have which is too much by

1

Of four terms, we have, which is too small by &c. If a, the fraction will be , to which is equal the series 1-} + ¦ − & + & − 213 + 729, 1+} &c. continued to infinity. Taking only two terms, we have, which is too small by: three terms are equal to 3, which is too much by Four terms are equal to 9, which is too small by and so on.

1

36.

676. The fraction 17 may be resolved by another method into an infinite series, namely, by dividing 1 by a + 1, thus:

1

Thus we find also that the fraction is equal to the infinite series

as

a+1

1 1 &c. If, then, we make a = 1, we shall have the series 1−1 +1 −1+1 −1, &c. =}, as before; and if a=2, we shall have the series -1 + 1 − 18 + 32 −34, &c. = f.

C

677. So, generally, by resolving the general fraction into an infinite series, we have a+b

a aa a3 a4

From which it appears that we may compare + with the series

C

c bc bbc b3c

a+b

+

&c.

to infinity. Let a=2, b=4, c=3; we shall then have +=+2=}=}=} −3+6−12, &c.

678. If a=10, b=1, and c=11, we have a+b=10+1=1=}8-100+1000-10000 only one term of this series, we have, which is too great by; if we take two terms, we have, which is too small by T; if three terms, 100, which is too great by Too, &c. divisor contain more than two terms, the division may, in the same manner, still be continued to infinity. Thus, suppose the fraction a+aa proposed; the infinite series to which it is equal would be found as follows:

1

Hence we have the equationa+aa =1+a—a3 — a4 + a + a7 — a9-a10, &c.; in which, if we make a=1, we have 1=1+1 −1 −1 +1 +1 − 1 − 1 + 1 + 1, &c., which series contains twice that of 1 - 1 + 1 − 1 + 1, &c. found above; and as that has been found equal to, it is to be expected that we should find, or 1 for the value of that just determined. If a={ we shall have the equation ==—— 18+ ¿4—1£s + st2, &c. If a= we shall have the equation ==1+-7+&c., of which series, if the four leading terms be taken, we have, which is only less than 4. Again, if a=3, we have 4=;=1+3−− +, &c., a series which is equal to the preceding one, and subtracting one from the other, -+, &c. must =0. The four terms added together make

63

679. The method of resolving generally all fractions into infinite series which has been above given, is often of great utility, and though it never ceases, an infinite series may have a determinate value. Many discoveries of the highest importance have been derived from this branch of mathematics, which, on that account, well deserves the study and comprehension of the reader.

THE SQUARES OF COMPOUND QUANTITIES.

680. To find the square of a compound quantity, it is only necessary to multiply by itself, and it is the square required; thus the square of a + b is found in the following

manner:

a+b
a+b

aa + ab

ab+bb aa+2ab+bb

From which we learn that the square of any number comprising two terms consists, first, of the squares of each term, namely aa and bb, and twice the product of the two terms, that is 2ab. In figures, suppose a = 12 and b = 4, that is, let it be required to find the square of 16, we have 144 +96 +16, or 256. This formula then gives us the power of finding the squares of numbers, however great, if we divide them into two parts. Thus, to find the square of 49, recollecting that this number is equal to 40+9, its square is =1600+ 720 + 81 =2401. From the same cause it is evident that the square of a + 1 will be aa + 2a + 1; and since the square of a is aa, we find the square of a +1 by adding to it 2a + 1, wherein it is observable that this 2a+1 is the sum of the two roots a and a + 1. Hence, the square of 10 being 100, that of 11 will be 100+ 21 (that is, 100 + 2 × 10+ 1). The square of 49 being 2401, that of 50 is 2401 +99=2500; the square of 50 being 2500, that of 51 = 2500+101 =2601, &c.

681. The square of a compound quantity, as a + b, is thus represented (a+b)2. We have then (a+b)2= =aa+2ab+bb, whence result the following equations : —

(a + 1)2=aa + 2a + 1; (a + 2)2=ua + 4a+4; (a+3)2=aa + 6a+9;
(a + 4)2=aa + 8a+ 16, &c.

682. If the root is a-b, the square of it is aa-2ab+ bb, which also contains the squares of the two terms, but in such a manner, that from their sum must be taken twice the product of those terms; thus, let a=10 and b=-1, the square of 9 will be found to be

100-20+1=81.

683. Having then the equation (a−b)2=aa-2ab+bb, we shall have (a−1)2 = aa— 2a + 1. From which it is evident that the square of a-1 is found by subtracting from aa the sum of the two roots a and a-1, that is, 2a-1. Let a 50, then aa=2500; and a-1=49; hence 492-2500-99=2401.

=

684. The rule is moreover confirmed and illustrated by fractions; for, taking as the root 3+ (which make 1), the square will be

++, that is, 1.

And further, the square of - (or of 1) will be }-{+}=%

685. If the root consists of a greater number of terms, the square is determined in a similar manner; take, for instance, the square of a + b + c.

