Prove this recursively defined sequence goes to 0

[Ryker]

Homework Statement

Show that 2^{n^{1001}} |a_{n} - a_{\infty}| \rightarrow 0 as n \rightarrow \infty.

Here, a_n is defined recursively by a_{1} = 1, a_{n+1} = \frac{1}{2}(a_{n}+\frac{x}{a_{n}}).

I already know that a_{\infty} = \sqrt{x}.

Homework Equations

We are given a hint to consider (y_{n}) = \frac{a_{n} - a_{\infty}}{a_{n} + a_{\infty}}.

The Attempt at a Solution

I considered the hint, and by defining the sequence in the hint to be (y_n) I got that y_n+1 = y_n². However, I don't know how to proceed and show that the above sequence goes to zero. I know a_{n} - a_{\infty} \rightarrow 0 by definition, but I don't see how that y_n thing helps. Any thoughts?

 

[Ryker]

Anyone?

 

[Ryker]

OK, one final bump, and then I give up. But from the lack of responses I see I'm not the only one really puzzled by this problem.

Basically, I need to find a way to show (a_n) as defined above converges faster than (2ⁿ)¹⁰⁰¹ diverges.

 

[Mr.Miyagi]

You haven't mentioned the domain of the functions in (a_n). It should probably not contain 0, as I think the quantity in question doesn't converge for x=0.

I haven't checked it, but I assume that y_{n+1}=y_n^2 is correct. This implies that y_{n}=y_1^{2^n}. Moreover, we can calculate that |y_1|&lt; 1 from the definition of (y_n) and the fact that x\neq 0. Now \begin{eqnarray}<br /> 2^{n^{1001}} |a_{n} - a_{\infty}| &amp;=&amp; 2^{n^{1001}} |a_{n} + a_{\infty}|||y_n| \nonumber \\<br /> &amp;\leq&amp; 2^{n^{1001}} |2\sqrt{x}||y_n| \nonumber \\<br /> &amp;=&amp; 2^{n^{1001}} |2\sqrt{x}||y_1^{2^n}| \nonumber \\<br /> &amp;=&amp; 2^{n^{1001}} |2\sqrt{x}||y_1|^{2^n}<br /> \end{eqnarray}.
Can you carry on from here to prove that this must converge to 0 pointwise? And if you can, is the convergence uniform?

 

[Ryker]

Hi Mr. Miyagi, and thanks for the response! Even though the homework due date has now already past, I appreciate the input. And in fact, after four hours of wracking my brain on how to do this, I did indeed finally come to the same conclusion as you did, and essentially used the fact that exponentials rise faster than polynomials. I feel really stupid for needing such a long time to figure that out, but eh...

Oh, and the domain is the real numbers, but x must be greater than zero.