Of course these results compare unfavourably with the well-known formula a a+10 v2dx= }} [v2+Va+10 + 4 (Va+i+Ya+3+...+ Va+9) +2(Va+2 + 2 (Var2+...+ Va+s)], which is 4P+20P,... in defect, but they have the advantage of involving fewer terms. 3. Taking the summation-formula in its general form, and adding the two sums for k and (1−k), we have h (8) 1/4 [(Va+kk+Vp-kh) + (Va+h+kh + Vp-h-kh) +••• (b+kh-h+ Va-kh+h)] = 2 Many new approximate summation-formulæ may be built up from this equation by keeping hor k constant, or by varying both, but as fractional values of the variable will usually be involved the results are not of much interest. Many of the results already given may, however, be obtained very readily. For example if h=6 and k=0, 2, 3, formula (2) is obtained; while h = 6, k=0, 4, and will give Hardy's formula. Again, Weddle's formula is given by taking the set of corresponding values k = 0, 1, 3 = 3}. h=1, 2, The numerical values of the coefficients required in the calculations can be obtained from equations derived from the fact that pr (4) is the coefficient of h' in hekh 2r-1 [ør (§) +pr (1)] = pr (1) = A, and therefore 4, (4) = − A, | 2r ̈1 [ør (4) +ør (3)] = pr() and therefore, if r is even, 2r-' [ør (¿) +ør (3)] = Þr (¿) and therefore pr (¿) = − (1 − 1) Þr (3) When is even the value of p, (3) is given by the equation 6 1 64 If k=4, p。(k) = -3.44, and consequently all the formulæ obtained corresponding to (1) to (7a), where k was zero, will have errors of less than 4th of the corresponding errors. The formula corresponding to (1), Weddle's formula, is obtained by taking h=1, 2, and 3 in (8), and the result is 0 da 23 ô [5 (v4 + Vş + Vq + ... + vz + vz + vz +...+ V28) − 4 (v2 + v§ +.....+ v¥) + (v4 + v + v¥ +vy)], A 31 1 6 with an error of – 33.44 (v,“ – v ̧‹3).... If we write this as -24 6! v_dx = } [5 (v, + v ̧ + v ̧ + ... + v12) − 4 (v, + V ̧ + V10 + ... + V„) 5 64 (v2, (5) — v (3)) .... It will be seen that this error is almost equal to that in Weddle's formula when the integral is expressed in terms of 13 terms, namely, v ̧dx= § [(v + v ̧ + v2+...+ v1⁄2) + 5 (v ̧ + v ̧ +v10+...+ v„) A rather remarkable result is obtained by taking h=10 and k=,, and, and slightly modifying the coefficients in the resulting formula. v_dx = } [13 (v,+v ̧+ v ̧+ v ̧) − 20 (v,+v,)], ON SOME METHODS OF CALCULATING LOGARITHMS WITHOUT THE USE OF SERIES. By J. W. L. GLAISHER. Introduction, §1. $1. THE present paper consists of four parts. Part I. relates to a method of calculating logarithms by means of a table of successive square roots of 10. Such a table was actually constructed by Briggs, and the method is one which he might have used to advantage. The method resolves any given number into factors contained in the table and as, in its fundamental form, this is effected by means of a series of multiplications, I have called it the method of multiplications.' Part II. relates to the method of geometrical means. This method does not require the previous calculation of any table, the logarithm being obtained entirely from first principles. Nothing is assumed except that the logarithm of the geometrical mean of two numbers is the arithmetical mean of their logarithms. Part III. relates mainly to the history of the method of geometrical means. I have not been able to ascertain with whom the method originated, but I have given some account of its history since 1670. It is possible that it is one of the methods which Napier had in his mind, and which he intended to describe. This Part also contains an account of two methods which were similar in character to the method of multiplications. Part IV., which has been added since the first three parts were written, contains the table to which Napier's first method of constructing decimal logarithms would have led him if the operations which he proposed could have been performed. This table, it is shown, can be applied in the same manner as Briggs's table of the successive square roots of 10 to the calculation of logarithms by the method of multiplications described in Part I. PART I. THE METHOD OF MULTIPLICATIONS, §§ 2-19. Briggs's table of successive square roots of 10, § 2. § 2. One of the methods by which Briggs calculated his fundamental logarithms of prime numbers required the repeated extraction of the square root of 10 a great number of times. The values of these successive roots were given by him in Chapter vi. (p. 10) of his Arithmetica Logarithmica (1624). They form a table which contains the 2"-th root of 10 from n=1 to n = 54 to a varying number of decimal places ranging from 26 to 32. The method also required the formation of a similar series of repeated square roots of the number whose logarithm was sought, carried to as many significant figures as were required in the logarithm. I do not know whether it has been remarked that this second series of extractions of square roots could have been avoided, and that the logarithm of any number could have been derived from Briggs's table of the successive square roots of 10 by simple multiplications or divisions. The proposed process consists in resolving the number whose logarithm is sought into factors all of which are contained in Briggs's table. Notation, §3. 1 2n §3. For convenience [n] will be used to denote and (») will be used to denote 10"]. Thus (1), (2), (3), (4), ... denote 10, 10, 10, 10, Also (n)"1= = 10-["], so .... 1 (n) that (1)-1, (2)-1, (3)-1, ... denote 10, 10, 10, Briggs's table gives the numerical values of (n). quantity [n] is the logarithm of (n) to base 10.* Explanation of the proposed method, §§ 4—8. .... The §4. For the purpose of explaining the proposed method suppose that only six places of decimals are required in the *Briggs regarded logarithms as integers, so that his logarithm of (n) containing figures (besides the characteristic) is the integral portion of 10" [n]. logarithm. Briggs's table of the successive square roots of 10 reduced to seven decimals becomes: Taking as an example the calculation of the logarithm of 5, the process consists in first selecting from the table the largest number less than 5, which is (1). This number is to be multiplied by the largest number in the table for which the product is less than 5. This number is (3) and the product is 4.216 965. This product is to be multiplied by the largest number in the table for which the product will still be less than 5; this number is (4) and the resulting product is 4.869 675. The largest multiplier in the table which will still keep the resulting product less than 5 is (7), this product being 4.958 068. The next multiplier is (9), producing 4.980 416; and proceeding in this manner we find that 5 = (1) × (3) × (4) × (7) × (9) × (10) × (11) × (13) × (14) × (15) × (16) × (17) × (19) × (20) × (23) × (25), the corresponding logarithm being obtained by adding [1], [3], [4], ... that is by adding the values of 2-1, 2-3, 2−4, .... §5. The process is exhibited in the following table, which shows the intermediate products and their logarithms. |