5. To find the number of positive and negative roots that may be contained in 3 5 – 8 = 0. Completing the equation by the addition of the infinitesimal 0.x, it becomes 3x3- 5x2+0x-8=0; also completing the equation by the addition of 0.x, it becomes 3-5a2 0.x-8=0. Now, according to the first of these complete equations, the roots of the proposed equation may all be positive; while, according to the second complete equation, two of them must be negative; consequently, on account of the incompatibility of the preceding conclusions, two of the roots must be imaginary. 6. To find how many positive and negative roots may be in 25 50 = 0. 7x4 Because the equation may be completed in the forms a 72*+0x3-0.x2 + 0.x — 500 and x 0.04 0.x2 -0.x 500, it follows that it must have four imaginary roots; and it is easy to perceive that it has one real positive root. It is hence clear that any equation which wants 2m - 1 terms between terms having like signs, must have at least 2m imaginary roots. 7. To find how many imaginary roots there are in the equation 76 - 9′3 — 80 = 0. Because the equation wants four terms between terms which have like signs, it is evident, as in the last question, that it has four imaginary roots; and it is easy to perceive that the equation has two real roots, one of which is positive and the other negative. Hence, any equation which wants 2m terms between terms which have like signs, must have at least 2m imaginary roots. 8. To find how many imaginary roots there are in 8 20x+92 0. = Writing 820x-0.x +92 for the equation, then, since two terms are wanting between 20 and -0.x, terms which have like signs, it follows that the equation has (at least) two imaginary roots. Hence any equation which wants 2m+1 terms between terms having unlike signs, must have at least 2m imaginary roots. 9. To find how many imaginary roots there are in 10x 15x+72 0. Writing 10-1520.x + 72 = 0 for the equation, then according to what has been shown, since the equation wants three terms between 15 and 0.x, terms which have like signs, it results that the equation must at least have four imaginary roots. Hence, any equation which wants 2m terms between terms which have unlike signs, must at least have 2m imaginary roots. 10. To show that y1 — 19y3 + 132y2 — 302y + 200 = 0 has a pair of imaginary roots. = 19 Putting y=x+, we get an equation of the form a1 + Aa2+ etc. 0, in which A is positive; consequently, this equation has a pair of imaginary roots, since it wants a term between terms having like signs; and of course the proposed equation must have a pair of imaginary roots. 11. To show that y5y+13y2 — 17y + 10 = 0 has a pair of imaginary roots. 5 Putting y = x + and proceeding as in the last question, 4' it is easy to show that the proposed equation has at least a pair of imaginary roots. Indeed, since the equation is equivalent to 2 5 + 27y2 17y + 12 = 0, and that 273 - 17y+ 12 is positive 4 27y2 for all real values of y, it results that all the roots of the tion are imaginary. equa Resuming (4), given in (7), we shall have the series ƒ(x), ƒ'(x), ƒ''(x),ƒ'''(x), . . . . N, (1), in which f(x) = 0 denotes an algebraic equation, f'(x), ƒ''(x), etc., being the first, second, etc., derived functions of f(x), and N is the last derived function, which is evidently independent of x. If m represents the number of variations of signs among the terms of the series (1) which correspond to the (real) value p of x, and m' the number of variations corresponding to the value q of x, then the number of real roots of the equation f(x)=0, which lie between P and q, can not exceed the number of units in the difference of the numbers m and m'. In the demonstration of the theorem we shall suppose that p is greater than q, in such a sense that pq+r, r being positive; and that x increases by very small increments from the value q to p, and as it increases passes through the real roots of f(x) = 0, which lie between q and p. According to what has heretofore been shown, it follows that the signs of f(x) and f'(x) will form a variation just before a passes through any real root of f(x)= 0, which becomes a permanence just after the passage; consequently, a variation of signs between the first two terms of the series (1) will be lost by the passage of a through any real root of f(x) = 0. Because f'(x) is the first derived function of f'(x), it follows, as above, that as x increases and passes through a root of f(x) = 0, a variation between the signs of f(x) and f'(x) will be lost, which