let $f(t)=t$ be an analytic function which has a fixed-point at $t$. The multiplier $\lambda (t) = f'(t)$ is just the 1st derivative evaluated at the fixed-point. The standard nomenclature is that when $\| \lambda (t) \|=1$ the fixed-point is said to be indifferent, or neutral. However, it does not necessarily mean that trajectories can't be attracted or repelled by this point, only that its 1st derivative equals 1.

Wouldn't "indeterminate" be a better name than "indifferent" ?

The function $(1 / ( 1 - \| \lambda (t) \| ) )$ is ill-defined when $\lambda(t) = 1$


Perhaps a matter of opinion, but "indeterminate" should mean that its type cannot be determined, while "indifferent" should mean that its type can be any. Right?

From this point of view, although it is only language, it seems better to use "indifferent" since clearly the type of a fixed point will definitely be well defined, for each specific dynamics.

On the other hand, it also common to use "parabolic" (and in my opinion much better) in opposition to "hyperbolic". The latter (usually) means that $|f'(t)|\ne1$.

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    $\begingroup$ I agree with your interpretation of indeterminate vs indifferent. I think that parabolic is typically reserved for the case where $f'(t)$ is a root of unity, though, which is a stronger assumption that $|f'(t)|=1$. As such, parabolic points are definitely not indeterminate, as the Leau-Fatou flower theorem describes their dynamics quite precisely. $\endgroup$ – Mark McClure Mar 4 '17 at 23:32
  • $\begingroup$ I see, now that you put it that way, indifferent does make more sense. since if one is indifferent, they could go either way. indeterminate would imply it cant be determined as you say. on an indifferent fixed-point the trajectories at that point are only changing by at most a constant independent of t. at a superattractive point it isnt changing by anything since the derivative is exactly 0 $\endgroup$ – crow Mar 5 '17 at 0:09
  • $\begingroup$ i understand the case of indifferent fixed-points is much more complicated than attractive or repulsive . $\endgroup$ – crow Mar 5 '17 at 0:12

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