Superior Limit of Positive Roots. 305. Problem. To find a superior limit of the positive roots. Solution. The sum of all the negative terms being equal to the sum of, all the positive terms, must exceed each positive term. Let, then, —S be the greatest negative coefficient of the equation of the nth degree, and m the exponent of the highest negative term; the sum of the negative terms, neglecting their signs, must evidently be less than that of the series +Sxm, S+Sx+Sx2 + &c. ... + Sxm, for each term of this series is greater than the corresponding negative term of the equation. But this series is a geometrical progression of which S is the first term, Szm the last term, and x the ratio; so that its sum is, by example 3, of art. 261, and must be greater than any positive term, as x^, or x − 1 < x and (x — 1)n—m−1 < xn-m−1, Limits of Negative Roots. If we, then, denote by L this superior limit of the positive roots, we have n-m L=1+ √S; that is, a superior limit of the positive roots is unity, increased by that root of the greatest negative coefficient, whose index is equal to the excess of the degree of the equation above the exponent of the first negative term. 306. Problem. To find an inferior limit of the positive roots. Solution. Substitute in the given equation for x, the value 1 x= y and find, by the preceding article, a superior limit of the positive values of y, after the equation is reduced to the usual form; and denote this limit by L'. > L' is an inferior limit of the positive roots of the given equation. 307. Problem. To find the limits of the negative roots of an equation. Solution. Substitute for x Limits of Real Roots. and the positive roots of the equation thus formed are the negative roots of the given equation; and, therefore, the limits of its positive roots become, by changing their signs, the required limits. 308. Corollary. By the substitution of different numbers for p and q, in arts. 290 or 294, the limits between which each root is obtained can be narrowed to any extent which may be desired, until they may be adopted as the first approximations to the roots in the method of art. 179. Thus, it is easy to obtain the first left hand significant figure. 309. EXAMPLES. 1. Find the left hand significant figure of the real roots of the equation 5 x3 6 x + 2 = 0. Solution. First. In this case, 6 is the greatest negative coefficient and - 6x is the first negative term, so that, by art. 305, 1+ √6=3·5 is a superior limit of the positive roots. To find the limit of the negative roots, let x=- y, and the equation becomes, by reversing its signs, so that 5y36 y 2 =0; −(1+√6)=—3·5 is the superior limit of the negative roots, and the roots are all contained between 4 and 4. so that the row of signs when x = 4 is so that the equation has three real roots. The row of signs when x = 0 is 十, 一, 一, 十 ; so that two of the roots are positive and one is negative. The substitution of positive integers, gives for the rows of signs when x = =1 so that both the positive roots are contained between 0 and 1. The substitution of the positive decimals 0.1, 0.2, 0.3, &c., gives the following rows of signs. x = 0.1 x=0.2 x = 0·3 x=0·4 x = 0.5 x = 0.6 x = 0.7 x = 0.8 x = 0·9 | +, +, +, so that one real root is contained between 0-3 and 0.4, and the other between 0.8 and 09; their first approximate val ues are, then, 0.3 and 0.8. The substitution of the negative integers gives, in the same way root. Limits of Real Roots. —1, for an approximate value of the negative 2. Find the left hand significant figures of the roots of 3. Find the first approximation to the roots of the equa tion x5 — 15 x3 + 132 x2 + 36 x + 396 = 0. Ans. 1,-1,- 5. 4. Find, by Stern's theorem, the greatest possible number of real roots which the equation g10 — 10 8 — 2+x−11=0 can have between +1 and Solution. In this case we have, by art. 294, x10-10 x8 — x2 + x — 11. น 80 x74x3+1 560 x6 12 x2 UIV 30240 x5 3360 x5 16800 x4 24 x 67200 x3 24 -+-+-+, -, -, +, -,+; so that the number of these roots cannot exceed 8. |