ally approach to a state of equality, as their number is increased. P1 etc., or limited by them, it results that as the number of Qi clear that the number of converging ratios will be finite; but if А A is irrational, the number of converging ratios must be A' infinite, and such that they will ultimately differ from by A' a difference which will be less than any given difference. 6. If A and A' have a common divisor, since the quotients a, b, c, etc., are independent of the divisor, it follows that the converging fractions will also be independent of it; conseA quently, if the number of fractions is finite (or rational), A A' the last of them must be the ratio when reduced to its A" lowest terms. Hence, if A A" when reduced to its lowest U S T terms, is expressed by and that is the fraction which immediately precedes it, it results from (5) that we shall have the equation SV – TU ±1, (7); in which must be used for when there is an even number of converging fractions, and must be used for when there is an odd number of converging fractions. If C represents the greatest common divisor of A and A', then if we multiply (7) by C (since CV A' and CU = A) we get A'S AT = ± C, (S); consequently, if C = 1, the fraction (or ratio) which immediately precedes it. R S' 7. It is impossible to insert a rational fraction, such as between two successive converging fractions, such as P, Р Q and unless its denominator S is greater than Q (the QI greater denominator of the fractions, between which the fraction is inserted). For, by the supposition, we must have 1 1 P P1 == , or (rejecting the signs) we Q S QS 1 must have Q1 PS-QR (9). Because P, Q, R, and S are integers, it is clear that PS QR can not (numerically) be less than 1; consequently, it is evident that the inequality (9) can not be satisfied unless S is greater than Q1. tions which converge toward or to the same degree of ap continued fraction; observing that an expression which is partly integral and partly fractional, or wholly fractional, and is such that the denominator of the fractional part is a mixed number, and the denominator of the fractional part of the denominator is also a mixed number, and so on, to any required extent, is called a continued fraction. If we take only the integral part of the continued fraction, we have 1 4 for the first approximation to A Hence, our method of approximating to the ratio is substantially the same as the ordinary method, by using continued fractions. 9. To perceive the use of what has been done, take the following consequently, the quotients a, b, c, d are severally represented by 1, 4, 1, 4. Hence, the first two converging frac tions become quently, 1 + ab 1 + 1 x 4 5 α 1 5 4 + 5 × 4 24 = ; conse4 are the remaining fractions. Because the last fraction is equal to the given fraction, it follows that the given fraction can not be reduced to lower terms; and since there are an even number of fractions, we have 6 × 24 - 5 x 29 = - 1. 2d. To find the approximate values of 372 246 Here a, b, c, and d are represented by 1, 1, 1, and 20, and 1 2 3 62 372 thence = are the sought fractions. Because 1' 1' 2' 41 246 there are an even number of fractions, we have 3 × 41 2 × 62 = 1; and because 6 is the greatest common divisor of 246 and 372, we shall have 3 × 246 2 × 372= 6. 1728 3d. To find the approximate values of 1892 Here a, b, c, etc., are represented by 0, 1, 10, 1, 1, 6, and 0 1 10 11 21 137 432 1728 3, and we have 1' 1' 11' 12' 23' 150' 473 1892 = Because there are an odd number of fractions, we have 473 x 137 432 x 150 = 1; and because 4 is the greatest common divisor of 1728 and 1892, we also have 1892 × 137 - 1728 × 150 = 4. 4th. Given, 3.1415926535, etc. = the ratio of the circum 1 ference of a circle to its diameter, to find the approximate ratios. Here the numbers corresponding to a, b, c, d, etc., are 3, 7, 15, 1, 292, 1, 1, etc.; consequently, 7' 106' 113' 3 22 333 355 103993 , etc., are the sought ratios. The most celebrated of 33102' these ratios are and 7 des and Metius. 22 5th. Given, 25 - 31y=1, to find positive integral values of x and y which will satisfy the equation. 31 If we find the approximate ratios of we get 1, 4, and 6 25' for the numbers which correspond to a, b, c; consequently, 1 5 we have and for the ratios. Since the number of 1' 4' 25 31 ratios is odd, we have 25 × 531 x 4 = 1; or, using m to stand for any positive integer, we have 25 x (5 + 31m) — 31 × (4 + 25m) = 1; consequently, we may put = 5 + 31m, and y=4+ 25m. If we put 0, 1, 2, etc., successively, for m, we get x=5, y = 4; x = 36, y = 29; x = 67, y=54, and so on, for the corresponding values of x and y; conse quently, the question admits of an unlimited number of answers. 6th. Given, 41-56y= 1, to find positive integral values of x and y which will satisfy the equation. Here, since 1 3 4 11 15 fractions, we have 41 x 1556 × 111, or 41 × (56m -15) 56 x (41m11) 1. Putting 1, 2, etc., for m, we get 56m - 15 = 41 and 41m 11 = 30, and so on. Remark.-If A and B are positive integers prime to each other, it follows that we may, in like manner, find an unlimited number of positive integers for and y which will satisfy the equation Ax - By = 1. 7th. To find the sum or (more properly) the generating function of the series 1 + 2x + 3x2 + 4+ etc. If we can find the generating function of the fraction it is clear that, by taking its 1 1 + 2x + 3x2 + 4x3 + etc.' reciprocal, we shall get the generating function of the proposed series. If we divide the numerator of the fraction by its denominator, we shall get 1 for the quotient, and -2x-3x2-4x3etc., for the remainder. Then, dividing 1 + 2x + 3x2 + 4 + etc., by the preceding remainder, we get for the quotient, and 3x3 2 1 2x x 2 +24+ + etc., for the remainder; and dividing the 525 - last divisor by the last remainder, we have 4 for the quotient, and a2 + 2x3 + 3x2 + etc., for the remainder; also, dividing the last divisor by the last remainder, we have 1 2x for the quotient, and 0 for the remainder; consequently, the divisions terminate. Hence, the quotients a, b, c, d are here represented by 1, |