Let $f$ be a polynomial with a super attracting fixed point at $x=0$. Can the immediate basin of attraction of the fixed point contain other roots? If so, please provide a specific example with the immediate basin of attraction. If not, why?
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$\begingroup$ How about $z^3-z^2$? $\endgroup$– Mark McClureCommented Feb 24, 2020 at 20:20
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$\begingroup$ @Mark is zero an attracting fixed point? Is, say, $z=1/2$ attracted to zero? $\endgroup$– Gerry MyersonCommented Feb 24, 2020 at 21:17
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$\begingroup$ @Mark is 1 in the immediate basin of attraction of 0? It is not obvious $\endgroup$– user152801Commented Feb 24, 2020 at 21:35
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$\begingroup$ what does super attracting mean ? $\endgroup$– Piyush GroverCommented Feb 24, 2020 at 21:39
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$\begingroup$ @Piyush A fixed point is super-attracting if the derivative is 0. It means the points in the basin of attraction approach specifically at an exponential rate. In this case, for $z$ in the basin of attraction,$ |f(z)-0|\approx b|z-0|^p$ for some constant p. $\endgroup$– user152801Commented Feb 24, 2020 at 21:48
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1 Answer
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Let $f(z)=az^2(z-1)$. Zero is superattracting. Now choose $a$ so small that $|f(z)|<|z|/2$ for $|z|<2$. Then the root $z_0=1$ is in the immediate domain of attraction.
answered Feb 25, 2020 at 11:02