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Fixed point iteration is $p_n={p_{n-1}\over{2}}+{1\over p_{n-1}}$ $n=1,2,...$. As it is fixed point iteration $p_n=g(p_{n-1})$. Hence, $g(x)={x\over 2}+{1\over{x}}$.Can someone give me a head start and that could start this problem. I tried induction but got some really complicated expressions which I cannot simplify. I tried proving $\sqrt{2}<p_n<p_{n-1}$ by induction when $p_0$ is not $\sqrt{2}$. Maybe I am doing something wrong in my induction method.

Don
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  • if i recall correctly, the even terms of the sequence are monotonic, and the odd terms of the sequence are monotonic – mathworker21 Sep 14 '19 at 14:07
  • Carry out the first few steps in some examples. Try for example some cases with $p_0 <0, 0 < p_0 < \sqrt{2}, p_0 = \sqrt{2}, p_0 > \sqrt{2}$. – GEdgar Sep 14 '19 at 14:08
  • If you look at the link I posted as a duplicate, you will have an idea of how many times this question was asked ;) – rtybase Sep 14 '19 at 14:11
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    Thank you rtybase – Don Sep 14 '19 at 14:34

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Hints: if the iteration converges the limit $l$ must verify $l = g(l)$; is false the the sequence converges for all $p_0$ (if $p_0 = 0$...); but $g$ is contractive near...

Martín-Blas Pérez Pinilla
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