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Let $\alpha, \gamma$ be real numbers such that $0<\alpha<1$ and $\gamma>0$. Consider the sequence of real numbers given by $$ \begin{cases} x_0\ne 0&\\ x_{k+1}=x_k\left(1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right) \quad (k\in \mathbb{N}).& \end{cases} $$ Suppose that $x_k\ne 0$ for all $k\in \mathbb{N}$. Prove that :

  • The sequence $\{x_k\}_{k\in\mathbb{N}}$ does not converge.

  • The sequence $\{|x_k|\}_{k\in\mathbb{N}}$ converges to $[(1/2)(1+\alpha)\gamma]^{1/(1-\alpha)}.$

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Hint: Start with the second part, observing that $|x_{n+1}|=|x_n|-\gamma(1+\alpha)|x_n|^\alpha$. (And is there a typo? This seems like converging $\to0$ at most.) –  Hagen von Eitzen Sep 13 '12 at 23:06
    
@ Hagen von Eitzen How can we have $|x_{n+1}|=|x_n|-\gamma(1+\alpha)|x_n|^{\alpha}$? –  blindman Sep 13 '12 at 23:13
    
That should be $|x_{n+1}| = \left| |x_n| - \gamma (1+\alpha) |x_n|^\alpha \right|$. If this is $f(|x_n|)$, note that $f(t) < t$ for $t > t^*$ and $f(t) > t$ for $0 < t < t^*$, where $t^* = (\gamma (1+\alpha)/2)^{1/(1-\alpha)}$. –  Robert Israel Sep 13 '12 at 23:30
    
@Robert Israel: Thank you for your comment. How can we continue with your hint? –  blindman Sep 13 '12 at 23:39
    
Actually the statement is not true. There is a sequence of initial values for which some $x_k = 0$, and then it stays $0$: $y_1 = (\gamma (1 + \alpha))^{1/(1-\alpha)}$ for which $f(y_1) = 0$, $y_2$ for which $f(y_2) = y_1$, etc. –  Robert Israel Sep 14 '12 at 2:26

2 Answers 2

Let $L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$.

Suppose that $|x_k| < L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{\alpha/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}(\gamma(1+\alpha)/2)^{1/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$

from which we conclude that $|x_k|^{\alpha}(\gamma(1+\alpha)) < 2L$. Also $|x_k| < L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$ implies that $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}>2$. Then $|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|= |x_k|\left(\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}-1\right)>|x_k|$. If $|x_{k+1}| >L$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) < 2L$ we manipulate to get $|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L< L-|x_k|$ and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L< L-|x_k|$. So we conclude that $|x_{k+1}|-L < L - |x_k|$ and then from $|x_{k+1}| >L$ and $|x_k| < L$ we know that $||x_{k+1}|-L| < ||x_k|-L|$.

Supppose that $|x_k| > L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$ then $|x_k|^{\alpha}(\gamma(1+\alpha)) > 2L$. Also $0<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<2$ and $|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|<|x_k|$. If $|x_{k+1}| < L$ and $1<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<2$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) > 2L$ we manipulate to get $|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L> L-|x_k|$ and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L> L-|x_k|$. So we conclude that $|x_{k+1}|-L > L - |x_k|$ and $L-|x_{k+1}| > |x_k|-L$ then from $|x_{k+1}| <L$ and $|x_k| > L$ we know that $||x_{k+1}|-L| < ||x_k|-L|$. If $|x_{k+1}| <L$ and $0<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<1$ then $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<1$ implies that $|x_k| > 2^{1/(1-\alpha)}L\ge 2L$. And from $|x_{k+1}|<L$ we know $||x_{k+1}|-L|<L< |x_k| - L = ||x_k| - L|$.

In summary we have shown that if $|x_k| < L$ then $|x_{k+1}| >|x_k|$. If $|x_k| < L$ and $|x_{k+1}| <L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| < L$ and $|x_{k+1}| >L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| > L$ then $|x_{k+1}| <|x_k|$. If $|x_k| > L$ and $|x_{k+1}| >L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| > L$ and $|x_{k+1}| <L$ then $||x_{k+1}|-L|<||x_k|-L|$. No matter what the case it is true that $||x_{k+1}|-L|<||x_k|-L|$. Therefore the sequence $|x_k|-L$ converges. Showing that the limit of $|x_k|$ must be $L$ is easy.

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We only need to prove the second claim for the following reason.

If $|x_k|\rightarrow (\gamma (1+\alpha)/2)^{1/(1-\alpha)}$, $$\frac{x_{k+1}}{x_k}=\left(1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right) \rightarrow-1.$$ Thus ${x_{k}}$ will be an oscillating sequence.

I will write it later about how to prove the second claim.

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Dear Sir, How can we continue? –  blindman Sep 27 '12 at 4:45
2  
@ZhiyongWang Your "will write it later" seems to be long overdue. –  Lord_Farin Sep 17 '13 at 7:53

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