# Sect

### root square quantity term terms required powers power positive cube

SECT. II. - INVOLUTION AND EVOLUTION.

and - a3 denote the same powers of the root a .

quantities - 1 ' a" - 1 &c., may be respectively expressed a thus a-1, (4-2, (4-3, &c.; and considered as negative powers of the root a.

This method of expressing the fractions 1 1 1 - a' - a" - a" as powers of the root a, but with negative indices, is a consequence of the rule which has been given for the division of powers ; for we consider 1 a - as the quotient arising from the division of any power of a by the next higher power ; for example, from the division of the 2d by the 3d, and so we have - 1 - a = but since pov:ers of the same quantity are a a " Involution.

Case 1. When the quantity is simple.

Rule. Multiply the exponents of the letters by the index of the power required, and raise the coefficient to the same power.

Xote. If the sign of the quantity be + , all its powers will be positive ; but if it be - , then all its powers whose exponents are even numbers are positive, and all its powers whose exponents are odd numbers are negative.

lience it appears that the powers of a + x differ from the powers of a - x only in this respect, that in the former the signs of the terms are all positive, but in the latter they are positive and negative alternately.

Besides the method of finding the powers of a compound quantity by multiplication, which we have just now explained, there is another more general, as well as more expeditious, by which a quantity may be raised to any power whatever without the trouble of finding any of the inferior powers, namely, by means of what is commonly called the binomial theorem, to be proved hereafter. This theorem may be expressed as follows : - Let a +x be a binomial quantity, which is to be raised to any power denoted bv the nuaiber n, then (a + x)"-= This series will always terminate when 77, is any whole positive, number, by reason of some one of the factors n - 1, n - 2, &c., becoming = 0; but if n be either a negative or fractional number, the series will consist of an infinite number of terms. As, however, we mean to treat in this section only of the powers of quantities when their exponents are whole positive numbers, we shall make no further remarks upon any other. The nth power of a - x will not differ from the same power of a +x, except in the signs of the terms which compose it, for it will stand thus : (a - x)"= and – alternately.

Let it be required, for instance, to raise a+ x to the fifth 1-; ower.

Here n, the exponent of the power, being 5, the first term a" of the general theorem will be equal to a5, the = 5a4x, the third w(n 2 „ - 1 x 2cea 2 = . - 1) 1'),/,',-62, the fourth 1 3 n(n-1) (n•– 2)(0–,x3-1x 2 x 3 a2x3 = = 5axl, and the sixth and last n – 1) (n– 2) (9i - 3) (a - 4) 5x4x3x2x1 acx5 = x5 the remaining terms of the general theorem all vanish, by reason of the factor n - 5 = 0 by which each of them is nuiltiplied, so that we get (a + x)5 = a5 + 5a4x + 10a3x2 + 1 0: + 5ax4 + x5.- If the quantity to be involved consists of more than tv,-,) terms, as, if p + q - r were to be raised to the second power, put p = a and q - r =b, then (p + q - r)2 = (a + b)2 = a2 +2ab +b2 = p2 +2p(q - r) + (q - r)2, but 2p(q - r) = 2p7 - '2pr, and by the general theorem (q - 02= q2 - r2, therefore we get (p + q - = p2 + 2pq - 2pr 2 - 2qr + r2 ; and by a similar method of proceeding a quantity consisting of four or more terms may be raised to any power.

Ex. 1. From the value of (a + x)4 found in example 4, to find that of (a + b + c)4. From example 4 we write at once, by symmetry, (a+ b + c)4 = a4 + 4a3b + 6a2b2+R + b4 + 4a3c + 6a2c2 + c4 + 4 b3a + 6b2c2 where Ti is the series of remaining terms denoting the three following forms, a2bc, b2ac, c2ab. Now when a, b, c are each unity, there are 81 terms (viz. 34). But the number of terms already written down (4a3b being considered as 4 terms, &c.) is 45. The quantity 11 must consequently make up the other 36 terms, .•. it can be nothing else than 12a2bc +12b2ac+12c2ab.

Ex. 2. (p -1- q + r)2 = p2 + q2 + r2 + 2 (pq + qr + rp). Cor.Ifp +q+r = 0 ; thenp2+ q2+r2 +2(pq+qr +rp)=- O. Case 1. a - b+b - c+c - a= 0, gives Case 2. a(b - c)+b(c - a) + c(a - b) - 0, gives Ex. 3. Prove that (x2 - yz)3+ (y2 - xz)3+(z2 - xy)3 - 3(x2 - yz)(y2 - xz)(z2 - xy) is a complete square.

