n 220. Convergents. The value obtained by taking only the first 1 quotients in a continued fraction is called the nth convergent of the fraction. When there is no whole number preceding the fractional part of the continued fraction the first convergent is zero. Thus in or p1 = α1, 91 = 1. This indicates that the form of the nth convergent is Pn anPn-1+PR-2 = In anIn-1 + In-2 (1) This is in fact the case, as we proceed to show by complete induction. We have already established form (1) for n = - 2 and n = 3. We assume it for n = m, and will show that its validity for n = m +1 follows. The (m + 1)th convergent differs from the mth only in the fact that am + fraction in place of am In (1) 1 appears in the continued replace n by m, and a, by Am + 1 Pm-1 +Pm-2 Im-1+ Im-2 (am+1am+1) Pm-1+am+1Pm-2 am+1(amPm-1+Pm-2)+Pm-1 am+1Pm+Pm-1, am+12m +2m-1 which is form (1). EXERCISES 1. Express the following as continued fractions, and find the convergents. (a) 0. Solution: By the method already explained, we find that 1 The first convergent is evidently 0, the second is 1, and the third is 2. Find the value of the following by finding the successive convergents. 57 (f) 1999. 221. Recurring continued fractions. We have seen that every terminating continued fraction represents a rational number, and conversely. We now discuss the character of the numbers represented by the simplest infinite continued fractions. A recurring continued fraction is one in which from a certain point on a group of denominators is repeated in the same order. are recurring continued fractions if the denominators are assumed to repeat indefinitely as indicated. That a repeating continued fraction actually represents a number we shall establish in § 223. Unless this fact is proven, one runs the risk of dealing with symbols which have no meaning. If for certain continued fractions the successive convergents increase without limit, or take on erratic values that approach no limit, it is important to discover the fact. All the fractions that we discuss actually represent numbers, as we shall see. We shall consider only continued fractions in which every denominator has a positive sign. THEOREM. Every recurring continued fraction is the root of a quadratic equation. Let, for instance, x= 1 1 1 1 1 1 Evidently the part of the fraction after the first denominator c may be represented by x, and we have thus virtually the terminating fraction 1 1 1 (ab + 1) x2 +[c (ab + 1) + a − b]x — be − 1 = 0, which is a quadratic equation whose root is x, the value of the continued fraction. Since this equation has a negative number for its constant term it has one positive and one negative root. The continued fraction must represent the positive root, since we assume that the letters a, b, c represent positive integers. The quadratic equation whose root is a recurring continued fraction with positive denominators will always have one positive and one negative root. The equation will be quadratic, however, whatever the signs of the denominators may be. The proof may be extended to the case where there are any number of recurring denominators or any number of denominators before the recurrence sets in. Since every real irrational root of a quadratic equation is a surd, our result is equivalent to the statement that every recurring continued fraction may be expressed as a surd. EXERCISES Of what quadratic equations are the following roots? Express the continued fraction as a surd. Since x2 is negative, x1 must be the surd that is represented by the continued fraction. |