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Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$

Let $(x_{n})$ be a sequence of rational numbers defined recursively : $x_0=1$, $x_{n+1}=\frac{1}{2}(x_{n}+\frac{2}{x_{n}})$.
Prove that that sequence is Cauchy without using any form of the Completeness property

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Nameless
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  • Why wouldn't you use the completeness property? – Pedro Apr 01 '12 at 16:14
  • How exactly are we supposed to 'use' the Cauchy criterion? The only interesting thing about the Cauchy criterion is that any sequence satisfying the Cauchy criterion must converge, but this fact is equivalent to the Completeness Property of the Real Numbers. – John Gowers Apr 01 '12 at 16:15
  • These questions seem to be related: Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ and Finding the limit of a recursive sequence. – Martin Sleziak Apr 01 '12 at 16:19
  • The Cauchy criterion implies the completeness of $\mathbb R$ as given any sequence $(x_n)$ we have $\lim\limits_{n\to \infty}x_n\text{ exists }\iff \sum\limits_{n=1}^\infty (x_{n+1}-x_n)\text{ exists}$, and if $(x_n)$ is Cauchy the latter satisfies the Cauchy criterion. – Alex Becker Apr 01 '12 at 16:27
  • After reading the comments I realised you were right and changed the question – Nameless Apr 01 '12 at 16:39
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    One answer to "Prove this sequence is Cauchy" is that some sequences are Cauchy sequences but no sequence is Cauchy. Cauchy was a person. I once met a mathematician who insisted on usages consistent with that position. – Michael Hardy Apr 01 '12 at 17:18
  • @MichaelHardy I have to agree. It always distresses me when people say that sequences are Cauchy, or sets are Ramsey etc. However, I think that that is standard usage now. – John Gowers Apr 01 '12 at 18:44
  • @Michael, Donkey_2009: Alas nowadays we say that a set has a property of being Ramsey, or a sequence has the property of being a Cauchy sequence. Much like an apple may have the property of being red, or a group being abelian. Does it bother you when someone says "This apple is red"? – Asaf Karagila Apr 01 '12 at 19:30
  • .....and spaces are Hausdorff. – Michael Hardy Apr 01 '12 at 21:40
  • @Asaf : If the color had been named after Prof. Red, the situation might be different. – Michael Hardy Apr 01 '12 at 21:40
  • In the phrase 'Cauchy sequence' the word 'Cauchy' is not an adjective, but is a noun being used as a modifier. You might go into a shop to buy mountain bikes, but would you not be bothered if the shopkeeper pointed to a row of bikes and said, "All these bikes are mountain"? – John Gowers Apr 01 '12 at 23:44
  • @Asaf Coming back to the apple analogy - we have cooking apples and eating apples, but if someone told you that their apples were eating, what would you think? – John Gowers Apr 02 '12 at 15:06
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    @Donkey_2009: Genetic experiment gone wrong; Chernobyl Apples; "Feed me, Seymour!". – Asaf Karagila Apr 02 '12 at 15:33

3 Answers3

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For $x_n>0$,

$$ x_{n+1}=\frac12\left(x_n+\frac{2}{x_n}\right)=\frac12\left(\sqrt{x_n}-\sqrt{\frac{2}{x_n}}\right)^2+\sqrt{2}\ge\sqrt{2}\tag{1} $$
Edit: a good point was raised regarding the existence of the square roots above. Instead, we can use $$ x_{n+1}^2=\frac14\left(x_n+\frac{2}{x_n}\right)^2=\frac14\left(x_n-\frac{2}{x_n}\right)^2+2\ge2\tag{1} $$ Thus, each $x_n$ after the first must satisfy $x_n^2\ge2$. This implies $$ x_{n+1}-x_n=\frac{2-x_n^2}{2x_n}\le0\tag{2} $$ Since $\{x_n\}$ is a decreasing sequence, bounded below, it is a Cauchy sequence.

Moved to this answer, which required extra work. This seems to indicate that the questions are not equivalent.

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robjohn
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  • 'Since {xn} is a decreasing sequence, bounded below, is converges'. I think that counts as using a form of the Completeness Property. However, it is true that a decreasing sequence bounded below is always a Cauchy sequence, even over sets such as the rational numbers where the Completeness Property does not hold. – John Gowers Apr 01 '12 at 23:59
  • @Donkey_2009: thanks, I missed the part about not using the Completeness Property when I first answered. I have been gone until now, so I have just removed the claim of convergence. – robjohn Apr 02 '12 at 00:09
  • Another slight issue - how do you know that $\sqrt{2}$ exists without invoking the Completeness Property for the Real Numbers? – John Gowers Apr 02 '12 at 20:01
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    @Donkey_2009: Thanks again. $\sqrt{x_n}$ also poses a problem. I think the repaired equation $(1)$ above removes all assumptions about the Completeness Property. – robjohn Apr 02 '12 at 21:08
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You might wish to show recursively that, for every $n\geqslant0$, $1\leqslant x_n\leqslant3/2$ and $$ x_{n+1}^2-2=\frac{(x_n^2-2)^2}{4x_n^2}\quad\text{hence}\quad 0\leqslant x_{n+1}^2-2\leqslant\frac14(x_n^2-2)^2 . $$ This proves that, for every $n\geqslant1$, $2\leqslant x_n^2\leqslant2+1/4^{n}$, which implies the result you are after. (Naturally, the rate of convergence is much faster but this control seems to be sufficient for your purpose.)

