LOCATION OF THE ROOTS. 568. If the variable x changes continuously from a to b the function f(x) will change continuously from f(a) to f(b). Let c and c+h be any two values of x lying between a and b. We have h2 hn ƒ(c+h) − f(c)=hƒ' (c) + ƒ'" (c) + ... + H ƒ^(c) ; 12 and by taking h small enough the difference between f(c+h) and f(c) can be made as small as we please; hence to a small change in the variable x'there corresponds a small change in the function f(x), and therefore as x changes gradually from a to b, the function f(x) changes gradually from f(a) to f(b). 569. It is important to notice that we have not proved that f(x) always increases from ƒ(a) to ƒ(b), or decreases from f(a) to f(b), but that it passes from one value to the other without any sudden change; sometimes it may be increasing and at other times it may be decreasing. 570. If f(a) and f(b) are of contrary signs then one root of the equation f(x)=0 must lie between a and b. As x changes gradually from a to b, the function f(x) changes gradually from f(a) to f(b), and therefore must pass through all intermediate values; but since ƒ(a) and ƒ(b) have contrary signs the value zero must lie between them; that is, ƒ(x)=0 for some value of x between a and b. It does not follow that f(x)=0 has only one root between a and b; neither does it follow that if ƒ(a) and ƒ(b) have the same sign f(x)=0 has no root between a and b. 571. Every equation of an odd degree has at least one real root whose sign is opposite to that of its last term. In the function f(x) substitute for x the values +∞0, 0, successively, then ƒ(+∞)=+∞, ƒ(0)=Pn ƒ(-∞) = -∞. 81 If pn is positive, then f(x)=0 has a root lying between 0 and -∞, and if p, is negative f(x)=0 has a root lying between 0 and +∞. 572. 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 f(+∞)=+∞, ƒ(0)=Pn ƒ(-∞) = +∞ ; but Pn is negative; hence f(x)=0 has a root lying between 0 and +, and a root lying between 0 and -8. 573. 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 b; 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. ... K Suppose that a is greater than b, and that a, ß, y, 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-α)(x-ẞ) (x−y) ... (x−к); then Hence f(x)=(x-a)(x − ẞ)(x − y) ... (x−k)P(x). f(a)=(a− a)(a−ẞ) (a− y) ... (a−k)Þ(a). - 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. 570], which is contrary to the hypothesis. Hence if ƒ(a) and ƒ(1) have contrary signs, the expressions (a-a) (a-B) (ay)... (a−k), must have contrary signs. Also the factors in the first expression 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 a, ß, y, ...K must be odd. Similarly if f(a) and ƒ(b) have the same sign the number of factors must be even. In this case the given condition is satisfied if a, ß, y, ... 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 L. f. 1. Find the successive derived functions of 2x1 — x3 — 2x2+5x−1. Solve the following equations which have equal roots: 2. x4-9x2+4x+12=0. 3. x2-6x+12x2-10x+3=0. 4. x5-13x+67x3-171x2+216x-108=0. 5. x5-x+4x2-3x+2=0. 6. 8x4+4x3-18x2+11x-2=0. 7. Show that the equation 10x3−17x2+x+6=0 has a root between 0 and -1. 8. Show that the equation x1-5x3-3x2+35x-70-0 has a root between 2 and 3 and one between -2 and -3. 9. Show that the equation x-12x2+12x-3=0 has a root between 3 and 4 and another between 2 and 3. 10. Show that x5+5x-20x2-19x-2=0 has a root between 2 and 3, and a root between -4 and -5. STURM'S THEOREM AND METHOD. 574. In 1829, Sturm, a Swiss mathematician, discovered a method of determining completely the number and situation of the real roots of an equation. 575. Let f(x) be an equation from which the equal roots have been removed, and let f(x) be the first derived function. Now divide f(x) by f(x) and denote the remainder with its signs changed by f(x). Divide f(x) by ƒ2(x) and continue the operation, which is that of finding the H.C.F. of f(x) and ƒ¡(x), except that the sign of every remainder is changed before it is used as a divisor, until a remainder is obtained independent of x; the sign of this remainder must also be changed. The expressions f(x), f(x), ƒ2(x), ... Sturm's Functions. fn(x) are called Let Q1, Q2, ... Qn-1, denote the successive quotients obtained; then the steps in the operation may be represented as follows: f(x) =Q1 fi(x) −ƒ1⁄2(x), f(x) = Q2 ƒ2(x) −ƒ1⁄2(x), 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)−ƒ4(x) if f(x)=0 we have f2(x)=-ƒ1(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 c be a value of x which makes some function except the first vanish: for example fr(x) so that ƒ,(c)=0. Now when x=c, fr-1(x) and fr+1(x) have contrary signs and thus just beforex=c and also just after x=c the three functions fr-1(x), fr(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 exceptƒ(x) vanish. · (2) Let c be a root of the equation f(x)=0 so that f(c)=0. Let h be any positive quantity. 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)=hfi(c), and as h is a positive quantity, f(c+h) and ƒ¡(c) have the same sign. That is, the function just after x passes a root has the same sign as f(x) at a root. In a like manner we may show that f(c-h) = −hf1(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 f(x)=0. There is at no time a gain in the number of variations of sign, hence the theorem is established. 576. 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. 577. When + or - is substituted for x, the sign of any function will be that of the highest power of x in that function. 578. Example. Determine the number and situation of the real roots of x3-3x2-9x-4=0. Here fi(x)=3x2+6x−9. Now any positive factor may be introduced or removed in finding f2(x), fs(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)3x3+9x2-27x-12(x+1 Hence the number of real roots is 3, of which one is positive and two are negative. |