Page:Elementary algebra (1896).djvu/422

404 Rh Explanation. With the first and second quotients take the first and second convergents, which are readily determined. Thus, in this example, 2 is the first convergent, and $$2 + \frac{1}{6}$$ or $$\frac{13}{6}$$ the second convergent. The numerator of the third convergent, 15, equals the numerator of the preceding convergent, 13, multiplied by 1, the third quotient, plus 2, the numerator of the convergent next preceding but one. The denominator is formed in a similar manner: thus 7 = 1 × 6 + 1.

The fifth convergent = $$\frac{11(28)+15}{11(13)+7} = \frac{323} {150}$$

503. If the fraction is a proper fraction, we may consider zero as the first convergent, and proceed as follows:

Reduce $$\frac{84}{227}$$, to a continued fraction, and calculate the successive convergents.

Proceeding as in Art. 499,

227)84(0 00 84)227(2 168 59)84(1 59 25)59(

We obtain $$0+ \frac{1}{2 +} \frac{1}{1 +} \frac{1}{2 +} \frac{1}{ 2+} \frac{1}{1 +} \frac{1}{ 3+} \frac{1}{2}$$.

The successive quotients are 0, 2, 1, 2, 2, 1, 3, 2.

Writing $$\frac{0}{1}$$ for the first convergent and arranging the work as show in the example of the preceding article, we have Quotients 0, 2, 1, 2, 2, 1, 3, 2. Convergents $$\frac{0}{1}, \frac{1}{2}, \frac{1}{3}, \frac{3}{8}, \frac{7}{19}, \frac{10}{27}, \frac{37}{100}, \frac{84}{227}.$$

504. It will be convenient to call a_n the nth partial quotient; the complete quotient at this stage being

$$a_n + \frac{1}{a_{n+1} +} \frac{1}{a_{n+2} +} \ldots$$

We shall usually denote the complete quotient at any stage by k.