# Math Help - numerical method

1. ## numerical method

Show that by putting $x_{n+1} = x_{n} = x$ that if the sequence defined by $x_{n+1} = \frac{1}{2}( x_{n} + \frac{N}{x_{n}})$

converges, it will converge to $\sqrt{N}$
Usually the questions tell you what $x_{0} =$ some number, but not sure how to start with this one, any help appreciated

thank you

2. Originally Posted by Tweety
Usually the questions tell you what $x_{0} =$ some number, but not sure how to start with this one, any help appreciated

thank you
Would it shock you if I told you that the way to start is to do what they tell you to do? Replace both $x_n$ and $x_{n+1}$ in the formula by x.

That gives you $x= \frac{1}{2}\left(x+ \frac{N}{x}\right)$ solve that for x.

IF this sequence converges, then $\lim x_{n+1}= \frac{1}{2}\left(\lim x_n+ \frac{N}{\lim x_n}\right)$ and each of those limits is of the same sequence and so is equal to x. This is, by the way, a well known iterative method for finding square roots. In fact, it is often known as the "Babylonian method"!
Methods of computing square roots - Wikipedia, the free encyclopedia

3. okay thanks,

When I solve for x I get $x = \frac{1}{2}(x+ \frac{N}{x})$

$x = \frac{1}{2}(\frac{Nx^{2}}{x})$

$x = \frac{1}{2}(Nx)$

$2x = Nx$

x=0
Is this correct?

4. Originally Posted by Tweety
okay thanks,

When I solve for x I get $x = \frac{1}{2}(x+ \frac{N}{x})$

$x = \frac{1}{2}(\frac{Nx^{2}}{x})$
You need practice adding fractions! $x+ \frac{N}{x}= \frac{x^2}{x}+ \frac{N}{x}= \frac{x^2+ N}{x}$

The simpler thing to do is just multiply the entire equation, $x= \frac{1}{2}(x+ \frac{N}{x})$, by 2x.
$x = \frac{1}{2}(Nx)$

$2x = Nx$

x=0
Is this correct?
Clearly not, since you were told that this sequence was supposed to converge to $\sqrt{N}$!

By the way, there is a slight inaccuracy in the problem, related to the fact that they did not give you $x_0$.

For any $x_0> 0$, this recursive sequence converges to $\sqrt{N}$.

But if $x_0< 0$, it converges to $-\sqrt{N}$.

(And, of course, if N< 0, the sequence does not converge.)