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I am following the book Iteration of Rational Functions by Alan F. Beardon. In this book, On page number 2, they are characterizing the behaviour of a sequence of iterates of the fixed point.

It says that if z is a point close to the fixed point $\zeta$ of the rational function $R($z$)$, then, approximately, $|$$R($z$)-$$\zeta$|=$|$$R($z$)-$$R($$\zeta$)|=|$R$$'$(z)||$z$$-$$\zeta$|

Clearly, From this, I can see that points close to attracting fixed points move even closer to it when we apply $R$, while points close to the repelling fixed point tend to move away from it.

Further, the following line is written in the book and I am not able to understand it-

"If z is close to (but not equal to ) the repelling fixed point $\zeta$, itnially it is repelled away from it but it may return to the vicinity of $\zeta$ (or even to $\zeta$ itself) at a later stage".

According to me if we take a sequenc of iterates of a point which is close the repelling fixed point $\zeta$ then sequence will be always repelled away to the $\zeta$.

Please someone help me regarding this. Also, please suggest me any software in which I can visualize this kind of iterations of rational functions.

N_Set
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  • Can you include more contexts into the post, like the definition of the repelling fixed point, and more importantly your doubt about it. – Arctic Char Jul 19 '22 at 16:46
  • Repelling fixed point is the fixed point whose modulus of the derivative at that fixed point is strictly greater than the $1$. Also, For your kind information, I have posted my doubt clearly and also someone posted the answer of it. – N_Set Jul 19 '22 at 18:15

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Consider the iteration of $f(z)=4(z-z^3)$ for values of $z$ a bit bigger than the repelling fixed point at zero. It's not too hard to see that these iterates are generally increasing and that, if you pick your starting point just right, then you'll hit $z=1$ exactly. Of course, $f(1)=0$ so that you land on the repelling fixed point at the next step.

As a concrete example, if $z_0 \approx 0.067709$ is the smallest positive root of $$ 4 \left(4 \left(z-z^3\right)-64 \left(z-z^3\right)^3\right)=1, $$ then $z_0$ is in the component of $|f'(z)|>1$ about zero, yet $f^3(z_0)=0$

Mark McClure
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    Okay, I got it. So we can say that the fact, a point close to the repelling fixed point, moves away from it when we apply the function, is true for only finite number of iterations? – N_Set Jul 19 '22 at 18:11