The expression will assume symmetry if (x2 - yz)(y2 - xz) (z2 - xy), instead of being multiplied by 3, be repeated three times, each being connected with one of the cubes in turn; this gives - + ;22 __ szt„,z)).1 (,;22 __ yx2,122 i (r2 i yx,z)) (( :22 - xx yll + ,2 _ xy) 1 z.,2 _ xy 2 _ x2 _ yz)(y2 _ xz = X2 - 7/Z)X ( X3 + y3 + z3 - 3xyz) 1 + &c., &c.

= (x3+ y3+ z3 - 3xyz)(x3 + y3+,,,3_ 3xyz) .

Ex. 4. Prove that (a2 + b2+ 02)3+ 2(ab + bc + ca)3 - 3(a2 + b2 + c2)(ab + be + ca)2 = (a3 + 63 + c3 - 3abc)2 .

Combine each of the cubes with each of the products in succession, and reduce, as in the last example.

Ex. 5. To find the condition that px2 + 2qxy + ry2 may be incapable of changing its sign through any change of sign or value of x and y. It is evident that p and r must have the same sign. Suppose it positive. By multiplying by p, the quantity may be thrown into the form (px + qy)2 + (pr - q2)y2, which is the sum of two positive quantities provided pr>22. The condition required is, therefore, pr > q2 _; or as a particular case pr = q2.

Ex. 6. To find the condition that ax2 + by2+cz2+ 2Pyz + 2Qzx + 2Rxy may be incapable of changing its sign through any change of sign or value of x, y, z.

We will suppose a, b, c to be all positive, in which ease the whole result is also positive.

If we nniltiply the whole by a, we may write it under the form of a square and a supplement, viz., (ax+ Qz +llyr 4. (ac - Q2)z2 + (ab - 112)y2+ 2(aP - QIl)yz.

Now as the first term of this expression is a square, it is essentially positive. Hence the required condition can be satisfied only by rendering the remainder positive.

It follows that ac >Q2, ab >112, and (Example 5) (ac - Q2)(ab - R2)>(„p - QIt)2, i.e., «bc + 2PQR >aP2 + bQ2 + cR2 .

If we had begun by throwing the expression into the form of (by + Ps +1-19:)' + &c., a resulting condition would have been be >P2. The four conditions are consequently ab>R2, ac > Q2, bc >P2 , abc+ 2PQR >aP2+ LQ2+ cll".

Results of this kind are of the utmost value in the higher analysis.

Evolution,.

To denote that the root of any quantity is to be taken, the sign „/ (called the radical sign) is placed before it, and a small number placed over the sign to express the denomination of the root. Thus ,;,/a denotes the square root of a, „3„Ict its cube root, /a its fourth root, and in general, Zla its nth root. The number placed over the radical sign is called the index or exponent of the root, and is usually omitted in expressing the square root: thus, either „2,Ft or ,/a- denotes the square root of a.

Case 1. When roots of simple quantities are to be found.

Rule. Divide the exponents of the letters by the index of the root required, and prefix the root of the numeral coefficient; the result will be the root required.

Note 1. The root of any positive quantity may be either positive or negative, it' the index of the root be an even number; but if it be an odd number, the root can be positive only.

The root of a negative quantity is also negative when the index of the root is an odd number.

But if the quantity be negative, and the index of the root even, then no root can be assigned.

Ex. Required the cube root of 125a'x9.

here the index of the root is 3, and the root of the coefficient 5, therefore :/125a6x9=5a2x3, the root required; and iu like manner the cube root of - 125a6x° is found to be - 5a2x3.

The root of a fraction is found by extracting the root of both numerator and denominator. Thus the square 4a2x4. 2ax2 root - - • 9b2y6 35y3 I. To extract the square root.

Range the terms of the quantity according to the powers of one of the letters, as in division.

Find the square root of the first term for the first part of the root sought, subtract its square from the given quantity, and divide the remainder by double the part already found, and the quotient is the second term of the root.

Add the second part to double the first, and multiply their sum by the second part ; subtract tho product from the remainder, and if nothing remain, the square root is obtained. But if there is a remainder, it must be divided by the double of the parts already found, and the quotient will give the third term of the root, and so on.