Did
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We're not allowed to use the Completeness Axiom, so we're going to have to formulate the proof so that all terms involved are rational. So I'm going to start by showing that $x_n^2\to 2$.

$$x_{n+1}^2=\left (\frac{1}{2}\left (x_n+\frac{2}{x_n}\right )\right )^2=\frac{1}{4}\left (x_n^2+\frac{4}{x_n^2}+4\right )=\frac{x_n^2}{4}+\frac{1}{x_n^2}+1$$

We now rewrite $x_n^2$ as $2(1+\delta_n)$ and get:

$$ \begin{align} 2(1+\delta_{n+1}) &=\frac{1+\delta_n}{2}+\frac{1}{2(1+\delta_n)}+1\\ &=\frac{1}{2}+\frac{\delta_n}{2}+\frac{1}{2}-\frac{\delta_n}{2(1+\delta_n)}+1\\ &=2+\frac{\delta_n^2}{2(1+\delta_n)}\\ \end{align} $$

So,

$$\delta_{n+1}=\frac{\delta_n^2}{4(1+\delta_n)}$$

Now in all of this I've assumed that $\delta_n \neq -1$, whereas $\delta_0$ is in fact $-1$. This doesn't matter: we'll start from $\delta_1$, and it should be fairly clear that $\delta_n$ is positive for all $n\ge 1$.

$$x_1^2=\frac{1}{4}+1+1=2\left (1+\frac{1}{8}\right )\textrm{ so }\delta_1=\frac{1}{8}$$

Now for simplicity's sake we note that $0<\delta_n<1$ for all $n\ge 1$, so

$$\delta_{n+1}=\frac{\delta_n^2}{4(1+\delta_n)}<\frac{\delta_n}{4}.$$

Now $\delta_1=\frac{1}{8}$, so it is obvious that $\delta_n<\frac{1}{2}\times\left (\frac{1}{4}\right )^n$ for all $n\ge 1$ (though you can come up with a formal proof using induction, if you like). This sequence clearly tends to $0$ (and you can prove that using the definition of convergence if you like). Therefore, $x_n^2=2(1+\delta_n)\to 2(1+0)=2$.

Now we know that $x_n>0$ for all $n\ge 1$, and we (I hope) know the result that for convergent sequences $a_n\to a$ and $b_n\to b$, $a_n\times b_n\to a\times b$.

Now suppose that $(x_n)$ diverges. But then $(x_n^2)$ also diverges, which is a contradiction. So $(x_n)$ converges to a limit, which we shall call $x$. Now $x_n^2\to x^2$, so $x^2=2$. Clearly $x$ is positive, so $x=\sqrt{2}$.

I see no reason to use any form of the Completeness Axiom for the Real Numbers. This proof of the statement holds in the algebraic numbers and $\mathbb{Q}\cup \{\sqrt{2}\}$ as well as in the reals, whereas the Completeness Axiom only holds in the Real Numbers, so using it weakens the proof.

John Gowers
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  • A kind of big proof, when compared to Didier's one but thanks anyway. – Nameless Apr 01 '12 at 17:09
  • ..aaand the question has now changed completely. Since we've shown that $x_n\to \sqrt{2}$, we can now say that $|x_{n+1}-x_n|=|\frac{1}{x_n}-\frac{x_n}{2}|\to \frac{1}{\sqrt{2}}-\frac{\sqrt{2}}{2}=0$. So the sequence is Cauchy. And we didn't use the Completeness Axiom, but we went rather a long way round. – John Gowers Apr 01 '12 at 17:10
  • And I'm pretty sure that my proof is exactly equivalent to Dider's, which wasn't up when I started writing... :-( – John Gowers Apr 01 '12 at 17:11
  • I'm astonished that no one has noticed the contradiction in my post. I actually implicitly assumed completeness when I supposed that $x_n$ must either tend to a limit or diverge. Oh well.... – John Gowers Apr 01 '12 at 18:43
  • Well with Dieder's proof we have that $x_n^2\to 2$ and so that it is Cauchy (a Cauchy sequence if you prefer) and we have only used the fact that $\mathbb{R}$ is archimedean. – Nameless Apr 01 '12 at 19:07
  • No - we both proved that $x_n^2\to 2$, but that does not imply that $x_n\to \sqrt{2}$ or that $(x_n)$ is a Cauchy sequence. For example, if we were working over the rational numbers rather than the real numbers (which we might as well be doing since we're not allowed to use completeness), $x_n^2$ would still tend to $2$, but $x_n$ would not converge, as for any $x$ in $\mathbb{Q}$ we can find some $\varepsilon$ such that for all $N$ there is some $n>N$ with $|x_n-x|\ge \varepsilon$ - just take $\varepsilon$ to be some rational number less than $|x-\sqrt{2}|$. – John Gowers Apr 01 '12 at 23:40