If p is positive, then f(x) = 0 has a root lying between O and∞, and if p, is negative f(x) = 0 has a root lying between 0 and +∞. 601. Every equation which is of an even degree and has its last term negative has at least two real roots, one positive and one negative. For in this case Pn ƒ(+∞)=+∞, ƒ(0)=Pn_ƒ(− ∞) = +∞ ; but p, is negative; hence f(x) = 0 has a root lying between O and +∞, and a root lying between 0 and 602. If the expressions f(a) and f(b) have contrary signs, an odd number of roots of f(x) = 0 will lie between a and 6; and if f(a) and f(b) have the same sign, either no root or an even number of roots will lie between a and b. ... Suppose that a is greater than b, and that c, d, e, k represent all the roots of f(x) = 0, which lie between a and b. Let (x) be the quotient when f(x) is divided by the product (x — c) (x − d) (x − e) ..... (x − k); then ... ƒ (x) = (x − c) (x — d) (x − e) ··· (x − k) & (x). Hence f(a) = (a− c) (a — d) (a − e) ..... (a — k) $ (a). f(b) = (b − c) (b − d) (b − e) ... (b − k) ¢ (b). Now (a) and (b) must be of the same sign, for otherwise a root of the equation (x) = 0, and therefore of f(x) = 0, would lie between a and b [Art. 599], which is contrary to the hypothesis. Hence if ƒ (a) and ƒ(b) have contrary signs, the expressions must have contrary signs. Also the factors in the first expressions are all positive, and the factors in the second are all negative; hence the number of factors must be odd, that is, the number of roots c, d, e, •k must be odd. Similarly if f(a) and f(b) have the same sign the num ... ber of factors must be even. dition is satisfied if c, d, e, ... In this case the given con k are all greater than a, or less than b; thus it does not necessarily follow that f(x)=0 has a root between a and b. EXAMPLES XLVIII. f. 1. Find the successive derived functions of 2x4-x3-2x2+5x-1. Solve the following equations which have equal roots : 2. x4 9x2+4x+12= 0. 3. x46x8 + 12x2 - 10x + 3 = 0. 4. x5 - 13x2 + 67 x3 171 x2+216 x 108 = 0. 5. x5 3x + 2 = 0. 7. Show that the equation 10 x3 — 17 x2 + x + 60 has a root between 0 and 1. 8. Show that the equation x4 – 5 x3 + 3x2 + 35 x − 70 = 0 has a root between 2 and 3, and one between 2 and -3. 9. Show that the equation x4 12 x2 + 12 x 30 has a root between 3 and 4, and another between 2 and 3. 10. Show that x5 +5 x 2 and 3, and a root between 20 x2 · 19x20 has a root between 4 and 5. STURM'S THEOREM AND METHOD. 603. In 1829, Sturm, a Swiss mathematician, discovered a method of determining completely the number and situation of the real roots of an equation. 604. Let f(x) be an equation from which the equal roots have been removed, and let fi(x) be the first derived function. Now divide f(x) by fi(x), and denote the remainder with its signs changed by f(x). Divide f(x) by f(x) and continue the operation, which is that of finding the H.C.F. of f(x) and fi(x), except that the signs in every remainder are changed before it is used as a divisor, until a remainder is obtained independent of x; the signs in this remainder must also be changed. No other changes of sign are allowed. ... The expressions f(x), fi(x), f2(x), fn(x) are called Sturm's Functions. Let Q1, Q2,... Qn-1 denote the successive quotients obtained; then the steps in the operation may be represented as follows: fn-2(x)=Qn-1fn-1(x) — ƒn (x). From these equalities we obtain the following: (1) Two consecutive functions cannot vanish for the same value of x. For if they could, all the succeeding functions would vanish, including f(x), which is impossible, as it is independent of x. (2) When any function except the first vanishes for a particular value of x, the two adjacent functions have opposite signs. Thus in f(x)=Qƒ3(x)—ƒ1(x) if ƒ (x)= 0, we have ƒ2(x)=— ƒ4(x). We may now state Sturm's Theorem. If in Sturm's Functions we substitute for x any particular value a and note the number of variations of sign; then assign to x a greater value b, and again note the number of variations of sign; the number of variations lost is equal to the number of real roots of f(x) which lie between a and b. (1) Let be a value of x which makes some function except the first vanish; for example, f(x), so that f(c)=0. Now when x = c, fr-1(x), and fr+1(2) have contrary signs, and thus just before x = c and also just after x = c, the three functions f(x), f(x), fr+1(x) have one permanence of sign and one variation of sign, hence no change occurs in the. number of variations when x passes through a value which makes a function except f(x) vanish. (2) Let c be a root of the equation f(x)=0 so that ƒ(c)=0. Leth be any positive quantity. Now h2 f(c+h)=ƒ(c)+hf' (c) + "/2ƒ''(c) +......, [Art. 594.] and as c is a root of the equation f(x) = 0, ƒ(c) = 0, hence If h be taken very small, we may disregard the terms containing its higher powers and obtain f(c + h) = hf'(c), and as h is a positive quantity, f(c + h) and f'(c) have the same sign. That is, the function just after x passes a root has the same sign as ƒ'(x) at a root. In a like manner we may show that f(ch) = — hf'(c), or that the function just before x passes a root has a sign opposite to f'(x) at a root. Thus as x increases, Sturm's Functions lose one variation of sign only when x passes through a root of the equation ƒ(x) = 0. There is at no time a gain in the number of variations of sign, hence the theorem is established. 605. In determining the whole number of real roots of an equation f(x) = 0 we first substitute ∞ and then +∞ for x in Sturm's Functions: the difference in the number of variations of sign in the two cases gives the whole number of real roots. By substituting and 0 for x we may determine the number of negative real roots, and the substitution of +∞ and 0 for x gives the number of positive real roots. 606. When +∞ or ∞ is substituted for x, the sign of any function will be that of the highest power of x in that function. 607. Let us determine the number and situation of the real roots of x3-3x2 - 9x-4=0. Here f(x)=3x2+6x-9. Now any positive factor may be introduced or removed in finding f(x), f(x), etc., for the sign of the result is not affected by so doing; hence multiplying the original equation by 3, we have 3x2+6x-9)3x2+9x2 - 27x-12(x+1 Hence the number of real roots is 3, of which one is positive and two are negative. To determine the situation of these roots we substitute different numbers, commencing at 0 and working in each direction, thus: |