When we wish to obtain root correct to a limited number of places, we may save much of the labor of the operation by cutting off all figures beyond a certain decimal. Thus if, in the example above, we cut off all beyond five decimal places in the successive dividends, and all beyond four decimal places in the divisors, it will not affect the first six decimal places in the root. Ex. 3. Find the roots of the equation x-12x2+12x=3. Ex. 4. Find the roots of the equation Ex. 5. Find the roots of the equation +0.934685, +3.308424, Ans. +3.824325, +1.879508, +7.053058. Ex. 6. Required the fourth root of 18339659776. Ans. 368. Ex. 7. Required the fifth root of 26286674882643. Ans. 483. Ex. 8. There is a number consisting of four digits such that the sum of the first and second is 9; the sum of the first and third is 10; the sum of the first and fourth is 11; and if the product of the four digits be increased by 36 times the product of the first and third, the sum will be equal to 3024 diminished by 300 times the first digit. Required the number. 469. Newton's Method of Approximation. Let x3+Bx2+Сx=V be an equation to be solved. Find, by trial, a number, r, nearly equal to the root sought, and let r+h denote the exact value of the root, so that h is a small fraction which is to be determined. Substitute r+h for x in the given equation, and there will result a new equation containing only h and known quantities. Now, since his supposed to be a small fraction, h2 and h3 will be small compared with h; and if we reject the terms which contain the second and third powers of h, we shall have, approximately, h= p3+Br2+Cr-V -3r2-2Br-C This correction applied to the assumed root gives a closer approximation to the value of x. Repeat the operation with this corrected value of r, and a second correction will be obtained which will give a nearer value of the root; and, by successive repetitions, the value of the root may be obtained to any required degree of accuracy. The value of h may, however, be found more briefly by observing that the numerator is the first member of the equation after V has been transposed and x changed to r; and the denominator is the first derived function of the numerator with a negative sign, Art. 450. EXAMPLES. 1. Find a root of the equation 3+2x2+3x=50. For the numerator of the value of h, we have We find, by trial, that x is nearly equal to 3. If we substi tute 3 for r, we shall have Hence x=2.9 nearly. If we substitute this new value of r, we shall find the value of h to be +.00228. Hence x 2.90228. If we repeat the operation with this last value of r, we shall find the value of h to be +.0000034. Hence x=2.9022834. 2. Find a root of the equation 25—6x=10. Hence x=1.86 nearly. If we assume r=1.86, we shall find the value of h to be -.021. Hence x=1.839 nearly. If we assume r=1.839, we shall find the value of h to be +.00001266. 3. Find a root of the equation x3-9x=10. 4. Find a root of the equation x3+9x2+4x=80. Ans. x 2.4721359. 470. Approximation by Double Position.—Find, by trial, two numbers, r and r', as near as possible to the true value of x; substitute them successively for x in the given equation, and let E and E' represent the errors which result from these substitutions. We assume that the errors of the results are proportional to the errors of the assumed numbers. This supposition is not entirely correct; but if we employ numbers near to the true values, the error of this supposition is generally not very great, and the error becomes less and less the further we carry the approximation. We have then E: E':: x-r:x-r'. Whence, Art. 305, E-E': r'-r:: E:x-r; that is, As the difference of the errors is to the difference of the two assumed numbers, so is either error to the correction required in the corresponding assumed number. This correction, being added to the assumed number when it is too small, or subtracted when too great, will give a near approximation to the true root. This result, and some other number, may now be used as new values of r and ' for obtain. ing a still nearer approximation, and so on. It is generally most convenient to assume two numbers which differ only by unity in the last figure on the right, or one of the values of r already used, together with the approximate root, may be employed for the two assumed numbers. This method of approximation is applicable to many equations which can not be solved by either of the preceding methods. EXAMPLES. 1. Find one root of the equation x3+x2+x-100=0. When 4 and 5 are substituted for x in this equation, the results are 16 and +55. Hence Therefore 55+16:5-4:: 16:22. We now assume the two values 4.2 and 4.3, and, substituting them for x in the given equation, we obtain the results -4.072 and +2.297. Hence 4.072+2.297: 4.3-4.2 :: 2.297:.036. Therefore x=4.264 nearly. Assuming again the two values 4.264 and 4.265, and substituting them for x, we obtain the results -.027552 and +.036535. Hence Therefore .064087:.001::.027552:.0004299. 2. Find one root of the equation x3+2x2-23x-70=0. Ans. x=5.13458. Ans. x 10.2610. 3. Find one root of the equation x3-3x2-75x-10000=0. 4. Find one root of the equation 25+3x+2x3-3x2-2x-2=0. Ans. x=1.059109. 471. The different Roots of Unity.—The equation "—a would appear to have but one root, that is, x=Va; but, by Art. 436, it must have n roots; that is, the nth root of a must have n different values. Unity must therefore have two square roots, three cube roots, four fourth roots, five fifth roots, six sixth * roots, and so on. Ex. 1. Find the two roots of the equation x2=1. Since one root of this equation is x=1, the proposed equation must be divisible by x-1; and dividing, we obtain x2+x+1=0. Now the roots of this equation are Hence the required roots are +1, 1(−1+√−3), and ₫(—1—√−3), which are the cube roots of unity; and these results may be easily verified. Ex. 3. Find the four roots of the equation x1=1. The square root of this equation is x2=+1, or =— -1. Hence the required roots are +1, −1, +√−1, −√—1. Ex. 4. Find the five roots of the equation x=1. Since one root of this equation is x=1, the proposed equa tion must be divisible by x-1; and dividing, we obtain which, being substituted in equation (1), gives v2+v-1=0. This equation, solved by the usual method, gives v=−1+1√5, or v———√5. |