Steps / Historical AI

Leonhard Euler · 1770 (1828 English ed.) · Internet Archive

Elements of Algebra

All 627 leaves · leaf 560

Transcription

CHAP. II. ADDITIONS. 523 Now, if we consider the values of the terms $r^0, r', r'', r'''$, &c. found in the two cases, we shall perceive that the least of these terms is equal to $-3$; in the first case, it is the term $r'''$, to which the values $p'''$ and $q'''$ answer; and, in the second case, it is the term $r^{iv}$, to which the values $p^{iv}$ and $q^{iv}$ answer. Whence it follows, that the least value, which the given quantity can receive, is $-3$; and, in order to have the values of $p$ and $q$, which answer to it, we shall take, in the first case, the numbers $\mu, \mu', \mu''$, namely, $7, 1$, and $1$, and shall form with them the *principal* converging fractions $\frac{7}{1}, \frac{8}{1}, \frac{15}{2}$; the third fraction will, therefore, be $\frac{p'''}{q'''}$, so that we shall have $p''' = 15$, and $q''' = 2$; that is to say, the values required will be $p = 15$, and $q = 2$. In the second case, we shall take the numbers $\mu, \mu', \mu'', \mu'''$, namely, $5, 1, 1, 3$, which will give these fractions, $\frac{5}{1}, \frac{6}{1}, \frac{11}{2}, \frac{39}{7}$; so that we shall have $p^{iv} = 39$, and $q^{iv} = 7$; therefore $p = 39$, and $q = 7$. The values which we have just found for $p$ and $q$, in the case of the minimum, are also the least possible; but if we choose, we may likewise successively find others greater: for it is evident, that the same term, $-3$, will always return at the end of every interval of six terms; so that, in the first case, we shall have $r''' = -3, r^{ix} = -3, r^{xv} = -3$, &c. and, in the second, $r^{iv} = -3, r^x = -3, r^{xvi} = -3$, &c. Therefore, in the first case, the satisfactory values of $p$ and $q$ will be these; $p''', q''', p^{ix}, q^{ix}, p^{xv}, q^{xv}$, &c.; and, in the second case, $p^{iv}, q^{iv}, p^x, q^x, p^{xvi}, q^{xvi}$, &c. Now, the values of $\mu, \mu', \mu''$, &c. are in the first case $7, 1, 1, 5, 3, 2, 1$; $1, 1, 5, 3, 2, 1$; $1, 1, 5, 3$, &c. to infinity, because $\mu^{vii} = \mu'$, and $\mu^{viii} = \mu''$, &c. so that we shall only have to form, by the method of Art. 20, the fractions, $7, 1, 1, 5, 3, 2, 1, 1, 1, 5, \ldots$ $\frac{7}{1}, \frac{8}{1}, \frac{15}{2}, \frac{83}{11}, \frac{264}{35}, \frac{611}{81}, \frac{875}{116}, \frac{1486}{197}, \frac{2361}{313}, \frac{13291}{1762}, \ldots$ And we may take for $p$ the numerators of the third, ninth, &c. and for $q$ the corresponding denominators: we shall therefore have $p = 15, q = 2$, or $p = 2361, q = 313$, &c. In the second case, the values of $\mu, \mu', \mu''$, &c. will be $5, 1, 1, 3, 5, 1, 1, 1, 2; 3, 5, 1, 1, 1, 2$, &c. because $\mu^{ix} = \mu^{iii}, \mu^x = \mu^{iv}$, &c. We shall, therefore, form these fractions, $5, 1, 1, 3, 5, 1, 1, 1, 2, 3, \ldots$ $\frac{5}{1}, \frac{6}{1}, \frac{11}{2}, \frac{39}{7}, \frac{206}{37}, \frac{245}{44}, \frac{451}{81}, \frac{696}{125}, \frac{1843}{331}, \frac{6225}{1118}, \ldots$
confidence 0.70 53 expressions source scan at the Internet Archive

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