$\sqrt{11}$ as rational number within $10^{-4}$?

Question

Use continued fractions to find a rational number which approximates $\sqrt{11}$ to within $10^{−4}$.

I know how to solve for continued fractions like this: $$\sqrt{11}=3+x$$ $$11=9+6x+x^2$$ $$11=9+(6+x)x$$ $$2=x(6+x)$$ $$x=\frac{2}{6+x}=\frac{1}{3+\frac{x}2}=\frac{1}{3+\frac{1}{6+x}}$$

therefore, $$\sqrt{11}=[3;\overline{3,6}]$$

How do I find out where to terminate the fraction so as to obtain close value with error less than the mentioned limit

Answer 1

It's a fact that if $m/n$ is a convergent in the continued fraction expansion of $x$, and $m$ and $n$ are relatively prime, then $| x - m/n | < 1/n^2$. This allows you to bound the error in your approximation.

Answer 2

Theorem: Let $p_k/q_k$ is the $k$-th convergent of a positive real $x$ then

$$ \Bigg|x - \frac{p_k}{q_k}\Bigg| \le \frac{1}{q_k q_{k+1}} < \frac{1}{q_k^2} $$

So to approximate within $10^{-4}$, $q_k > 100$ sufficient not always necessary because in general you can get the same accuracy with a smaller denominator if $q_k q_{k+1} > 10^4$.