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Find the limit of the sequence represented by the following non linear difference equation and determine their rate of convergence $$x_{n+1}=\frac{1}{2}\bigg{(}x_{n}+\frac{16}{x_n}\bigg{)}.$$

Is there anything wrong in this problem? since there is no intial value$(x_0)$, Is it possible to find rate of convergence? Give me some hint?

Do I need to consider to cases when $x_{n}>0 ,x_{n}<0 \space\forall n \in \mathbb{N} $

ALMEra
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  • You might study what happens for $x_0$ in each of the intervals $(-\infty, -4)$, $[-4,-4]$, $(-4,0)$, $(0,4)$, $[4,4]$, and $(4,\infty)$... – Eric Towers Nov 25 '21 at 17:49
  • @EricTowers since it converge $\pm 4$ is it necessary? don't we need to calculate $\lim_{n \to \infty} \frac{|x_{n+1} +4|}{{|x_{n}+4|}}$ and$ \lim_{n \to \infty} \frac{|x_{n+1} -4|}{|{x_{n}-4|}}$ – ALMEra Nov 25 '21 at 17:54
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    By the way, this is the Babylonian Method https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method – b00n heT Nov 25 '21 at 17:57
  • See also the related or duplicate https://math.stackexchange.com/questions/2455260/prove-fixed-point-iteration, https://math.stackexchange.com/questions/2289703/finding-the-square-root-newton-raphson. (One would think there is more to find, but apparently that is only for convergence, not the quadratic convergence order.) – Lutz Lehmann Nov 26 '21 at 09:20
  • @LutzLehmann but there is no way to find the rate converges in those duplicates – ALMEra Nov 26 '21 at 10:26
  • $e_{n+1}\approx Ke_n^2\implies Ke_n\approx (Ke_0)^{2^n}$ indicates a quadratic rate or order of convergence. – Lutz Lehmann Nov 26 '21 at 10:29

3 Answers3

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If a recursion of the form $x_{n+1} = g(x_n)$ is convergent to $z$ (with regular enough $g$) and $$ g'(z)=\cdots = g^{(p-1)}(z) =0, g^{(p)}(z) \ne 0 $$

then $(x_n)$ converges with order $p$.

In this case, since the only possible values for $z$ are $\pm 4$, and noting that $g'(\pm 4)=0, g''(\mp 4) \ne 0$, we conclude that when $x_0$ is chosen in a way that $x_n$ converges, the convergence is quadratic.

In this case, it is possible to establish that when $x_0>0$ the sequence converges quadratically to 4 and, when $x_0<0$, the sequence converges quadratically to $-4$.

Convergence when $x_0>0$.

In this case, since $g(x_0) > 4$ when $x_0 < 4$, we can assume without loss of generality that $x_0 \ge 4$. Applying the fixed point theorem to $g$ on the set $[4, +\infty[$, we get convergence for any $x_0 >0$.

PierreCarre
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  • How do we find the rate of convergence? – ALMEra Nov 26 '21 at 10:21
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    What you can find is the order of convergence. In this case, since $p=2$, it converges with order $p$, in the sense that $|e_{n+1}|\approx K |e_n|^2$. This is most likely what you are expected to provide. – PierreCarre Nov 26 '21 at 13:53
  • [+1] Interesting. I see that you do not enter more into the details than me when you say "when $x_0$ is chosen in a way that $x_n$ converges" ;)... – Jean Marie Nov 26 '21 at 17:33
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    @JeanMarie In the end I did address convergence when $x_0>$... – PierreCarre Nov 27 '21 at 22:09
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First of all, let us assume without loss of generality that $x_0>0$ (implying that all further $x_n$ are themselves $>0$). Indeed if $x_0<0$, we have completely symmetrical results.

In fact, this sequence can be "casted" into a Newton's method convergence:

$$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}=x_n-\dfrac{x_n^2-16}{2x_n}$$

with $f$ defined by $f(x)=x^2-16$ (roots of $f(x)=0$ are precisely $\pm 4$).

