Question

Show that $\frac{1}{q_1}-\frac{x}{q_1{q}_2}+\frac{x^2}{q_2{q}_3}-\frac{x^3}{q_3{q}_4}+\cdots$ is equal to the continued fraction $\frac{1}{a_1+}\frac{x}{a_2+}\frac{x}{a_3+}\frac{x}{a_4+}\cdots$, where $q_1,q_2,q_3,\dots$ are the denominators of the successive convergents.

Answer 1

$\Rightarrow$ We know that

$\frac{1}{a_1}-\frac{x}{a_2}+\frac{x^2}{a_3}-\frac{x^3}{a_4}+....=\frac{1}{a_1+}\frac{a_1^{2}x}{a_2-a_{1}x+}\frac{a_2^{2}x}{a_3-a_{2}x+}.....$

Hence, $\frac{1}{q_1}-\frac{x}{q_1{q}_2}+\frac{x^2}{q_2{q}_3}-\frac{x^3}{q_3{q}_4}+.....$

$=\frac{1}{q_1+}\frac{q_1^2}{q_1{q}_2-q_{1}x+}\frac{q_1^2{q}_1^{2}x}{q_2{q}_3-q_1{q}_{2}x}.....$

$=\frac{1}{q_1+}\frac{x}{\frac{q_2-x}{q_1}+}\frac{x}{\frac{q_3-q_{1}x}{q_2}+}\frac{x}{\frac{q_4-q_{2}x}{q_3}+}.....$

$=\frac{1}{a_1+}\frac{x}{a_2+}\frac{x}{a_3+}\frac{x}{a_4+}....$

Since $q_1=a_1,q_2=a_1{a}_2+x,q_3=a_3{q}_2+x{q}_1$

and so on

$\therefore q_1,q_2....$ are denominators of successive convergents