= Because a, b, and c are represented by -4, -3, and 6, (5) becomes 23-822-8z9, one of whose roots equals 9. Since 3 and 3 are the values of m, which correspond to a 3 + V 3 3 2 = 9, we (from (6)), get 2, 1, 2 for the roots of the equation in x; consequently, from y = x+ 4. To find the roots of y8y + 24y- 36y+27=0. Putting y+2, the equation becomes a-4x+3=0; consequently, (5) gives 23 122 = 16, which has 4 for one of its roots. Because 2 and 2 are the values of m which correspond to z = 4, we readily get 1, 1, −1 + √−2, − 1 − √-2 for the roots of the equation in x; consequently, 3, 3, 1+ 1-2, 1-1-2 are the roots of the given equation. SECTION XVIII. LIMITS OF THE REAL ROOTS OF AN EQUATION OF THE nth DEGREE. -2 (1.) USING + An-1-1 + A--9+.... + N = 0, (1) to stand for an equation of the nth degree, which is supposed to have real roots; then, if some of the coefficients are positive (regarding N as being the coefficient of ao), while others are negative, if we use -b to stand for the greatest negative coefficient, b + 1 shall be greater than the greatest positive root of (1), or b + 1 shall be what is called a superior limit to the positive roots of (1). For it is clear that " - bxn-1. bxn-2 b can not be greater than the first member of (1), and since (b + 1)"— b(b + 1)n−1 — b(b + 1)”—2 — etc. = (b + 1)"−1 — b(b + 1)n—2— etc.(b+1)-2-b(b+1)-3 etc. etc. etc. 1, it follows if we put b + 1, or any greater number for x in (1), that the = = = result can not be less than 1; consequently, since the great est positive root, when put for a, reduces the first member of (1) to 0, it results that b+1 must be greater than the greatest positive root. Thus, 5 is a superior limit to the positive roots of the equation a4x3 + 3x2-x-3=0. (2.) On the same supposition, if M is the greatest positive coefficient which precedes the first negative term, and if N' is the greatest negative coefficient, then N' M +1 shall be a superior limit to the positive roots of (1). For if we divide the coefficients by M, and disregard the terms which precede the term whose coefficient is M, then, Thus, 3 is a superior limit to the positive roots of a3 + 2x1 + 6x3 7x2+9x12 = 0. -m (3.) Using to represent the greatest negative coefficient in (1), and 2”-” to express the power of x in the first negative term, then shall +1 be a superior limit to the positive roots of (1). For since an =xn bl x m · b(xn−m + xn−m−1 + ¿n-m−2 + + x + 1) .... is manifestly not greater than the first member of (1), if we determine a so as to make this expression positive, it is manifest that the value of x so found, and all greater values, will be superior limits to the positive roots of (1). Because the preceding condition is equivalent to the b Xn -xm-1 inequality x"> = if we put bp” or r+1= x 1 +1, and substitute r+ 1 for x, we shall reduce the in equality to (r+ 1)"> (. E ) + m-1 × [(r + 1)” — (r + 1)m−1]; which being evidently true, it results that +1 is a supe rior limit to the positive roots of (1). Thus, 3 is a superior limit to the positive roots of the equation + 2x3-3x2 - 4x+6=0; since b=-4, n = 4, and n m = 2. = (4.) If M is the greatest positive coefficient which precedes the first negative term in (1), and if m' is the difference of the indices of the corresponding powers of x; then, if — N' is the greatest negative coefficient in the equation, it is evident, from what has been shown in (3), that m' N' M + 1 will be a superior limit to the positive roots of (1). Thus, 3 is a superior limit to the positive roots of the equation + 8x+4x1-5x3-7x2-9x320, because M = 8,N'-32, and that 5 is the first negative coefficient, and that 532 = m'. (5.) If for a we put — in (1), (and change the signs of all the terms in the result when n is an odd number), we shall clearly obtain an equation such, that its positive roots, when their signs are changed, will be the negative roots of (1). Hence, having found a superior limit to the positive roots of (1), and then put -x for x in (1), and obtained a superior limit to the positive roots in the resulting equation, by changing the sign we shall get what is called an inferior limit to the negative roots of (1); consequently, in this way we shall (generally) find a positive and negative number, between which all the real roots of (1) must lie, the numbers thus found being called limits to the real roots of the equation. EXAMPLES. 