Page:Encyclopædia Britannica, Ninth Edition, v. 14.djvu/802

then by seventeen or a less number of significant figures) the logarithms were proportional to these significant figures. He then by means of a simple proportion deduced that log (1.00000 00000 00000 1) = 0.00000 00000 00000 04342 94481 90325 1804, so that, a quantity 1.00000 00000 00000 x (where x consists of not more than seventeen figures) having been obtained by repeated extraction of the square root of a given number, the logarithm of 1.00000 00000 00000 x could then be found by multiplying x by .00000 00000 00000 04342 . . …

To find the logarithm of 2, Briggs raised it to the tenth power, viz., 1024, and extracted the square root of 1.024 forty-seven times, the result being 1.00000 00000 00000 16851 60570 53949 77. Multiplying the significant figures by 4342… he obtained the logarithm of this quantity, viz., O.OOOOO 00000 00000 07318 55936 90623 9336, which multiplied by 2⁴⁷ gave 0.01029 99566 39811 95265 277444, the logarithm of 1.024, true to 17 or 18 places. Adding the characteristic 3, and dividing by 10, he found (since 2 is the tenth root of 1024) log 2 = .30102 99956 63981 195. Briggs calculated in a similar manner log 6, and thence deduced log 3.

It will be observed that in the first process the value of the modulus is in fact calculated from the formula

h/10^h − 1 = 1/log_e10,

the value of h being 1/2⁵⁴, and in the second process log₁₀ 2 is in effect calculated from the formula

$\log _{10}2=\left(2^{\frac {10}{2^{4}7}}\right)\times {\frac {1}{\log _{e}10}}\times {\frac {2^{4}7}{10}}$.

Briggs also gave methods of forming the mean proportionals or square roots by differences; and the general method of constructing logarithmic tables by means of differences is due to him.

The following calculation of log 5 is given as an example of the application of a method of mean proportionals. The process consists in taking the geometric mean of numbers above and below 5, the object being to at length arrive at 5.000000. To every geometric mean in the column of numbers there corresponds the arithmetical mean in the column of logarithms. The numbers are denoted by A, B, C, &c., in order to indicate their mode of formation.

	Numbers.		Logarithms.
A =	1	.000000	0.000000
B =	10	.000000	1.000000
C = √(AB) =	3	.162277	0.500000
D = √(BC) =	5	.623413	0.750000
E = √(CD) =	4	.216964	0.625000
F = √(DE) =	4	.869674	0.687500
G = √(DF) =	5	.232991	0.718750
H = √(FG) =	5	.048065	0.703125
I = √(FH) =	4	.958069	0.6953125
K = √(HI) =	5	.002865	0.6992187
L = √(HK) =	4	.980416	0.6972656
M = √(KL) =	4	.991627	0.6982421
N = √(KM) =	4	.997242	0.6987304
O = √(KN) =	5	.000052	0.6989745
P = √(NO) =	4	.998647	0.6988525
Q = √(OP) =	4	.999350	0.6989135
R = √(OQ) =	4	.999701	0.6989440
S = √(OR) =	4	.999876	0.6989592
T = √(OS) =	4	.999963	0.6989668
V = √(OT) =	5	.000008	0.6989707
W = √(TV) =	4	.999984	0.6989687
X = √(WV) =	4	.999997	0.6989697
Y = √(VX) =	5	.000003	0.6989702
Z = √(XY) =	5	.000000	0.6989700

Great attention was devoted to the methods of calculating logarithms during the 17th and 18th centuries. The earlier methods proposed were, like those of Briggs, purely arithmetical, and for a long time logarithms were regarded from the point of view indicated by their name, that is to say, as depending on the theory of compounded ratios. The introduction of infinite series into mathematics effected a great change in the modes of calculation and the treatment of the subject. Besides Napier and Briggs, special reference should be made to Kepler (Chilias, 1624) and Mercator (Logarithmotechnia, 1668), whose methods were arithmetical, and to Newton, Gregory, Halley, and Cotes, who employed series. A full and valuable account of these methods is given in Hutton’s “Construction of Logarithms,” which occurs in the introduction to the early editions of his Mathematical Tables, and also forms tract 21 of his Mathematical Tracts (vol. i., 1812). Many of the early works on logarithms were reprinted in the Scriptores logarithmici of Baron Maseres (6 vols. 4to, 1791-1807).

