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additional terms two on the left, and two on the right—of the harmonical progression

40

30

24

20

Taking the reciprocals of the given numbers, we obtain the arithmetical progression

40

30 131
24

=

=

The common difference being, we extend this progression as follows: (left-hand side) 10-120 89; 80 130 120 (right-hand side) 20+128=120; 120+120=120=15. The arithmetical progression, in its extended form, being thus found to be

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40. 30 24 20 130 15 we obtain the harmonical progression, in its extended form, by taking the reciprocals of these fractions:

15.

120 60 40 30 24 20 17 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 together, vary in length as the numbers

&c.

I 122424 These numbers, it will be seen, form a harmonical progression, and are the reciprocals of the series of "natural" numbers

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273. If, between any two numbers, regarded as extremes, there were inserted (I.) a harmonical mean, (II.) a geometrical mean, and (III.) 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) 57; the geometrical mean, (√4×9=√36=)

4+9

=)

6; and the arithmetical mean, (4+9=) 61;—and 51, 6, and 6 are in continued proportion:

2

51: 66: 61

If the extremes were x and y, the harmonical mean would

2xy

be the geometrical mean, √xy; and the arithmetical

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2xy

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;-and it is obvious that √xy, and

2

x+y'

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would form three consecutive terms of a geometrical progression :

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LOGARITHMS.

274. 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

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above any geometrical progression whose first term is unity. Thus, selecting 7 for common ratio, we have

7

8

&c.

Ο Ι 2 3 4 6 1 7 49 343 2,401 16,807 117,649 823,543 5,764,801 &c. 16,807 These two sets of numbers afford us a ready means of, for instance, (a) multiplying 49 by 16,807; (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 49 and 16,807 stand 2 and 5, respectively; the sum of 2 and 5 is 7; and below 7 stands 823,543-the product of 49 and 16,807.

(b) Above 823,543 and 2,401 stand 7 and 4, respectively; the difference between 7 and 4 is 3; and below 3 stands 343the 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 arithmetical progression are the indices of those powers, respectively: 1=7° (§ 222); 7=71; 49=72; 343-73; 2,401=7*; &c.:

(a) 49× 16,807-72X75-77-823,543
(b) 823,543÷2,401=77÷71=73=343
(c) 2,4012 (74)=78-5,764,801

(d)

117,649=√76—72=49.

It will be observed that the number of sevens which occur, as factors, in any particular term of the geometrical progression is indicated by the corresponding term of the arithmetical progression. This explains the word LOGARITHM, which literally means "number of [common] ratios." In the case under con

sideration, the logarithm of 1 is o; of 7, 1; of 49, 2; of 343. 3; &c.-the common ratio, 7, being recognised as the BASE of the table.

We now proceed to construct a different table, by taking (say) 4 for common ratio, instead of 7:

OI 2 3 4 5 6 7 8 9 ΙΟ &c. 1 4 16 64 256 1,024 4,096 16,384 65,536 262,144 1,048,576 &c. A glance at this table enables us to say that, for instance, (a) the product of 256 and 4,096 is 1,048,576; (b) the division of 262,144 by 16,384 gives 16 for quotient; (c) the third power of 16 is 4,096; and (d) the square root of 1,048,576 is 1,024:

(a) Above 256 and 4,096 stand 4 and 6, respectively; the sum of 4 and 6 is 10; and below 10 stands 1,048,576—the product of 256 and 4,096.

(b) Above 262,144 and 16,384 stand 9 and 7, respectively; the difference between 9 and 7 is 2; and below 2 stands 16– the quotient obtained when 262,144 is divided by 16,384.

(c) Above 16 stands 2; the treble of 2 is 6; and below 6 stands 4,096—the cube of 16.

(d) Above 1,048,576 stands 10; the half of 10 is 5; and below 5 stands 1,024-the square root of 1,048,576.

This is explained by the fact that the terms of the geometrical progression are all powers of 4, and that the terms of the arithmetical progression are the indices of those powers, respectively: 1=4o; 4=4'; 16=42; 64=43; 256=41; &c.:

(a) 256X4,096=4X46-410=1,048,576

(b) 262,144÷16,384=4°÷47=42=16
(c) 163=(42)3=46=4.096

(d) √1,048,576=√/410=45=1,024.

In this last table, each term of the arithmetical progression indicates the "number of [common] ratios"-that is, the number of fours-which occur, as factors, in the corresponding term of the geometrical progression. So that the logarithm of I is o; of 4, 1; of 16, 2; of 64, 3; &c.-the common ratio, 4, being recognised as the "base" of the table.

It must be borne in mind that two different tables-or "systems"-of logarithms cannot be employed indiscriminately. As an illustration of this, let us endeavour, by means of the two tables now before us, to find (a) the product of 343 and 256; also, (b) the number of times 16 is contained in 5,764,801 :

5

6

7

8

&c.

