Question

Express the following surds as continued fractions, and find the fourth convergent to $\sqrt{a^2+\frac{2a}{b}}$.

Answer 1

$\sqrt{a^2+\frac{2a}{b}}=\frac{\sqrt{a^2{b}^2+2ab}}{b}=a+\frac{\sqrt{a^2{b}^2+2ab}-ab}{b}$

$=a+\frac{2a}{\sqrt{a^2{b}^2+2ab}+ab}$

$\frac{\sqrt{a^2{b}^2+2ab}+ab}{2a}=b+\frac{\sqrt{a^2{b}^2+2ab}-ab}{2a}$

$=b+\frac{b}{\sqrt{a^2{b}^2+2ab}+ab}$

$\frac{\sqrt{a^2{b}^2+2ab}+ab}{b}=2a+\frac{\sqrt{a^2{b}^2+2ab}-ab}{b}$

$=2a+\frac{2a}{\sqrt{a^2{b}^2+2ab}+ab}$

$\sqrt{a^2+\frac{2a}{b}}=a+\frac{1}{b}+\frac{1}{2a}+\frac{1}{b}+\frac{1}{2a}+....$

Convergents are

$\frac{a}{1},\frac{ab+1}{b},\frac{2a^{2}b+3a}{2ab+1}$

$\frac{2a^2{b}^2+4ab+1}{2a{b}^2+2b}$