Hence, the half sum added to the half difference gives the greater of the two; and the half difference subtracted from the half sum gives the less of the two. It may be noticed that the method of doing this question. is substantially the same as the Rule of Double Position in Arithmetic. See Rutherford's edition of Hutton's Mathematics, p. 79. Ex. 3. Given 5, the square root of 25, and 5.1961, the approximate square root of 27, to find the approximate square root of 26. Here, using 25, 26, and 27 as different values of x, and their square roots as the corresponding values of X, we shall 9.0980126-5; consequently, 26 5.0980 nearly. Since 126 5.099, we perceive that our process gives the root correctly to two decimal places. = Ex. 4. Given 1289= 17, and 1/291 = 17.058722, to find √290. = By proceeding as in the preceding question, we get √290 = 17.02936 very nearly; since 1290 17.02938, it appears that our root does not differ from the true root by more than two units in the fifth decimal place; so that we have found the root more correctly than in the preceding question. The reason why we have found the root more correctly in this than in the preceding question clearly is, that the numbers 289, 290, 291 are nearer to a ratio of equality than the numbers 25, 26, 27. Remarks. Hence we perceive the reason for correcting (numbers or) quantities found nearly from tables, such as logarithmic, astronomical, etc., tables, and then taking proportional parts for the corrections. Thus, having 2.414973 and 2.416641, the logarithms of 260 and 261, by proceeding as in the preceding questions, we get 2.415307 very nearly for the logarithm of 260.2, which agrees with the logarithm as found directly from the tables. Ex. 5. To find a number whose square added to three times the number makes 20. Let a represent the number, then, using 3.2 and 3.3 as particular values of x, we get (3.2)2 + 3 x 3.2 = 19.84 and (3.3)2 + 3 × 3.3 = 20.79 for the corresponding particular H 9.5 values of X; consequently, we get ; and since h = = 0.16 0.0168, which gives a = 3.2168 nearly; for (3.2168) 9.5 = + 3 x 3.216819.998 +, which differs but little from 20. If 3.2168 and 3.217 are used as particular values of x, we shall find a more nearly, and so on to any extent. Remark. Since this question virtually requires such a value of as satisfies the equation 2+3x= 20, we perceive how to apply our process to the solution of equations. Ex. 6. Supposing X to have unlike signs for the particular values a and b of x, and that b is greater than a, we propose to show that there is at least one value of x which is greater than a and less than b, such that the corresponding value of X equals 0. Representing the values of X which correspond to a and b by A and B, and supposing a to increase from a to b by the successive additions of the very small positive (numbers or) quantities h, i, k, l, etc., and that H, I, K, L, etc., are the very small variations of the corresponding values of X, then we shall clearly have a +h+i+k+ etc. b, (1), and A+II+I+K+ etc. B,(2). = = = If we subtract A from the two members of (2), we get H + I + K + L+ etc. BA, (3), in which (since A and B are supposed to have unlike signs) B and A have like signs; consequently, the first member of the equation equals the absolute sum of A and B. Hence, if we form the successive sums H, H+I, H+I+K, H+I+K+ L, and so on, it is clear that A must either equal one of these sums, or be greater than one of two that are successive, and - less than the remaining one of the two. Hence, by taking , i, k, etc., sufficiently small, we shall clearly have H+I + K + .... + T′ = - A, or A+ H + I + K + . . . . T' = 0, while the sum of the remaining terms of the first member of (3) equals B. .... Because X = A + H+I+....+T'0 corresponds to x = a + h + i + . . . . +t', which is clearly between a and b, or greater than a and less than b, it is evident that a value of a can be found as proposed. Remarks.-1st. If ba is very small, while A and B are very great, it is evident from (3) that H, I, K, etc., can not be very small, unless h, i, k, etc., are extremely small. 