ΟΙ 2 3 4 I 7 49 343 2,401 16,807 117,649 823,543 5,764,801 &c. 1 4 16 64 256 1,024 4,096 16,384 65,536 &c.

(a) The logarithms of 343 and 256 are 3 and 4, respectively -but not to the same base; the sum of 3 and 4 is 7, the logarithm of 823,543 (to the base 7), and also of 16,384 (to the base 4); neither 823,543 nor 16,384, however, is the product of 343 and 256:

343 X 256-73 × 41; and 73×41 is neither 77 nor 47-i.e., is neither 823,543 nor 16,384.

(b) The logarithms of 16 and 5,764,801 are 2 and 8, respectively-but not to the same base; the difference between 2 and 8 is 6, the logarithm of 117,649 (to the base 7), and also of 4,096 (to the base 4); neither 117,649 nor 4,096, however, expresses the number of times 16 is contained in 5,764,801:

5,764,801÷16=78÷42; and 7o÷42 is neither 76 nor 4o—i.e., is neither 117,649 nor 4,096.

In constructing a table of logarithms, therefore, we must confine ourselves to some one base: moreover, we must devise some means of determining the logarithms of all the numbers intermediate between every two powers of the base. Neither of the preceding tables, for instance, is of any practical utility -the first exhibiting the logarithm of none of the numbers between 1 and 7, between 7 and 49, between 49 and 343, &c.; whilst the second table exhibits the logarithm of none of the numbers between 1 and 4, between 4 and 16, between 16 and 64, &c.

We now proceed to consider how this defect can be remedied. We have already seen (p. 274) that, of any three consecutive terms of a geometrical progression, the middle term is the square root of the product of the other two; and (p. 269) that, any three consecutive terms of an arithmetical progression, the middle term is half the sum of the other two. These facts suggest a means by which either of the foregoing tables could be completed. Let us take the table whose base is 4:

of

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&c. 256 &c.

Now, if 16 and 2, for instance, were removed from this table, we could replace both numbers by inserting a geometrical mean between 4 and 64, and an arithmetical mean

between 1 and 3 : √4×64=√256=16; !+3—4—2. Again,

2

2

if 64 and 3 were removed, we could replace both by inserting a geometrical mean between 16 and 256, and an arithmetical mean between 2 and 4: √16×256=√4096=64; 2+4_6

2

=

2

3.

In like manner, by inserting a geometrical mean between I and 4, and an arithmetical mean between o and 1, we find that the logarithm of 2 is; by inserting a geometrical mean

between 4 and 16, and an arithmetical mean between 1 and 2, we find that the logarithm of 8 is ; &c.: √I×4=√4=2; 1+2== 3; &c. :*

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2

=

O

I

=}; √4×16=√64=8;

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2

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So that the continual insertion of geometrical means in the lower, and of arithmetical means in the upper series, would ultimately complete the table, by exhibiting the logarithms, not merely of 1, 4, 16, 64, &c., but of all the intermediate numbers also. The arithmetical mean between o and 2, for instance, would be the logarithm of the geometrical mean between

and

2; the arithmetical mean between and I would be the logarithm of the geometrical mean between 2 and 4; and so on.t

The base of the table of logarithms now in universal use is neither 7 nor 4, but 10, which, as we shall presently find, is by far the most appropriate number that could have been selected. The following portion of the table involves no calculation :

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Here the logarithm of 1 is o; of 10, I; of 100, 2; of 1,000, 3; &c. Because o, 1, 2, 3, &c. are the indices of the powers which 1, 10, 100, 1,000, &c., respectively, are of the base 10:1=10°; 10=101; 100=102; 1,000=103; &c. The table could be completed upon the principle just explained. For instance a being the geometrical mean between 1 and 10, and a' the arithmetical mean between o and I, a' would be the logarithm of a; b being the geometrical mean between I and a, and b' the arithmetical mean between o and a', b' would be the logarithm of b; c being the geometrical mean between 1 and b,

* It is evident that and are the logarithms of 2 and 8, respectively -to the base 4. Because when 2 is expressed as a power of 4, the index is; and when 8 is so expressed, the index is 3:2-4; 8=4.

Here is a general demonstration of the fact that the arithmetical mean between the logarithms of any two numbers is the logarithm of the geometrical mean between the numbers themselves. Let x and z be two numbers, whose logarithms are a and c, respectively; let y be the geometrical mean between x and z, and b the arithmetical mean between a and c: to prove that b is the logarithm of y.

Putting B for the base of the table, we have (the logarithm of x being a) x=B", and (the logarithm of z being c) z=Bc; xxz=Ba × Bc= Bate; i.e., (xxz being=y2, and a+c=2b,) y2=Bab; y=Bb; b= logarithm of y.

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