Question

Express the following surds as continued fractions, and find the fourth convergent to $\sqrt{a^2+1}$.

Answer 1

We can write $\sqrt{a^2+1}=a+(\sqrt{a^2+1}-a)=a+\frac{1}{\sqrt{a^2+1}+a}$

$\Rightarrow \sqrt{a^2+1}+a=2a+(\sqrt{a^2+1}-a),2a+\frac{1}{\sqrt{a^2+1}+a}$

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

Convergents are:-

$\frac{a}{1},\frac{2a^2+1}{2a},\frac{4a^3+3a}{4a^2+1},\frac{8a^4+8a^2+1}{8a^3+4a}$

Thus fourth convergent is $\frac{8a^4+8a^2+1}{8a^3+4a}$.