1. To find the limits of the roots of +2x2-3x-16=0. Here 16+1= 5 is a superior limit to the positive roots; and changing into, or (which is the same) changing the signs of the second and fourth terms, we get - 2 3x+160; and since 3 + 1 = 4 is a superior limit to the positive roots of this, it follows that 4 is an inferior limit to the negative roots of the given equation, and its real roots lie between 4 and 5. = 0. 2. To find the limits of the roots of + 2x2 + x + 5 Because the equation has no negative terms, it clearly has no positive roots; and putting for x, we get 6 for the limit of the negative roots, and the real roots lie between 6 and 0. -- 3. To find the limits of the roots of a3 2x2+4x-7= 0. The number 7+18 is a limit to the positive roots; and putting for a, the equation becomes + 2x2 + 4x + 7 0, which has no positive roots, and of course the given equation has no negative roots, and the roots lie between 0 and 8. = easily reduced to an equation of the form y" + A'n_1?”-1 + A'n-2y-2 + .... + N' = 0, (2), whose roots are clearly the reciprocals of the roots of (1). Representing the limits of the roots of (2) by a and b, it is evident that 1 1 a can not be greater than the least positive root of (1), and will be what is called an inferior limit to a 1 the positive roots of (1); and in like manner not being numerically greater than the numerically least negative root of (1), will be what is called a superior limit to the negative roots of (1). Hence, limits can be assigned to the real positive and negative roots of any equation which has such roots. EXAMPLES. 1. To find limits to the roots of a3. 4x10. Since 4 is the greatest negative coefficient, (3) gives 14 +1=3= a superior limit to the positive roots, and putting - for x, we get (in like manner) — 3 for an inferior limit of the negative roots; consequently, the roots lie between 1 Putting for x in the equation, it is easily reduced to y3+ 4y2 - 10, and 2 is easily seen to be a superior limit to the positive roots; and in like manner (by putting — y for y) we get 5 for the inferior limit of the negative roots; LIMITS OF THE ROOTS OF EQUATIONS. 1 1 consequently, and 2 5 are inferior and superior limits to the positive and negative roots of the given equation. 2. To find limits to the roots of 3x+6x3 5 = 0. Dividing the terms by 3, the equation is reduced to the 4 5 proper form, 24+ 2x03 — 4x2 + 32-3 = 0; then, from (2), (3), or (4) we get 3 for a superior limit to the positive roots; and putting - for x, or changing the signs of the second 5 for an inferior limit to the and fourth terms, we get negative roots; consequently, the real roots lie between -5 and 3. 1 Putting for x, we easily reduce the equation to y - У 5 2.2 and -(1+) for limits; and of course 1 ÷ 2 . 2 and 4 /3 - 1 ÷ (1 + √√3) are inferior and superior limits to the posi tive and negative roots of the given equation. Hence, we easily find that the equation has two real roots, one of which lies between 1 and 2, and the other between - 3 and (7.) Representing an equation of the nth degree by Ax2 + A1+ Ax2 + .... + A-1 + An = ƒ(x) = 0, (3), supposing A to be positive, while the other coefficients may be positive or negative, according to the nature of the case; then, if we denote the first, second, third, etc., derived functions of f(x) by ƒ'(x), ƒ''(x), ƒ'"'(x), etc., we shall get the equations A-1 + (n − 1)A ̧ãл−2 + (n − 2)Axn−3+ 1)Ax2¬2 + (n − 1) +2A+ A-1 = ƒ'(x) = 0, n(n (n-2) A+ (n − 2) (n − 3)A+.... + 3 × 2A-3 + 2An-2 = ƒ''(x) = 0, and so on to n(n − 1) (n − 2) × × 2 × 1A = N, a term which does not contain x. Now, if x is so assumed that all the terms of the series |