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  1. {
  2. "cells": [
  3. {
  4. "cell_type": "markdown",
  5. "metadata": {},
  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,
  35. "metadata": {},
  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,
  61. "metadata": {},
  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,
  72. "metadata": {},
  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",
  93. "metadata": {},
  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,
  101. "metadata": {},
  102. "outputs": [],
  103. "source": [
  104. "X = H_1[3:, 3:]"
  105. ]
  106. },
  107. {
  108. "cell_type": "code",
  109. "execution_count": 5,
  110. "metadata": {},
  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",
  128. "metadata": {},
  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,
  136. "metadata": {},
  137. "outputs": [],
  138. "source": []
  139. }
  140. ],
  141. "metadata": {
  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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