c'd -} c-d (a+b) = 5 (a+b) — — * 3 = (11) (a3⁄4+a2b3+a3b33⁄4+ab+a3¿a3+b3) × (a3—b3) —a3—bo. (12) (x2+x*y*+y3)× (x*—y1)=x*—y3. 3 (13) (x*+y3)×(x−1+y ̄1)=x3y—$+2+x ̄*y*. 3 5 1 27 99 3316 27 13 (18) m3p1q3r•×p—Hqr ̄‡m2×p»q=m"p"?q¥ 2 (20) (z3+6z1a3+9a3). √3⁄41⁄23. (VZ+3√α3)=(Z*+3a3)3 . }{√5k3. It may be asked here whether the rules for the calculus of exponents apply to incommensurable and imaginary exponents. With regard to incommensurable exponents, it may be said that they have not absolutely of themselves any signification, and that, in order to give them one, it is necessary to conceive them in imagination, replaced by their approximate commensurable values. A formula, therefore, into which incommensurable exponents enter, should be considered as representing the limit toward which the values deduced from it tend by the substitution of commensurable numbers for the exponents, differing from them by as small a quantity as we choose to assign; in this way we perceive that the proposed expression will represent exactly this same limit, when the same operations shall have been executed upon the in commensurable exponents which it contains, as would be if they were commensurable. Thus, for example, m and n being incommensurable quantities, we shall always have amXan=a+ For, if m' and n' represent their approximate commensurable values, we have am' Xan=am+w. The first members of this equality tend toward the same limit as the second. But am Xan represents the limit of the one, and am+n that of the other; hence, amXa1=am+". With regard to imaginary exponents, there is necessary here, as every where, a tacit admission that the general relations of real quantities, represented by letters, hold good when these letters are replaced by symbols of quantities which are imaginary. This subject will be better understood after the student has been over that of extraction of roots by approximation. 78. Having thus discussed the formation of powers, and the extraction of roots in monomial quantities, we shall how direct our attention to polynomials; and, in the first place, let it be required to determine the square of x+a; then, (x+a)=(x+a)x(x+a) =x2+2xa+a2 by rules of multiplication. By inspection of this result, it is perceived that the square of a binomial contains the square of each term together with twice the product of the two. Next, let it be required to form the square of a trinomial (x+a+b). Let us represent, for a moment, the two terms, x+a, by the single letter z. Then, Therefore, substituting for z2 and 2zb their values in (1), we find (x+a+b)2= x2+a2+b2+2xa+2xb+2ab. Hence it appears that the square of a trinomial is composed of the sum of the squares of all the terms, together with the sum of twice the products of all the terms multiplied together two and two. We shall now prove that this law of formation extends to all polynomials, whatever may be the number of terms. In order to demonstrate this, let us suppose that it is true for a polynomial consisting of n terms, and then endeavor to ascertain whether it will hold good for a polynomial composed of (n+1) terms. Let x+a+b+c+----+k+l be a polynomial consisting of n+1 terms, and let us represent the sum of the first n terms by the single letter z; then (x+a+b+c+---+k+l) =(z+1), or, putting for z its value, =(x+a+b+c+---+k)2+2(x+a+b +c+---+k)2+1o. But the first part of this expression, being the square of a polynomial consisting of n terms, is, by hypothesis, composed of the sum of the squares of all the terms, together with twice the sum of the products of all the terms multiplied two and two; the second part of the above expression is equal to twice the sum of the products of all the first n terms of the proposed polynomial, multiplied by the (n+1)" term l; and the third part is the square of the (n+1)" term l. Hence, if the law of formation already enounced holds good for a polynomial composed of n terms, it will hold good for a polynomial composed of (n+1) terms. But we have seen above that it does hold good for a polynomial composed of three terms; therefore it must hold for a polynomial composed of four terms, and therefore for a polynomial of five terms, and so on in succession. Therefore the law is general, and we have the following RULE FOR THE FORMATION OF THE SQUARE OF A POLYNOMIAL. The square of any polynomial is composed of the sum of the squares of all the terms, together with twice the sum of the products of all the terms multiplied together two and two. According to this rule, we shall have, (1) (a+b+c+d+e)2=a2+b2+c2+d2+e2+2ab+2ac+2ad+2ae+2bc +2bd2be+2cd+2ce+2de. (2) (a-b-c+d)2=a2+b2+c2+d2-2ab-2ac+2ad+2bc-2bd-2cd. If any of the terms of the proposed polynomial be affected with exponents or coefficients, we must square these monomials according to the rules already established. (3) (2a-4bc)=4a+16bc-16ab2c3. (4) (3a2—2ab+4b2)2=9a*+4a2b2+16ba—12a3b +24a2b-16ab3 =9a-12a3b+28a2b-16ab3+16b4, arranging according to powers of a, and reducing. (5) (5a b-4abc6bc2-3a2c)2=25a+b2+16a2bc2+36bc1+9a*c2 -40a3b3c+60a2bc2-30a1bc -48ab2c+24a3bc-36a2bc3. =25a1b2 —40a3b°c+76a2b2c2-48ab2c3 +36bc-30abc+24a3bc2 79. Let us now pass on to the extraction of the square root of algebraic quantities. Let P be the polynomial whose root is required, and let R represent the root which for the moment we suppose to be determined; let us also suppose the two polynomials, P and R, to be arranged according to the powers of some one of the letters which they contain; a, for example. If we reflect upon the law just given of the formation of the square of a polynomial, it will be seen that the first two terms of the polynomial P, when thus arranged, are formed without reduction, and will enable us at once to determine the first two terms of the root sought; for, 1o. The square of the first term of R must involve a, affected with an exponent greater than any that is to be found in the other terms which compose the square of R; because this exponent is double the highest exponent of a in R, and must be greater than the double of any lower exponent, or than the result produced by adding it to one of the lower exponents, or by adding any two of them together. 