as that of the logarithm of 12456; and so on for the other inferior numbers. (528.) The series [3] may be expressed in a different form. Let l'(n+1)—l'n=D, then the formula becomes D=2M{ 1 + +...} when n is 100 or upwards, the second term is = '00000004, and the first will give the value of D correct to the 7th decimal place, and (100+1)='100+ D, or '1012-00432134 7' This logarithm, however, is not quite correct in the 8th decimal place, for if the second term of the series be added, then D00432138, and omitting the last figure 7' 1012-0043214 When n is 1200 or greater, 2 n may be taken instead of 2n+1, and then M n Thus, for n = 10000, D=·000043427 and hence 1000110000+ D=4.000043427 So l'10002ľ 10001 + D= 4·000086854 ... In a similar manner, the logarithms of 10003, 10004, 10005, may be calculated. If the logarithms were required to be correct only to 7 decimal places, then the value of D for n = 10000 is 0000434, and D would have the same value for several successive numbers, so that by merely adding D to 10001, the sum is '10002; and adding D to the latter, the sum is l 10003, and so on. When the number is still higher, as n = 99840, then the value of Dis·000004349, and for the value of n = 99860, D has exactly the same value, so that for all numbers between these two, D has this constant value, even to the 9th decimal place; and therefore when '99840 is known, the logarithm of each of the successive numbers up to 99860 is found, by merely adding this value of D to the logarithm of the preceding number. If the logarithms be limited to a smaller number of places, the difference will be constant over a greater interval. Thus, when the logarithms are extended to only seven places, the difference D is – 44 for a. the successive numbers between 97600 and 99800. When the logarithms of prime numbers are computed, those of composite numbers are found, by adding the logarithms of their component factors (501.) EXAMPLE. Given the logarithms of 2 and 7, find that of 14. So the log. of 41′22=21′2; 1'8 = 1'23 = 31′2. EXERCISES. 1. Given the logarithms of 2=0·3010300 and of 3 0-4771213, find those of 4, 6, 8, 9, and 12. Ans. 0-6020600, 0.7781513, 0·9030900, 0·9542425, 1·0791812. 2. Calculate by the formula in (528) the common logarithm of 11. Ans. 1.0413927. SERIES. THE DIFFERENTIAL METHOD. (529.) The use of the differential method is to find the successive differences of the terms of a series, or any term of the series, or the sum of a finite number of its terms. The successive differences are divided into different orders. The differences of the first order are the differences of the terms of the series; those of the second order are the differences of the terms of the first order; those of the third are the differences of the terms of the second order; and so The differences in every case are found by subtracting a term from the succeeding term.' on. Thus, if the series be 1, 8, 27, 64, 125, 216, ... then 7, 19, 37, 61, 91, is the 1st order of differences and ... (530.) Prob. I. To find the first term of any order of differences. Let the series be a, b, c, d, e, then the different orders of differences are, 1st order b-a, cd, dc,... e-2b+a, d-2c+b, e-2d+c,... The coefficients of the different orders of differences are evidently, for the and so on; being evidently the terms of the successive powers of 1-1, or the coefficients of the terms of the powers of a binomial x y; and hence the coefficients of the nth order are n 1 — n + " (n − 1) __ n (n − 1) (n — 2) 2 2x3 The last term is +1 when n is an even number, and 1 when it is odd. It is also evident that the last term is the coefficient of a in the first of the nth order of differences; the last term but one is the coefficient of b; the last but two the coefficient of c, and so on; and these coefficients are the same in reverse order as in the direct order (486); therefore, if dn represent the first difference of the nth order, and di, da, da, ... those of the first, second, third, ... orders, then, when n is an even number, (531.) It is evident from the coefficients, that when n1, the value of d has only two terms, for then n-1=0; when n=2, this value has only three terms, for then n2=0; and so on. EXAMPLES. 1. Find the first term of the second order of differences of the series 12, 22, 32, 42, ... or 1, 4, 9, 16, 25, ... Here n = 2; hence take three terms of the first value of dn, and take a = 1, 64, and c = :9; therefore c-1-2x4+ = 1-8+9=2 The accuracy of the result may be proved by finding the differences by subtraction; thus, 1, 4, 9, 16, 25, ... and hence 2 is the first difference of the second order. 2. Find the first term of the fourth order of differences of the series 13, 23, 33, 43, 53, or 1, 8, 27, 64, 125, . ... ... Here n = 4; hence take five terms of the value of dn, : 1, b = 8, c = 27, d = 64, e = 125, and hence d4 = and a= 4x3 1-4x8+ -x27—. 2 4x3x2 4x3x2x1 x64+ X 125 2x3x4 =1—32+162 — 256 + 125=0 or the required difference is 0. EXERCISES. 1. Find the first term of the second order of differences of the series 1, 3, 6, 10, 15, 21, ... Ans. 1. 2. Find the first term of the third order of differences of the series 1, 6, 20, 50, 105, 196, ... Ans. 7. 3. Find the first term of the third order of differences of the series 1, 5, 15, 35, 70, ... Ans. 4. 4. Find the first term of the eighth order of differences of the series 1, 3, 9, 27, 81, ... (532.) Prob. II. To find any term of a series. ... Ans. 256. Let d1, d2, dз, represent as above the first terms of the different orders of differences, then by the preceding formula (530), and so on. It is evident from inspection that the coefficients of the different orders of differences in the value of any of the terms, as of e the fifth term, are the coefficients' of the terms of a binomial involved to a power, whose exponent is one less than the number denoting the place of the terms; and hence the nth term is d3+... |