Proving that a sequence is contractive.

[daon]

Let \(\displaystyle x_{n+1} = \sqrt{25-2x_n}\) and \(\displaystyle x_1 > 0\)

I need to show that this sequence is contractive and am having trouble.

\(\displaystyle x_n\) is contractive \(\displaystyle \Leftrightarrow\) \(\displaystyle |x_{n+1} - x_{n+2}| < r |x_n - x_{n+1}|\), where \(\displaystyle 0 < r < 1\). I have tried plugging in the square roots but am unable to get close to what it needs to be.

The way I'm attempting it is by starting with \(\displaystyle |x_{n+1}-x_{n+2}|\) and trying to get a series of less-than/less-than-or-equals to attain the above. I've done a few of these and they were more or less straight-forward, but I'm feeling there's a trick of some kind I can't find.

Thanks in advance,
-Daon

 

[pka]

Daon, I am frankly concerned about the size of x<SUB>1</SUB>. But, say it is not too big. Then
\(\displaystyle \begin{array}{rcl}
\left| {x_{n + 2} - x_{n + 1} } \right| & = & \\
& = & \left( {\left| {\sqrt {25 - 2x_{n + 1} } - \sqrt {25 - 2x_n } } \right|} \right)\frac{{\sqrt {25 - 2x_{n + 1} } + \sqrt {25 - 2x_n } }}{{\sqrt {25 - 2x_{n + 1} } + \sqrt {25 - 2x_n } }} \\
& = & \frac{{\left| {2x_n - 2x_{n + 1} } \right|}}{{\sqrt {25 - 2x_{n + 1} } + \sqrt {25 - 2x_n } }} \\
& \le & \frac{{\left| {2x_n - 2x_{n + 1} } \right|}}{{\sqrt {25} + \sqrt {25} }} \\
& \le & \frac{1}{5}\left| {x_n - x_{n + 1} } \right| \\
\end{array}.\)

 

[daon]

Thanks pka. I see your point about it not being to big, but I believe we are supposed to assume it was well-defined, so x_n <= 25/2. I better ask him just in case.

 

[daon]

Actually pka, I have a question about one of your steps:
\(\displaystyle \frac{{\left| {2x_n - 2x_{n + 1} } \right|}}{{\sqrt {25 - 2x_{n + 1} } + \sqrt {25 - 2x_n } }}
\le \frac{{\left| {2x_n - 2x_{n + 1} } \right|}}{{\sqrt {25} + \sqrt {25} }}\)

Doesn't \(\displaystyle \sqrt{25-2x_{n}} < \sqrt{25} \Rightarrow \frac{1}{\sqrt{25-2x_{n+1}}} > \frac{1}{\sqrt{25}\)?

 

[marcmtlca]

seems that way to me too...

 

[pka]

Yes I fear you are correct.
Right now I do not see a way around that.
But I am sure that is what has to be done.

PS
I had to think about it. I am so use to doing this with +2x<SUB>n</SUB>.
By induction show that 3.5<x<SUB>n</SUB><5, for 3<n
Then the above will work.