Given $a_1=\frac{1}{2}(a_0+\frac{A}{a_0})$, $a_2=\frac{1}{2}(a_1+\frac{A}{a_1})$, $a_{n+1}=\frac{1}{2}(a_n+\frac{A}{a_n})$ for $n\ge2$ where $a>0$ and $A>0$; prove that $${a_n-\sqrt{A} \over a_n+\sqrt{A}} = \Big({a_1-\sqrt{A} \over a_1 + \sqrt{A}}\Big)^{2^{n-1}}$$

I tried to do this:

There are two steps to solve this: 1. prove that this true for $n=2$, then 2. prove that it is true for $n+1$ if it is true for $n$.

The first step is very simple, and I know that how how to do it but I am facing.some problem in the second step.

• What is $a$? Why not simplify the recursion to $a_{n+1}=\frac12(a_n+\frac A{a_n})$ for $n\ge0$? Also, I suggest to use $n=1$ as base case for the proof by induction - it is even more trivial. – Hagen von Eitzen Dec 29 '13 at 15:46
• i know that is the basis of induction but it is given in my homework assignment as it is – Vibha Dec 29 '13 at 15:47
• ohh i had solved it i think your suggestion really works thankyou – Vibha Dec 29 '13 at 15:48
• "What is $a$?" This should read $a_0>0$. With $a_0<0$, this recursion converges to $-\sqrt A$. – Lutz Lehmann Dec 29 '13 at 15:56

$$\frac{a_{n+1}-\sqrt{A}}{ a_{n+1}+\sqrt{A}} = \Big(\frac{a_n-\sqrt{A} }{ a_n + \sqrt{A}}\Big)^2?$$
Then all that remains is to show that the recursion formula $\theta_{n+1}=\theta_n^2$, that is, repeated squaring, leads to $\theta_n=\theta_1^{2^{n-1}}$.
• recursion, see recursion. You compute $a_{n+1}$ from $a_n$. The formula that expresses this computation is a recursion formula. – Lutz Lehmann Dec 29 '13 at 15:55