We perceive that the first term of the series must be Therefore, assume 1 + x X 2x2 + 6x3 = Ax1+Вx0 + Сx+Dx2+ Ex3 + Fx1+.... Clearing of fractions and transposing, Ax+ Bx+ C│x2+ D│x3 + E│x2+ F|25 0= -1 -2A -2B -20 -2D -2E Substituting these values in the assumed development, and observing that the term containing C will disappear because NOTE. It is not necessary to transpose the terms to one member; for if neither member is zero, we have simply to equate the coefficients of the like powers of x in the two members, according to the third property of identical equations. The method of Indeterminate Coefficients is applicable to a great variety of examples, but always with this provision, viz.: That we determine by inspection what power of the variable will be contained in the first term of the expansion, and make the first term of the assumed development correspond to the known fact. If the assumed development commence with a power of the variable higher than it should, the fact will be indicated by an absurdity in one of the resulting equations. If, however, the assumed development commence with a power of the variable lower than is necessary, no absurdity will arise; but the redundant terms will disappear by reason of the coefficients reducing to Ans. 1 + x + 3x2 + 9x3 + 27x1 + 81x3 +..ca 3. Develop x2 into a series. Ans. 1 + 3x + 4x2 + 7x3 + 11x1 + 18x5 +.... Ans. 1+ 2x + 8x2 + 28x3 + 100x + 356x5 + 2 Ans. x + 9x2 + 32x3 + 92xa + 240x3 +.... 5. Develop into a series. Ans. 1-2x2 + x4 + 4x6 — 11x8 + 10x10 + 13x12. Ans. 1+(1-2a)x-(2a-3a2)x2+(3a2-4a3)x3— .... ... .; then square - 2 2.4 2.4.6 NOTE.-Assume √1-x= A + Bx + Cx2 + Dx3 + both members, and the equations for the coefficients will be readily obtained. .... into a series 9. Develop 1 + 3x + 5x2 + 7x3 + 9x1 + ... of rational terms. 3x 11x2 23x3 179x4 10. Develop Ans. 1 + + + + 2 8 16 128 1-2x2+3x-4x6 +5x8 — 6x10+. 1+ x2+ x2+ 26 + x8 + 210 +.... Ans. 1-3x2 + 5x1 7x69x8 11x10 +.... into a series. REVERSION OF SERIES. 383. The Reversion of a Series is the process of finding the value of the unknown quantity in the series, expressed in terms of another unknown quantity. 1. Given y = ax + bx2 + cx3 + dx1 + ex3 + ...., to find the value of x in terms containing the ascending powers of y. In this equation, x and y are two indeterminate quantities, and either may have any value whatever without altering the form of the series. We may therefore apply the method of Indeterminate Coefficients. Assume x = Ay + By2 + Cy3 + Dy1 + Ey3 + . . . . .... We may now find by involution the values of x2, x3, x1, x3, etc., carrying each result only to the term containing y5. Then substituting for x, x2, x3, etc., in the given equation, we shall have, after transposing y, This is an identical equation, being true for all values of y. And if we place the coefficients of the different powers of y separately equal to zero (368, IV), and reduce the resulting equations, we shall obtain the values of the assumed coefficients and the powers of If we substitute these values of A, B, C, etc., in (1), we shall have the value of x in terms of a, b, c . . . ., y; that is, the given series will be reversed. 2. Given y = ax + bx3 + cx5 + dx2 + exo + value of x in terms of y. Assume to find the x= Ay+By+ Cy5+ Dy1 + Ey3 +....... (1). Proceeding as before, we shall obtain In the preceding examples the letters a, b, c, . . . ., represent any coefficients whatever. Hence, in reverting any series in either of these forms, we may determine the values of the assumed coefficients by an application of formula (F), or (G). 3. Revert the series y = x + 2x2 + 4x3 + 8xa + .... Assume x= Ay + By + Cy3 + Dy1 +........ If we now substitute in formula (F), we shall obtain Hence, a=1, b = 2, c = 4, d = 8, A=1, B2, C=4, D=-8. x=y-2y2+ 4y3 — 8y1 +...., Ans. EXAMPLES FOR PRACTICE. 1. Revert the series y = x + x2 + x3 + x2 + 205 + · .... 2. Revert the series y = x + 3x2 + 5x3 + 7xa + 925 +.... Ans. xy-3y2 + 13y3 — 67y+381y5. Ans. y=x+ + + + + 1.2 1.2.3 1.2.3.4 1-2-3-4-5 4. Revert the series y=x-x3 + x5 − x2 + x9 — x11 + Ans. x = y + y3 + Qy5 + 5y3 + 14y3 + 5. Revert the series y = 2x + 3x3 + 4x3 + 5x2 +........ .... .... 6. Revert the series x = 2y+4y2 + by3 + 8y1 + 10y+ .... 384. One of the principal objects in reverting a series is, to obtain the approximate value of the unknown quantity when the sum of the series is known. Thus, |