Note 1.-When a negative root is to be found, change the signs of the alternate terms of the equation, and proceed as for a positive root. Note 2.-When three or four decimal places in the root are obtained, the operation may be contracted, and much labor saved, as will be seen in the following examples: EXAMPLES. (1) Find all the roots of the cubic equation x3−7x+7=0. By Sturm's theorem, the several functions are (Note, p. 328), V = x3-7x+7 V1 =3x2-7 V1 =+ Hence, for r=+∞ the signs are ++++ no variation, -+-+ three variations; therefore the equation has three real roots, one negative, and two positive. To determine the initial figures of these roots, we have hence there are two roots between 1 and 2, and one between -3 and —4. But in order to ascertain the first figures in the decimal parts of the two roots situated between 1 and 2, we shall transform the preceding functions into others, in which the value of x is diminished by unity. Thus, for the function V we have this operation: And transforming the others in the same way, we obtain the functions V'y3+3y2-4y+1; V',=3y2+6y-4; V'2-2y-1; V'3=+. y=1 then the signs are ++ two variations, Let Therefore, the initial figures of the three roots are 1·3, 1·6, and —3. The rest of the process, with a repetition of the above, is exhibited and afterward explained below. The process here is similar to that on p. 318. The numbers marked with stars are the coefficients of the equation having the reduced roots. Thus, *3, *4, and *1 are the coefficients of the equation whose roots are 1 less than those of the proposed equation. The right-hand 3 of *33 is the 3 tenths added in the next step of the process, which has for its object to reduce the roots by .3. The coefficients of the resulting equation are *39, —*193, and *97. Now, instead of going on in this manner to obtain the following figures, 568, &c., of the root, the method of proceeding changes; the 193, which is the penultimate coefficient, becomes a trial divisor, by which dividing the absolute term 97, which is .0097, the divisor being 1.93, the quotient is 5, the next figure of the root, which is .05. This 5 is annexed to the *39, and we proceed as before; that is, multiply the *395 in the first column by this 5, producing 1975 in the second column, and by addition, 1·7325, and so on. To show that the quotient figure 5 is obtained by means of the trial divisor, observe that the 1.7325 is nearly equal to the *1.93 above, and that the 088625 in the third column, which is the product of 1-7325 by the 05, is nearly equal to the *-097 above; hence the quotient of *·097 by 1·93 is nearly this same ·05. The further we proceed, the more accurate this process becomes, for the first figure of each number in the first column being units, this, multiplied by the decimal figure found in the root, which is thousandths, tens of thousandths, and so on; that is, soon a very small fraction gives thousandths, ten of thousandths, and so on, or a very small fraction, for the product; and, the first figure in the numbers of the second column being also units, these numbers are not much affected by the addition of the above-named products.* When the number of decimal places in the numbers of the third column becomes equal to the number of decimal places required in the root, it will not be necessary to obtain any more in the third column; and as each new decimal figure in the root, multiplied by the number in the second column, would make one more place in the third, it will be necessary to cut off one figure in the second column, and, for a similar reason, two figures in the first column. As soon as the figures are all cut off in the first column, the process becomes simply one of division, the divisor and dividend rapidly diminishing. We have thus found one root x=1.356895867......., and the coefficients of the successive transformed equations are indicated by the asterisks in each column. To find another, we have the following: For the negative root, change the signs of the second and fourth terms. To show this in a more general way, let ax"+Bx"¬1+Bx"....+Вn-1+B=0 be one of the depressed equations which is to furnish the next decimal place of the root of the proposed equation; the value of x in this depressed equation will of course be a very small fraction; hence the higher powers of it may, without much error, be neglected. The depressed equation thus reduces to nearly; that is, it may be obtained by dividing the ultimate by the penultimate coefficient. Hence the three roots of the proposed cubic equation are (2) Find the roots of the equation 23+113-102x+181=0. We have already found the roots to be nearly 3·21, 3·22, and —17. Example 4, p. 330.) (See In a similar manner, the two remaining roots will be found to be and (3) Given the real roots. Here we have x=3.22952121 x=-17.44264896. +3+x2+3x-100=0, to find the number and situation of V = x2+ x3+ x2+3x-100 V1=4r3+3x+2x+3 V1=-5x-34x+1603 V3=-1132x+6059 Let x=∞ then signs are +−−+− three variations, hence two roots are real and two imaginary; and the real roots must have contrary signs, for the last term of the equation is negative. To find the sit In this example the function V1 vanishes for x=-1, and for the same value of x the functions V and V, have contrary signs, agreeably to Lemma 2, and writing + or for 0 gives the same number of variations. initial figures of the root are, therefore, 2 and 3. Y -- The |