# Tag Info

23

This is exactly the dynamics studied by V. I. Arnold, which exhibits what is known as Arnold's tongues. See this link.

21

Nice question! Here's what I can show. Let $X$ be a smooth closed manifold. Then: (1) If $\chi(X) = 0$, then $X$ is not $n$-POP for any $n$. (2) If $\chi(X) \neq 0$ and $X$ is orientable, then $X$ is $\text{lcm}(1, 2, ... n)$-POP with respect to maps $f : X \to X$ of nonzero degree, where $n = \text{max}(b_0 + b_2 + ..., b_1 + b_3 + ...)$ (where $b_i$ is ...

10

I want to leave a few elementary comments, maybe they will be helpful. The question asks about recurrence relation $$u_{n+1}= \lfloor u_n \rfloor (u_n − \lfloor u_n \rfloor + 1)$$ Suppose you write rational $u$ in terms of natural numbers $\frac{pq+r}{p}$. Then $\lfloor u \rfloor = q$. The recurrence relation is now $$u_{n+1} = q_n (\frac{pq_n+r_n}{p} − ... 10 Given an entire function f\colon\mathbb{C}\to\mathbb{C}, the escaping set, I(f), is the set of z\in\mathbb{C} such that f^n(z)\to\infty. Per the Wikipedia article, the escaping set of a non-linear entire function is nonempty. The reference for this is On the iteration of entire functions by Eremenko. 6 To be precise, orb\big(\frac{1}{2}\big)=\mathbb{Z}\big[\frac{1}{2}\big] \cap [0,1). Since you say \mathbb{Z} rather than \mathbb{N} you mean the orbit of y to include the solutions t of f^k(t)=y for k \in \mathbb{Z}. You say that the orbit of \frac13 is contained in$$ \Big\{\frac{m}{3\times 2^n}\mid n \in \mathbb{Z},\ 0 \leq m\lt 3\...

6

Polynomial maps $f(z)$ for which there is a general formula for the $n$-th iterate are called integrable. Besides polynomials of degree $1$, there are two types of them: a) those which are conjugate (by an affine map) with $z^d$ and b) those which are conjugate with $T_d$, where $T_d$ is the Chebyshev polynomial, defined by $$\cos dx=T_d(\cos x).$$ For ...

6

As noted in the question's comments by Aleksei Kulikov, a necessary and sufficient condition is given by the following: Theorem 1 A real continuous function f is a runaway function iff $f(x)=x$ has no solution for $x\in \mathbb{R}$. To prove this we need the following lemma: Lemma Let $f$ be continuous on $\mathbb{R}$ and $f(x)>x$ for all $x \in \mathbb{R}... 5 Explicit solutions for arbitrary$r$exist in various forms: Logistic map: an analytical solution (1995) represents the solution as a power of a transfer matrix. An explicit solution for the logistic map (1999) gives a functional integral solution. A note on exact solutions of the logistic map (2020, paywall) gives the solution in terms of a power series. 5 No, even if we assume$\nu$to be invariant under$\phi$. Let$X = \{0,1\}^\mathbb{Z}$be the set of two-way infinite binary sequences with the prodiscrete topology, and let$\phi$be the left shift on$X$. Let$\nu = (\mu_1 + \mu_2)/2$where$\mu_1$is the uniform Bernoulli measure on$X$and$\mu_2$is an atomless$\phi$-invariant probability measure on ... 4 The following is proved by F. Brock Fuller in "The Existence of Periodic Points," Annals of Mathematics, Vol. 57, 1953, pp. 229-230: Theorem. Let X be a compact simplicial complex with Betti numbers$B_i$, and nonzero Euler characteristic. If a continuous map$f: X\rightarrow X$induces isomorphisms of homology groups, then$f$has a point whose period ... 4 The right part of$f$being not relevant, we consider the diffeo$f:[0,1/2]\to[0,1]$,$f(x)=x+2^\alpha x^{1+\alpha}$, whose inverse map$g:[0,1]\to[1,1/2]$is strictly increasing with unique fixed point$0$, where it has the expansion $$g(x)=x\big(1-2^\alpha x^{\alpha}+o(x^{\alpha})\big).$$ (Incidentally, note that there is a power series expansion valid ... 3 Let$p_i$be at distance$r_i > 2$from the origin, WLOG assume$p_i = (r_i, 0)$. Let$q_i = (\cos \alpha, \sin \alpha)$, and$v_i = (\cos \beta, \sin \beta)$be a direction vector of$M_i$. By choice$\alpha$and$\beta$are independent and equidistributed on$[0, 2 \pi)$. Simple calculations show that$||p_{i + 1}||^2 =: r_{i + 1}^2 = r_i^2 + 4r_i \sin \...

