Contractive Sequences problem

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daon

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Jan 27, 2006
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Okay. I am to prove that \(\displaystyle b_n\),defined below, is contractive:

\(\displaystyle b_1 = \frac{4}{5}\)
\(\displaystyle b_{n+1} = \frac{1}{2}(b_n + \frac{2}{b_n})\)

Okay, here's my thoughts... A while ago I worked on the general case of this squence being convergent. So, I know that \(\displaystyle b_n \ge \sqrt{2}, \,\, \forall n \ge 2\). I also know this sequence is non-increasing if that would help.

Now, I have gotten as far as showing that:
\(\displaystyle \L |b_{n+1} - b_{n+2}| \,\, \le \,\, (\frac{1}{2} + \frac{1}{b_nb_{n+1}})|b_n - b_{n+1}|\)

Now, by the definition of contractive, I need that for \(\displaystyle \L 0 \,\, < \,\, r \,\, < \,\, 1\)
\(\displaystyle \L |b_{n+1} - b_{n+2}| \,\, \le \,\, r \cdot |b_n - b_{n+1}|\).

Thus, I need to show that there is a positive real number C < 1 such that:
\(\displaystyle \L (\frac{1}{2} + \frac{1}{b_nb_{n+1}}) \,\, \le \,\, C \,\, < \,\, 1\).

I figure that it would be fine to show that:
\(\displaystyle \L \frac{1}{b_nb_{n+1}} \,\, < \,\, \frac{1}{2} \,\, \,\, or \,\, \,\, b_nb_{n+1} \,\, > \,\, 2\).

Since I only know that:
\(\displaystyle \L b_n \,\, \ge \,\, \sqrt{2}\)

I can only get that
\(\displaystyle \L b_nb_{n+1} \,\, \ge \,\, \sqrt{2}\sqrt{2} = 2\) (but I need strictly greater than.)

Any ideas?
 
Its funny how after I take the time to write out the problem presentably, I am likely to gain more insight.

I proved that \(\displaystyle n>2 \,\, \Rightarrow \,\, b_n\) is NON-INCREASING, when it would have helped to prove that for \(\displaystyle n>2 \,\, \Rightarrow \,\, b_n\) is DECREASING, which I have just done.

For this tells me that (for n>2):
\(\displaystyle b_n > b_{n+1} \,\, \Rightarrow \,\, b_n = b_{n+1} + \epsilon \,\,\) for some \(\displaystyle \epsilon \,\, \in \L \mathbb{R}^+\).

So that:
\(\displaystyle \L b_nb_{n+1} = (b_{n+1} + \epsilon )(b_{n+1}) \,\, \ge \,\, (\sqrt{2} + \epsilon)(\sqrt{2}) \,\, = \,\, 2 + \sqrt{2} \epsilon \,\, > \,\, 2\)

I think thats correct.
Thanks,
-Daon
 
daon, I have done real analysis a long time but never seen a definition of contractive sequence. However, what you have done makes sense. I will say that the sequence does converge to \(\displaystyle \sqrt 2\) and does so very rapidily.
 
pka said:
daon, I have done real analysis a long time but never seen a definition of contractive sequence. However, what you have done makes sense. I will say that the sequence does converge to \(\displaystyle \sqrt 2\) and does so very rapidily.
Click to expand...

I don't doubt that. Searching for examples of contractive sequences on the internet proved very toublesome. The only resource (besides my book) Ive found was: http://everything2.com/index.pl?node_id=1766070

My teacher introduces contractiveness in the section on Cauchy sequences. The full-blown definition is:

A sequence \(\displaystyle {a_n}\) in \(\displaystyle R^m\) is a said to be contractive if there is a 0 < r < 1 such that for all n, \(\displaystyle |a_{n} - a_{n+1}| \le r|a_{n-1} - a_n|\)

And with use of facts from Cauchy sequences he proves: Every contractive sequence converges.