One of the essential properties of Newton's method is that it has a quadratic convergence. It means that

$$e_{n+1}=Ke_n^2 \ \ \text{where} \ \ e_n=|x_{n+1}-4| \ \ \text{and} \ \ K=\dfrac{f''(L)}{2f'(L)}=\dfrac18.$$

(where $L$ is the limit, with $L=4$ here).

for a certain $K$, which gives a fantastic speed of convergence (under the condition to be in a good vicinity of the root).

Therefore, the computation of $\lim_{n \to \infty} \frac{e_{n+1}}{e_{n}}$ i.e.

$$\lim_{n \to \infty} \frac{|x_{n+1}-4|}{{|x_{n}-4|}}$$

will give $0$ but gives no information, plainly because the convergence isn't linear as is the case for example for a fixed point method.

Jean Marie
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  • I am using a definition "A sequence is said to converge with order $k(>1)$ if $\exists C>0$ and $N>0$ $|x_{n+1}-\alpha| \leq C|x_n-\alpha| ^k$, Can't we find such a $K$? – ALMEra Nov 25 '21 at 19:23
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    I haven't been precise enough: I should have said $e_{n+1} \approx Ke_n^2$ with $K=\dfrac{f''(L)}{2f'(L)}$ where the limit $L=2$ here. But I don't completely stick to your definition : what is usually taken for a convergence with order $k$ is $|x_{n+1}-\alpha| \approx C|x_n-\alpha|^k$... – Jean Marie Nov 25 '21 at 19:29
  • I am new to this subject I am using above definition, If you know about this definition can you please tell me what it is really saying – ALMEra Nov 25 '21 at 19:36
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    I find $K=2/(2 (2L))=1/8$. – Jean Marie Nov 25 '21 at 19:36
  • The definition of $f(x) \approx g(x)$ in a vicinity of $L$ (here the limit) is $\lim_{x \to L}\dfrac{f(x)}{g(x)}=1$. – Jean Marie Nov 25 '21 at 19:38
  • Thank you very much for your help. My definition and your definition is totally different therefore I have no idea – ALMEra Nov 25 '21 at 19:50
  • Here is a site where you will find how to use your definition with $\le$ signs (which is fact not so different from mine...): https://www.cantorsparadise.com/a-modern-look-at-square-roots-in-the-babylonian-way-ccd48a5e8716 (you will find it on the 4th or 5th screen) – Jean Marie Nov 25 '21 at 19:53
  • Thank you very much it is interesting page for me. if there is no difference is rate of convergence $1/8$? – ALMEra Nov 25 '21 at 19:56
  • There is no reason that it's not the same, but once again I wouldn't call it a "rate" of convergence ("rate" should be reserved to the case $\Delta Y/\Delta X$ whereas here it is a $(\Delta Y)^2/\Delta X$. – Jean Marie Nov 25 '21 at 20:11
  • Thank you very much – ALMEra Nov 25 '21 at 20:33
  • @JeanMarie I get the idea, but you don't actually show that the sequence converges when $x_0>0$. Newton's method is not always convergent, and does not always converge with order two. – PierreCarre Nov 26 '21 at 09:19
  • @PierreCarre You are right. It is why I have advise the OP to take a look here to have a proof that convergence takes place whatever $x_0 \ne 0$, which is a rare case in the framework of Newton's method. About the fact that the order can be different from two, isn't it in very particular cases such as when $f'(L)=0$ ? – Jean Marie Nov 26 '21 at 17:33
  • @JeanMarie Yes, the order can be lower when $f'(L)=0$, but it can also be higher in certain conditions, namely when $(x-f(x)/f'(x))''_{|x=z} = 0$. – PierreCarre Nov 27 '21 at 22:12
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We will restrict ourselves to $x_0>0$. Let's rewrite it like: $$x_{n+1} = \frac12\left(x_n+\frac{k^2}{x_n}\right)$$

In your case, $k=4$. You can see that at the limit, you have: $$\lim_{n->\infty} x_n = A = \frac12\left(A+\frac{k^2}{A}\right)$$ so: $$A = k$$

Now, if you compute $\frac{x_{n+1}-x_n}{x_n}$, you get: $$\frac12\left(\frac{k^2}{x_n^2}-1\right)$$

So you have quadratic convergence as $x_n \to k$.

PC1
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