Operation. 251)764(3 753 11)251(22 22 31 22 9)11(1 9 2)9(4 8 1)2(2 2 0 The partial fractions are,,.,; therefore, we 56. Let us now endeavor to reverse the foregoing process; that is, let us seek the vulgar fraction which is equivalent to a continued fraction. all but the first partial fraction, its value will become 1. Again, omitting all but the first and second partial fractions, we find 1 2+1 3 7 1 Again, including one more partial fraction, we obtain 2+1 3+1 4. Our successive values, obtained in this way, are,, 13, and. 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 3; 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 1; 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, 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, 7. 68 This last value is the true value of the continued frac tion, 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 symbol as a fraction; then write this symbol, and the first partial fraction, for the first two 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 numerator and denominator of the next preceding approximate value, and the result will be the succeeding approximate value. Thus continue to multiply the last approximate 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. In this example, we find the successive approximate values to be f, 1, 4, 11, 74%, and 27. |