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.

  • $\begingroup$ 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. $\endgroup$ – Hagen von Eitzen Dec 29 '13 at 15:46
  • $\begingroup$ i know that is the basis of induction but it is given in my homework assignment as it is $\endgroup$ – Vibha Dec 29 '13 at 15:47
  • $\begingroup$ ohh i had solved it i think your suggestion really works thankyou $\endgroup$ – Vibha Dec 29 '13 at 15:48
  • $\begingroup$ "What is $a$?" This should read $a_0>0$. With $a_0<0$, this recursion converges to $-\sqrt A$. $\endgroup$ – Lutz Lehmann Dec 29 '13 at 15:56

Did you show that the recursion formula leads to

$$ \frac{a_{n+1}-\sqrt{A}}{ a_{n+1}+\sqrt{A}} = \Big(\frac{a_n-\sqrt{A} }{ a_n + \sqrt{A}}\Big)^2?$$

Then all that remains is to show that the recursion formula $\theta_{n+1}=\theta_n^2$, that is, repeated squaring, leads to $\theta_n=\theta_1^{2^{n-1}}$.

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  • $\begingroup$ i dont know what is recursion formula $\endgroup$ – Vibha Dec 29 '13 at 15:44
  • $\begingroup$ recursion, see recursion. You compute $a_{n+1}$ from $a_n$. The formula that expresses this computation is a recursion formula. $\endgroup$ – Lutz Lehmann Dec 29 '13 at 15:55

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