686. It will be perceived that the product includes the square of each term of the roots, and, besides that, the double products of those terms multiplied two by two. To exhibit this in figures, divide the number 345 into three parts, 300+ 40+5; its square will then be composed of the following parts: —

119025, which is equal to the product of 345 × 345.

687. Though some of the terms of the root be negative, the rule still holds good, only that we must be careful in prefixing the signs to the double products. For instance, the square of a-b-c being aa+bb + cc -2ab2ac + 2bc, if the number 345 be represented by 40050-5, we shall have

688. For the extraction of the roots of compound quantities, and the rule by which the operation is guided, we must consider with attention the square of the root a + b, which is aa+2ab+bb. It will be seen that it is composed of several terms; and that, therefore, the root will comprise more than one term, and that if the square be written so that the powers of one of the letters, as a, may continually diminish, the first term will be the first square of the root; and as, in the instance adduced, the first term of the square is aa, it is certain that the first term of the root is a. Having thus found the first term of the root, that is, a, we have to consider the rest of the square, namely 2ab+ bb, and endeavour to ascertain from it the second part of the root, which is b. Now as the remainder 2ab+bb, which may be represented by the product (2a + b)b, has two factors, 2a + b and b, it is evident that the latter b, which is the second part of the root, will be found by dividing the remainder 2ab+bb by 2a+b. The quotient then arising from the division of the above remainder by 2a + b, is the second term of the root required. In the division, it is to be observed that 2a is the double of the first term a, which is already determined, and as the second term is yet unknown, though for the present its place must be left empty, the division may be attempted, since in it we attend only to the first term 2a. When, however, the quotient is found, which is here b, it must be put in the empty place, by which the division is rendered complete. The operation is thus represented:

689. In a similar manner may be found the square root of other compound quantities, provided they are squares, as may be seen by the following examples.

(I.) aa + 6ab + 9bb (a+3b (II.) 4aa ·4ab+bb (2a-b (III.) 9pp +24pq + 16qq (3p+ 4q

690. If a remainder occurs after division, it proves that the root is composed of more than two terms. In which case, the two terms already found are considered as forming the first part, and we must try to obtain the other from the remainder in the same manner as the second part of the root was found. The following examples will show the mode of

(III.) a® - 6a3b + 15a1bb — 20a b3 + 15aaba — 6ab5 + b6 (a3 — Saab + Sabb—b3

a6

2a3-3aab)-6a5b+15a4bb

-6ab9a4bb

2a3-6aab+3abb)6a4bb — 20a3b3 + 15aab1

6a4bb18a3b3+ 9aab4

2a3-6aab+6abb-b3) — 2a3b3 + 6aab*. 6ab5+ b5

-2a3b3+6aab1 — 6ab5 + b6

691. From this method of extracting the square root, it is easy to reduce the rule given for that purpose in common books of arithmetic; as in the following examples in numbers.

692. If there be a remainder after the whole operation, it proves that the number is not an exact square, and therefore its root cannot be assigned. When this is the case, the radical sign before mentioned must be written before the quantity, and the quantity itself is placed between parentheses, or under a line. Thus, the square root of aa+bb is represented by

✔ (aa+bb), or by √ aa+bb, and √(1−xx) or √1−xx expresses the square root of Instead of the radical sign the fractional exponent may be used, thus (aa + bb)ŝ

to aa + bb − equally represent the square root of aa + bb.

CALCULATION OF IRRATIONAL QUANTITIES.

a+ √a Thus the The sum

693. The addition of two or more irrational quantities is performed by writing all the terms in succession, each with its proper sign. They may often be abbreviated thus: for we may write 2√a, and √a- √a=0, because the terms destroy one another. quantities 3+ √2 and 1 + √2 added together, make 4 + 2√2 or (590.) 4 + √⁄8. of 5+ √3 and 4-3 is 9, and that of 2√3+3√2 and 3-2 is 3√3+2√2. 694. Subtraction is equally simple, since we have only to change the signs of the numbers to be subtracted, and then add them together, thus:

Subtract from 4- √2+2√3 −3√5+4√6

The numbers 1+2√2−2√3−5√5+6√4

3−3√3+4√3+2√5+2√6

695. In multiplication it must be remembered that a multiplied by ✔a produces a, and that if the numbers which follow the sign ✔ are different, as a and b, we have ab for the product of a multiplied by √b. From due observance of this, the following examples are easily calculated.

The rules apply also to imaginary quantities, and it will only be necessary to mention that ✔-a multiplied by a produces

-a.

696. To find the cube of −1+√ −3, we take the square of that number, and multiply such square by the number, as follows:

697. To divide surds, we have only to express the quantity in the form of a fraction, which may then be changed into another expression having a rational denominator. Thus, for instance, if the denominator be a + b, and it as well as the numerator be multiplied by a√b, the new denominator will be aa-b, in which no radical sign occurs. Let it, then, be required to divide 3 + 2√2 by 1 + √2, we have Multiplying both terms of the fraction by 1-√2, we have for the numerator,

3+22
1+√/2

3√2-4-√2−1;