57 2 90 -- p3 72 90 — 64 - 16-4 6 Whence2= .10, 372 +2r + 1 48 +8+1 And consequently, x = 4.1, nearly. Again, if 4.1 be substituted in the place of r, in the last equation, we shall have 90 - 83 - pa - -_ * --- ተጎ 90-68.921-16.81 - 4.1 =.00283; 3y2 + 2r + 1 50.43 +8.2 +1 And consequently, x = 4.1 +.00283 == 4.10283 for a second approximation. And if the first four figures, 4,102, of this number be again substituted for r, in the same equation, a still nearer value of the root will be obtained; and so on, as far as may be thought necessary. 2. Given x2 + 20x = 100, to find the value of x by approximation. Ans. X = 4.1421356. 3. Given 23 + 9x2 + 4x 80, to find the value of x by approximation. Ans. x = = 2.4721359. 4. Given 24 38x3 + 210x2 + 538x + 289 = 0, to find the value of x by approximation. Ans. X = 30.53565375. 5. Given 2015 + 6x4 - 103 - 112x2 - 207x +110 = 0, to find the value of x by approximation. Ans. 4.46410161. The roots of equations, of all orders, can also be determined, to any degree of exactness, by means of the following easy rule of double position ; which, though it has not been generally employed for this purpose, will be found in some respects superior to the former, as it can be applied, at once to any unreduced equation consisting of surds, or compound quantities, as readily as if it had been brought to its usual form RULE 2.--Find, by trial, two numbers as near the true root as possible, and substitute them in the given equation instead of the unknown quantity, noting the results that are obtained from each. Then, as the difference of these results is to the difference of the two assumed numbers, so is the difference between the true result, given by the question, and either of the former, to the correction of the number belonging to the result used ; which correction being added to that number when it is too little, or subtracted from it when it is too great, will give the root required nearly. And if the number thus determined, and the nearest of the two former, or any other that appears to be more accurate, be now taken as the assumed roots, and the operation be repeated as before, a new value of the unknown quantity will be obtained still more correct than the first; and so on, proceeding in this manner as far as may be judged necessary.* EXAMPLES. 1. Given x3 + x2 + x = 100, to find an approximate value of Here it is soon found by a few trials, that the value of x lies between 4 and 5. Hence, by taking these as the two assumed numbers, the operation will stand as follows: First Sup. Second Sup. 25 125 5 16 And consequently x = 4 + .225 = .4.225, nearly. * The above rule for Double Position, which is much more simple and commodious than the one commonly employed for this purpose, is the same as that which was first given at p. 311 of the octavo edition of my_Arithmetic, published in 1810. To this we may further add, that when one of the roots of an equation has been found, either by this method or the former, the rest may be determined as follows: Bring all the terms to the lefthand side of the equation, and divide the whole expression, so formed, by the difference between the unknown quantity (2) and the root first found; and the resulting equation will then be depressed by a degree lower than the given one. Find a root of this equation, by approximation, as in the first instance, and the number so obtained will be a second root of the original equation. Then by means of this root, and the unknown quantity, depress the second equation a degree lower, and thence find a third root; and so on, till the equation is reduced to a quadratic; when the two roots of this, together with the former, will be the roots of the equation required. Thus in the equation x3 15x2 +-63.0 = 50, the first root is found by approximation to be 1.02804. Hence, x— 1.02804).3 —- 15x2 +-63.2 -- 50 (.C2 — 13.971954 +48.63627 =0. And the two roots of the quadratic equation, X2 — 13,971953 48.63627, found in the usual way, are 6.57653 and 7.39543. So that the three roots of the given cubic equation 23- - 15x2+63x 50, are 1.02804, 6.57653, and 7.39543, their sum being. = 15, the coefficient of the second term of the equation, as it ought to be when they are right. Again, if 4.2 and 4.3 be taken as the two assumed numbers the operation will stand thus :First Sup. Second Sup 4.2 4.3 17.64 22 18.49 74.088 79.507 X 2013 95.928 102.297 95.928 Results 4.3 102.297 Therefore 6.369 .1 2.297 .036. And consequently, * = 4.3 - .036 = 4.264, nearly. Again, let 4.264 and 4.265 be the two assumed numbers; then First Sup. Second Sup. 4.264 4.265 18.181696 18.190225 77.526752. 