Since

log(e)(1 + x) = x - ½x² + (1/3)x³ - ¼x^4 + &c.,

we have, by changing the sign of x,

log(e)(1 - x) = -x - ½x² - (1/3)x³ - ¼x^4 - &c.;

whence

1 + x

log(e) ----- = 2(x + (1/3)x³ + (1/5)x^5 + &c.),

1 - x

p - q

and, therefore, replacing x by -----,

p + q

_ _

p | p - q /p - q\³ /p - q\^5 |

log(e) --- = 2 | ----- + (1/3)( ----- ) + (1/5)( ----- ) + &c. |,

q |_ p + q \p + q/ \p + q/ _|

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,

_ _

| p - 1 /p - 1\³ /p - 1\^5 |

log(e) p = 2 | ----- + (1/3)( ----- ) + (1/5)( ----- ) + &c. |,

|_ p + 1 \p + 1/ \p + 1/ _|

and by putting q = p + 1,

_ _

| 1 1 1 |

log(e)(p + 1) - log(e)(p) = 2 | ------ + (1/3)--------- + (1/5)----------- + &c. |;

|_ 2p + 1 (2p + 1)³ (2p + 1)^5 _|

the former of these equations gives a convergent series for log(e)p,

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(e)(p/q) we may deduce the following very

convergent series for log(e)2, log(e)3 and log(e)5, viz.:--

log(e)2 = 2( 7P + 5Q + 3R),

log(e)3 = 2(11P + 8Q + 5R),

log(e)5 = 2(16P + 12Q + 7R),

where

1 1 1

P = -- + (1/3) · ------ + (1/5) · ------ + &c.

31 (31)^3 (31)^5

1 1 1

Q = -- + (1/3) · ------ + (1/5) · ------ + &c.

49 (49)^3 (49)^5

1 1 1

R = --- + (1/3) · ------- + (1/5) · ------- + &c.

161 (161)^3 (161)^5

The following still more convenient formulae for the calculation of

log(e)2, log(e)3, &c. were given by J. Couch Adams in the _Proc. Roy.

Soc._, 1878, 27, p. 91. If

10 / 1 \ 25 / 4 \

a = log -- = -log ( 1 - -- ), b = log -- = -log ( 1 - --- ),

9 \ 10 / 24 \ 100 /

81 / 1 \ 50 / 2 \

c = log -- = log ( 1 + -- ), d = log -- = -log ( 1 - --- ),

80 \ 80 / 49 \ 100 /

126 / 8 \

e = log --- = log ( 1 + ---- ),

125 \ 1000 /

then

log 2 = 7a - 2b + 3c, log 3 = 11a - 3b + 5c, log 5 = 16a - 4b + 7c,

and

log 7 = ½(39a - 10b + 17c - d) or = 19a - 4b + 8c + e,

and we have the equation of condition,

a - 2b + c = d + 2e.

By means of these formulae Adams calculated the values of log(e)2,

log(e)3, log(e)5, and log(e)7 to 276 places of decimals, and deduced

the value of log(e)10 and its reciprocal M, the modulus of the

Briggian system of logarithms. The value of the modulus found by Adams

is

Mo = 0.43429 44819 03251 82765 11289

18916 60508 22943 97005 80366

65661 14453 78316 58646 49208

87077 47292 24949 33843 17483

18706 10674 47663 03733 64167

92871 58963 90656 92210 64662

81226 58521 27086 56867 03295

93370 86965 88266 88331 16360

77384 90514 28443 48666 76864

65860 85135 56148 21234 87653

43543 43573 17253 83562 21868

25

which is true certainly to 272, and probably to 273, places (_Proc.

Roy. Soc._, 1886, 42, p. 22, where also the values of the other

logarithms are given).

If the logarithms are to be Briggian all the series in the preceding

formulae must be multiplied by M, the modulus; thus,

log(10) (1 + x) = M (x - ½x² + (1/3)x³ - ¼x^4 + &c.),

and so on.

As has been stated, Abraham Sharp's table contains 61-decimal

Briggian logarithms of primes up to 1100, so that the logarithms of

all composite numbers whose greatest prime factor does not exceed this

number may be found by simple addition; and Wolfram's table gives

48-decimal hyperbolic logarithms of primes up to 10,009. By means of

these tables and of a factor table we may very readily obtain the

Briggian logarithm of a number to 61 or a less number of places or of

its hyperbolic logarithm to 48 or a less number of places in the

following manner. Suppose the hyperbolic logarithm of the prime number

43,867 required. Multiplying by 50, we have 50 × 43,867 = 2,193,350,

and on looking in Burckhardt's _Table des diviseurs_ for a number near

to this which shall have no prime factor greater than 10,009, it

appears that

2,193,349 = 23 × 47 × 2029;

thus

43,867 = (1/50)(23 × 47 × 2029 + 1),

and therefore

log(e) 43,867 = log(e) 23 + log(e) 47 + log(e) 2029 - log(e) 50

1 1 1

+ --------- - ½ ------------ + (1/3) ---------- - &c.

2,193,349 (2,193,349)² (193,349)³

Encyclopaedia Britannica, 11th Edition, "Logarithm" to "Lord Advocate" Volume 16, Slice 8 · 第 12 / 67 题