IV. A NEWTONIAN METHOD. 113. It remains to exemplify a fourth method of proceeding which may often be employed with advantage, and moreover is of historical interest, as having been employed by Newton. Assume a series for the expansion 2 0 } Then form a differential equation in the way indicated in several of the examples in the preceding chapter. Substitute the series in the differential equation and equate the coefficients of like powers of x on each side of the equation. We shall thus get equations enough to find all the coefficients except one or two of the first which may be easily obtained from the values of f(0) and ƒ’(0). Ex. 1. Expand aa in this manner. therefore, substituting from (1) and (3) in the differential equation (2), A1+2αqx+3ɑzx2+...log.a(a+α1x+α 2 x2 + ...). Hence, comparing coefficients, a1=a logea, 2α2=α1logea, 3agagloga, etc. Now, let ... therefore and + Anx2+an+1 x2+1+an+2x2+2+. Y1=α1+2α2x+...+namx2-1+(n+1)an+] +(n+2)αn+2x2+1+....., An + 1 xn Y2=2α2+...+n(n-1)a,xn-2 + (n+1)nan+12m-1 +(n+2)(n+1)an+2x+.... ... Picking out the coefficient of " in the equation (which may be done without actual substitution) we have (n+2)(n+1)αn+2 − n (n − 1)an=na„; N2 (n + 1)(n+2) a。=ƒ(0)=(sin−10)2, (2) and if we consider sin-1x to be the smallest positive angle whose A slightly different method of proceeding is indicated in the following example. x2 Ex. 3. Let y=sin (m sin -1x)=αo+α1x +α 22! Differentiating again, and dividing by 2y1, we have Differentiating this n times by Leibnitz's Theorem Now, a。=(y)x=0=sin (m sin-10)=0, (2) ..(3) (assuming that sin-1 is the smallest positive angle whose sine is x) α3 = − (m2 — 12)α1 = — m(m2 — 12), (m2 - 32)α3=m(m2 - 12) (m2 - 32), α= (m2 — 52)α=-m(m2 - 12) (m2 - 32) (m2-52), The corresponding series for cos (m sin-1x) is cos (m sin-1x)=1 2 — 22) (m2 — 42) x6 +.... m2x2 + m2 (m2 — 22) μ4 __ m2(m2 — 22) (m2 — 42) + 2! 4! x4 If we write = sin these series become 6! In this manner find all the coefficients of the Binomial Theorem. 2. If y=sin-1x=αj+A1X+A2x2+A3X3 +......., and in this manner deduce the expansion given in Ex. 2, Art. 112. 3. If y=(tan-1x)2=αo+α1x+αqx2 +ɑzx3 + ..., prove that (n+2)(n+1)an+2+2n2a, + (n-2) (n − 1)an-2=0. CONTINUITY. 114. DEF. A function is said to be continuous between any two values of the independent variable involved if, as that variable is made to assume successively all intermediate values from the one assigned value to the other, the function does not suddenly change its value, but changes so that for any indefinitely small change in the variable there is never a change of finite magnitude in the value of the function. 115. Suppose the function to be p(x). Trace the curve y=p(x) between the ordinates AL(x=a) and BM(x=b). Then if we find that as x increases through some value, as ON (Fig. 15), the ordinate p(x) suddenly changes from NP to NQ without going through the intermediate values, the function is said to be discontinuous for the value x=ON of the independent variable. 116. Similarly, we may represent geometrically the dy discontinuity of a differential coefficient. For dx repre sents the tangent of the angle which the tangent line to the curve makes with the axis of x. If, therefore, as the point P travels along the curve the tangent suddenly changes its position (as, for example, from PT to PT in the figure), without going through the intermediate positions, there is a discontinuity in the value of dy dx 117. PROP. If any function of x, say p(x), vanish when x=a and when x=b and is finite and continuous, as |