Continuous fractions theorem

Question

$$k_n h_{n-1} - k_{n-1} h_n = (-1)^n$$

Can you guys explain to me about the continued fractions and about this particular theorem? What does this mean?
(I have some general info about this, but it is kind of vague for me)

Answer 1

Take a continued fraction (infinite or finite). Truncate the calculations after step $n-1$ to arrive at a rational number $\frac{k_{n-1}}{h_{n-1}}$ and truncate it after step $n$ to arrive at the rational number $\frac{k_n}{h_n}$. These rational numbers will satisfy $k_n h_{n-1} - k_{n-1} h_n = (-1)^n$ which is also equivalent to $\frac{k_n}{h_n} - \frac{k_{n-1}}{h_{n-1}}=\frac{(-1)^n}{h_n h_{n-1}}$.

This is important for a few reasons. Firstly, your equation is a linear diophantine equation. If I wanted to solve $83x-36y=1$, then we can begin by looking at the continued fraction expansion of $\frac{83}{36}=[2;3,3,1,2]$ and finding some partial convergents. We see that $[2;3,3,1]=\frac{30}{13}$ so we try $83(13)-36(30)=-1$ which actually wasn't what we wanted to solve initially....so the trick is to actually use $\frac{83}{36}=[2;3,3,1,1,1]$ and try to use the partial convergent of $[2;3,3,1,1]=\frac{53}{23}$. This leads to $83(23)-36(53)=1$ which works. So solving $ax-by=\pm1$ is solvable by looking at the continued fraction expansion of $\frac{a}{b}$.

The second equation shows the difference between successive partial convergents becomes increasingly small (as you move farther into the continued fraction expansion, the $q_n$ and $q_{n-1}$ terms will get larger). This is used to prove that infinite continued fractions actually converge to their limit and that for an irrational real number $x=[a_0;a_1,a_2,a_3,a_4,...,a_n,...]$, you'll always have $\left|x-\frac{k_n}{h_n}\right| < \frac{1}{h_n h_{n-1}}$. This is useful for finding close rational approximations to irrational numbers. For example, $\pi=[3;7,15,1,292,1,1,1,2,...]$. Stopping at $\pi \approx [3;7,15,1]=\frac{335}{113}$, you'll see that this approximation gives an error of $\left|\pi-\frac{335}{113}\right| < \frac{1}{113 \cdot 33102}$. The $33102$ came from the next stopping convergent of $[3;7,15,1,292]$. So if you find an irrational number with a big number in the continued fraction expansion, you can find a very good rational approximation using this method.