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Using the Newton's method the function $f(x)=\frac{1}{x^2}-a$ calculate approximation of $\frac{1}{\sqrt a}$.
a) Is this method locally convergent squarely or cubically?
b) Find $x_0$ such that Newton's iteration is convergent to $\frac{1}{\sqrt a}$

I know that if we have convergent squarely then exist $C\ge0$ such that $|x_{n+1}-x^*|\le C |x_n-x^*|^2$ and if we have convergent cubically then exist $C\ge0$ such that $|x_{n+1}-x^*|\le C |x_n-x^*|^3$.

So I have: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{\frac{1}{x_n^2}-a}{-2x_n^{-3}}=x_n+(\frac{1}{x_n^2}-a)\frac{x_n^3}{2}=x_n+\frac{x_n}{2}-\frac{ax_n^3}{2}=\frac{x_n}{2}(3-ax_n^2)$$ $$|x_{n+1}-x^*|=|\frac{x_n}{2}(3-ax_n^2)-\frac{1}{\sqrt a}|$$ But I have a problem with further estimation...

When it comes to finding $x_0$ I completely don't know what conditions should be met

newuser458
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2 Answers2

1

If we let $x_n=\frac1{\sqrt a}(1+r_n)$ where $r-n$ is the relative error in the $n^{\text{th}}$ iterate, then $$x_{n+1}=\frac1{2\sqrt a}(1+r_n)(3-(1+r_n)^2)=\frac1{\sqrt a}\left(1-\frac32r_n^2-\frac12r_n^3\right)=\frac1{\sqrt a}(1+r_{n+1})$$ So we are trying to find $C$ such that $$\frac{|r_{n+1}|}{\sqrt a}=\frac{\left|\frac32r_n^2+\frac12r_n^3\right|}{\sqrt a}\le C\frac{r_n^2}a$$ So any $C>\frac32\sqrt a$ should do. Since $$\lim_{r_n\rightarrow0}\frac{\frac{\left|r_{n+1}\right|}{\sqrt a}}{\left(\frac{\left|r_n\right|}{\sqrt a}\right)^2}=\frac32$$ the algorithm can't be cubically convergent.

The second question is more interesting. If $x_0=0$ then we get stuck at $0$, while if $x_0=\frac{\sqrt5}{\sqrt a}$ then we bounce back and forth between $\pm x_0$, while if $x_0>\frac{\sqrt5}{\sqrt a}$ out iterates grow at each step. For $0<x_0<\frac{c_0}{\sqrt a}=\frac{\sqrt3}{\sqrt a}$ we converge to $\frac1{\sqrt a}$ but if $x_0=\frac{c_0}{\sqrt a}$ then $x_1=0$ so we get stuck again. Looking at the problem in more detail we see that if we let $c_{k+1}=\sqrt[3]{c_k+\sqrt{c_k^2-1}}+\sqrt[3]{c_k-\sqrt{c_k^2-1}}$ then if $x_0=c_k$ we eventually converge to $0$; if $c_{2k}<x_0<c_{2k+1}$ we converge to $-\frac1{\sqrt a}$ and if $c_{2k+1}<x_0<c_{2k+2}$ we converge to $\frac1{\sqrt a}$. Whatever $x_0$ converges to, $-x_0$ will converge to its additive inverse. Note that $$\lim_{k\rightarrow\infty}c_k=\sqrt5$$

user5713492
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  • I have a few questions: 1) since $\lim_{r_n \to 0} ...=\frac{3}{2}$ so the algorithm can't be cubically convergent but can be quadratically convergent, yes? 2) What is $c_k$ in your solution? – newuser458 Mar 31 '20 at 11:58
  • Yes, it is quadratically convergent. $c_{-1}=0$; $$\begin{align}c_0=&\sqrt[3]{c_0+\sqrt{c_0^2-1}}+\sqrt[3]{c_0-\sqrt{c_0^2-1}}=\sqrt[3]{0-\sqrt{-1}}+\sqrt[3]{0-\sqrt{-1}}\&=\sqrt[3]{i}+\sqrt[3]{-i}=\left(\frac{\sqrt3}2+\frac i2\right)+\left(\frac{\sqrt3}2-\frac i2\right)=\sqrt3\end{align}$$ And further $c_{k+1}=\sqrt[3]{c_k+\sqrt{c_k^2-1}}+\sqrt[3]{c_k-\sqrt{c_k^2-1}}$. That's sort of explained above.The sequence ${c_k}$ separates regions that converge to $\frac1{\sqrt a}$ and to $-\frac1{\sqrt a}$. – user5713492 Mar 31 '20 at 15:06
  • Oh I misspoke slightly: it's $\left{\frac{c_k}{\sqrt a}\right}$ that separates the regions of convergence. I think I have a proof that $$\lim_{k\rightarrow\infty}6^k\left(1-\frac{c_k}{\sqrt5}\right)=C\approx0.23932103507792341419821$$ – user5713492 Mar 31 '20 at 18:59
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Regarding b), you can use the convergence criterion for Netwon's method. I assume a proof of this can be found in your textbook or course notes, but here is a related SE answer in case you're curious.

First pick an interval that contains your root, say $[\frac{1}{2}, 1]$

$f'(x) = \frac{-2}{x^3}$

$f''(x) = \frac{6}{x^4}$

$M = max |\frac{f''(x)}{2f'(x)} | = max|\frac{-3}{2x}|\ , \ x\in[\frac 1 2, 1]$

$M = 3$

So you can pick any $x \in (\frac 1 {\sqrt 2}- \frac 1 3, \frac 1 {\sqrt 2} + \frac 1 3)$

Eleftheria Chatziargyriou
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