2d. If the particular values a and b of a can be found, such that the corresponding values A and B of X have unlike signs, it results (from what has been done) that there is at least one particular value of x between a and b which will reduce X to 0, or which will satisfy the equation X = 0. Hence, if we define the roots of an equation to be those values of the unknown letter (or the independent variable) which satisfy the equation, we perceive how we may (often) proceed in order to find them. Thus, if we put a2 7x+ 5 for X, and use 6 and 7 for a and b, we get - 1 and 5 for the corresponding values of A and B; consequently, the equation 2 7x+50 has a root, which is greater than 6 and less than 7. Also, if we use 0.8 and 1 for a and b, we get 0.04 and 1 for the corresponding values of A and B; consequently, the equation -7x+ 50 has another root, which lies between 0.8 and 1. - To find these roots more nearly, we remark that B - A is the variation of a 7x+5, which results from the variation b-a of x, and that to find the root x we have x2 — 7x + 5 = 0 for the corresponding value of a 7x+5; consequently, 0A is the variation of 2 75, which results from changing x into x-a; so that we shall have B-A = b -a If, in this equation, we put 1, 5, 6, and 7, severally for A, B, and b, we shall get 6 = 1 ; or we shall have x 6 x=6 = 61 6 6.16+ for the value of the first of the preceding roots, correct to one decimal place. If 6.1 and 6.2 are put for a and b in a2 -7x+5, the corresponding values of A and B will be found to be 0.49 and 0.04; consequently, if we put 0.49, 0.04, 6.1, and 6.2 for A, B, a, and b, and proceed as before, we shall have 6.192 + for a still more correct value of the root; and by continuing the preceding process sufficiently far, the root may be found to any required number of decimal places. If 0.04, 1, 0.8, and 1 are severally used for A, B, a, and b, we shall in like manner get 0.807+ for the second of the preceding roots. -2 - 1 x + An 3d. Taking the equation of the nth degree, " + Â ̧æn−1 + A‚æ2¬2 + A3x2¬3 + . . . . + An it = may be noticed, if n is an odd number, that the equation must have at least one real root, which has a sign contrary to An, the absolute term of the equation; and that if n is an even number, and the absolute term is negative, the equation must have at least two real roots, one of which is positive and the other negative. For if n is odd, and we put x = 0, the function + A-1 etc., is reduced to An; and if we give a contrary sign to An, and suppose to be very great, it is clear that the sign of the function + A-1+ etc., will be the same as that of its first term a", which will be the same as the sign of x, because n is an odd number; consequently, the sign of the function will be contrary to that of An. Hence, because the function " + A1un−1 + Agen−1 + .... + An changes its sign between 0 and a very great and of a contrary sign to An, it follows, from what has been shown, that the given equation has at least one real root which lies between 0 and a very great and of a contrary sign to An; consequently, if An is negative, the equation has a positive root, and if An is positive, the equation has a negative root. Again, if n is an even number and An is negative, then, by putting = 0, the function will be reduced to An, a negative result; but if we give a the same sign as An, or a sign which is contrary to it, and make a very great, the sign of the function will be the same as that of its first term a", which + An will be positive, because n is an even number; consequently, because the function "+ A12-1 + A22-2 + .... changes its sign between x = 0 and x very great and of the same sign as An, and between x = 0 and a very great and of a contrary sign to An, it follows that the given equation has a pair of roots, one of which is positive and the other negative. Thus, the equation + 3x2 - 4 = 0, has a positive root; the equation — 7x+5= 0, has a negative root; and the equation a+4x-5=0, has a pair of roots, one of which is positive and the other negative. - Ex. 7.-Let X be such a function of a that when a, a +h, a + h + k are severally put for x, it shall become A', 0 (or naught), and K; then supposing hand k to be positive and very small, we propose to show that A' and K will have unlike signs. Let I express the variation of X = A' which results from changing x = a into x + h = a + h; then, by the question, we shall have A' + H = 0, or II A', so that H and A' is clearly the variation = have unlike signs. Again, since K of X = A' + H = 0, which results from changing x + h = a + h into x + h + k = a + h + k, and that h and k are very small, it follows (from (4), given at p. 229) that we shall have H K h = k very nearly; consequently, since h and k have like signs, H and K must also have like signs, and since H and A' have unlike signs, K and A' must also have unlike signs, which is in accordance with what we proposed to show. H K Remarks.-1st. If the quotients and are finite, it is h k evident (because A' is the value of X just before it is reduced to 0, and K its value just after) that the sign of the function X is changed when it passes through 0, in consequence of the (supposed) increase of its independent variable x. Hence, supposing a to increase from being less than any root of the equation X = 0, and to pass through the successive roots (or to become equal to them) as it is increased; then if H K the quotients and are finite for all the roots, it follows from what has been done, that X and H K or will have unh k |