420 lines
11 KiB
Plaintext
420 lines
11 KiB
Plaintext
{
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"cells": [
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Zadanie 4.6"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<h3> Zadanie 6 </h3>\n",
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"\n",
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"Rozwiąż układ równań $Ax=b$ metodą przybliżoną, gdzie\n",
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"\n",
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"$$A=\\left(\\begin{array}{rrr}\n",
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"2 & 4 & 6 \\\\\n",
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"8 & 10 & 12 \\\\\n",
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"14 & 16 & 18 \\\\\n",
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"20 & 22 & 24 \\\\\n",
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"26 & 28 & 31\n",
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"\\end{array}\\right)$$\n",
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"\n",
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"oraz $b=(-1,0,1,0,1)$."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 126,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"A =\n",
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"[[ 2 4 6]\n",
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" [ 8 10 12]\n",
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" [14 16 18]\n",
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" [20 22 24]\n",
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" [26 28 31]] \n",
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"\n",
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"b =\n",
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"[-1 0 1 0 1] \n",
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"\n",
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"A^T * A =\n",
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"Matrix([[1340, 1480, 1646], [1480, 1640, 1828], [1646, 1828, 2041]]) \n",
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"\n",
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"Macierz A^T*A jest kwadratowa, więc rozwiązanie istnieje\n",
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"\n",
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"(A^T * A)^(-1)*A.transpose() =\n",
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"[[0.050000 -0.233333 -0.516667 -0.800000 1.000000]\n",
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" [-0.600000 0.216667 1.033333 1.850000 -2.000000]\n",
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" [0.500000 0.000000 -0.500000 -1.000000 1.000000]] \n",
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"\n",
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"Rozwiązanie:\n",
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"\n",
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"x = (A^T * A)^(-1) * A^T * b =\n",
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"[0.433333 -0.366667 -0.000000] \n",
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"\n",
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"\n"
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]
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}
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],
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"source": [
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"import numpy as np\n",
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"from sympy import symbols, Matrix, vector\n",
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"from numpy.linalg import eig\n",
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"\n",
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"m = np.array([[2, 4, 6],\n",
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" [8, 10, 12],\n",
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" [14, 16, 18],\n",
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" [20, 22, 24],\n",
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" [26, 28, 31]])\n",
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"\n",
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"A=Matrix(m)\n",
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"\n",
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"np.set_printoptions(formatter={'float_kind':'{:f}'.format})\n",
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"\n",
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"print('A =')\n",
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"print(m, '\\n')\n",
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"\n",
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"print('b =')\n",
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"b=np.array([-1,0,1,0,1])\n",
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"print(b, '\\n')\n",
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"\n",
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"print('A^T * A =')\n",
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"print(A.transpose()*A, '\\n')\n",
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"print('Macierz A^T*A jest kwadratowa, więc rozwiązanie istnieje\\n')\n",
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"\n",
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"#print('(A^T * A)^(-1) =')\n",
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"#print(np.linalg.inv(m.transpose() @ m), '\\n')\n",
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"\n",
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"print('(A^T * A)^(-1)*A.transpose() =')\n",
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"print(np.linalg.inv(m.transpose() @ m) @ m.transpose(), '\\n')\n",
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"\n",
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"print('Rozwiązanie:\\n')\n",
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"#u=(A.transpose()*A)^(-1)*A.transpose()*b\n",
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"x = np.linalg.inv(m.transpose() @ m) @ m.transpose() @ b\n",
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"print('x = (A^T * A)^(-1) * A^T * b =')\n",
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"print(x, '\\n\\n')\n",
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"\n",
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"# SPRAWDZENIE\n",
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"x, residuals, _, _ = np.linalg.lstsq(m, b, rcond=None)\n",
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"#print(\"Przybliżone rozwiązanie:\")\n",
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"#print(x)"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Zadanie 4.7"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<h3> Zadanie 7</h3>\n",
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"\n",
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"Przybliż funkcją $f(t)=a+be^{t}$ zbiór punktów $(1,1)$, $(2,3)$, $(4,5)$ metodą z zadania 6.\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 127,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Przybliżone parametry:\n",
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"a = 1.641485981693081\n",
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"b = 0.06298603382678646\n"
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]
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}
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],
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"source": [
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"import numpy as np\n",
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"from scipy.optimize import curve_fit\n",
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"\n",
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"# Definicja funkcji, której chcemy dokonać przybliżenia\n",
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"def func(t, a, b):\n",
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" return a + b * np.exp(t)\n",
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"\n",
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"# Dane punktów\n",
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"t_data = np.array([1, 2, 4])\n",
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"y_data = np.array([1, 3, 5])\n",
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"\n",
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"# Przybliżanie funkcji do danych punktów\n",
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"params, _ = curve_fit(func, t_data, y_data)\n",
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"\n",
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"# Rozwiązania parametrów a i b\n",
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"a = params[0]\n",
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"b = params[1]\n",
