# Limit of a sequence

• May 24th 2008, 10:41 PM
arbolis
Limit of a sequence
I'm not able to find a limit of a sequence.
It's related to the Babylonian's method to approximate a square root. In our case, we have $a>0$, and we want to approximate $\sqrt a$. Thanks to the Newton's method I've found the Babylonian's method. We have $x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}$. We can start the method for any $x_0>0$. I also showed that the sequence $x_n$ is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by $\sqrt a$. Now I must prove that the sequence ${x_n}$ converges to $\sqrt a$ when $n$ tends to $+\infty$ and this is precisely what I'm asking you to help me.
• May 25th 2008, 12:05 AM
Moo
Hello,

Quote:

Originally Posted by arbolis
I'm not able to find a limit of a sequence.
It's related to the Babylonian's method to approximate a square root. In our case, we have $a>0$, and we want to approximate $\sqrt a$. Thanks to the Newton's method I've found the Babylonian's method. We have $x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}$. We can start the method for any $x_0>0$. I also showed that the sequence $x_n$ is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by $\sqrt a$. Now I must prove that the sequence ${x_n}$ converges to $\sqrt a$ when $n$ tends to $+\infty$ and this is precisely what I'm asking you to help me.

:D

$\lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} \frac{x_n+\frac{a}{x_n}}{2} \ (1)$

Let L be the limit of the sequence.
$\lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L$

Therefore, from (1), we get :

$L=\frac{L+\frac aL}{2}$

...
Conclude ? (Nod)
• May 25th 2008, 08:46 AM
arbolis
Thanks so much!! I got $L^2=a$, so $L=\sqrt a$ because $a>0$. I feel I'm not used to find limit of sequences of this type. You helped me a lot. (Hi)
I'll try to demonstrate that $x_n$ is bounded below by $\sqrt a$.
• May 25th 2008, 09:58 AM
ThePerfectHacker
Quote:

Originally Posted by Moo
Hello,

:D

$\lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} \frac{x_n+\frac{a}{x_n}}{2} \ (1)$

Let L be the limit of the sequence.
$\lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L$

Therefore, from (1), we get :

$L=\frac{L+\frac aL}{2}$

...
Conclude ? (Nod)

Everything else is perfect, you just need to prove that the sequence is convergent.
• May 25th 2008, 09:59 AM
Moo
Quote:

Originally Posted by ThePerfectHacker
Everything else is perfect, you just need to prove that the sequence is convergent.

"Decreasing and bounded below" :)

If I read correctly, he said he had to prove the limit after showing that it was decreasing and bounded.
• May 25th 2008, 10:07 AM
arbolis
I confused Theperfecthacker because I said " I'll try to demonstrate that http://www.mathhelpforum.com/math-he...1720c05d-1.gif is bounded below by http://www.mathhelpforum.com/math-he...3ee61184-1.gif." I assumed the sequence was bounded below, but I had to prove it and I wasn't able. (Now I'll try).