2o. Twice the product of the first term of R by the second must contain a, affected with an exponent greater than any to be found in the succeeding terms; for it will be the sum of the highest, and the next to the highest exponent of a in R. It follows from this, that if P be a perfect square, I. The first term must be a perfect square; and the square root of this term, when extracted according to the rule for monomials (Art. 49), is the first term of R. II. The second term must be divisible by twice the first term of R thus found, and the quotient will be the second term of R. III. In order to obtain the remaining terms of R, square the two terms of R already determined, and subtract the result from P; we thus obtain a new polynomial, P', which contains twice the product of the first term of R by the third term, together with a series of other terms. But twice the product of the first term of R by the third must contain a, affected with an exponent greater than any that is to be found in the succeeding terms, and hence this double product must form the first term of P'.* IV. The first term of P' must be divisible by twice the first term of R, and the quotient will be the third term of R. V. In order to obtain the remaining terms of R, square the three terms of the root already determined, and subtract the result from the original polynomial P; we thus obtain a new polynomial, P", concerning which we may reason precisely in the same manner as for P', and continuing to repeat the operation until we find no remainder, we shall arrive at the root required. The above observations may be collected and imbodied in the following RULE FOR THE EXTRACTION OF THE SQUARE ROOT OF ALGEBRAIC POLYNOMIALS. 1o. Arrange the polynomial according to the powers of some one letter. 2o. Extract the square root of the first term according to the rule for monomials, and the result will be the first term of the root required. 3o. Square the first term of the root thus determined, and subtract it from the original polynomial. 4°. Double the first term of the root, and divide by it the first term of the remainder, and annex the result (which will be the second term of the root), with its proper sign, to the divisor. 5°. Multiply the whole of this divisor by the second term of the root, and subtract the product from the first remainder. 6o. Divide this second remainder by twice the sum of the first two terms of the root already found, and annex the result (which will be the third term of the root), with its proper sign, to the divisor. 7°. Multiply the whole of this divisor by the third term of the root, and subtract the product from the second remainder; continue the operation in this manner until the whole root is ascertained. The above process will be readily understood by attending to the following examples: EXAMPLE 1. Extract the square root of 10x1-10x3—12x5+5x2+9x¤—2x+1. Or, arranging according to the powers of r, The square of the second term of R usually contains the same exponent of the letter of arrangement, but this is already subtracted from P, and not left in P'. In practice, this operation is dispensed with by following the precepts 5o, 7°, in the following rule, which evidently come to the same thing. 9x12 10x-10x+5x-2x+1|3x3-2x2+x-1 6x3-2x2-12x+ 10 -10.x3+5x2-2x+1 Having arranged the polynomial according to powers of x, we first extract the square root of 926, the first term; this gives 33 for the first term of the root required; this we place on the right hand of the polynomial, as in division; squaring this quantity, and subtracting it from the whole polynomial, we obtain for a first remainder, 12x+10xa—10x3+5xo-2x+1; we now double 3x3, and place it as a divisor on the left of this remainder, and dividing by it -12, the first term of the remainder, we obtain the quotient -2x2 (the second term of the root sought), which we annex, with its proper sign, to the double root 623; multiplying the whole of this quantity, 6x3-2x2, by —2x2 (which produces twice the product of the first term of the root by the second, together with the square of the second), and subtracting the product from the first remainder, we obtain for a second remainder, 6xa—10x3+5x2−2x+1. Next, doubling 3r3-2x2, the two terms of the root thus found, and dividing 6x4, the first term of the new remainder, by 6x3, the first term of the double root, we obtain r for a quotient (which is the third term of the root sought), and annex it to the double root 6x3—4x2, multiplying the whole of this quantity 6x3-4x2+x by x (which produces twice the first by the third, twice the second by the third, and the square of the third), and subtracting the product from the second remainder, we obtain a third remainder, -6x+4x2−2x+1; we now double 3x3-2x2+x, the three terms of the root already found, and dividing -6x3, the first term of the new remainder, by 623, the first term of the double root, we obtain -1 for the quotient (which is the fourth term of the root sought), and annex it to the double root 6x-4x2+2x; multiplying the whole of this quantity 6x3-4x2+2x-1 by -1, and subtracting it from the third remainder, we find 0 for a new remainder, which shows that the root required is 3-2x+x-1. |