and omit all but the first partial fraction, its value will become =. Again, omitting all but the first and second partial frac tions, we find 1 2+1 Again, including one more partial fraction, we obtain 2+1 30 . 3+1 When we include the whole, we find 2+1 tind I 5. Our successive values obtained in this way, are , , 43, and photo These values may be derived in the following manner : Take the first partial fraction for the first value, multiply both numerator and denominator by the denominator of the next partial fraction, and we get ]; if we increase this denominator by 1, it will give the second value . Again, multiplying numerator and denominator by the denominator of the next partial fraction, we get 17; if we increase this numerator by the numerator of the last value, also increase the denominator by the denominator of the last value, we get ts, which is the third value. Again, multiplying both numerator and denominator of this value, by the denominator of the next partial fraction, and to the respective products add the numerator and denominator of the preceding value, we obtain the last value, 198 This last value is the true value of the continued fraction, whilst the other values are successive approximations. From what has been said we derive the following rule, for finding the vulgar fraction equivalent to a continued fraction. RULE. Consider the symbole as a fraction; then write this symbol and the firsl partial fraction for the two first terms of the approximate values. Multiply the numerator and denominator of the second approximate value by the denominator of the next partial fraction, and to the respective products add the numer. ator and denominator of the next preceding approximate value, and the result will be the succeeding approximate value. Thus continue to multiply the last approximative value, by the denominator of the succeeding partial fraction, and to the products add the numerator and denominator of the preceding approximate value, the result will be the succeeding approximate value. Examples. 1. What vulgar fraction is equivalent to the continued fraction 3+1 In this example we find the successive approximate values to be f, 3, 4, it, non, and 447. 2. What are the approximative values of the continued frac. Ans. s, $, is, and att 3. What are the approximative values of the continued frac tion 2+1 2+1 Ans. i, j, k, s, 4. What are the approximative values of the continued frac. 5. What are the approximative values of the continued frac. tion " 9 +1 8+1 2 ? Ans. f, t, 145, 14, 15, 145, 2755, Win the highest 57. We will now show the application of the foregoing principles of continued fractions, by the solution of several practical questions. 1. Express approximately the fractional part of 24 hours, by which the solar year, of 365 days 5 hours 48 minutes and 48 seconds, exceeds 365 days. 5 hours 48 minutes 48 seconds=20928 seconds. 24 hours=36400 seconds. Therefore the true value of the fraction required is th=15%. Now, converting is into a continued fraction, by rule un. der Art. 55, we get 109_1 34 450 4+1 7+1 converted into its approximative values (by rule under Art. 56) gives s, š, zboglipi 15%. The fraction 4, agrees with the correction introduced into the Calendar by Julius Cæsar, by means of bissectile or leap. year. The fraction is the correction used by the Persian Astronomers, who add 8 days in every 33 years, by having seven regular leap-years, and then deferring the eighth until five years. 2. The French metre is 39.371 inches. Required the ap. proximative ratio of the English foot to the metre. In this example, the true ratio is 24. Operating with this fraction as in the last example, we find some of the first approximative values to be ), M , ze 32 . Hence, the foot is to the melre, as 3 to 10, nearly; a more correct ratio is as 32 to 105. . . 3. What are some of the approximative values of the ratio of the diameter of a circle to its circumference? If we take the value of the circumference of the circle whose diameter is 1, to ten decimals, we have the vulgar fraction Tih 1882055, given to find its approximative values. Proceeding with this, as in the former examples, we find some of the first approximative values to be $, ža, }}}, &c. 58. Continued fractions have been the means of obtain. ing elegant approximations to the roots of surds. The method of converting a surd into a continued fraction is too complicated for us to explain in this place. The square root of 2, or, what is the same thing, the ratio of the diagonal of a square to its side, is found to cqual the fol. |