Writing down only the signs which occur in the operation, we have The double sign + occurs whenever there are two terms with different signs to be added. On examining the product it follows that: I. The ambiguous sign occurs whenever + follows -- and – follows — in the original sequence of signs. II. The signs before and after an ambiguity, or set of ambiguities, are unlike. III. A change of sign is introduced at the end. Take the most unfavorable case, that in which all the ambiguities in signs are taken as continuations; then it follows from II that the number of changes in signs is the same whether the upper or lower sign is taken; e.g., take the upper sign, then the number of changes of sign can not be less than the number in + + — — — —H — — —H — + — which has the same arrangement of signs as the original polynomial excepting a change of signs at the end. Suppose now that a polynomial is formed of the factors corresponding to the negative and imaginary roots of an equation; the result of multiplying this product by each of the factors ac—a, a -b, a -c, etc., corresponding to the positive roots a, b, c, etc., is that at least one variation in sign for each root is introduced; therefore an equation can not have more positive roots than variations in sign. 817. Descartes's Rule of Signs for Negative Roots.-If ~ a2 is substituted for a in the equation f(z) = 0, the resulting equation f(–2)=0 has the same roots as the equation f(r)=0, excepting that their signs are changed. This follows from the identical equation f(r) = (x –al) (c. — a,) . . . . (a — a,) from which we deduce f(– r) = (–1)" (a + ai) (or + as) . . . . (or + a.). Hence the roots of f(– r) = 0 are – a, , – as , . . . . – an: Therefore the negative roots of f(a) = 0 are the positive roots of f(– ar) = 0, and hence follows Descartes's rule for negative roots: The number of negative roots of the equation f(z) = 0 can not be greater than the number of variations of signs in f(– ar). ExAMPLE. —The equation a' + 15 a” + 7.c – 11 = 0 has one variation of sign and can not have more than one positive root. Again f(–2)=z'-- 15 w” – 72 – 11 = 0 has one variation of sign and therefore f(x) can not have more than one negative root. 818. Determination of the Existence of Imaginary Roots by means of Descartes's Rule. In case the sum of the maximum number of positive and negative roots is less than the degree of the equation we are sure that the equation has imaginary roots. ExAMPLE. –The equation ac' -- 152* + 72 – 11 = 0 has, 3817, at most one positive root and,. 3817, at most one negative root. Hence the given equation can not have more than two real roots, and therefore must have at least two imaginary roots. f'(a), f"(a), f"(z), . . . . . are called respectively the first, second, and third derivatives of f(r) with respect to ar. It follows that f'(a) is obtained from f(r) by multiplying each term in f(r) by the exponent of a in that term and diminishing the exponent of a by 1. ExAMPLE. – Find the derivatives of 3 a." – 2 w" – 5 r + 7. * The notation f(z), fosz), foo(a) ..., becomes inconvenient when the number * of accents is large, and hence f"(a) is used for the coefficient of o repeated r—2 times, and so on. Therefore, f(r)=0 has or has not equal roots according as f(z) and f'(a) have or have not the common factor, a - a, or some power of ac – a. 821. Hence it follows that the equal roots of f(r)=0 are given by finding the greatest common divisor of f(r)=0 and f'(r)=0, and placing it equal to zero, then solving the resulting equation. 822. Continuity of a Rational Integral Function of x. —If f(r) is a rational integral polynomial in ac, and ac is made to vary by infinitesimal increments (4637) from a to a larger number b, we shall prove that f(z) at the same time varies by infinitesimal increments; in this case it is said that f(a) varies continuously with ac. Let c and c + h be any two values of a lying between a and b. Hence we have, 4819, Since f(r), f'(a), f'(a), . . . are all finite for a = e, the limit of the second member of equation (1) is 0 for h > 0 (4668). Therefore, the }% [f(c + h) — f(e).] = 0. Hence, to infinitesimal changes in a there correspond infinitesimal changes in f(r), and as a changes continuously from a to b, the function f(a) changes continuously from f(a) to f(b). 823. When f"(c) is positive, f(a) increases with a ; and when f(c) is negative, f(a) diminishes as a increases. This theorem may be made more perspicuous by aid of the graphical representation of the following example. - ; } for which the corresponding val- 824. Theorem.—If f(a) and f(b) have contrary signs then the equation f(z) = 0 must have at least one real root situated between a and b. For, since f(x) changes continuously from f(a) to f(b), and therefore passes through all intermediary values as a changes continuously from a to b, and since by hypothesis one of the signs of f(a) and f(b) is positive and the other negative, it follows that f(z), for some value of a between a and b, must take the value zero situated between Caution.—It does not follow that f(z) = 0 has but one root between a and b; nor does it also follow, in case f(a) and f(b) have the same sign, that f(z) = 0 has no root between a and b. |