0.4, we get 0.63+ and -0.63 for its roots; consequently, one of the roots of the equation must lie between these numbers; hence, we easily get 0.3 for the first figures of this root. And because one of the remaining roots must be less than 0.63, and the other greater than 0.63 +, we soon find - 1.2 and 0.8 for the first figures of these roots. Hence, the roots can easily be found by Horner's method. 1 Another solution. Substitute for a in the given equation, y and it will be readily reduced to 2y3 - 6y+5=0; and as the first derived equation gives y3 - 2y = 0, whose roots are 0 and 2, it follows that one of the roots of the equation in y must be less than 0 or negative; another root must lie between 0 and 2, and the remaining root must be greater than 2. The first figures of the root which lies between 0 and 2 are soon found to be 1.2; and 0.8, 2.6 are the first figures of the remaining roots. Hence, the roots are easily found by Horner's method, or by any of the common methods of approximation. 1 - Because x = it follows that if we divide 1 by the roots, y or take the reciprocals of the roots, we shall get the roots or values of a which satisfy the given equation. - Remark. It will be noticed that the roots of the equation in x, which are numerically less than 1, are in the equation in y represented by roots which are numerically greater than 1. It will also be observed that the positive root which is without the limits 0.63+ and 0.63 in the equation in x, is represented by a root in the equation in y, which is within the limits 2 and 0. It is hence clear that it will often be useful to proceed in like manner to find the reciprocals of the roots of equations. Ex. 6. Supposing the roots of 8x+14x2+4x-8 = 0 to be real, it is required to find how many of them lie between 0 and 3. are Here the first, second, and third derived equations become If we put 0 and 3 for x in the given equation, the results it follows that an even number of the roots must lie between 0 and 3. And putting 0 and 3 for x in the first derived equation, the results are 1 and 5; which having opposite signs, it follows that an odd number of the roots of the (first derived) equation lie between 0 and 3. Also, putting 0 and 3 for x in the second derived equation, the results are 7 and -2; which having opposite signs, it follows that only one root of the (second derived) equation, lies between 0 and 3. Hence, because the root of the second derived equation lies between two roots of the first derived equation, it is clear that there is only one root of this equation which lies between 0 and 3; consequently, since this root lies between two roots of the given equation, it follows that the given equation has only two roots, which lie between 0 and 3. Indeed, the roots of the given equation are - 0.732050, 0.763932, 2.732050, and 5.236068, nearly; two of which lie between 0 and 3. Hence we perceive how we may proceed in order to find the number of real roots of a given equation, which may lie between two given numbers. (49.) We will now proceed to complete the indicated expansion of X' (see (3), p. 229), and to show its use in the development of functions. 1. Resuming the equation, we have X' = X + Ah + Â ̧h3 + A2h3 + Agh* + etc., (1); in which X, A, A1, A2, etc., are supposed to be functions of x, and X' is the value of X which results from changing x into x + h, supposing x and h to be indeterminates. If the value of X which corresponds to +h+k is represented by X", it is clear, from (1), that we shall have X" = X + A(h + k) + A1(h + k)2 + A2(h +k)3+ Ag(h + k)*+ etc., or developing the powers of h + k according to the ascending powers of k, we shall get X" = X + A(h + k) + A1(h2 + 2hk + etc.) + А(h3 + 3h2 + etc.) + etc., (2). Again, since X, A, A1, A2, Ag, etc., are functions of x, if we change a into a +k they become X+ Ak + A1h2 + Áh2 + etc., A+ A'k+A"k+ etc., A, + A'k+A" + etc., A + A'‚k + A"2 + etc., and so on; in which X + Ak+ 2 etc., is clearly deduced from the right member of (1) by changing into k, and A' is the first derived function of A, A', that of A, and so on; hence, if we change in (1) into +k, it is clear that X' will be changed into X", since it is the value of X which corresponds to x+h+k; consequently, substituting the values of X, A, A1, etc., resulting from changing a into a +k, we shall have X" = X + Ak + A1k2 + etc. + (A + A'k + etc.)h + (A1 + A'k + etc.)h2 + (A2 + A ́‚k + etc.)h3 + etc. = X + A(h + k) + A1(h2 + k2)+ A(h+k+ etc. + (A'h + A',h2 + A'gh3 + A'ght + etc.)k+ (A”h + A”‚h2 + A′′1⁄2h3+ etc.)k2 + (A'''h + A'''‚h2 + etc.)k3 + · etc., (3). Because the first members of (2) and (3) are equal, it is clear that their second members must also be equal, and it is evident that they must be identical, so as to leave h and k arbitrary; and