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I don't understand one step of proving of the Babylonian Method for $a=2,x_1=1$, and

$$x_{n+1} =\frac{1}{2}( x_n + \frac{a}{x_n}). \quad (n=1,2,\ldots)$$

$$x_{n+1} - \sqrt{2} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - \sqrt{2} = \frac{(x_n^2-2\sqrt{2} x_n+2)}{2x_n} = \frac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$$

For all values of $x_n>0$. Conclude $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.

After using $x_n \geq \sqrt{2}$ we find

$$x_{n+1} - x_n = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - x_n = \frac{x_n^2+2-2x_n^2}{2x_n} =\frac{2-x_n^2}{2x_n} \leq 0,$$

which means the sequence $(x_n)$ is monotonically decreasing. Now we know $(x_n)$ is bounded below by $\sqrt{2}$ and $(x_n)$ is monotonically decreasing so $(x_n)$ has a limit.

My problem is I don't understand how could we say $x_n \ge \sqrt{2}$ by using $\dfrac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$?

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if a>1 and $x_1=a$ it will be easy to explain why $x_n≥√2$ so $ (2-x_n^2)/2x_n ≤0$ – Shamill Mar 24 '12 at 17:37
Could you please edit you question and (1) use \sqrt{2} instead of and \ge instead of (2) Could you please state the problem clearly: what is given, what is to be proven (3) Write the step you don't understand. (4) Refrain from bullets. (5) Add new lines, use new paragraphs when needed. – user2468 Mar 24 '12 at 17:46
@Shamill, note that for all values of $x_n > 0$, $\frac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$, this is because numerator is positive (square function is always positive) and denominator is positive because it is given $x_n > 0$. – quartz Mar 24 '12 at 18:36
I find the injunction to Conclude $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$ highly misleading since $x_{n+1}\geqslant\sqrt2$ for every positive $x_n$, irrespectively of whether $x_n\geqslant\sqrt2$ or $0\lt x_n\lt\sqrt2$. – Did Apr 2 '12 at 14:00
up vote 2 down vote accepted

There seems to be confusion on what you think they claim, and what they actually claim. The step you are talking about is as follows.

Conclude that $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.

What is claimed here is that if $x_n \geq \sqrt{2}$, then $x_{n+1} \geq \sqrt{2}$. We know this holds, since, as shown in the previous step,

$$x_{n+1} - \sqrt{2} = \frac{(x_n - \sqrt{2})^2}{2 x_n} \geq 0. \quad \quad (\text{if } x_n > 0)$$

So a stronger statement actually holds: $x_n > 0 \Rightarrow x_{n+1} \geq \sqrt{2}$. Since $x_n \geq \sqrt{2}$ implies $x_n \geq 0$, this then also implies $x_{n+1} \geq \sqrt{2}$.

However, this does not mean that $x_n \geq \sqrt{2}$ for any $n$! This only means that if $x_{n_0} \geq \sqrt{2}$ for some $n_0 \in \mathbb{N}$, then we can use induction to show that $x_n \geq \sqrt{2}$ for all $n \geq n_0$.

So probably you would now also like to find such an $n_0$. Since $x_n > 0 \Rightarrow x_{n+1} \geq \sqrt{2}$ as shown above, and $x_1 = 1 > 0$, it follows that $x_2 \geq \sqrt{2}$. And indeed, $x_2 = 2 \geq \sqrt{2}$. So then, by induction we get $x_n \geq \sqrt{2}$ for all $n \geq 2$.

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thank you very much, I had to see " However, this does not mean that $x_n \geq \sqrt{2}$ for any $n$! This only means that if $x_{n_0} \geq \sqrt{2}$ for some $n_0 \in \mathbb{N}$, then we can use induction to show that $x_n \geq \sqrt{2}$ for all $n \geq n_0$. " – Shamill Mar 24 '12 at 19:05
@Shamill: You are welcome :) – TMM Mar 24 '12 at 19:09

you have concluded that for $x_n > 0$,$$x_{n+1} - \sqrt{2} = {{((x_n−\sqrt{2})^2)}\over{(2x_n)}}≥0. $$ This means $$x_{n+1} - \sqrt{2} ≥ 0$$ Now since n is arbitrary, any value of n will satisfy this equation , put $n = n -1$ in this equation, you will get $$x_{n} - \sqrt{2} ≥0$$

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sequence $(x_n)$ is for n=1,2,.. so we can only say this if $n\ge2$. if we ignore n=1 so $x_1=1$ we can explain $x_n$ is monotonically decreasing, if we dont ignore it goes like $x_1=1$, $x_2=3/2$, $x_3=17/12$, $x_4=577/408$, ... that means $x_1$<$x_2$ and $x_2>x_3>x_4>...$ – Shamill Mar 24 '12 at 18:51
@Shamill, your observation is right that if $n\ge2$, then the sequence is monotonically decreasing. – quartz Mar 24 '12 at 19:04

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