23 77.581310 x2 99.972448 Results 100.036535 Therefore, 100.036535 4.265 100 99.972448 4.264 99.972448 .064087 : .001 :: .027552 : 0004299 And consequently, x = 4.264 +.0004299 = 4.2644299, very nearly. 2. Given (12 - 15) + x V x=90, to find an approximate value of x. Here, by a few trials, it will be soon found, that the value of ą lies between 10 and 11; which let, therefore, be the two assumed numbers, agreeably to the directions given in the rule. Then, Second Sup. V 36.482 64.5 1. And consequently, x = : 11 .482 -= 10.518. Again, let 10.5 and 10.6 be the two assumed numbers Then, Second Sup. 34.511099 6.615483 .1 : .341883:.0051679 And consequently, x = 10.6 - .0051679 = 10.5948321, very nearly. EXAMPLES FOR PRACTICE.' 1. Given 23 + 10x2 + 5x 2600, to find a near approximate value of x. Ans. x == 11.00673. 2. Given 2x4 - 16x3 + 40x2 30x + 1 = 0, to find a near value of x. Ans. x = 1.284724, 3. Given 25 + 2x4 + 3x + 4x2 + 5x == 54321, to find the value of x. Ans. 8.414455. 4. Given V (7x2 + 4x) + V (20x2 - 10x) = 28, to find the value of x. Ans. 4.510661. 5. Given ✓ [1440,2 (x2 + 20)] + ✓ [196m2 — (x2 + 24)4] : 114, to find the value of x. Ans. 7.123883. OF EXPONENTIAL EQUATIONS. 1 An exponential quantity is that which is to be raised to some unknown power, or which has a variable quantity for its index; as, ai, at, and or , &c. And an exponential equation is that which is formed between any expression of this kind and some other quantity whose value is known; as, Que = b, deri Where it is to be observed, that the first of these equations, when converted into logarithms, is the same as log. b x log. a = b, or x = ; and the second equation ariza log. 11 is the same as x log. x In the latter of which cases, the value of the unknown a, &c. log. a. log. a, quantity x may be determined, to any degree of exactness, by the method of double position, as follows : RULE.—Find, by trial, as in the rule before laid down, two numbers as near the number sought as possible, and substitute them in the given equation x log. x = instead of the unknown quantity, noting the results obtained from each. Then, as the difference of these results is to the difference of the two assumed numbers, so is the difference between the true result , given in the question, and either of the former, to the correction of the number belonging to the result used; which correction, being added to that number when it is too little, or subtracted from it when it is too great, will give the root required, nearly. And if the number thus determined, and the nearest of the two former, or any other that appears to be nearer, be taken the assumed roots, and the operation be repeated as before, a new value of the unknown quantity will be obtained still more correct than the first; and so on, proceeding in this manner as far as may be thought necessary. EXAMPLES. 1. Given 220 = 100 to find an approximate value of x. Here, by the above formula, we have & log. a log. 100 = 2. And since x is readily found, by a few trials, to be nearly in the middle between 3 and 4, but rather nearer the latter than the former, let 3.5 and 3.6 be taken for the two assumed numbers. Then log. 3.5 = .5440680, which, being multiplied by 3.5, gives 1.904238 first result; And log. 3.6 =.5563025, which, being multiplied by 3.6, gives 2.002689 for the second result. Whence, 2.002689.. 3.6 .. 2.002689 1.904238.. 3.5., 2. .098451 : .1 :: 002689 : .00273 for the first correction; which, taken from 3.6, leaves 3.59727, nearly. And as this value is found, by trial, to be rather too small, let 3.59727 and 3.59728 be taken as the two assumed numbers Then log. 3.59728 0.555974243134677 to 15 places. The log. 3.59727 = 0.555973035847267 to 15 places, |