it is clear that they will be identical, if we put the coefficient of the simple power of k in the second member of (2) equal to that of k in the second member of (3); consequently, we must have A + 2Ah + 3A,h2 + 4Ah etc. A + A'h + Ah + A'gh3 + etc., (4). Since must be arbitrary, it is clear that the coefficient of any power of h in the first member of (4) must equal the coefficient of the same power of h in the second member; consequently, we must have the equations A = A, 2A1 = A', 3A, = A'1, 4A, = A'2, 5A4 = A',, and so on (5). Because A, the coefficient of the simple power of h in the right member of (1), is called the first derived function of X, we may express it by X1; and because A' is the first derived function of A, we may call it the second derived function of X, and denote it by X,; and in like manner we may call the first derived function of A' the third derived function of X, and represent it by X,; and so on. = Hence, because A'= X2, the second of (5) becomes 2A1 = X1, which gives A1 = 1⁄2 ; and thence we have A1 = = X2 1.2 2 X3 1.2' and the third of (5) is reduced to 3A,= X, 1.2' or we get ; and in a similar way we get from the fourth it will be reduced to X' = X + X1 . h + the values of A, A1, A2, Ag, etc., in the right member of (1), 1 1.2.3 1.2.3.4 h+ etc. (a), whose law of continuation is manifest; (a), which is usually called Taylor's theorem, is the development which we proposed to find. It is clear that X1 can be obtained from X by changing into a +h, and putting the coefficient of the simple power of h in the expansion equal to X1; and if we change x into + h in X1, the coefficient of the simple power of h in the expansion will clearly equal X.; and in a similar way, if we change x in X, into x + h, the coefficient of the simple power of h in the expansion will equal X,; and so on. 2 To show the use of (a), take the following EXAMPLES. Ex. 1.-To find the expansion of a when x is changed into a +h, or to develop (+) according to the ascending powers of h. Here X is represented by a1, and allowing a + 4x3h to be the first two terms of the expansion of (x + h)1, we shall have X1 = 4a3, and allowing the first two terms of the expansion of (x + h)3 to be a3 + 3x2h, we shall have 4(x + h)3 = 4x3 + 12x2h + ; consequently, 122, the coefficient of the simple power of h, will equal X.; and from 12(x + h)2 12x2+24xh + we get X, 24; and from 24(x + h) = 24x +24h we get X, 24; and since X, does not contain a, it is clear that X5, X6, etc., are (each) equal to 0. = = Hence, putting (x + h), x1, 4x3, 12, 24x, and 24, severally, for X', X, X1, X2, X3, X4, in (α), we shall get (x + h)* = 20+ 4x3h + 6x h2 + 4xh3 + h', as required. = Ex. 2.-To develop 3x-5x+2x-7, when x is changed into x + h. Since 9-10x+2, 18x10, 18, 0, 0, etc., are easily seen to be the first, second, third, etc., derived functions, we get (from (a)) 3(x + h)3 − 5(x + h)2 + 2(x + h) — 7 = 3x3 5x2 + 2x−7+ (9x2 - 10x + 2)h + (9x — 5)h2 + 3h3, as required. Ex. 3. To develop -2, when x is changed into x + h. Here the first, second, etc., derived functions are 4x3-6x2, 122-12x, 24x-12, 24, 0, 0, etc.; consequently, we get (x + h)-2(x + h)3 = x* — 2x3 + (4x3 — 6x3)h + (6x2 — 6x)h2 + (4x − 2)h3 + h1, as required. Ex. 4. Supposing the first derived function of a to be no-1 for any value of n, it is required to develop (x + h)" according to the ascending powers of h. - — — Here the first, second, etc., derived functions are non-1, n(n - 1)x2-2, n(n − 1) (n − 2)xn−3, n(n − 1) (n − 2) (n − 3) 11, and so on; consequently, from (a) we get (x + h)n = x2 n(n − 1) (n − 2) xn−8p3 + etc., n(n - 1) as required. 1 2 3 2. According to what has heretofore been shown, we may apply (a) to expand any function of x + h, according to the ascending powers of h, when particular values are assigned to a and h, provided X, X1, X2, etc., do not any of them become infinite; should any of these quantities, as X,, become infinite, then the expansion will be correct no further than to the term which contains X3. If for x, in (a), we put the particular value 0, while h is arbitrary; then if we represent the values of X, X1, X2, etc., which correspond to x=0 by (X), (X1), (X2), etc., (a) will become X' = (X) + (X1) . h + (X2) h2+ etc.; or since we 1. 2 may clearly (here) use x for h, we shall get X' = (X) + MacLaurin's Theorem; where it will be noticed that X' is a function of x and known quantities, and that (X) is the value of X' which corresponds to x=0; (X1), (X2), etc., being the values of the first, second, etc., derived functions which correspond to x = 0. MacLaurin's Theorem is often very useful in the expansion of functions. Thus, if we wish to expand (b + x)TM according to the ascending powers of x, we have, as in Taylor's Theorem, (b + x + h)m = (b + x)TM + m (b + x)m−1h + |