1

Let $$x_{n+1}= \dfrac{1}{2}\left ( x_n+\dfrac{R}{x_n} \right )$$

  1. Interpret this relation in terms of quadratic convergence. State your condition for such quadratic convergence clearly.

  2. Prove that after the first step, the iteration is monotonical decreasing.

Here is my attempt:

  1. We need the condition that $x_n\neq 0$,for all $n=0,1,2,...$ Then, I represent it as follows:
    $$x_{n+1}^2 - R =\left ( \dfrac{x_n^2 -R}{2x_n} \right )^2$$

We put $f(x) =x^2-R$, then, $f'(x) \neq 0 $ when $R\neq 0$.

Then, by Newton's method theorem : If $f$ , $f'$, and $f''$ are continuous in a neighborhood of a root $r$ of $f$ and if $f'(r) \neq 0$, then there is a positive $\eta$ with the following property: If the initial point in Newton’s method satisfies $|r - x_0|\geq \eta$, then all subsequent points $x_n$ satisfy the same inequality, converge to $r$, and do so quadratically; that is, $$|r − x_{n+1}| \geq c(\eta)|r − x_n|^2$$

Hence, in this case, $x_n$ converges quadratically if $x_0^2 \in [R-\eta, R+\eta]$, then we get $$ \sqrt{R-\eta} \leq x_0 \leq \sqrt{R+\eta} \quad \text{ or} \quad -\sqrt{R+\eta}\leq x_0 \leq -\sqrt{R-\eta}$$

Could you please check my answer or give me some suggestion to solve it? Thank you so much.

  1. I have $x_{n+1}-x_n= \dfrac{R-x_n^2 }{2x_n} $, it would be easy if we have the condition that $x_0> 0$,but in this case, I am not sure whether $x_n$ decreases when $x_0 <0$. Could you please show me how to solve it?

Many thanks for any help.

anna
  • 65
  • R is a constant – anna Sep 21 '20 at 22:14
  • Your root is $r = \sqrt{R}$, so express $(x_{n+1} - \sqrt{R})$ in terms of $(x_n - \sqrt{R})$. – Christopher A. Wong Sep 21 '20 at 23:06
  • @ChristopherA.Wong I tried but I could not, could you please give me more detail to express like that? and what about the root $ -\sqrt{R}$? – anna Sep 21 '20 at 23:39
  • Which root is obtained would depend on the sign of $x_0$.

    Try this. Let $\delta_{n+1} = x_{n+1} - \sqrt{R}$, and $\delta_n = x_n - \sqrt{R}$. Then $\delta_{n+1} = (1/2)(x_n + R/x_n) - \sqrt{R} = (1/2)(\delta_n + \sqrt{R} + R/(\delta_n + \sqrt{R})) - \sqrt{R}$. Now algebraically simplify.

     – Christopher A. Wong Sep 22 '20 at 01:39
  • @ChristopherA.Wong Should I divide it into 2 case: $x_0>0 $ and $x_0<0$? For the first case, based on your advice, I got that $\delta_{n+1}=\frac{\delta_n^2}{2(\delta_n+\sqrt{R})}$, i.e., $x_{n+1}-\sqrt{R}=\dfrac{(x_n-\sqrt{R})^2}{2x_n}$, do I need to evaluate the denominator here? I mean something like $x_n \geq$ a constant – anna Sep 22 '20 at 02:05
  • Correct you'll have two cases, I only covered the positive case. I agree the denominator must be addressed; the denominator is $x_n$; so you must first establish that the iteration converges; in which case the denominator is bounded by a constant. – Christopher A. Wong Sep 22 '20 at 02:24
  • I got it, I could proceed with this case. Could you kindly give me some advice about the negative case? I am not sure that it is the same with the positive case. – anna Sep 22 '20 at 02:32
  • 1
    You get a simplification of the expression in the approach of @ChristopherA.Wong if you use the fraction of the distance to the two roots, $$\frac{x_{n+1}-\sqrt{R}}{x_{n+1}+\sqrt{R}}=\left(\frac{x_{n}-\sqrt{R}}{x_{n}+\sqrt{R}}\right)^2.$$ – Lutz Lehmann Sep 22 '20 at 06:04
  • Would you consider this question a duplicate of https://math.stackexchange.com/questions/2289703/finding-the-square-root-newton-raphson? Are all of your questions treated in the answers there? See also the similar topics https://math.stackexchange.com/questions/2386351, https://math.stackexchange.com/questions/3009864/babylonian-method – Lutz Lehmann Sep 22 '20 at 14:21
  • @LutzLehmann I saw these topics, but I could not understand in a case $x_0 <0$, because I think it follows that $x_n<0$, and then this sequence does not decrease after the first step. Could you please help me clarify this? – anna Sep 22 '20 at 15:35
  • The easiest way is to see that the iteration is symmetric to zero, so that if $x_0<0$, the sequence $(-x_n)$ obeys the recursion equation on the positive half axis. This should reduce sign errors/errors due to "sign intuition". Indeed, the negative sequences should increase after the first step and approach $-\sqrt{R}$ from below. – Lutz Lehmann Sep 22 '20 at 15:53

0 Answers0