Page:Elementary algebra (1896).djvu/425

407 Rh 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 $$a$$.