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# 数学代写|丢番图逼近代写Diophantine approximation代考|MAT6932 Ammann-Beenker tilings

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## 数学代写|丢番图逼近代写DIOPHANTINE APPROXIMATION代考|Ammann-Beenker tilings

Collections of vertices of Ammann-Beenker tilings can be obtained as canonical cut and project sets, using the two dimensional subspace $E$ of $\mathbb{R}^4$ defined by
$$E=\left{\left(x, L_1(x), L_2(x)\right): x \in \mathbb{R}^2\right},$$
with
$$L_1(x)=\frac{\sqrt{2}}{2}\left(x_1+x_2\right) \quad \text { and } \quad L_2(x)=\frac{\sqrt{2}}{2}\left(x_1-x_2\right) .$$
Although we cannot directly appeal to either Theorem $4.2 .2$ or Corollary 4.3.3, we will explain how the machinery we have developed can be used to easily show that these sets are LR.

The canonical window $\mathcal{W}$ in $F_\rho$ is the regular octagon with vertices at
$$\left(\frac{1+\sqrt{2}}{2} \pm \frac{1+\sqrt{2}}{2}, \frac{1}{2} \pm \frac{1}{2}\right) \text { and }\left(\frac{1+\sqrt{2}}{2} \pm \frac{1}{2}, \frac{1}{2} \pm \frac{1+\sqrt{2}}{2}\right)$$
By Lemma 4.1.1, every patch of size $r$ corresponds to a finite collection of connected components of $\operatorname{nsing}(r)$. Therefore to demonstrate that a canonical cut and project set formed using $E$ is LR, it is enough to show that the there is a constant $C>0$ with the property that, for all sufficiently large $r$, the orbit of any nonsingular point $w \in F_\rho$, under the action of the collection of integers
$$\rho^{-1}(C r \Omega) \cap \mathbb{Z}^k,$$
intersects every connected component of $\operatorname{nsing}(r)$.

## 数学代写|丢番图逼近代写DIOPHANTINE APPROXIMATION代考|Penrose tilings

This example is similar to the previous one, but it also gives an indication of how to apply our techniques in cases when the physical space does not act minimally on $\mathbb{T}^k$. Let $\zeta=\exp (2 \pi i / 5)$ and let $Y$ be a canonical cut and project set defined using the two dimensional subspace $E$ of $\mathbb{R}^5$ generated by the vectors
$$\left(1, \operatorname{Re}(\zeta), \operatorname{Re}\left(\zeta^2\right), \operatorname{Re}\left(\zeta^3\right), \operatorname{Re}\left(\zeta^4\right)\right)$$
and
$$\left(0, \operatorname{Im}(\zeta), \operatorname{Im}\left(\zeta^2\right), \operatorname{Im}\left(\zeta^3\right), \operatorname{Im}\left(\zeta^4\right)\right) \text {. }$$
Well known results of de Bruijn [10] and Robinson [25] show that the set $Y$ is the image under a linear transformation of the collection of vertices of a Penrose tiling, and in fact that all Penrose tilings can be obtained in a similar way from cut and project sets. The fact that $Y$ is LR can be deduced directly from the definition of the Penrose tiling as a primitive substitution. However, as in the previous example, we will show how to prove this starting from the definition of $Y$ as a cut and project set.

## 数学代写|丟番图逼近代写DIOPHANTINE APPROXIMATION代考|Ammann-Beenker tilings

、left 缺少或无法识别的分隔符

$$L_1(x)=\frac{\sqrt{2}}{2}\left(x_1+x_2\right) \quad \text { and } \quad L_2(x)=\frac{\sqrt{2}}{2}\left(x_1-x_2\right) .$$

$$\left(\frac{1+\sqrt{2}}{2} \pm \frac{1+\sqrt{2}}{2}, \frac{1}{2} \pm \frac{1}{2}\right) \text { and }\left(\frac{1+\sqrt{2}}{2} \pm \frac{1}{2}, \frac{1}{2} \pm \frac{1+\sqrt{2}}{2}\right)$$

## 数学代写|丢番图逼近代写DIOPHANTINE APPROXIMATION代考|Penrose tilings

$$\rho^{-1}(C r \Omega) \cap \mathbb{Z}^k,$$

$$\left(1, \operatorname{Re}(\zeta), \operatorname{Re}\left(\zeta^2\right), \operatorname{Re}\left(\zeta^3\right), \operatorname{Re}\left(\zeta^4\right)\right)$$

$$\left(0, \operatorname{Im}(\zeta), \operatorname{Im}\left(\zeta^2\right), \operatorname{Im}\left(\zeta^3\right), \operatorname{Im}\left(\zeta^4\right)\right)$$
de Bruijn [10] 和 Robinson [25] 的众所周知的结果表明该集合 $Y$ 是彭罗斯平铺顶点集合线性亲换下的图像，事实上，所有彭罗 斯平铺都阿以用类似的方式从切割和投影集中获得。事实上 $Y$ 是 LR 可以直接从 Penrose tiling 的定义中中推导出来作为原始晴换。 然而，就像前面的例子一样，我们将展示如何从定义开始证明这一点 $Y$ 作为剪辑和项目集。

## MATLAB代写

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