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Consider a smooth map $f: \mathbb{R} \rightarrow \mathbb{R}$ with an attracting fixed point $F$. Then, we have

  • if $f'(F) \ne 0$, $F$ is a "simple" attracting fixed point,
  • if $f'(F) = 0$, $F$ is a super-attracting fixed point,
  • if $f^{(k)}(F) = 0$ for $1 \le k \le n$, we can say $F$ is super-attracting of order $n$,
  • if $f^{(k)}(F) = 0$ for all $k \ge 1$, we could call $F$ a mega-attracting fixed point (I don't think that's a "standard" piece of terminology, but I like it!).

Examples:

  • $f(x) = \frac{\sin(x)}{2}$ has a simple attracting fixed point at $0$,
  • $f(x) = x^2$ has a super-attracting fixed point at $0$,
  • $f(x) = x^k$, $k > 1$, has a super-attracting fixed point of order $k-1$ at $0$,
  • $f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & \mbox{if } x \ne 0 \\ 0, & \mbox{if } x = 0\end{cases}$ has a mega-attracting fixed point at $0$, yet is not a constant function.

What are the upper and lower bounds on the "rate" of attraction for each class, and what's the proof? I suppose the absolute maximum "strength" of attraction of a mega-attracting fixed point, and thus of all fixed points, is the strength of the mega-attractor of a constant function, but what about the minimums, and what about the other classes?

Also, is there a generalization of these concepts to continuous flows, and to more complicated attracting sets?

Rodrigo de Azevedo
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The_Sympathizer
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    The trouble with this question is that the OP has not given any definition of "rate" or "strength" of attraction, making it difficult to say anything at all in the way of an answer. – Lee Mosher Sep 18 '14 at 22:41
  • @Lee Mosher: Good point. I'll have to think about this. – The_Sympathizer Sep 18 '14 at 22:52
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    That's also the meaning in standard numerical analysis. It boils down to the same definition which allows you to say, for example, that certain numerical methods (such as Newton's) converge "linearly" or "quadratically" or "exponentially" or whatever else and it reduces to how many digits of accuracy you get per iteration. A "linear" rate of convergence then, would be $O(n)$, whereas a quadratic rate would be $O(n^2)$, etc. –  Sep 20 '14 at 11:44
  • Just as a point if interest, google some stuff on Lyapunov exponents. – snar Sep 21 '14 at 02:13

1 Answers1

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As ioannis galidakis noted, the "rate of attraction" has a concrete meaning in numerical analysis, which is the asymptotics of the difference $|x_n-F|$ where $x_n$ is a sequence such that $x_n\to F$ and $x_n=f(x_{n-1})$ for $n>1$. In practical terms it helps to focus on the logarithm of difference, $$D_n=|\log |x_n-F||$$ which roughly corresponds to the number of digits of $x_n$ that are the same as the digits of $F$. (This is of obvious interest when $F$ is a root of an equation we are trying to solve.)

  • Simple attracting fixed point has $|x_n-F| \approx |f'(F)||x_{n-1}-F|$. Hence, $D_n$ grows linearly with $n$.
  • Super-attracting fixed point of order $k$ has $|x_n-F| \approx |c_{k+1}| |x_{n-1}-F|^{k+1}$ where $c_{k+1}$ is the first nonzero term of the Taylor expansion of $f(x)-F$. Hence, $D_{n}$ grows exponentially, $D_n\approx (k+1)^n$.
  • Mega-attracting: according to the above, $D_n$ grows faster than every exponential, i.e., has superexponential growth.

There is no upper bound on the growth of $D_n$ in the mega-attracting case. Indeed, given a rapidly convergent monotone sequence $x_n\to 0$, we can construct a $C^\infty$ function such that $f(x_n)=x_{n+1}$ for all $n$. (Use disjoint smooth bumps of height $x_{n+1}$ supported near each $x_n$. For the derivative of order $k$ to have limit zero at the origin, we need $(x_n-x_{n+1})^k x_{n+1} \to 0$ as $n\to\infty$. This is satisfied for every $k$ provided $x_n$ goes to zero rapidly enough.)