Recursion relation convergence

hasan_researc
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Homework Statement



Consider the sequence \left x_{n}\{\right\} defined by the recursion relation,

x_{n+1} = \frac{1}{2} \left( x_{n} + \frac{2}{x_{n}} \right)

where x0 > 0.

Use the fact that "if a sequence of real numbers is monotonically decreasing and
bounded from below, then it converges" to prove that the sequence converges.

Show that it converges to \sqrt{2}.

Homework Equations





The Attempt at a Solution




No idea!
Any help would be greatly appreciated. :smile:
 
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No ideas? How about the idea of doing what the hint says?

1) Prove the sequence is increasing. Use "proof by induction" to show that x_{n+1}\ge x_n for all n.

2) Prove that the sequence has an upper bound. Again, use "proof by induction to show that x_n< 2 for all n.
 
This is what I have found so far:

x_{n+1} < x_{n}
\frac{1}{2}\left( x_{n} + \frac{2}{x_{n}} \right) < 2x_{n}
x_{n} > \sqrt{2}

But I have n't used induction to show that the sequence is bounded below by \sqrt{2} :confused:

Now, I have to show that the sequence is monotonically decreasing (by induction, as you say).

But how I prove the x_{n+1} \leq x_{n} for n = 1 (to begin with)?:confused:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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