(n Approximate Values of Continued Fractions. 320. Theorem. The numerator of any approximate value, as the nth, is obtained from the numerators of the two preceding approximate values, the 1)st, and the (n-2)nd, by multiplying the 1)st numerator by the nth denominator contained in the given continued fraction, and adding to the result the numerator of the (n-2)nd approximate value. (n The denominator of the nth approximate value is obtained in the same way from the two preceding denominators. Demonstration. Let the (n-3)rd, (n—2)nd, (n−1)st, and nth approximate values be, respectively, and let the (n 1)st and the nth denominators, contained in the given continued fraction, be p and q. We shall suppose the proposition demonstrated for the (n-1)st approximate value, and shall prove that it can thence be continued to the nth value; that is, we shall suppose it proved that M PL+ K pL'+K Now it is plain, from the remarks at the end of the preceding article, that the nth value is deduced from the (n—1)st, 1 by changing p into p+; which change, being made in the preceding value, gives Difference between successive Approximate Values. Hence we have, by substituting N MpLK, Mp L'K'; N M'q + L' that is, the value required to satisfy the theorem. If, therefore, it can be shown that the proposition is true for any approximate value, it follows that it must be true for every succeeding value. But the comparison of the values given in the preceding article shows that it is true for the third value, and therefore for every succeeding value. 321. Theorem. If two succeeding approximate values are reduced to a common denominator equal to the product of their denominators, the difference of their numerators is unity. Demonstration. Let the (n-2)nd, (n—1)st, and nth approximate values be the difference between the (n-2)nd and (n-1)st is and that between the (n − 1)st and nth is of both which differences the numerators are the same; and, therefore, this is always the case. Approximate Value compared with True Value. Now the first and second approximate values, as given in art. 319, are, when reduced to a common denominator, the difference of the numerators of which is 1; and, therefore, unity must always be the difference of two such nu merators. 322. Theorem. The approximate values of a coninued fraction are alternately larger and smaller han its true value, the first being larger, the second maller, and so on alternately. Demonstration. Since, in the preceding demonstration, the subtraction of the (n-1)st value from the (n-2)nd, gave a fraction having the same numerator as that obtained, by its subtraction from the nth; we see that if the (n-1)st value is larger than the (n-2)nd, it must also be larger than the nth; and if the (n-1)st is smaller than the (n-2)nd, it is also smaller than the nth. But the true value is, by art. 319, nearer the (n-1)st value than the (n-2)nd, and nearer the nth than the (n − 1)st; so that when the (n-1)st value is larger than the (n-2)nd, the true value must likewise be larger than the (n -2)nd, and smaller than the (n-1)st, and so on alternately; but when the (n-1)st value is smaller than the (n-2)nd, the true value must be smaller than the (n-2)nd, and larger than the (n-1)st, and so on alternately. Now the first value is, by the preceding article, larger than the second, and therefore the true value is smaller than the first, larger than the second, and so on alternately. 323. Theorem. Each approximate value of a continued fraction differs from the true value by a Transformation of a Quantity to a Continued Fraction. quantity less than the fraction whose numerator is unity, and whose denominator is the square of the denominator of this approximate value. Demonstration. Let the denominator of the two successive approximate values be M' and N'; N' must, by art. 320, be larger than M'; and the difference between these two values must be 1 MN But, by the preceding article, the true value is contained between these two approximate values, and therefore differs from either of them by a quantity less than their difference. Now, since we have M' < N', so that the true value must differ from the approximate value, whose denominator is M', by a quantity less than 1 M'2' that is, less than a fraction whose numerator is 1, and de nominator M12. 324. Problem. To transform any quantity into a continued fraction. Solution. Let X be the quantity to be transformed. Find the greatest integer contained in X, and denote it by A, and denote the excess of X above 1 A by the fraction; and we have |