3

See Mathworld, where it is stated, "While some quadratic maps are solvable in closed form, most are not." Examples with a closed-form solution are $p_n=p_{n-1}^2$, $p_n=p_{n-1}^2+1$ with $p_0=1$, $p_n=p_{n-1}^2-p_{n-1}+1$ with $p_1=2$, and of course anything that you can reduce to these forms by completing the square.

3

Let $\Sigma:=(\mathbf{x},\mathbf{y},B)$ be an initial seed where $\mathbf{x}:=\{x_1,\ldots,x_n\},\mathbf{y}:=\{y_1,\ldots,y_n\}$ and let $\mathbf{c} := \{x_{n+1},\ldots,x_m\}$ denote the set of frozen variables. By the Laurent phenomenon, we know that every cluster variable in the seed pattern generated by $\Sigma$ is a Laurent polynomial in the cluster ...

3

To have here the invariant distribution with cdf $F$ given by $F(x)=x^2$ for $x\in[0,1]$, all that is needed is a change of variables. More generally, let $F$ be the cdf of any non-atomic distribution supported on an interval $I$ in $\mathbb R$. Let $F^{-1}$ denote the inverse of the restriction of $F$ to $I$. Let $U$ be a random variable uniformly ...

2

As said by Sangchui Lee, we are interested in $r_i = |p_i|$ and we assume that the reflections induce enougth random rotation such that the "bands" of radius $r$ just reveal the number of time $r_i$ was closed to $r$. And we have $$| p_{i+1}| =|p_i -2\langle q_i, n_i \rangle n_i|$$ We define then a random march $X_i$ defined by $X_0=0$ and X_{i+1}=X_i-2\... 2 (Too long for a comment) Let n_i denote the unit normal vector to the mirror line M_i. Then we can provide a simple recursive formula p_{i+1} = p_i - 2\langle p_i-q_i, n_i \rangle n_i. Since p_i - \langle p_i, n_i \rangle n_i is orthogonal to n_i, it follows that \begin{align*} \| p_{i+1} \| &= \| p_i - 2\langle q_i, n_i \rangle n_i \|. \end{... 2 No. Let \phi be the left shift on the set X = \{0,1\}^\mathbb{Z} of bi-infinite binary sequences with the prodiscrete topology, and let V = \{ x \in X : x_0 = 0 \} be the set of sequences that have 0 at the central coordinate. Then X is Hausdorff (even metrizable and compact), \phi is transitive (even mixing) and V is nonempty and open (even ... 2 In the linear case (for a bounded operator T on a Banach space), the analogous construction for equivalent norms, possibly with a different ratio in place of 1/2, is often named adapted norm (see e.g. Global stability of dynamical systems, by Michael Shub). Its main feature is that gives T an operator norm arbitrarily close to its spectral radius. For ... 2 The questions you are asking are fundamental to the theory of one-dimensional dynamical systems. I would suggest starting with an introductory textbook, such as An Introduction to Chaotic Dynamical Systems by Devaney. Books with more in-depth results include Iterated Maps of the Interval as Dynamical Systems by Collet and Eckmann, and One-Dimensional ... 2 This is an extremely well studied problem. The canonical reference is Bougerol and LaCroix's book: Bougerol, Philippe; Lacroix, Jean, Products of random matrices with applications to Schrödinger operators, Progress in Probability and Statistics, Vol. 8. Boston - Basel - Stuttgart: Birkhäuser. X, 283 p. DM 88.00 (1985). ZBL0572.60001. But the paper by ... 2 In base 2 the solutions are A306259: Composite numbers k such that 2^{(k(k-1))} \equiv 1 \pmod{ k^2} From a comment: It contains all Fermat pseudoprimes to base 2, A001567. The sequence starts: 21, 105, 165, 205, 231, 273, 301, 341, 385, 465 There are many solutions in other bases, the following pari/gp program computes them: c=0;a=2;for(p=1,10^4,if(... 2 Direct calculation shows ((T\Delta)^2 v)_{0,0} = 2\sqrt{2}. This implies that((T\Delta)^{2n}v)_{0,0} \ge (2\sqrt{2})^n for $n = 1,2,\ldots$. You might be interested in studying the iterates of $P_{\lambda} = \frac{1}{4}\lambda T\Delta = \lambda T\Delta'$ where $\Delta' = \frac{1}{4}\Delta$ and $\lambda > 0$ (I prefer $\Delta'$ to $\Delta$ since it ...

