dy tion (142), and as is explicitly a function of x and z, it is dz du convenient for the purposes of differentiation to consider the du dy dr product of two functions, viz. of (y) and of ; whence, differentiating, we have d2u d du dy $ (y) dz § dy du +(y)} dy + dx dz dy day (y) dx dz dz let = 0, in which case, as before, y = z, dy = dz, and u = f(z); Again, considering to involve a product of two functions, du viz. {+(y)} and dy, the former of which is explicitly a function of y only, and the latter is an explicit function of both ☛ and z, and differentiating and substituting from (143), dxn and since the order of differentiation may be reversed by virtue of Art. 79, dnu dn-2 d S du = don dzn-2 dx dy dy | dz S' Sdy dy du + dy dz dx dy dy {+(y)}" }, by virtue of equation (143) ; If therefore the formulæ are true for n-1, they are true for n; they are true when n = 3, therefore they are true when n = 4, and therefore are true for all positive integral values of n, which are the only cases in which it is necessary for us to find them. Substituting then, in equation (141), the values above determined, we have If, having given y=z+xp(y), the problem is to determine y, then f(y) = y, and f(z) = 2; and the above formula becomes In applying the above theorems to particular examples, it is most convenient first to substitute the specific forms of the functions, and subsequently to replace the variables by their specific values. 91.] Ex. 1. Given a-by+cy = 0; to find y. a series which is identical with that arising from the develop Ex. 2. Given y3-ay+b=0; to find y". b 1 On comparing the given equation y=+y3 with the typical form, we have a d x3 X3 + {nzn−1; 1 dz x2 d2 1.2 dz2 x2 = zn+nzn+2 +n(n+5)zn+4. +n(n+8)(n+a) +6 + 1 1.2 b21 n (n+5) b1 1 { 1 + n + a2 a Ex. 3. Given y = a+be"; 1.2 a4 a2 to find loge y. Ex. 4. Given y = a + e sin y; to find cos y and sin 2y. cos y = cos a — (sin a)2 € – 3 (sin a)2 cos a — 1 e2 92.] By the preceding theorem of Lagrange any function of x, say f(x), may be expanded, when certain conditions are fulfilled, in ascending powers of any other function (x); so that, as by Maclaurin's Theorem a function is expanded in ascending powers of x, by this Theorem it is expanded in ascending powers of another function of x. Let the form of the required series be our object is to determine the coefficients A0, A1, A2, which are independent of x. Let x-a be a factor of (x); which renders 4(x) = 0; and let other factors; so that ... that is, let a be a value of x 1 4(x) be the product of all the (149) (150) .'. x = a + ((x) × √(x). It will be convenient to replace (x) by t ; and thus (150) becomes x = a + t√(x). Now this form is clearly the same as (142); and when t=0, x = a; and our object is to expand f(x) in ascending powers of t; therefore replacing y by x and z by a in (146), we have f(x) = [f(x)] + [d.f(x) y(x)]{ + โร d (d.f(x) 12 dx dx 1.2 wherein the square brackets indicate that particular values of the |