can not be recovered, unless a permanence exists between the signs of f(x) and f(x) before the passage, which is changed into a variation, in consequence of the change of the sign of f'(x); consequently, the variations among the signs of the terms of (1) can not be increased by the passage of a through a real root of f'(x) = 0. Because a similar reasoning and conclusion is applicable to each of the terms of (1) which lie between f(x) and the constant N, it follows that any changes of signs among these terms, resulting from the supposed increase of x, can not increase the number of variations in (1). Because a variation of signs between f(r) and f'(x) in (1) is lost by the passage of x, during its increase, through each real root of f(x) = 0, it results that the theorem is true; and that whether f(x) = 0 has or has not any equal roots. To perceive the use of the theorem, take the following EXAMPLES. 1. To find the number of real roots of a3 =0 which may lie between x = 0 and x = 2. 3x2+3x-1 and 6; 6, Here the successive terms of (1) become a3-3x2 + 3x — 1, 3x2 6x + 3, 6x and = 0 and P = 2. Putting 0 for x in the preceding series, we get the signs −, +, −, +, three variations; and putting 2 for æ, the signs become +, +, +, +, no variation; consequently, m and m' are here represented by 0 and 3. Consequently, since m' - m3, it follows that the proposed equation may have three real roots, which may lie between 0 and 2. Indeed, because the equation is equivalent to (x-1). (x-1). (x-1)=0, it has three equal real roots, expressed by x 1, x 1, a 1, which lie between 0 and 2. = = = 2. To find how many real roots of a 2x-1=0 may lie between - 2 and 2. Here (1) become a 2-1, 3-2, 6x, and 6; and putting 2 for a, their signs are -, +, -, +, three variations; also, putting 2 for x, the signs are +, +, t, t, no variation; consequently, the roots of the equation must all lie between 2 and 2. 2 and 2. Indeed, since the (proposed) equation is equivalent to (x + 1). (x2 — x − 1) = (x + 1) . 2 2 =0, it is easy to perceive that 2 and 2 are inferior and superior limits to the roots. 3. To find how many roots of x3 — 4x2 between - 4 and 5. 9x+360 lie Here (1) become a -4x-9x+36, 3x-8x-9, 6x-8, and 6; and putting -4 for X, the signs are +, +, three variations; also, putting 5 for x, the signs are +, +, +,+, no variation; consequently, 4 and 5 are inferior and superior limits to the roots. Indeed, the roots are - 3, 3, and 4. - Remarks. If we suppose the equation - 4a2 — 9x+ 360 has equal roots with contrary signs, then, putting ≈ and for a pair of these roots, it is clear that the equa tions 3 4x2 9x+ 360 and -3 4x2+9x+36 = 0 must be coexistent; consequently, we must have ∞3 — 9x=0 or a2 = 9 and 4a36, which concur in giving -3 and 3 for a pair of equal roots having opposite signs; and dividing - the equation by -9, the quotient is x 4, and of course 4 is the remaining root. Hence, if we unite all the terms of an equation which contain the odd powers of the unknown letter, according to their signs, into one sum, then the greatest common divisor of the sum thus obtained and the remaining terms of the equation, when put equal to 0, will give the equal roots, with opposite signs, in the equation. Thus, in the equation - 9x+6x+15x3 — 12x2 — 7x + 60, we find a 2x2 + 1 = (-1) to be the greatest 9x5+ 15×3. 12x2+6; common divisor of consequently, from a = 7x, and 6x 2a+1=0 we get a = 1, x = − 1, and a 1, x=1, for the equal roots having opposite signs. 4. To find how many of the roots of a1 - 480 lie between 5 and 5. 13x2 + 16x Here (1) become a -X08 26x16, 12x2-6x-26, 24-6, and 6; for x in these, the signs are +, -, +, -, +, four variations; also, putting 5 for a, the signs are +, +, +, t, t, no variation; consequently, -5 and 5 are inferior and superior limits to the roots. Because a3 16x and a 13x2 48 have x2 16 for a factor, it follows that 4 and 4 are roots of the equation; and it is hence easy to show that the remaining roots are imaginary. 5. Find the number of roots of a 1800, which may lie between 1 and 6. 6. Find the number of roots of 53 may lie between 0 and 4x3-29x2 + 156x Ans. Three. 6x+2=0, which Ans. One. 1. Representing any algebraic equation by Ax" + A1æn−1 + A geo-2 + .... + An-1°c + A, 0, (1), and its first member by V; then multiplying each term of (1) by the corresponding index of c, and subtracting 1 from the index, we get nAx2-1+ (n-1)A ̧æ” ̄2 + (n − 2) Asic13 + . . . . + 2A-90 + An-1, (2), for the first derived function, which we denote. by V1. -1 .... |