Natural logarithm of 2

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

$\ln 2 \approx 0.69314718055...$

as shown in the first line of the table below. The logarithm in other bases is obtained with the formula

$\log_b 2 = \frac{\ln 2}{\ln b}.$

The common logarithm in particular is ( A007524)

$\log_{10} 2 \approx 0.301029995663981195.$

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

$\log_2 10=1/\log_{10} 2 \approx 3.321928095$ ( A020862).

number	approximate natural logarithm	OEIS
2	0.693147180559945309417232121458	A002162
3	1.09861228866810969139524523692	A002391
4	1.38629436111989061883446424292	A016627
5	1.60943791243410037460075933323	A016628
6	1.79175946922805500081247735838	A016629
7	1.94591014905531330510535274344	A016630
8	2.07944154167983592825169636437	A016631
9	2.19722457733621938279049047384	A016632
10	2.30258509299404568401799145468	A002392

Series representations

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} = \ln 2.$

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

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

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

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

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

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

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

$\ln 2 = \sum_{k\ge 1} \frac{1}{k2^k}.$

$\ln 2 = \sum_{k\ge 1}\left(\frac{1}{3^k}+\frac{1}{4^k}\right)\frac{1}{k}.$

$\ln 2 = \frac{2}{3}+\sum_{k\ge 1}\left(\frac{1}{2k}+\frac{1}{4k+1}+\frac{1}{8k+4}+\frac{1}{16k+12}\right)\frac{1}{16^k}.$

$\ln 2 = \frac{2}{3} \sum_{k\ge 0} \frac{1}{(2k+1)9^k}.$

($\gamma$ is the Euler–Mascheroni constant and $\zeta$ Riemann's zeta function).

Some Bailey–Borwein–Plouffe (BBP)-type representations fall also into this category.

Representation as integrals

$\int_0^1 \frac{dx}{1+x} = \ln 2,\text{ or, equivalently, }\int_1^2 \frac{dx}{x} = \ln 2.$

$\int_1^\infty \frac{dx}{(1+x^2)(1+x)^2} = \frac{1}{4}(1-\ln 2).$

$\int_0^\infty \frac{dx}{1+e^{nx}} = \frac{1}{n}\ln 2; \int_0^\infty \frac{dx}{3+e^{nx}} = \frac{2}{3n}\ln 2.$

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

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

$\int_0^1 \ln\frac{x^2-1}{x\ln x}dx=-1+\ln 2+\gamma.$

$\int_0^{\pi/3} \tan x \, dx=2\int_0^{\pi/4} \tan x \, dx=\ln 2.$

$\int_{-\pi/4}^{\pi/4} \ln(\sin x+\cos x)\,dx=-\frac{\pi}{4}\ln 2.$

$\int_0^1 x^2\ln(1+x)\,dx=\frac{2}{3}\ln 2-\frac{5}{18}.$

$\int_0^1 x\ln(1+x)\ln(1-x)\,dx=\frac{1}{4}-\ln 2.$

$\int_0^1 x^3\ln(1+x)\ln(1-x)\,dx=\frac{13}{96}-\frac{2}{3}\ln 2.$

$\int_0^1 \frac{\ln x}{(1+x)^2}\,dx = -\ln 2.$

$\int_0^1 \frac{\ln(1+x)-x}{x^2}\,dx=1-2\ln2.$

$\int_0^1 \frac{dx}{x(1-\ln x)(1-2\ln x)} = \ln 2.$

$\int_1^\infty \frac{\ln\ln x}{x^3}\,dx = -\frac{1}{2}(\gamma+\ln 2).$

($\gamma$ is the Euler–Mascheroni constant).

Other representations

The Pierce expansion is  A091846

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

The Engel expansion is  A059180

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

The cotangent expansion is  A081785

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

As an infinite sum of fractions:

$\log 2 = \frac{1}{1} -\frac{1}{2} +\frac{1}{3} -\frac{1}{4} +\frac{1}{5} -\cdots.$

This generalized continued fraction:

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

also expressible as

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

Bootstrapping other logarithms

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

$c=2^i3^j5^k7^\ell\cdots\rightarrow \ln c=i\ln 2+j\ln 3+k\ln 5+\ell\ln 7+\cdots$

Apart from the logarithms of 2, 3, 5 and 7 shown above, this employs

prime	approximate natural logarithm	OEIS
11	2.39789527279837054406194357797	A016634
13	2.56494935746153673605348744157	A016636
17	2.83321334405621608024953461787	A016640
19	2.94443897916644046000902743189	A016642
23	3.13549421592914969080675283181	A016646
29	3.36729582998647402718327203236	A016652
31	3.43398720448514624592916432454	A016654
37	3.61091791264422444436809567103	A016660
41	3.71357206670430780386676337304	A016664
43	3.76120011569356242347284251335	A016666
47	3.85014760171005858682095066977	A016670
53	3.97029191355212183414446913903	A016676
59	4.07753744390571945061605037372	A016682
61	4.11087386417331124875138910343	A016684
67	4.20469261939096605967007199636	A016690
71	4.26267987704131542132945453251	A016694
73	4.29045944114839112909210885744	A016696
79	4.36944785246702149417294554148	A016702
83	4.41884060779659792347547222329	A016706
89	4.48863636973213983831781554067	A016712
97	4.57471097850338282211672162170	A016720

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 $\ln \sqrt[n]{c} = \frac{1}{n}\ln c$.

The logarithm of 2 is useful in the sense that the powers of 2 are rather densely distributed; finding powers $2^i$ 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

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

$s\ln p-t\ln q = \ln\left(1+\frac{d}{q^t}\right) = \sum_{m=1}^\infty (-1)^{m+1}\frac{(d/q^t)^m}{m}.$

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 $3^2=2^3+1$, for example, generates

$2\ln 3 = 3\ln 2-\sum_{k\ge 1}\frac{(-1)^k}{k8^k}.$

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...

References

• 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. Nat. Acac. Sci. U. S. A. 26: 205–212. MR 0001523. 
• Sweeney, Dura W. (1963). "On the computation of Euler's constant". Mathematics of Computation 17. MR 0160308. 
• Chamberland, Marc (2003). "Binary BBP-formulae for logarithms and generalized Gaussian–Mersenne primes". Journal of Integer Sequences 6: 03.3.7. MR 2046407. 
• Gourévitch, Boris; Guillera Goyanes, Jesus (2007). "Construction of binomial sums for π and polylogarithmic constants inspired by BBP formulas". 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

• Weisstein, Eric W., "Natural logarithm of 2", MathWorld.
• table of natural logarithms, PlanetMath.org.

• Gourdon, Xavier; Sebah, Pascal. "The logarithm constant:log 2".