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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.)2012-09-13
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    @ Hagen von Eitzen How can we have $|x_{n+1}|=|x_n|-\gamma(1+\alpha)|x_n|^{\alpha}$?2012-09-13
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    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 $t^* = (\gamma (1+\alpha)/2)^{1/(1-\alpha)}$.2012-09-13
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    @Robert Israel: Thank you for your comment. How can we continue with your hint?2012-09-13
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    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.2012-09-14
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    @Robert Israel: Thank you for your careful comment. I revised the question.2012-09-14

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