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Can $x \rightarrow \sqrt{x}$ be considered as one dimensional discrete dynamical system? As I see it will have the fixed points $x^{*} = 0,1$.

I am not sure about the stability of $x^* = 0$ as $|f'(x^*)|$ is not defined where $f(x) = \sqrt{x}$. The fixed point at $x=1$ is stable as $|f'(1)|<1$. So all positive real numbers except zero will converge to the fixed point $x^* = 1$.

BAYMAX
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    Since "all positive real numbers except zero will converge to the fixed point $x^* = 1$", we conclude that $x^* = 0$ is... – John B Jun 16 '20 at 23:32
  • Yeah, unstable...but maybe there is some other test using which we can show it analytically? – BAYMAX Jun 16 '20 at 23:50
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    You can note that $\sqrt x>x$ for $x\in(0,1)$. – John B Jun 16 '20 at 23:52
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    A fixed point repels if, in any small-enough deleted neighborhood, the derivative is greater than $1$. In this case, choose any neighborhood $(0,\epsilon)$ for $0 < \epsilon < 1/4$. Conclude. – Brevan Ellefsen Jun 16 '20 at 23:52
  • Oh cool! thanks@JohnB @Brevan Ellefsen! – BAYMAX Jun 16 '20 at 23:55

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