几何迭代 Markov链

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旧官网的cover image(封面图)
如何构造下面的 cover image 4 呢? 这些四边形都相似, 相邻两个四边形有一条公共边.

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p = (A_1 - B_0) / (A_0 - B_0) q = (B_1 - A_1) / (B_0 - A_1) Sequence(((p - 1) (p q)^n (A_0 - B_0) - B_0 + p ((q - 1) A_0 + B_0)) / (p q - 1), n, 2, 40) Sequence((-B_0 + p ((q - 1) (A_0 + (p q)^n (-A_0 + B_0)) + B_0)) / (p q - 1), n, 2, 40)

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\begin{cases}A_{n+1}=p(A_n-B_n)+B_n\\
B_{n+1}=q (B_n-A_{n+1})+A_{n+1}=B_n+(1-q)p(A_n-B_n)\end{cases}即\[\pmatrix{A_{n+1}\\B_{n+1}}=\pmatrix{p&1-p\\p-pq&1-p+pq}\pmatrix{A_n\\B_n}\]相似对角化,\[\left(
\begin{array}{cc}
1 & p-1 \\
1 & p-p q \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & 0 \\
0 & p q \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & p-1 \\
1 & p-p q \\
\end{array}
\right)^{-1}\]
$n$次方,\[\left(
\begin{array}{cc}
1 & p-1 \\
1 & p-p q \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & 0 \\
0 & (p q)^n \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & p-1 \\
1 & p-p q \\
\end{array}
\right)^{-1}\]
当$|pq|<1$时收敛到\[\left(
\begin{array}{cc}
1 & p-1 \\
1 & p-p q \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & p-1 \\
1 & p-p q \\
\end{array}
\right)^{-1}=\left(
\begin{array}{cc}
\frac{p-p q}{1-p q} & \frac{1-p}{1-p q} \\
\frac{p-p q}{1-p q} & \frac{1-p}{1-p q} \\
\end{array}
\right)\]因此$A_n$、$B_n$收敛到同一个点 $\frac{p-p q}{1-p q}A_0+\frac{1-p}{1-p q}B_0$
将$p = \frac{A_1 - B_0}{A_0 - B_0},q = \frac{B_1 - A_1}{B_0 - A_1}$代入化简得$$\frac{A_0 B_1-A_1 B_0}{A_0-A_1-B_0+B_1}$$

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以$(A_0,B_0)(A_1,B_1)$、$(A_0,A_1)(B_0,B_1)$为初始点会收敛到同一个点, 这个点是“顺相似中心”(两对复数的等比分点)
见Wikipedia - Spiral similarity

Solution with complex numbers
If we express \(A, B, C,\) and \(D\) as points on the complex plane with corresponding complex numbers \(a, b, c,\) and \(d\), we can solve for the expression of the spiral similarity which takes \(A\) to \(C\) and \(B\) to \(D\). Note that \(T(a) = x_0+\alpha(a-x_0)\) and \(T(b) = x_0+\alpha(b-x_0)\), so \(\frac{T(b)-T(a)}{b-a} = \alpha\). Since \(T(a) = c\) and \(T(b) = d\), we plug in to obtain \(\alpha = \frac{d-c}{b-a}\), from which we obtain\[x_0 = \frac{ad-bc}{a+d-b-c}\]
Pairs of spiral similarities
...Thus $X$ is also the center of the spiral similarity which takes $\overline {AC}$ to $\overline {BD}$.

Fundamental Theorem of Directly Similar Figures
顺相似变换基本定理

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四个实数$a,b,c,d$的等比分点:
$A,B,C,D$的横坐标为$a,b,c,d$, 且$AB∥CD$, 则$AC,BD$的交点的横坐标为
\[\frac{ad-bc}{a+d-b-c}\]