Ex. Required the square root of x4 - 2x + x„ - - - + - • To understand the reason of the rule for findimg the square root of a compound quantity, it is only necessary to involve any quantity, as a + b + c, to the second power, and observe the composition of its square; for we have (a + b + c)2= a2 2ab + b' + 2ac+ 2bc + e2; but 2ab d-b2 = (2a+ b)b and 2ac+ 2bc + c2 = (2a +2b + c)c, therefore, (a + b + c)2 = a2+ (2a + b)b + (2a + 2b + c)c; and from this expression the manner of deriving the rule is obviom.

As an illustration of the common rule for extracting the square root of any proposed number, we shall suppose that the root of 59049 is required.

Accordingly we have (a + b + c)2 = 59049, and from hence we are to find the values of a, b, and c.

59049(200= aience 243 is the runt a2 = 200 x 200 = 40000 40 = b 3=c required.

2a =400119049 2a+b=41°1 2a + 2b + c= 483 1449 = (2a + 2b + c)c The reason of the preceding rule is evident from the composition of a cube ; for if any quantity, as a + b + c, be raised to the third power, we have (a + b + c)3 = a'+(3a2 + 3ab + b2)b + (3(a + b)2 + 3(a+ b)c + c2)c, and by considerII. To extract the cube root. Cube roe Range the terms of the quantity according to the powers of seine one of the letters.

Find the root of the first term, for the first part of the root sought ; subtract its cube from the whole quantity, and divide the remainder by three times the square of the part already found, and the quotient is the second part of the root.

Add together three times the square of the part of the root already found, three times the product of that part and the second part of the root, and the square of the second part; multiply the sum by the second part, and subtract the product from the first remainder, and if nothing remain, the root is obtained ; but if there is a remainder, it must be divided by three times the square of the sum of the parts already found, and the quotient is a third term of the root, and so on, till the whole root is obtained.

Ex. Required the cube root of a3 30a; 3ax2 + x3 .

ing in what manner the terms a, b, and c are deduced from this expression for the cube of their sum, we also see the reason for the common rule for extracting the cube root in numbers. Let it be required to find the cube root of 13312053, where the root will evidently consist of three figures ; let us suppose it to be represented by a +6 + c, and the operation for finding the numerical values of these quantities may stand as follows : -

III. To extract any other root.

Range the quantity of which the root is to be found, according to the powers of one of its letters, and extract the root of the first term; that will be the first member of the root required.

Involve the first member of the root to a power less by unity than the number that denominates the root required, and multiply the power that arises by the number itself ; divide the second term of the given quantity by the product, and the quotient shall give the second member of the root required.

Find the remaining members of the root in the same manner by considering those already found as making one term.

square root of 0+ x2 will be found to be a + -+ 16(45 128a7 + &c., and the cube root of a3+0 will stand x3 x6 5x2 10x12 thus, a + - 3a2-9a6+ 81a8 243a11+ &c. But as the ex- traction of roots in the form of series can be more easily performed by the aid of the binomial theorem, we shall refer the reader to the section where this subject is resumed.

Fx. 1. Write down the square root of x4 - 20+ 3 - x2 - 2 Since the square contains 5 terms, the root must contain 3. Of these the first is x2 on account of x4, the second - x on account of 2x3, and the third 4 - on account of 16. But as the last term but one of the square is -, and the last term but one of the root also - , the last term of the root must be +.

x2 - x 4 - is the root required.

Ex. 2. Extract the square root of 250+160- Gxy (50 + 4y2) +490,0. We must first arrange the square in terms of some one quantity (say x).

The first term of the square is 25x4, which gives 5x2 as the first term of the root. The second term of the square, - 300y gives - 3xy as the second term of the root. The last term 1Gy4 gives 4y2; which, since the last term but one is -, leads to the root 5x2- 3xy + 42/2.

Ex. 3. Extract the cube root of 8x6 - 36x5 + 660 - 63x3+ 33x2 -9x +1 .

Since there are seven terms in the cube, there must bo three terms in the root. The first is 2x2, the second - 3x, the third 1, as will be seen at once by examining the cube of p - q +1, viz., p3 -3p2q + - 3q +1 .

These examples have been solved by the assumption that the root is capable of extraction without leaving a remainder. When this is not the case, or when there is no certainty that it is so, the only resource is to work the example through, abbreviating the process by the aid of detached coefficients.

Ex. 4. Extract the square root of 40+ 12x5y+ 5x4y2 - 2x3y3 + 7x2y4- 2xy5 + v6. The work is written thus :