Thanks for your help,
-daon
 
daon said:
And with use of facts from Cauchy sequences he proves: Every contractive sequence converges.
Click to expand...
Be careful there. You must be in a complete metric space for that to be true.
 
The idea of a matric space has been spoken, but not yet introduced in my class. I think it is safe to assume what you said. In our book it states that in future classes such as topology we'll see more of them.

In any case, I thought \(\displaystyle \mathbb{R}^m\) was a complete metric space? If it is, the idea of a contractive sequence deals only with sequences in \(\displaystyle \mathbb{R}^m\), so what you said is implied. But, again, I don't know...
 
daon said:
...
I proved that \(\displaystyle n>2 \,\, \Rightarrow \,\, b_n\) is NON-INCREASING, when it would have helped to prove that for \(\displaystyle n>2 \,\, \Rightarrow \,\, b_n\) is DECREASING, which I have just done.
Click to expand...

Okay, so I thought I proved it was strictly decreasing, but unfortunately I made a mistake. I've spent the last couple hours trying to show this to no avail. Its also very aggrivating since it seems so obvious!

My latest idea was to prove that b_n > \(\displaystyle \sqrt{2}\) for all n>2, by assuming (by way of contradiction) that b_n <= \(\displaystyle \sqrt{2}\) for some n, and since I've proven that b_n is non-increasing and b_n >= \(\displaystyle \sqrt{2}\) for all n>1, this imples that for some n, b_n = \(\displaystyle \sqrt{2}\) and for all m>n b_m\(\displaystyle =\sqrt{2}\).. This is so obviously false, but I'm stuck nevertheless...
Thanks,
-Daon
 
A thing that might work is assume that \sqrt(2)=b_n for some n then show a contradiction, then you can get the bn>sqrt(2).
 
Hmm.. Not quite sure on how to get a contradiction...

\(\displaystyle b_n = \sqrt{2}\) imples that \(\displaystyle b_{n+1}=\sqrt{2}\)... (so no help there)

Now assume that \(\displaystyle b_{n-1} > \sqrt{2}\) (I know this is true for at least n=3 and can be verified numerically for any n if you;re patient. I went all the way to n=6)

\(\displaystyle \L \sqrt{2} = b_n = \frac{b^2_{n-1} + 2}{2b_{n-1}} > \frac{(\sqrt{2})^2 + 2}{2b_{n-1}} = \frac{2}{b_{n-1}}\)

So,
\(\displaystyle \L b_{n-1} > \frac{2}{\sqrt{2}} = \sqrt{2}\) (DOH!!)

Now, I could also do this


\(\displaystyle \L \sqrt{2} = b_n = \frac{b^2_{n-1} + 2}{2b_{n-1}} < \frac{b^2_{n-1} + 2}{2\sqrt{2}} \,\, \Longrightarrow \,\, (\frac{b_{n-1}}{2})^2 > \frac{1}{2} \,\, \Longrightarrow \,\, b_{n-1} > \frac{2}{\sqrt{2}} = \sqrt{2}\) (DOHHH!!!)

I just keep going in circles... My class is over with but I'm still going to work on this and post the solution if I figureit out. But, if anyone else figures it out... for goodness sakes let me know!
 
I had gotten the contradiction yesterday when I posted it... I'll post it in a few minutes.
 
Assume \(\displaystyle b_{n+1}=\sqrt{2}\)

Then:

\(\displaystyle \sqrt{2}=\frac{b_n}{2}+\frac{1}{b_n}\)

This implies b_n=\sqrt{2}.

Thus if it is true for some n, all the n's before them are sqrt(2).

It does not hold for b_2. I think that should be enough.
 
marcmtlca said:
Assume \(\displaystyle b_{n+1}=\sqrt{2}\)

Then:

\(\displaystyle \sqrt{2}=\frac{b_n}{2}+\frac{1}{b_n}\)

This implies b_n=\sqrt{2}.

Thus if it is true for some n, all the n's before them are sqrt(2).

It does not hold for b_2. I think that should be enough.
Click to expand...

Wow, thanks.. worked beautifully.
 
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