(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider the positive integer p and any positive number [tex]\alpha[/tex] and the formula:

[tex]x_{n+1}=\frac{p-1}{p}x_n+\frac{\alpha}{p}x_n^{-p+1}[/tex]

and describe the behavior of the resulting sequences [tex]\{x_n\}[/tex]

2. Relevant equations

In the problem statement.

3. The attempt at a solution

So in the earlier problem of this chapter where p=2. We first choose [tex]x_1>\sqrt{\alpha}[/tex] and then show that the resulting sequence is monotonically decreasing and is bounded below by [tex]\sqrt{\alpha}[/tex]. Hence a limit exists and letting [tex]x_{n+1}[/tex] and [tex]x_n[/tex] be equal to that limit and solving we get the limit of the sequence.

Using a similar strategy I wanted to begin with [tex]x_1>\alpha^{\frac{1}{p}}[/tex] and show that the resulting sequence is monotonically decreasing and bounded below by [tex]\alpha^{\frac{1}{p}}[/tex] and then show that the sequence converges to [tex]\alpha^{\frac{1}{p}}[/tex].

I have proved that [tex]x_{n+1}<x_n[/tex] iff [tex]x_n>\alpha^{\frac{1}{p}}[/tex] so proving inductively that [tex]x_n>\alpha^{\frac{1}{p}}[/tex] for all n should do the job. But I can't prove that. This is where I am stuck.

I have used excel to plot the sequence for various values of [tex]\alpha[/tex] and p and each time the sequence monotonically decreases and converges to [tex]\alpha^{\frac{1}{p}[/tex] if we choose [tex]x_1>\alpha^{\frac{1}{p}}[/tex].

I'm just stuck at the point I mentioned earlier.

Any help would be appreciated. Thank you.

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# Homework Help: Question from Rudin

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