CONTINUED FRACTIONS. 343. To represent in a continued fraction*. Let b be contained p times in a, with a remainder c; again, let c be contained q times in b, with a remainder d, and so on; then we have b) a p c) b (q d) cr e &c. 344. COR. 1. An approximation may thus be made to the value of a fraction whose numerator and denominator are in too high terms; and the farther the division is continued, the nearer will the approximation be to the true value. a 1 345. COR. 2. This approximation is alternately less and greater than the true value. Thus p is less than; and p+is greater, because a part of the denominator of the fraction is omitted: 9+ α b 1 q is too great for the denominator, therefore p+ 1 is less than ; and so on. Although a 'continued fraction' (see Def. Art. 8) may be of the form p a It is obvious that, when is a proper fraction, p=0, and c=a, so α that the operation of converting into a continued fraction in this case commences with dividing b by a. DEF. The quantities p, q, r, &c., which are always positive integers, are called the Partial Quotients; and, p+: 1 +, p+−1, when reduced to simple fractions, are called the Converging Fractions, or Convergents, to α Ex. To find a fraction which shall be nearly equal to and in lower terms. 100000) 314159 (3 300000 14159) 100000 (7 99113 887) 14159 (15 314159 100000 887 5289 4435 854) 887 (1 854 33 &c. Here p=3, q=7, r=15, 8=1, &c. therefore The first approximation is 3, which is too little, the next is The proposed fraction expresses nearly the circumference of a circle whose diameter is 1; therefore the circumference is greater 346. To convert any continued fraction into a series of converging fractions. Let the continued fraction be (Art. 343) or 1 pq+1 gr+1 pqrs+ps+rs+pq+1 &c. grs+q+s (pq+1)r+p ̧ (pq+1.r+p)s+pq+1 in which the law of formation is observed to be as follows: Write down in one line the quotients p, q, r, s, &c., and the first and second converging fractions at sight; then the other fractions may be obtained thus: : For the 3rd, (num. 3rd quot.x num'. of 2nd fract.+ num'. of 1st fract. (num". =4th quot.x num'. of 3rd fract.+ num'. of 2nd fract. And generally, for the nth fraction in the series, Multiply the nth quotient by the numerator of the n-1th fraction and add the product to the numerator of the n-2]th fraction. This will give the numerator. th Multiply the nth quotient by the denominator of the n-1] fraction and add the product to the denominator of the n-2th fraction. give the denominator. To prove this generally: Let 91, 92, 93, 94, be any 4 successive quotients, and This will N, N, N, D1' D' D' NA D the corresponding convergents; and suppose the law above stated to hold for the 3rd fraction, so that N ̧=q,N,+N1, and _D ̧=q ̧D ̧+D1; then, since any convergent is obtained from the preceding one by merely bringing to account another quotient, it is clear that may be found 1 N instead of q, that is, Now it may be proved that these fractions are both in their lowest terms by the application of the following: LEMMA. If an integer be added to a fraction in its lowest terms, the result, when reduced to an improper fraction, will also be in its lowest terms. b ac+b == ; C C Let a be the integer, and the fraction; then a+ and if this is not in its lowest terms, its numerator and denominator have a common measure, which will therefore also measure (ac+b)-ac or b, thus making the fraction not in its lowest terms. b с It is important here to observe that this reduction to an improper fraction is to be effected by multiplying by the denominator. are all in their lowest terms, considered as arising from the continued fraction, and not as derived from each other. This notation, although at first sight somewhat complex, is very expressive, and greatly relieves the memory throughout the operation, the first letter of each word being used, q for quotient, N for Numerator, and D for Denominator, whilst the small figures indicate the position of each with respect to the others here employed. Thus, for instance, the final result in this proof may be as easily read and understood, as if it were written at full length in ordinary language. N N1 D Now when we derive from as we have written q2+ and multiplied by the denominator q, we clearly must get the same result, whether we retain the form q3+ until the end of the operation, or perform the multiplication by q, first. But under this latter mode of reducing the fraction the result we obtain is in its lowest terms, therefore it is also in its lowest terms in the other case: i. e. is in its lowest N1 q, N2+N2 q1D2+D2 terms. Also has been proved to be in its lowest terms, .. the nume N1=q,N2+N,, and D.=q,D2+D2, which proves that, if the law holds for any one convergent, it also holds for the next; and as we know it does hold for each of those near the beginning of the series, it follows that the law holds generally. (NOTE 4.) 84 Ex. To find a series of converging fractions for 227 347. To express Ja2+1 in the form of a Continued Fraction. a result which is easily remembered and applied to any proposed case. |