Page:Elementary algebra (1896).djvu/425

507. To find limits to the error made in taking any convergent for the continued fraction.

Let be three consecutive convergents, and let k denote the complete (n+ 2)th quotient ;

then

Now k is greater than 1, therefore the difference between the continued fraction $x$, and any convergent, ${\frac {p_{n}}{q_{n}}}$, is less than ${\frac {1}{q_{n}q_{n+1}}}$ and greater than ${\frac {1}{q_{n}(q_{n+1}+q_{n})}}$.

Again, since $q_{n+1}>q_{n}$, the error in taking ${\frac {p_{n}}{q_{n}}}$ instead of $x$ is less than ${\frac {1}{{q_{n}}^{2}}}$, and greater than ${\frac {1}{{2q_{n+1}}^{2}}}$.

508. From the last article it appears that the error in taking \frac{p_n}{q_n} instead of the continued fraction is less than ${\frac {1}{q_{n}q_{n+1}}}$ or, ${\frac {1}{q_{n}(a_{n+1}q_{n}+q_{n+1})}}$; that is, less than ${\frac {1}{{a_{n+1}q_{n}}^{2}}}$ hence the larger $a_{n+1}$ is, the nearer does ${\frac {p_{n}}{q_{n}}}$ approximate to the continued fraction; therefore, any convergent which immediately precedes a large quotient is a near approximation to the continued fraction.

Again, since the error is less than ${\frac {1}{{q_{n}}^{2}}}$ it follows that in order to find a convergent which will differ from the con- tinued fraction by less than a given quantity ${\frac {1}{a}}$ we have only to calculate the successive convergents up to ${\frac {p_{n}}{q_{n}}}$ where ${q_{n}}^{2}$ is greater than $a$.