In the following account only those formulæ and methods will be referred to which would now be used in the calcula tion of logarithms.

Since

$\log _{e}\left(1+x\right)=x-{\frac {1}{2}}x^{2}+{\frac {1}{3}}^{3}-{\frac {1}{4}}x^{4}+\And c.$,

we have, by changing the sign of x,

$\log _{e}\left(1-x\right)=-x-{\frac {1}{2}}x^{2}-{\frac {1}{3}}x^{3}-{\frac {1}{4}}x^{4}-\And c.$;

whence

$\log _{e}{\frac {1+x}{1-x}}=2\left(x+{\frac {1}{3}}x^{3}-{\frac {1}{5}}x^{5}+\And c.\right)$,

and, therefore, replacing x by p - q/p + q,

$\log _{e}{\frac {p}{q}}=\left\{{\frac {p-q}{p+q}}+{\frac {1}{3}}\left({\frac {p-q}{p+q}}\right)^{3}+{\frac {1}{5}}\left({\frac {p-q}{p+q}}\right)^{5}+\And c.\right\}$,

in which the series is always convergent, so that the formula affords a method of deducing the logarithm of one number from that of another.

As particular cases we have, by putting q = 1,

$\log _{e}p=2\left\{{\frac {p-1}{p+1}}+{\frac {1}{3}}\left({\frac {p-1}{p+1}}\right)^{3}+{\frac {1}{5}}\left({\frac {p-1}{p+1}}\right)^{5}+\And c.\right\}$,

and by putting q = p + 1,

$\log _{e}\left(p+1\right)-\log _{e}p=2\left\{{\frac {1}{2p+1}}+{\frac {1}{3}}{\frac {1}{\left(2p+1\right)^{3}}}+{\frac {1}{5}}{\frac {1}{\left(2p+1\right)^{5}}}+\And c.\right\}$;

the former of these equations gives a convergent series for log_ep, and the latter a very convergent series by means of which the logarithm of any number may be deduced from the logarithm of the preceding number.

From the formula for log_ep/q we may deduce the following very convergent series for log_e2, log_e3, and log_e5, viz.:—

log_e2 = 2(7P +5Q + 3R),
log_e3 = 2(11P + 8Q + 5R),
log_e5 = 2(16P + 12Q + 7R),

where

P = 1/31 + 1/3 · 1/(31)³ + 1/5 · 1/(31)⁵ + &c.
Q = 1/49 + 1/3 · 1/(49)³ + 1/5 · 1/(49)⁵ + &c.
R = 1/161 + 1/3 · 1/(161)³ + 1/5 · 1/(161)⁵ + &c.

The following still more convenient formulæ for the calculation of log_e 2, log_e 3, &c. are given by Professor J. C. Adams in the Proceedings of the Royal Society, vol. xxvii. (1878), p. 91. If

$a=\log {\frac {10}{9}}=\log \left(1-{\frac {1}{10}}\right)$,	$b=\log {\frac {25}{24}}=\log \left(1-{\frac {4}{100}}\right)$,
$c=\log {\frac {81}{80}}=\log \left(1+{\frac {1}{80}}\right)$,	$d=\log {\frac {50}{49}}=-\log \left(1-{\frac {2}{100}}\right)$,
$e=\log {\frac {126}{125}}=\log \left(1+{\frac {8}{1000}}\right)$,

then

$\log 2=7a-2b+3c$,

$\log 3=11a-3b+5c$,

$\log 5=16a-4b+7c$,

and

$\log 7={\frac {1}{2}}\left(39a-10b+17c-d\right)$ or $=19a-4b+8c+d$,

and we have the equation of condition,

$a-2b+c=d+2c$,

By means of these formula Professor Adams has calculated the values of log_e 2, log_e 3, log_e 5, and log_e 7 to 260 places of decimals,