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# Untitled

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1. {
2.  "cells": [
3.   {
4.    "cell_type": "markdown",
6.    "source": [
7.     "We use the harmonic extensions\n",
8.     "\n",
9.     "$$\n", 10. "H_1 = \\sum_{i=1}^3 \\frac{1}{r} R_i^T D R_i\n", 11. "$$\n",
12.     "\n",
13.     "In the case of the Sierpinski gasket, it is well known that $r=\\frac{3}{5}$, \n",
14.     "$$\n", 15. "D=\\begin{pmatrix}\n", 16. "-2 & 1 & 1 \\\\\n", 17. "1 & -2 & 1 \\\\\n", 18. "1 & 1 & -2\n", 19. "\\end{pmatrix}\n", 20. "$$ \n",
21.     "and $R_i: \\ell(V_1) \\to \\ell(V_0)$ is given by\n",
22.     "\n",
23.     "$$\n", 24. "(R_i f) (v_j) = f(F_i(v_j)), \\quad \\text{ where }f\\in \\ell(V_1), v_j \\in V_0.\n", 25. "$$\n",
26.     "\n",
27.     "In our case, we let $v_1, v_2, v_3$ be the left, top, and right most points on $V_0$, and $v_4, v_5, v_6$ denote the points in $V_1$ between $v_1$ and $v_2$, $v_2$ and $v_3$, and $v_1$ and $v_3$ respectively.\n",
28.     "\n",
29.     "We compute $H_1$."
30.    ]
31.   },
32.   {
33.    "cell_type": "code",
34.    "execution_count": 1,
36.    "outputs": [],
37.    "source": [
38.     "import numpy as np\n",
39.     "import sympy as sym\n",
40.     "\n",
41.     "R_1 = np.array([[1, 0, 0, 0, 0, 0], \n",
42.     "                [0, 0, 0, 1, 0, 0], \n",
43.     "                [0, 0, 0, 0, 0, 1]])\n",
44.     "\n",
45.     "R_2 = np.array([[0, 0, 0, 1, 0, 0], \n",
46.     "                [0, 1, 0, 0, 0, 0], \n",
47.     "                [0, 0, 0, 0, 1, 0]])\n",
48.     "\n",
49.     "R_3 = np.array([[0, 0, 0, 0, 0, 1], \n",
50.     "                [0, 0, 0, 0, 1, 0], \n",
51.     "                [0, 0, 1, 0, 0, 0]])\n",
52.     "\n",
53.     "D = np.array([[-2, 1, 1], [1, -2, 1], [1, 1, -2]])\n",
54.     "\n",
55.     "r = sym.Rational(3, 5)"
56.    ]
57.   },
58.   {
59.    "cell_type": "code",
60.    "execution_count": 2,
62.    "outputs": [],
63.    "source": [
64.     "R = R_1 + R_2 + R_3\n",
65.     "\n",
66.     "H_1 = np.dot(np.dot(np.transpose(R), D), R) / r"
67.    ]
68.   },
69.   {
70.    "cell_type": "code",
71.    "execution_count": 3,
73.    "outputs": [
74.     {
75.      "name": "stdout",
76.      "output_type": "stream",
77.      "text": [
78.       "[[-10/3 5/3 5/3 -5/3 10/3 -5/3]\n",
79.       " [5/3 -10/3 5/3 -5/3 -5/3 10/3]\n",
80.       " [5/3 5/3 -10/3 10/3 -5/3 -5/3]\n",
81.       " [-5/3 -5/3 10/3 -10/3 5/3 5/3]\n",
82.       " [10/3 -5/3 -5/3 5/3 -10/3 5/3]\n",
83.       " [-5/3 10/3 -5/3 5/3 5/3 -10/3]]\n"
84.      ]
85.     }
86.    ],
87.    "source": [
88.     "print(H_1)"
89.    ]
90.   },
91.   {
92.    "cell_type": "markdown",
94.    "source": [
95.     "In our case, we need to compute $G = X_1^{-1}$, where $X_1$ is given by $H_1$ restricted to $V_1 - V_0$."
96.    ]
97.   },
98.   {
99.    "cell_type": "code",
100.    "execution_count": 4,
102.    "outputs": [],
103.    "source": [
104.     "X = H_1[3:, 3:]"
105.    ]
106.   },
107.   {
108.    "cell_type": "code",
109.    "execution_count": 5,
111.    "outputs": [
112.     {
113.      "name": "stdout",
114.      "output_type": "stream",
115.      "text": [
116.       "[[-10/3 5/3 5/3]\n",
117.       " [5/3 -10/3 5/3]\n",
118.       " [5/3 5/3 -10/3]]\n"
119.      ]
120.     }
121.    ],
122.    "source": [
123.     "print(X)"
124.    ]
125.   },
126.   {
127.    "cell_type": "markdown",
129.    "source": [
130.     "Which is noninvertible. In particular, $(1, 1, 1) \\in ker(X)$."
131.    ]
132.   },
133.   {
134.    "cell_type": "code",
135.    "execution_count": null,
137.    "outputs": [],
138.    "source": []
139.   }
140.  ],
142.   "kernelspec": {
143.    "display_name": "Python 3",
144.    "language": "python",
145.    "name": "python3"
146.   },
147.   "language_info": {
148.    "codemirror_mode": {
149.     "name": "ipython",
150.     "version": 3
151.    },
152.    "file_extension": ".py",
153.    "mimetype": "text/x-python",
154.    "name": "python",
155.    "nbconvert_exporter": "python",
156.    "pygments_lexer": "ipython3",
157.    "version": "3.7.2"
158.   }
159.  },
160.  "nbformat": 4,
161.  "nbformat_minor": 2
162. }
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