Natural logarithm of 2

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 (sequence A002162 in the OEIS) truncated at 30 decimal places is given by:

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

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

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

The common logarithm in particular is (OEIS: A007524)

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

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. It is also contained in the ring of algebraic periods.

${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots .}$This is the well-known "alternating harmonic series".

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

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

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

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

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

${\displaystyle \ln 2={\frac {2}{3}}\left(1+{\frac {2}{4^{3}-4}}+{\frac {2}{8^{3}-8}}+{\frac {2}{12^{3}-12}}+\dots \right).}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2}$using ${\displaystyle \lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-3n}}=\ln 2+{\frac {\pi }{6}}}$(sums of the reciprocals of decagonal numbers)

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

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

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

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

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

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

(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:

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

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

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

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:

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

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

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

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

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

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

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

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

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

The Pierce expansion is OEIS: A091846

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

The Engel expansion is OEIS: A059180

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

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:

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

also expressible as

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

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

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

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.

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

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

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

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

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

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

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.

• Rule of 72#Continuous compounding, in which ln 2 figures prominently
• Half-life#Formulas for half-life in exponential decay, in which ln 2 figures prominently
• Erdős–Moser equation: all solutions must come from a convergent of ln 2.

• 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.

• Weisstein, Eric W. "Natural logarithm of 2". MathWorld.
• Gourdon, Xavier; Sebah, Pascal. "The logarithm constant:log 2".