271. To insert a harmonical mean between two numbers: Divide the sum of the numbers into twice their product. Thus, taking any three consecutive terms of the harmonical progression 20, 120 60 40 30 24 we find that the middle term is obtained as quotient when twice the product of the other two is divided by their sum: 60=1 2 X 120 X 40 ; 40=2×60×30 ; &c. 60+30 Putting y for the harmonical mean between x and z, we have x: z:: x-y: y−z; xy-xz=xz―yz; xy+yz=xz+xz ; 272. By taking the reciprocals* of its terms, we convert a harmonical into an arithmetical progression, or an arithmetical into a harmonical progression- -as the case may be. Thus, taking the reciprocals of the terms of the harmonical progression 40 30 24 20, we obtain the arithmetical progression 120 60 On the other hand, taking the reciprocals of the terms of the arithmetical progression— Let x, y, and z be three consecutive terms of a harmonical progression, and let their reciprocals be x', y', and z', respectively: to prove that x', y', and z' are three consecutive * Any two numbers whose product is unity are said to be the "reciprocals" of one another. Thus, and are reciprocals of one another, being = 1; 7 and are also reciprocals of one another, 7X being = 1. So that, to find the reciprocal of a number, we divide unity by the number. The reciprocal of 51, for instance, or of, is (1) terms of an arithmetical progression; in other words Here we have (§ 271) xy+yz (§ 255), that '+'=2y. Next, let x', y', and 2' be three consecutive terms of an arithmetical progression, and let their reciprocals be x, y, and z, respectively: to prove that x, y, and z are three consecutive terms of a harmonical progression; in other 222 words (§ 271), that y=x+z Here we have (§ 255) x' z' zy' = x'+z'=zy'; (dividing by x'y'2') x'y'z+x'y'z ̄x′y'z' 273. To insert two or more harmonical means between a given pair of extremes: Between the reciprocals of the extremes insert as many arithmetical means as there are harmonical means to be determined; and the reciprocals of the arithmetical means will be the harmonical means required. This follows from § 272. As an illustration, let it be required to insert three harmonical means between 36 and 4. The reciprocals of the given numbers are% and Between these two fractions we insert (§ 254) three arithmetical means— the reciprocals of which are the harmonical means required As a second illustration, let it be required to set down four additional terms-two on the left, and two on the right-of the harmonical progression Taking the reciprocals of the given numbers, we obtain the arithmetical progression The common difference being ō, we extend this progression as follows: (left-hand side) -10=; -10=10; (right-hand side) +10=120; 130+10=180=1. The arithmetical progression, in its extended form, being thus found to be 110 % % 30 24 20 130 15, we obtain the harmonical progression, in its extended form, by taking the reciprocals of these fractions: 120 60 40 30 24 20 177 15. NOTE. A harmonical progression is said to be so called from the circumstance that musical strings of equal thickness and tension must, in order to produce harmony when sounded simultaneously, vary in length as the numbers These numbers, it will be seen, form a harmonical progression, and are the reciprocals of the series of "natural" numbers I 2 3 4 5 &c. 274. If, between any two numbers, regarded as extremes, there were inserted (a) a harmonical, (b) a geometrical, and (c) an arithmetical mean, those means taken in the order in which they have been mentioned-would form three consecutive terms of a geometrical progression. Between 4 and 9, for instance, the harmonical mean is 2 X 4X9 4+9 X=) 51; the geometrical mean, (√4×9 -√36=) 6; and the arithmetical mean, (++9=) 64;— 2 and 513, 6, and 61⁄2 are in continued proportion: 51:6:6:61. If the extremes were x and y, the harmonical mean would 2xy be x+y; the geometrical mean, the geometrical mean, √xy; and the arith x+y metical mean, and it is obvious that 2 2xy √xy, and x+y would form three consecutive terms of a geometrical 2 F E NOTE. The diagram in the margin exhibits at a glance the three MEANS-" arithmetical," "geometrical," and "harmonical"-between the straight lines AB and BD (of which AB is the longer). Upon the straight line AD, the sum of AB and BD, a semicircle is described; from the point B a perpendicular is erected to AD; from the point E, where this perpendicular meets the semicircumference, the radius EC is drawn-C being the point of bisection of AD; and on this radius is let fall, from B, the perpendicular BF. means-taken in the order in which they have been mentioned above-are EC, EB, and EF, respectively. As to the reason of this: those acquainted with Geometry cannot fail to see that EC is half the sum of AB and BD; that EB is the square root of the product of AB by BD; and that, ECX EF being=EB', the lines EF, EB, and EC form three consecutive terms of a geometrical progression A C EF: EB :: EB : EC.* * See §§ 255, 266, and 274. B D Then, the three LOGARITHMS. 275. By means of what is termed a "Table of LOGARITHMS we are able to substitute addition for multiplication, subtraction for division, multiplication for involution, and division for evolution. We can set about the construction of such a table by simply writing the arithmetical progression 1 2 3 4 5 &c. above any geometrical progression whose first term is unity. Thus, selecting 7 for common ratio, we have— 5 7 8 ΟΙ 2 3 4 6 &c. 1 7 49 343 2,401 16,807 117,649 823,543 5,764,801 &c. These two sets of numbers afford us a ready means of, for instance, (a) multiplying 16,807 by 49; (b) dividing 823,543 by 2,401; (c) raising 2,401 to the second power; and (d) extracting the cube root of 117,649: (a) Above 16,807 and 49 stand 5 and 2, respectively; the sum of 5 and 2 is 7; and below 7 stands 823,543the product of 16,807 by 49. (b) Above 823,543 and 2,401 stand 7 and 4, respectively; the difference between 7 and 4 is 3; and below 3 stands 343-the quotient resulting from the division of 823,543 by 2,401. (c) Above 2,401 stands 4; the double of 4 is 8; and below 8 stands 5,764,801-the square of 2,401. (d) Above 117,649 stands 6; one-third of 6 is 2 ; and below 2 stands 49-the cube root of 117,649. In order to understand the reason of this, we have merely to reflect that the terms of the geometrical progression are all powers of 7, and that the terms of the Throughout this chapter it is presumed that the pupil has such a table at hand. |