Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\dfrac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$?

share|improve this question
    
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
1  
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
add comment

2 Answers

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 $n \geq 2$.

share|improve this answer
    
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
add comment

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 $$x_{n} - \sqrt{2} ≥0$$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.