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I want to show that the sequence

$$ x_n=\left(\frac{3+\sqrt{5}}{2}\right)^n\cdot\begin{pmatrix}\frac{1+\sqrt{5}}{2}\\ 1\end{pmatrix}\,\mathrm{mod}\,1\in\mathbb{R}^2 $$ is dense in the unit square $\lbrack 0,1\rbrack^2$ (The $\mathrm{mod}\,1$ operator is taken componentwise). Unfortunately I'm not able to show this. Any kind of help is appreciated.

lbf_1994
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  • I don't think this is true. – Angina Seng Dec 21 '17 at 12:39
  • I think it is true. I generated the first several thousand points of the orbit and plotted them. They seem to cover the unit square. – lbf_1994 Dec 21 '17 at 12:45
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    Given that $(\frac{3+\sqrt5}{2})^n = (\frac{1+\sqrt5}{2})^{2n}$, $\lim_{n\to\infty} (\frac{1-\sqrt5}{2})^n = 0$, and $(\frac{1+\sqrt5}{2})^n + (\frac{1-\sqrt5}{2})^n \in \mathbb Z$, your sequence actually converges to $(0,0)$. Can you show us your plot and, even better, the code? – Hw Chu Dec 21 '17 at 12:50
  • I did the following: It holds that $Av=\lambda v$ where $A=\begin{pmatrix}2&1\\ 1&1\end{pmatrix}$, $v=\begin{pmatrix}\frac{1+\sqrt{5}}{2}\\ 1\end{pmatrix}$ and $\lambda = \frac{3+\sqrt{5}}{2}$, i.e. we have $x_{n+1}=A^{n+1}v\,\mathrm{mod}\,1=AA^nv\,\mathrm{mod}\,1=(A(A^nv\,\mathrm{mod}\,1))\,\mathrm{mod}\,1 = Ax_n\,\mathrm{mod}$. The second to last equality holds since $A$ is integer-valued. This gives us a numerically more stable scheme of generating the orbit as we avoid the exponential growth by going back to the unit square after each iteration. – lbf_1994 Dec 21 '17 at 13:13
  • This is now the code for generating the orbit: lambda = (3+sqrt(5))/2; v = [(1+sqrt(5))/2;1]; v=v/norm(v); n=100000; orbit=zeros(2,n); orbit(:,1)=v-floor(v); for i=2:n orbit(:,i) = lambda * orbit(:,i-1) -floor(lambda * orbit(:,i-1)); end plot(orbit(1,:), orbit(2,:), 'k.') – lbf_1994 Dec 21 '17 at 13:17
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    With $\varphi = \frac{1}{2}(1+\sqrt{5})$ and $\psi = \frac{1}{2}(1 - \sqrt{5}) = 1 - \varphi = -1/\varphi$, we have $$x_n = \begin{pmatrix} L_{2n+1} - \psi^{2n+1} \\ L_{2n} - \psi^{2n} \end{pmatrix} \bmod{1} = \begin{pmatrix} -\psi^{2n+1} \\ 1 - \psi^{2n} \end{pmatrix}$$ where $L_k$ is the $k^{\text{th}}$ Lucas number. Since $\lvert\psi\rvert < 1$, $x_n \to (0,1)$. – Daniel Fischer Dec 21 '17 at 13:20
  • @DanielFischer Yes, this is what I also found out numerically with PARI/GP after eliminating a parantheses-error. – Peter Dec 21 '17 at 13:24
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    @lbf_1994 I cannot quite understand from your code why you are dividing v by norm(v)... – Hw Chu Dec 21 '17 at 13:27
  • This is just such that the second component doesn't get mapped to 0 forever. – lbf_1994 Dec 21 '17 at 13:28
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    @lbf_1994 In your code, you have `v = v/norm(v);`, which is not in your question, and that changes things substantially. – Daniel Fischer Dec 21 '17 at 13:28
  • Oh yes you are right. I assumed it would hold that $x_n=A^nv\,\mathrm{mod}\,1$ is dense iff $x_n=\alpha A^nv\,\mathrm{mod}\,1$ is dense for any $\alpha\in\mathbb{R}$ as $v$ and $\alpha v$ lie in the same eigenspace. But apparently this is not the case. For my purposes it suffices to show that there exists an $\alpha\in\mathbb{R}$ such that the orbit $x_n=\alpha A^nv\,\mathrm{mod}\,1$ is dense. Sorry for the missleading question. I wasn't aware that the constant would change the behaviour. – lbf_1994 Dec 21 '17 at 13:38

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