greatest possible number of tenths, then the greatest possible number of hundredths, etc. The approximation to the root at any stage is the total distance the origin has been moved. 2. The process on page 591 illustrates the general method of extending a positive root to any number of figures after finding enough figures to separate it from other roots, if any, that are nearly equal to the root sought. The following process illustrates certain modifications of Horner's method for roots that are nearly equal, as explained below. Let x-12x + 38 x2+8x-112 = 0. EXPLANATION.-By Sturm's theorem it is found that the equation has a negative root and three positive roots, one between 2 and 3, and two between 5 and 6. The above process is concerned with the roots between 5 and 6. Decreasing the roots by 5, the first transformed equation is which has only two positive roots, both between 0 and 1, the least positive root having been passed, that is, changed to a negative root. This is indicated by the change of sign of the absolute term from to + and the loss of one variation of sign. Since the first transformed equation has two roots between 0 and 1, the rule that the next figure of the root is the greatest number of tenths by which the roots can be decreased without changing the sign of the absolute term is not sufficient to guard against passing the roots, for it is possible, by using too large a number, to pass both roots at the same time and still leave the absolute term positive in consequence of two changes in sign. The roots may be separated, however, by applying the rule that the equation loses one variation of sign for each positive root that is passed when the roots are decreased by a number greater than the root in question. If the roots of (1) are decreased by .4, as shown in the process, or by any less number of tenths, the absolute term remains positive, and besides, no variations are lost. Hence, both positive roots are greater than .4. But if the roots are decreased by .5 or by any greater number of tenths, two variations are lost. Hence, both positive roots are less than .5. That is, both positive roots of (1) lie between 4 and 5, and the corresponding roots of the given equation lie between 5.4 and 5.5. The second transformed equation 24 +9.6 23 + 18.56 22. 1.504 z+.0176 = 0 (2) has two roots between 0 and .10. Trying .01, .02, .03, ..., it is found that the absolute term changes sign and one variation is lost in passing from .01 to .02, and again in passing from .06 to .07. Hence, the roots of (2) are .01+ and .06+, and the corresponding roots of the given equation are 5.41+ and 5.46+. The roots are now separated and the above process may be continued to find either of them by the easier method given in 1. In finding the less root 5.414 +, the absolute terms will continue to be positive; in finding the greater root 5.464 +, they will be negative beginning with the third transformed equation. RULE FOR POSITIVE ROOTS. Find the first figure of the root by trial or by Sturm's theorem, and decrease the roots by this number. If the root is separated from other roots, find each succeeding figure of the root by decreasing the roots by the greatest number of the next lower order that will not cause the absolute term to change sign. If the root is not separated from other roots, first separate the roots by Sturm's theorem; or decrease the roots by the greatest number of the next lower order that will not cause the equation to lose more variations than there are positive roots that are less than the root sought. After several figures of the root have been found, the next figure may be found usually by neglecting powers of the unknown number higher than the first and solving the resulting simple equation; or, if the coefficient of the first power is small in comparison with the coefficients of the higher powers or is equal to zero, by neglecting the powers higher than the second and solving the resulting quadratic equation. ADV. ALG.-38 694. To avoid decimals and to abbreviate the work when many figures of the root are required, Horner's process is modified as explained below. Compare with the process on page 591. EXPLANATION. 1. Decimals are avoided by multiplying the roots of each transformed equation by 10. the numbers substituted for y, z, v, ... being regarded as 4, 1, 4, instead of .4, .01, .004, .. ... 2. After several figures of the root have been obtained (in this case beginning with the fourth transformed equation), nearly as many more are accurately found by a contracted process. For since dividing both members of an equation by the same known number does not change the roots, after the roots of the fourth transformed equation have been multiplied by 10, we may divide each coefficient by 1000; and if the decimal parts thus formed are cut off, the accuracy of the root will not be affected for several figures. Some allowance for the decimal cut off is made, when convenient. Thus, 102.420 × 2 is nearer 205 than 204 and is taken as 205. In practice the two operations of multiplying the roots by 10 and dividing the equation by 1000 are performed simultaneously by omitting to annex the ciphers and then cutting off the last figure of the coefficient of the first power, the last two figures of the coefficient of the second power, and so on. In the above process this practically destroys the leading coefficient and reduces the equation to the quadratic 102 w2 + 2096619 w — 4478056 = 0, next to the quadratic u2 + 209703 u 284408 0, and finally to the simple equation 20970 t74704 = 0. = 3.562+. Solving this equation by dividing 74704 by 20970, t 695. To find the value of a negative root, Horner's method may be applied to find the corresponding positive root of that transformed equation whose roots are numerically equal to the roots of the given equation, but opposite in sign. The resulting positive root with its sign changed will be the required negative root. 696. Horner's method may be used to find the principal nth root of any number. For example, the principal fifth root of 2 may be found by solving the equation 220 by Horner's method. EXAMPLES 14 0. = 1. Find by Horner's method the root lying between 1 and 2 of the equation 4 x3 − 3 x2 + x Test the result by multiplying the roots by 4 and substituting 4 times the root found. Find the imaginary roots. 2. The equation 8 x3- 29 x2 + 29 x − 21 = 0 has a root between 2 and 3. Find all the roots. 3. The number and situation of the real roots of the equation x1 — x3 — 4 x2 + 3x + 3 = 0 have been discussed in example 1, page 588. Find all the roots to the third decimal place. 4. Find all the roots of 1 6 x2+5x2 + 14 x 40 to the nearest third decimal place. 5. Find the fifth root of 330383.69407 by Horner's method. 697. Newton's method of approximation. If one and only one real root of the equation lies between a and b, and if a is less than this root by a number h that is small in comparison with a, then a + h is the root, and in (a + h)" + p(a + h)"1 + q(a + h)n-2 + ... + s(a + h) + t = 0 (2) the terms involving h2, h3, h4, Then, approximately, απ +pa"-1 ... may be neglected. +ga"-2 -3 +[na"-1+p(n − 1)a"-2 + q(n − 2)a”-3 + ... + 8]h = 0; − that is, § 639, f(a) +f'(a)h = 0, or h=— f(a) (3) Thus, to approximate to the root of x3- 2x − 50 lying between 2 and 3, denote the root by 2 + h as in (2), neglect the terms involving h2 and h3 in (2+ h)3 − 2 (2 + h) - 5 = 0, and solve the resulting equation 10h 10. The result is h=.1, approximately. -- Or proceed as follows, using the result given in (3) : f(x) = x3 − 2 x − 5 and ƒ'(x) = 3 x2 — 2. Next, using 2.1 for a, it is found that ƒ(2.1)= .061 and ƒ'(2.1) = 11.23. Hence, by (3), the next addition to the root is root is 2.1 .005, or 2.095, approximately. NOTE. This method possesses but little practical value. RECIPROCAL EQUATIONS 698. When each root of an equation is the reciprocal of some other root or of itself, the equation is called a Reciprocal, or Recurring, Equation. 699. Since the only numbers that are reciprocals of themselves are 1 and 1, there are four kinds of reciprocal equations : 1. Those whose roots in pairs are reciprocals of each other. 2. The first kind with the root 1 in addition. 3. The first kind with the root 1 in addition. 4. The first kind with the roots 1 and - 1 in addition. |