# Eikonal Blog

## 2010.02.12

### Some continued fractions

Filed under: mathematics — Tags: — sandokan65 @ 21:01
• $\sqrt{2} = 1+ \frac1{2+}\frac1{2+}\frac1{2+}\cdots = [1;2,2,2,\cdots] = [1;\dot{2}]$.
• $\sqrt{3} = [1;\dot{1},\dot{2}]$.
• $\sqrt{5} = [2; \dot{4}]$.
• $\sqrt{7} = [2; \dot{1},1,1,\dot{4}]$.
• $e = [2;1,2,1,1,4,1,1,6,1,1,8,\cdots]$.
• $\frac\pi4 = \frac1{1+} \frac{1^2}{2+} \frac{3^2}{2+} \frac{5^2}{2+} \frac{7^2}{2+} \frac{9^2}{2+} \cdots$.
• $[3;\dot{3}] = \frac{3+\sqrt{13}}2 = 3.302,77\cdots$.
• $[a;\dot{b}] = a+\frac{\sqrt{b^2+4}-b}2$.
• $[4;\dot{6}] = 1+\sqrt{10}$.
• $[a;\dot{b},\dot{c}] = ab + \frac{\sqrt{bc(bc+4)}-bc}2$.

Theorem: A simple infinite continued fraction $[a_0; a_1, a_2, \cdots]$ converges IFF $\sum_n a_n$ diverges.

Theorem:
Every real number $x$ has the unique simple continued fraction expression. That expression is:

• finite IFF $x \in {\Bbb Q}$,
• infinite and repeating IFF $x \not\in {\Bbb Q}$ but algebraic,
• infinite and non-repeating IFF $x$ is transcendental.