if a>0 and $\displaystyle x_{1}>0$

and $\displaystyle x_{n+1} = \frac{1}{2}(x_{n} +\frac{a}{x_{n}})$

find the limit of the sequence as n goes to infinity

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- Mar 16th 2009, 06:13 AMxalklimit of a sequence
if a>0 and $\displaystyle x_{1}>0$

and $\displaystyle x_{n+1} = \frac{1}{2}(x_{n} +\frac{a}{x_{n}})$

find the limit of the sequence as n goes to infinity - Mar 16th 2009, 06:17 AMJhevon
i assume you have already proven that a (finite) limit exists?

note that $\displaystyle \lim_{n \to \infty}x_n = \lim_{n \to \infty}x_{n + 1}$, call this limit $\displaystyle L$. Then we have

$\displaystyle \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \frac 12 \left( x_n +\frac a{x_n} \right)$

$\displaystyle \Rightarrow L = \frac 12 \left( L +\frac aL \right)$

Now find $\displaystyle L$ - Mar 16th 2009, 10:14 AMxalk
o.k thanks i have done that L= sqrt(a) or L= -sqrt(a)

Now which is the limit of the function ,because up to now we assumed that the sequence converges we have not proved that the sequence converges - Mar 16th 2009, 10:21 AMPlato
- Mar 16th 2009, 11:54 AMxalk
Thank you, but i am not so sure why we must take the positive sqrt of a as the limit of the function because all the terms of the sequence are positive .Is there a theorem or another problem ,saying that if all the terms of the sequence are +ve then the limit of the function must be +ve??

To prove convergence i tried hard with no result so a complete proof will be very mush appreciated it - Mar 16th 2009, 12:15 PMPlato
It is known that if a positive sequence converges its limit is non-negative.

- Mar 16th 2009, 12:49 PMxalk
can this be proved? if ,yes how??

- Mar 16th 2009, 01:03 PMPlato
If the limit of a sequence is negative then almost all of its terms must be negative.

Say $\displaystyle \left( {s_n } \right) \to L < 0$ then using the definition let $\displaystyle \varepsilon = \frac{{ - L}}{2}$.

$\displaystyle \left( {\forall n \geqslant N} \right)\left[ {\left| {s_n - L} \right| <\varepsilon } \right]\, \Rightarrow \,s_n < \frac{L}{2} < 0$ - Mar 16th 2009, 02:15 PMxalk
Good .So a proof of convergence is left

- Mar 16th 2009, 02:27 PMPlato
- Mar 16th 2009, 02:57 PMxalk
HOW??

- Mar 16th 2009, 03:14 PMPlato
- Mar 20th 2009, 05:30 PMxalk
By considering the difference :

$\displaystyle x_{n}-x_{n+1} = \frac{(x_{n}-\sqrt{a})(x_{n}+\sqrt{a})}{2x_{n}}$ we can study the monotony of the sequence .

But in this difference all terms are positive except the term $\displaystyle x_{n}-\sqrt{a}$ which can be positive or -ve for all values of nεN.

At the same time if we can prove whether the term $\displaystyle x_{n}-\sqrt{a}$ is +ve or -ve we will have proved whether the sequence is bounded above or bellow by sqrt(a).

So how do we that ,if anyone knows please help. - Mar 20th 2009, 05:34 PMThePerfectHacker
You can go here.

- Mar 21st 2009, 06:40 AMxalk
Thank you ,but i cannot see how you get:

$\displaystyle s_n^2 = x + (s_{n-1} - xs_{n-1})^2 $