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Does there exist an analytic function $f(t)$ which has a root $\alpha$ with multiplicity > 1 but which the multiplier $|N'_f(t)|$ of the Newton map $N_f(t)=t-f(t)/f'(t)$ is not equal to 1?

The multiplier of a fixed point $\alpha$ of a map $f (x)$ where $f (\alpha) = \alpha$ is equal to the absolute value of the derivative of the map evaluated at the point $\alpha$. \begin{equation} \lambda_f (\alpha) = | \dot{f} (\alpha) | \end{equation} If $\lambda_f (\alpha) < 1$ then $\alpha$ is a said to be an attractive fixed-point of the map $f (x)$. If $\lambda_f (\alpha) = 1$ then $\alpha$ is an indifferent fixed point, and if $\lambda_f (\alpha) > 1$ then $\alpha$ is a repelling fixed-point. When $\lambda_f (\alpha) = 0$ the fixed-point $\alpha$ is said to be super-attractive

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$f(z)$ is analytic and has a zero of multiplicity $m$ at $z=a$ means that for some function $h(z)$ analytic around $z=a$ : $$\log f(z) = h(z)+ m\log (z-a), \quad \frac{f'(z)}{f(z)} = h'(z)+\frac{m}{z-a}$$ $$N_f(z) =z- \frac{1}{h'(z)+\frac{m}{z-a}},\quad N_f'(z)= 1+\frac{h''(z)-\frac{m}{(z-a)^2}}{(h'(z)+\frac{m}{z-a})^2}$$ $$ \boxed{N_f'(a) = 1-\frac{1}{m}}$$

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Let $g(t)=f(t)/f'(t)$ where $f(t)=tanh(t-2)^2*tanh(t-4)$ then $g(t)$ has the same roots of $f(t)$ but with only simple roots. Here $g(t)={\frac { \left( \tanh \left( t-2 \right) \right) ^{2}\tanh \left( t-4 \right) }{2\,\tanh \left( t-2 \right) \tanh \left( t-4 \right) \left( 1- \left( \tanh \left( t-2 \right) \right) ^{2} \right) + \left( \tanh \left( t-2 \right) \right) ^{2} \left( 1- \left( \tanh \left( t-4 \right) \right) ^{2} \right) }} $ then the multiplier of the root of $g$ at $t=2$ is equal to $1/2$ so that the multiplicity of $f(2)$ is $m=1/(1-1/2)=1/(1/2)=2$ , a formula which cannot be applied to $f$ directly due to its multiple root being an indifferent fixed-point of $N_f(x)$

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  • $\begingroup$ what ? ${}{} {}{}$ $\endgroup$ – reuns Feb 27 '17 at 20:56
  • $\begingroup$ @user1952009 see p251 of pdfs.semanticscholar.org/a456/… $\endgroup$ – crow Feb 27 '17 at 21:04
  • $\begingroup$ the multiplier $l(a)$ of the fixed-point $a$ of the Newton map is the absolute value of the derivative of the map evaluated at the root $Z(a)=0$, which is a fixed-point of the Newton map $N_Z(a)=a$. The multiplicity $m(a)$ of the root $a$ to which the fixed-point corresponds is equal to $m(a)=1/(1-l(a))$ when $l(a) \neq 1$ . The zeros of $Z(t)/Z'(t)$ are the same as the zeros of $\frac{\frac{Z (t)}{\dot{Z} (t)}}{\frac{d}{dt} \frac{Z (t)}{\dot{Z} (t)}}$ $\endgroup$ – crow Feb 27 '17 at 21:15
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    $\begingroup$ $f(z) = z^2, N_f(z) = \frac{z}{2}, N_f'(z) = 1/2$ $\endgroup$ – reuns Feb 27 '17 at 21:49
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    $\begingroup$ $f(z) = e^{-1/z}$ has an essential singularity at $z=0$. You can't really think to it as a zero of order $\infty$ but $\lim_{z \to 0, Re(z) > 0} f(z) = 0$ while $\lim_{z \to 0, Re(z) < 0} f(z) = \infty$, and every essential singularity are of this kind $\endgroup$ – reuns Feb 27 '17 at 22:09
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Let $f(t)=tanh(t-2)^2*tanh(t-4)$ which has a double root at $t=2$ then the multiplier of the fixed-point at 2 is equal to 1/2 so its multiplicity is 1/(1-1/2)=2

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