b = 13.561 (ar. co.) 8.86771 c = 14.091 (ar. co.) 8.85106 (ar. co.) 8.85106 (ar.co.) 8.86771 A 26° 30′, ≥ B = 30° 39', ¦ C = 32° 51' A = 53° 0′, B=61° 18′, C 65° 42'. In the same way equations (152-154) would furnish a third method, (155-157) a fourth method, and (158) a fifth method. Given the three sides. 2. Given the three sides of a triangle equal to 17.856, 13.349, and 11.111; to solve the triangle. in which is an infinitesimal, into a series arranged according to powers of x. x i2 1.2 3 1.2.3 +(-1) (-2)+ &c. (160) But is infinite, and gives, therefore, i This quantity e is one of frequent occurrence in analysis, and is celebrated on account of its having been adopted by Napier as the base of his system of logarithms, which were called by him hyperbolic logarithms, but are known as the Naperian logarithms. The value of e is easily computed, from the consideration that it is the sum of the series (163), the first term of which is unity, and each succeeding term is obtained by dividing the preceding term by the number of the place of this preceding term. Logarithm of a number which is nearly equal to unity. The sixth place is neglected, in the sum of the decimals, as being inaccurate. 96. Corollary. The ath power of e is by (164 and 162) 98. Corollary. The logarithm of (166) is log. (1+i) i log. e, (166) which gives, by reversing the sign of i, = log. (1) i log. e. (167) (168) 99. Problem. To develop log. (1-x) into a series of terms arranged according to the powers of x. Solution. Let the series be denoted, as follows, log. (1-x)=A+A, x+A2 x2+&c... +4, x"+&c. (169) so that the number below the letter denotes the power of x of which the letter is the coefficient. First. To find the value of A; let which reduces (169) to x=0 log. 1A 0. (170) |