2

Let $A = V^{-1}([a, \infty))$. Both $A$ and $f^{-1}(A)$ are compact. Because $V$ is a strict Lyapunov function and $a \not\in V(Fix(f))$, it follows $\min_{f^{-1}(A)} V > a$. So you can set $\delta = \min_{f^{-1}(A)} V - a$, and you have $V(x) < a + \delta$ implies $V(f(x)) < a$.

2

There is a recursive and even polynomial upper bound. In the following I will denote the program string by $p$ and its length by $n$. First notice that the memory will always be of the form $0^i$, $0^i 1^j$ or $0^i 1^j 0^k$ (or symmetrically $1^i$, $1^i 0^j$ or $1^i 0^j 1^k$) for some $i,j,k>0$. This is a simple induction from the base case $01$, because ...

2

No. Any Markov operator is contracting in the total variation norm, whereas your function $F$ is subject to a much weaker condition of weak continuity. It is easy to construct a counterexample. For instance, take $X$ to be the two point set $\{0,1\}$, then the probability measures on $X$ are parameterized by a single parameter $t=\mu(1)\in [0,1]$. Take for $... 1$u_{0} \in \mathbb{Q} \quad u_{n+1}=\left[u_{n}\right]\left(u_{n}-\left[u_{n}\right]+1\right)$, then$\{u_{n}\}_{n=1}^{+\infty}$reach in integer.$\quad (*)$This can be proved if we prove,$p$prime,$t\in \mathbb{N}^{*}p^{t} u_{0} \in \mathbb{N}^{*}$,$\left\{u_{k}\right\}_{n=1}^{+\infty}$reach an integer. we call this as property$I(p,t)$. It is ... 1 Ah, I found the solution, it's just the mean value theorem. By the mean value theorem we know that there exist points$\eta_1, \eta_2such that \begin{align} Df(\eta_1) = \frac{|f(A)|}{|A|}, \quad Df(\eta_2) = \frac{|f(B)|}{|B|} \end{align} Thus we have \begin{align} \frac{|f(B)|}{|B|} \Big/ \frac{|f(A)|}{|A|} = \frac{|Df(\eta_2)|}{|Df(\eta_1)|} \leq e^M, \... 1 Writingy_n=2x_n$we get,$\frac{y_{n-1}}{y_n}-1=(y_n)^{\alpha}$and$\frac{y_{n}}{y_{n+1}}-1=(y_{n+1})^{\alpha}$. Dividing the first by second we get,$\frac{r_n-1}{r_{n+1}-1}=(r_{n+1})^{\alpha}$, where$r_n=\frac{y_{n-1}}{y_n}$. or,$\frac{t_{n+1}}{t_{n}}=(1+t_{n+1})^{-\alpha}....(1)$[Where,$r_n-1=t_n >0$]. Now,$f(n):=t_n$,$f'(n)=t_{n+1}-t_n\$. From ...

Only top voted, non community-wiki answers of a minimum length are eligible