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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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    what ? ${}{} {}{}$ – 2017-02-27
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    @user1952009 see p251 of https://pdfs.semanticscholar.org/a456/d7c92b85c03be1fd16a2bc7eaf9aa59f0100.pdf – 2017-02-27
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    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 $\frac{\frac{Z (t)}{\dot{Z} (t)}}{\frac{d}{dt} \frac{Z (t)}{\dot{Z} (t)}}$ – 2017-02-27
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    I see the confusion, the function $g(t)$ does not have multiple roots, but it does share the same set of distinct roots. the multipliers of its fixed-points give the multiplicities of the original function $f(t)$ – 2017-02-27
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    Yes see my answer nothing complicated here. My concern is why you think there is something interesting here – 2017-02-27
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    Thank you for your answer. No there is nothing interesting here. I was searching for a way to find the multiplicity of a root which corresponded to an indifferent fixed-point since the formula doesnt work for the case of indifferent fixed-points. I figured if I had an example of one I figured I could study the example. Another question would be, do all multiple roots correspond to indifferent fixed-points of the corresponding Newton map? – 2017-02-27
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    What is an indifferent fixed-point ? You are thinking to $Z(t),\zeta(s)$ so what do you mean ? And $Z(t)$ is real only on $t \in \mathbb{R}$ (in the same way that $\sin(t)$ is real for $t$ real) otherwise it is a complex analytic function – 2017-02-27
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    I've updated the question to include the definition, your answer is neat and informative but doesn't really answer the question. The function I posted, $g(t)$ doesnt have multiple roots, it has simple roots that coincide with the original function. That still doesn't answer my question, I cant think of a way to construct a function with multiple roots whose fixed-point multiplier is not equal to 1 – 2017-02-27
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    $f(z) = z^2, N_f(z) = \frac{z}{2}, N_f'(z) = 1/2$ – 2017-02-27
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    Oh neat, how simple... and easy to see that the multiplier/fixed-point correspondence works with that. – 2017-02-27
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    Sorry for the confusion. The original question was stupid, and I was getting confused looking at the sprawling Maple worksheets I'm working on. The example I posted actually fits the bill, just like the example @user1952009 posted...it seems like an indifferent fixed-point would correspond to a zero of infinite multiplicity (if such a thing is possible) – 2017-02-27
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    $f(z) = e^{-1/z}$ 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 – 2017-02-27
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    That's why I said your level in complex analysis is too low for studying $\zeta(s)$ seriously. You should start reading a complex analysis course, for example http://math.sfsu.edu/beck/papers/complex.pdf – 2017-02-27
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    It has nothing to do with my level of complex analysis knowledge, I posted here because I *knew* the results I was getting were non-sensical, because I had a syntax error in my Maple worksheet – 2017-02-27
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    It does not change the results of my conclusion that the (nontrivial) zeros of Z are simple. I will open another question for that – 2017-02-27
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    Come on... Don't dream you will prove something new on $\zeta(s)$ and $Z(t)$ ! As Peter Humphries said the simplicity of any function of the form $\zeta(s)h(s)$ with $h(s)$ analytic is unproven – 2017-02-27
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    Did you, or did you not read the pdf I posted? If you are so sure, prove it by showing how its wrong! – 2017-02-27
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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