can be given to this letter, is called an identical equation, in order to distinguish it from a common equation, that is, an equation which can only be satisfied by giving particular values to this letter. (42.) 189. The method of indeterminate coefficients requires that we should know, à priori, the form of the development with reference to the exponents of x. The development is generally supposed to be arranged according to the ascending powers of x, commencing with the power ao; sometimes, however, this form is not exact; in this case, the calculus detects the error in the supposition. For example, develope the expression 1 whence, by clearing the fraction, and arranging the terms, 0=−1+3Аx+3B | x2+3C | x3 +3D | xa + -1=0, 34-0, 3B-A=0..... Now the first equation,-1=0, is absurd, and indicates that the above form is not a suitable one for the expression 1 but if we put this expression under the form x 5 1 3 pose that X it will become, after the reductions are made, 3A+3B | x+3C | x2+3D | x2+ 0= which gives the equations 3A-1=0, 3B-A=0, 3C-B=0...., that is, the development contains a term affected with a nega tive exponent. 190. Demonstration of the Binomial Formula, by the method of indeterminate coefficients. xTM (1 + 2)", or Since (x+a) can be put under the form (1+ a xTM(1+y)", by making y=, it is only necessary to develope (1+y)", m being any quantity whatever. Ρ First, let m be equal to a positive number Assume q P (1+y)=1+Ay+By2+Cy3 + Dy1 + - - - - (1). [We are led to assume this form for the development, from the formation of the first entire powers, and by observing that when y=0, the first member reduces to 1, from which it follows that the part of the second member which is independent of y, must be equal to 1.] To determine the coefficients A, B, C, D for y in the equation (1); it becomes substitute z (1 + z)? = 1 + Az + Bz2 + Cz3 + Dza1 +- - - - -(2). It is evident that the values of A, B, C . . . . are the same in this equation as in the equation (1), since they are independent of any value given to y. Subtracting these two equations, member from member, we (1+y)? —(1+z)2 =A(y—z) + B(y2 — z2) + C(y3 —z 3) +D(y^—z1)+-----(3). Making (1+y)=u, and (1+z)'=v, there will result 1+y=u, 1+z=v; whence y-z-u-v, and the equation (3) becomes u” —vr=A(y—z)+B(y2 −z2)+C(y3 —z3)+D(y1 −z1)+..(4), or, dividing the first member by u-vq, and the second by its equal y-z, we have UP · VP __A(y—z)+B(y2—z2)+C(y3—z3) +D(ya —z1) +... y-z Now up is divisible by u-v, (No. 31), and gives the quotient, 1 UP 1+vUP 2+v2 up 3 +.....+ v»- 1‚ In like manner, (u'—v') : (u—v) gives .... Moreover, y-z, y2—z2, y3 —z3, y1—z1 . . . . divided by y-z, give the quotients 1, y + z, y2+yz+z2, y3 +yz2 + y2 z + z3• • • • ; therefore, the equation (4) becomes uv =A+B(y+2) +vu?-? + C(y2+yz+z2)+D(y3+yz2+y2z+z3)+ ... Making y=z in this last equation, whence u=v (since 12 (1+y)=u, and (1~2)=v), the first member reduces to ; and by putting in the place of u" its value (1+y),, or 1+Ay+By2+Cy3+....., and in place of u", its value 1+y, this first member becomes Moreover, the second member reduces to A+2By+3Cy2+4Dy3 + .... ; A+2By+3C | y2+4D | y3+5E | y1+.... + A +2B +3C Comparing the terms of the two members of this identical equation, we obtain the following equations: P. A=2B+A, 2B=4(−1); hence B= 2 4(-1) B(-2) ", B=3C+2B, 3C=B (¦—2); hence C= В 3 1. C=4D+3C, 4D=C(?, −3) hence D= (-3) ; and so on for the rest of the coefficients. 4 The law for the formation of the coefficients from each other, is manifest. Let N be that coefficient which has n coefficients before it, and M that which immediately precedes it. We evidently have By examining the preceding demonstration, it will be perceived that it applies to the case in which q=1; that is, in which the exponent is a whole number. As to the case in which m is equal to a negative fraction, p we proceed exactly in the same manner as above until we obtain the equation which corresponds to equation (4), viz. UP—v=A(y—2) + B(y2—z2) + C(y 3 — z 3) + . . . . Therefore, by dividing the first member by u'-v', and the second by its equal y-z, we have 1 u?—v” ___ A(y—z) + B(y2—z2) + then making y=z, whence uv, it becomes The remainder of the calculus is exactly similar to that of the preceding case. 191. Recurring Series. The development of algebraic fractions by the method of indeterminate coefficients, gives rise to certain series, called recurring series. It has been shown in No. 187, that the development of the in which each term is formed by multiplying that which precedes This property is not peculiar to the proposed fraction; it belongs to all rational algebraic fractions, and it consists in this, viz.: Every rational fraction involving x, when developed, gives a series of terms, each of which is equal to the algebraic sum of a certain number of preceding terms, multiplied respectively by certain constant quantities, which are the same for any term of the series. The collection of constant quantities, by which a certain number of the preceding terms should be multiplied, is called the scale of the series. In the preceding series, the scale is called a recurring series of the first order. b' a , and the series is |