stands for a term between the first and second of the given terms. When n is greater than 1 and less than 2, the intermediate term will lie between the second and third of the given terms, and so on. In general, the preceding formula will give the value of such intermediate terms. EXAMPLES. 1. Given the cube root of 60 equal to 3.914868, แ 61 "L 3.936497, to find the cube root of 60.25. Here D'+.021629, D′′——.000235, D'"+.000007, etc. a=3.914868, and n=.25. Substituting the value of n in the formula, we have Tn+1=a+D'—31⁄2D”+1 ́ ̧D””. The value of the 1st term is etc. +3.914868, Hence the cube root of 60.25 is 2. Find the cube root of 60.5. 3.920297. Ans. 3.925712. Ans. 3.931112. Ans. 3.927874. Ans. 3.922031. Development of Algebraic Expressions into Series. 373. An irreducible fraction may be converted into an infinite series by dividing the numerator by the denominator, according to the usual method of division. ·=1+x+x2+x3+x2+x3+, etc., to infinity. 1-x 1 1 = Ex. 2. Convert Suppose x=, we shall then have 1 1-x 1- 1 1+x Suppose x, we shall then have 1 -=2=1+1+1+++, etc. =÷=1+3+}+27+8+, etc. into an infinite series. Ans. 1-x+x2—x3 +x1 −x3+, etc. Ans. 1+2x+3x2+5x3+8x2+13x+, etc. 374. An algebraic expression which is not a perfect square may be developed into an infinite series by extracting its square root according to the method of Art. 198. Ex. 1. Develop the square root of 1+x into an infinite series. Hence the square root of 1+x is equal to х x2 XC3 ნეც. 1+ +2-8 + 16 128+, etc. Suppose x=1, we shall have 5 √2=1+1−3+1'8-198+, etc. T Ex. 2. Develop the square root of a2+x into an infinite series. X x2 X3 5x4 Ans. a+ + 2a 8a3 16a5 128a7 +, etc. Ex. 3. Develop the square root of a +x into an infinite series. Ex. 4. Develop the square root of a*-x into an infinite series. Ex. 5. Develop the square root of a2+x2 into an infinite series. Method of Undetermined Coefficients. 375. One of the most useful methods of developing algebraic expressions into series is the method of undetermined coefficients. It consists in assuming the required development in the form of a series with unknown coefficients, and afterward finding the value of these coefficients. This method is founded upon the properties of identical equations. 376. An identical equation is one in which the two members are identical, or may be reduced to identity by performing the operations indicated in them. As 377. It follows from the definition that an identical equation is satisfied by each and every value which may be assigned to a letter which it contains, provided that value is the same in both members of the equation. Every identical equation containing but one unknown quantity can be reduced to the form of A+B+Cx2+Dx3+, etc. A'+B'x+C'x2+D'x3+, etc. 378. If an equation of the form A+Bж+Сx2+, etc. = A'+B'x+C'x2+, etc., must be satisfied for each and every value given to x, then the coefficients of the like powers of x in the two members are equal each to each. For, since this equation must be satisfied for every value of x, it must be satisfied when x=0. But upon this supposition all the terms vanish except two, and we have A=A'. Suppressing these two equal terms, we have Bx+Cx2+, etc. B'x+C'x2+, etc. = Dividing each term by x, we obtain B+Cx+, etc. B'+C'x+, etc. Since this equation must be satisfied for every value of x, it must be satisfied when x=0. But upon this supposition B=B'. In the same manner we can prove that C=C', 379. Whenever we have an equation of the form M+Nx+Px2+Qx3+, etc. =0, which is true for every value of x, all the coefficients of x are equal to zero. For, if we transpose all the terms of the equation in the last article to the left-hand member, we shall have A—A'+(B–B′)x+(C−C')x2+(D−D′)1⁄23+, etc.=0. But it has been shown that A=A', B=B', etc.; whence A—A'=0, B—B'=0, etc. If we substitute M for A-A', and N for B-B', etc., the equation will be It is plain that this development is possible, for we may divide the numerator by the denominator, as explained in Art. 373. Let us, then, assume the identical equation |