operatorname {arccot}(0)-operatorname {arccot}(1)+operatorname {arccot}(5)-operatorname {arccot}(55)+operatorname {arccot}(14187)-cdots }).}$

The simple continued fraction expansion is OEIS: A016730

${displaystyle ln 2=left[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,…right]}$

,

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:

${displaystyle ln 2=left[0;1,2,3,1,5,{tfrac {2}{3}},7,{tfrac {1}{2}},9,{tfrac {2}{5}},…,2k-1,{frac {2}{k}},…right]}$

,^[1]

also expressible as

${displaystyle ln 2={cfrac {1}{1+{cfrac {1}{2+{cfrac {1}{3+{cfrac {2}{2+{cfrac {2}{5+{cfrac {3}{2+{cfrac {3}{7+{cfrac {4}{2+ddots }}}}}}}}}}}}}}}}={cfrac {2}{3-{cfrac {1^{2}}{9-{cfrac {2^{2}}{15-{cfrac {3^{2}}{21-ddots }}}}}}}}}$

Bootstrapping other logarithms[edit]

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

${displaystyle c=2^{i}3^{j}5^{k}7^{l}cdots rightarrow ln(c)=iln(2)+jln(3)+kln(5)+lln(7)+cdots }$

This employs

In a third layer, the logarithms of rational numbers r = a/b are computed with ln(r) = ln(a) − ln(b), and logarithms of roots via ln ⁿ√c = 1/n ln(c).

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

Example[edit]

If p^s = q^t + d with some small d, then p^s/q^t = 1 + d/q^t and therefore

${displaystyle sln p-tln q=ln left(1+{frac {d}{q^{t}}}right)=sum _{m=1}^{infty }(-1)^{m+1}{frac {({frac {d}{q^{t}}})^{m}}{m}}=sum _{n=0}^{infty }{frac {2}{2n+1}}{left({frac {d}{2q^{t}+d}}right)}^{2n+1}.}$

Selecting q = 2 represents ln p by ln 2 and a series of a parameter d/q^t that one wishes to keep small for quick convergence. Taking 3² = 2³ + 1, for example, generates

${displaystyle 2ln 3=3ln 2-sum _{kgeq 1}{frac {(-1)^{k}}{8^{k}k}}=3ln 2+sum _{n=0}^{infty }{frac {2}{2n+1}}{left({frac {1}{2cdot 8+1}}right)}^{2n+1}.}$

This is actually the third line in the following table of expansions of this type:

s	p	t	q	d/qt
1	3	1	2	1/2 = −0.50000000…
1	3	2	2	−1/4 = −0.25000000…
2	3	3	2	1/8 = −0.12500000…
5	3	8	2	−13/256 = −0.05078125…
12	3	19	2	7153/524288 = −0.01364326…
1	5	2	2	1/4 = −0.25000000…
3	5	7	2	−3/128 = −0.02343750…
1	7	2	2	3/4 = −0.75000000…
1	7	3	2	−1/8 = −0.12500000…
5	7	14	2	423/16384 = −0.02581787…
1	11	3	2	3/8 = −0.37500000…
2	11	7	2	−7/128 = −0.05468750…
11	11	38	2	10433763667/274877906944 = −0.03795781…
1	13	3	2	5/8 = −0.62500000…
1	13	4	2	−3/16 = −0.18750000…
3	13	11	2	149/2048 = −0.07275391…
7	13	26	2	−4360347/67108864 = −0.06497423…
10	13	37	2	419538377/137438953472 = −0.00305254…
1	17	4	2	1/16 = −0.06250000…
1	19	4	2	3/16 = −0.18750000…
4	19	17	2	−751/131072 = −0.00572968…
1	23	4	2	7/16 = −0.43750000…
1	23	5	2	−9/32 = −0.28125000…
2	23	9	2	17/512 = −0.03320312…
1	29	4	2	13/16 = −0.81250000…
1	29	5	2	−3/32 = −0.09375000…
7	29	34	2	70007125/17179869184 = −0.00407495…
1	31	5	2	−1/32 = −0.03125000…
1	37	5	2	5/32 = −0.15625000…
4	37	21	2	−222991/2097152 = −0.10633039…
5	37	26	2	2235093/67108864 = −0.03330548…
1	41	5	2	9/32 = −0.28125000…
2	41	11	2	−367/2048 = −0.17919922…
3	41	16	2	3385/65536 = −0.05165100…
1	43	5	2	11/32 = −0.34375000…
2	43	11	2	−199/2048 = −0.09716797…
5	43	27	2	12790715/134217728 = −0.09529825…
7	43	38	2	−3059295837/274877906944 = −0.01112965…

Starting from the natural logarithm of q = 10 one might use these parameters:

s	p	t	q	d/qt
10	2	3	10	3/125 = −0.02400000…
21	3	10	10	460353203/10000000000 = −0.04603532…
3	5	2	10	1/4 = −0.25000000…
10	5	7	10	−3/128 = −0.02343750…
6	7	5	10	17649/100000 = −0.17649000…
13	7	11	10	−3110989593/100000000000 = −0.03110990…
1	11	1	10	1/10 = −0.10000000…
1	13	1	10	3/10 = −0.30000000…
8	13	9	10	−184269279/1000000000 = −0.18426928…
9	13	10	10	604499373/10000000000 = −0.06044994…
1	17	1	10	7/10 = −0.70000000…
4	17	5	10	−16479/100000 = −0.16479000…
9	17	11	10	18587876497/100000000000 = −0.18587876…
3	19	4	10	−3141/10000 = −0.31410000…
4	19	5	10	30321/100000 = −0.30321000…
7	19	9	10	−106128261/1000000000 = −0.10612826…
2	23	3	10	−471/1000 = −0.47100000…
3	23	4	10	2167/10000 = −0.21670000…
2	29	3	10	−159/1000 = −0.15900000…
2	31	3	10	−39/1000 = −0.03900000…

Known digits[edit]

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^[2][3] of a natural number, except that of 1.

Date	Name	Number of digits
January 7, 2009	A.Yee & R.Chan	15,500,000,000
February 4, 2009	A.Yee & R.Chan	31,026,000,000
February 21, 2011	Alexander Yee	50,000,000,050
May 14, 2011	Shigeru Kondo	100,000,000,000
February 28, 2014	Shigeru Kondo	200,000,000,050
July 12, 2015	Ron Watkins	250,000,000,000
January 30, 2016	Ron Watkins	350,000,000,000
April 18, 2016	Ron Watkins	500,000,000,000
December 10, 2018	Michael Kwok	600,000,000,000
April 26, 2019	Jacob Riffee	1,000,000,000,000
August 19, 2020	Seungmin Kim[4][5]	1,200,000,000,100
September 9, 2021	William Echols[6][7]	1,500,000,000,000

See also[edit]

References[edit]

• Brent, Richard P. (1976). “Fast multiple-precision evaluation of elementary functions”. J. ACM. 23 (2): 242–251. doi:10.1145/321941.321944. MR 0395314. S2CID 6761843.
• 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. Bibcode:1940PNAS…26..205U. 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. Bibcode:2003JIntS…6…37C. MR 2046407. Archived from the original (PDF) on 2011-06-06. Retrieved 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.

External links[edit]