11. If p, q, be the rth convergent of 1 1 1 1 a + b + a + b + shew that Pan+2=P2+ bq,,, and that 12+2 = ap2 + (ab + 1)¶1⁄2n° 1 -- a + 2n 12. If p, q, be the rth convergent of the continued fraction 13. 1 1 1 1 shew that Pan+bP3, + (bc + 1) 9зn° +c+a+b+c+ If p, q, be the rth convergent of shew that Pan 92n− 1 − 92n P2n−1 2n - a"b". = P Q the first quotient being a, and the convergent preceding being be converted into a continued fraction, the last convergent will be (P-aQ)/(p-aq). ... +c+ b pq 355. To find the nth convergent of a continued fraction. We have in Art. 352 found a law connecting three successive convergents to a continued fraction, so that the convergents can always be determined in succession. In some cases an expression can be found for any convergent which does not involve the preceding convergents: the method of procedure will be seen from the following examples. Ex. 1. To find the nth convergent of the continued fraction - Pn− (2n+1) Pn-1= -- (2n − 3) {Pn− 1 − (2n − 1) Pn- 2}. Changing n into n-1 we have in succession Pn-1-(2n-1) Pn−2− − (2n − 5) {Pn−2 − (2n − 3) Pn−3}, n P3-7P2-3 {P2 - 5P1}. But, by inspection, p=1, p2=4; .'. P2 − 5p1 = − 1. 1 Hence Pn (2n+1) Pn−1 = (− 1)n – 1 (2n − 3) (2n − 5)...3.1. Since the denominators of convergents are formed according to the same law as the numerators, we have from the above In − (2n+1)In -1=(-1)"-23.5... (2n − 3) {q2-5q1}=0, since q3 and 42=15. Ex. 2. To find the nth convergent of the continued fraction = 1; 1+ 2+ 3+ 4+ The necessary transformations are given in Ex. 5, Art. 247. 356. Periodic continued fractions. When the elements of a continued fraction continually recur in the same order, the fraction is said to be a periodic continued fraction; and a periodic continued fraction is said to be simple or mixed according as the recurrence begins at the beginning or not. 357. To find the nth convergent of a periodic continued fraction with one recurring element. = 2-1 n-2 convergents after the second, we have pcpPn1+bP1· where b and c are constants, that is, are the same for all values of n. n-1 Now, if u1 +ux + Uzx2 +...+U„x2¬1+... be the recurring series formed by the expansion of A+ Bx 1 cx - bx2, the suc cessive coefficients after the second are connected by the law ucu,+bun-2 Hence, if A and B are so chosen that up, and up,, then will up, for all values of The necessary values of A and B are respectively p1 and p, - cp,, that is a and b. n. 2 = Hence the numerator of the nth convergent to the Similarly the denominator of the nth convergent is the coefficient of æ” ̄1 in the expansion of 1 + (¶1⁄2 − ¤Q1) x n-1 Ex. 1. Find the nth convergent of the continued fraction The numerator of the nth convergent is the coefficient of xn−1 Also In coefficient of x^-1 in the expansion of Ex. 2. Find the nth convergent of the continued fraction Since the last result is symmetrical in a and c, and also in b and d, it follows that P2n−1 − (a + c + bd) P2n−3+acp2n−5=0. Hence the relation Pn-(a+c+bd) Pn−2+acpn-4=0 holds good for all values of n. will be the coefficient of x-1 in the expansion of A+Bx+Cx2 + Dx3 1- (a+c+bd) x2 + acxa› provided the values of A, B, C, D are so chosen that the result holds good for the first four convergents. It will thus be found that p is the coefficient of x^-1 in the expansion of a+adx-acx2 1- (a+c+bd) x2 + acx1ˆ It will similarly be found that qn is the coefficient of x-1 in the expansion of b+ (bd+c) x - acx3 358. Convergency of continued fractions. When a continued fraction has an infinite number of elements it is of importance to determine whether it is convergent or When an expression can be found for the nth con not. |