containing the known quantities a, b, c, d, etc., and the powers of y. To revert this series, assume x=Ay+By2+Cy3+Dy1, etc. (2), in which the coefficients A, B, C are undetermined. Substituting these values in (2), and arranging, we have 0=Aa | x+Ab ¦ x2+ Acx3+ Adx++, etc. Since this is true, whatever be the value of x, and the coefficients of x, x2, x3, etc., will each =0, (Art. 314, Cor.), we have Ad+Bd2+2Bac+3Ca2b+Da1=0, .. D=—' =0, .. B: 262-ac =0, .. C= a5 a2d-5abc+563 ат 345. If the given series has a constant term prefixed thus, y=a'+ax+bx2+cx3+dx2+ assume y—a—z, and we have z=ax+bx2+cx3+dx+, etc. But this is the same as (1) in the preceding article, except that z stands in the place of y; hence, if z be substituted for y in [(3), Art. 344], the result will be the required development of x; and then, y-a' being substituted for 2, the result is 346. When the given series contains the odd powers of x, assume for x another series containing the odd powers of y. Thus, if y=ax+bx3+cx3+dx2+ to develope x in terms of y, assume x=Ay+By+Cy3+Dy1+..... Then, by substituting the values of y, y2, etc., derived from the former equation, in the latter, and equating the coefficients to zero, we find If both sides of the equation be expressed in a series, as ay+by2+cy3+, etc., a'x+b2x2+c'x3+, etc., and it be required to find y in terms of x, we must assume, as before, y=Ax+Bx2+Cx3+Dx1+, etc., and substitute the values of y, y2, y3, etc., derived from this last equation, in the proposed equation; we shall then, by equating the coefficients of the like powers of x, determine the values of A, B, C, etc., as before. The following exercises may be solved either by substituting the values of a, b, c, etc., in the equations obtained in the preceding articles, or by proceeding according to the methods by which those equations were obtained. 1. Given the series y=x-x2+x3—x2+ to find the value of x in terms of y. Ans. xy+y+y3+y++, etc. Find the value of x, in an infinite series in terms of y: 2. When y=x+x2+x2+, etc. Ans. x=y—y2+y3—y2+y3—, etc. 3. When y=2x+3x+4x+5x+, etc. 4. When y=1-2x+3x2. 128 Ans. x= —— 1 (y—1) + 3 (y—1)2—,% (y—1)3+, etc. 5. When y=x+}{x2+{x3+x}}x*+, etc. Ans. xy-y2+}y3—¡y‘+, etc. 6. When Ans. x= + y+ay2+by3+cy1...=gx+hx2+kx3+lx* . . . _y (ag2—h)y2 [bg1—kg—2h(ag2—h)]y3 + ... 9 XI. CONTINUED FRACTIONS: LOGARITHMS: EXPONENTIAL EQUATIONS: INTEREST, CONTINUED FRACTIONS. 347. A Continued Fraction is one whose denominator is continued by being itself a mixed number, and the denominator of the fractional part again continued as before, and so on; thus, in which a, b, c, d, etc., are positive whole numbers. Continued fractions are useful in approximating to the values of ratios expressed by large numbers, in resolving exponential equations, indeterminate equations of the first degree, etc. 348. To express a rational fraction in the form of a continued fraction. Let it be required to reduce 30 157 to a continued fraction. If we divide both terms of the fraction by the numerator, we 7 Again, if we divide both terms of the fraction less than the true value of the fraction. Hence, generally, By stopping at an odd reduction, and neglecting the fractional part, the result is too great; but by stopping at an even reduction, and neglecting the fractional part, the result is too small. are called converging fractions, because each one in succession gives a nearer value of the given expression. The fractions 11 1 a'b'è, etc., are called integral fractions. 350. To explain the manner in which the converging fractions are found from the integral fractions. By examining the third converging fraction, we find it is formed from the 1st, and 24, and from the 3d integral fraction, as follows: Num. 3d quot.Xnum. of 2d conv. fract.+num. of 1st conv. fract. Denom. 3d quot.Xden. of 24 conv. fract.+den. of 1st conv. fract. P Q R be the three To prove the general law of formation, let converging fractions corresponding to the three integral fractions |