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"\n",
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"# Wyświetlenie wyników\n",
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"print(\"Przybliżone parametry:\")\n",
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"print(\"a =\", a)\n",
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"print(\"b =\", b)\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 128,
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"metadata": {},
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"outputs": [
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{
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"ename": "TypeError",
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"evalue": "'module' object is not callable",
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"output_type": "error",
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"traceback": [
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"\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
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"\u001b[1;31mTypeError\u001b[0m Traceback (most recent call last)",
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"Cell \u001b[1;32mIn[128], line 9\u001b[0m\n\u001b[0;32m 5\u001b[0m m1\u001b[39m=\u001b[39mnp\u001b[39m.\u001b[39marray([[\u001b[39m1\u001b[39m,np\u001b[39m.\u001b[39mexp(\u001b[39m1.0\u001b[39m)],[\u001b[39m1\u001b[39m,np\u001b[39m.\u001b[39mexp(\u001b[39m2.0\u001b[39m)],[\u001b[39m1\u001b[39m,np\u001b[39m.\u001b[39mexp(\u001b[39m4.0\u001b[39m)]])\n\u001b[0;32m 7\u001b[0m \u001b[39m#a,b,t=var('a,b,t')\u001b[39;00m\n\u001b[1;32m----> 9\u001b[0m m1\u001b[39m*\u001b[39mvector([a,b])\u001b[39m-\u001b[39mvector([\u001b[39m1\u001b[39m,\u001b[39m3\u001b[39m,\u001b[39m5\u001b[39m])\n\u001b[0;32m 11\u001b[0m M1\u001b[39m=\u001b[39mm1\u001b[39m.\u001b[39mtranspose()\u001b[39m*\u001b[39mm1\n\u001b[0;32m 12\u001b[0m M1\u001b[39m.\u001b[39mdet()\n",
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"\u001b[1;31mTypeError\u001b[0m: 'module' object is not callable"
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]
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}
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],
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"source": [
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"import numpy as np\n",
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"\n",
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"zb1=np.array([[1,1],[2,3],[4,5]])\n",
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"\n",
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"m1=np.array([[1,np.exp(1.0)],[1,np.exp(2.0)],[1,np.exp(4.0)]])\n",
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"\n",
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"#a,b,t=var('a,b,t')\n",
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"\n",
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"m1*vector([a,b])-vector([1,3,5])\n",
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"\n",
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"M1=m1.transpose()*m1\n",
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"M1.det()\n",
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"\n",
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"M1^(-1)*m1.transpose()*vector([1,3,5])\n",
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"\n",
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"##########################################################\n",
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"\n",
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"def func(t, a, b):\n",
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" return a + b * np.exp(t)\n",
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"\n",
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"x = np.linalg.inv(m.transpose() @ m) @ m.transpose() @ b\n",
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"\n",
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"zbior=[(1,1),(2,3),(4,5)]\n",
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"print('zbior punktów = ', zbior)\n",
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"m=matrix(3,2,[1,exp(1.0),1,exp(2.0),1,exp(4.0)])\n",
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"\n",
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"a,b,t=var('a,b,t')\n",
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"\n",
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"m*vector([a,b])-vector([1,3,5])\n",
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"\n",
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"print('\\n (m^T * m)^-1 * m^T * vector =')\n",
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"z = (m.transpose()*m)^(-1)*m.transpose()*vector([1,3,5])\n",
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"print(z)\n",
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"\n",
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"plot(z[0] +z[1]*exp(t),(t,0,4))+sum([point(x) for x in zbior])"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Zadanie 4.9"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<h3> Zadanie 9 </h3>\n",
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"\n",
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"Znajdź bazę ortonormalnych wektorów własnych dla macierzy\n",
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"\n",
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"$$\\left(\\begin{array}{rrr}\n",
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"1 & 1 & 0 \\\\\n",
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"1 & 2 & 2 \\\\\n",
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"0 & 2 & 3\n",
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"\\end{array}\\right)$$"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Macierz A=\n",
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"[[1 1 0]\n",
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" [1 2 2]\n",
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" [0 2 3]]\n",
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"Baza ortonormalnych wektorów własnych:\n",
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"[[-0.593233 -0.786436 0.172027]\n",
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" [0.679313 -0.374362 0.631179]\n",
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" [-0.431981 0.491296 0.756320]]\n"
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]
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}
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],
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"source": [
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"import numpy as np\n",
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"from sympy import symbols, Matrix\n",
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"from numpy.linalg import eig\n",
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"\n",
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"#Definicja macierzy A\n",
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"A = np.array([[1,1,0],[1,2,2],[0,2,3]])\n",
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"#A = Matrix([[1,1,0],[1,2,2],[0,2,3]])\n",
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"print(\"Macierz A=\")\n",
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"print(A)\n",
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"\n",
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"def qr_eigval(matrix, epsilon=1e-10, max_iterations=1000):\n",
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"\n",
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" eigenv = np.diag(matrix)\n",
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"\n",
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" for _ in range(max_iterations):\n",
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"\n",
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" q, r = np.linalg.qr(matrix)\n",
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" matrix = np.dot(r, q)\n",
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" new_eigenv = np.diag(matrix)\n",
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"\n",
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" if np.allclose(eigenv, new_eigenv, atol=epsilon):\n",
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" break\n",
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" eigenv = new_eigenv\n",
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"\n",
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" return eigenv\n",
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"\n",
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"#Obliczenie wektorów własnych i wartości własnych\n",
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"#eigenvalues = qr_eigval(matrix)\n",
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"eigenvalues, eigenvectors = np.linalg.eig(A)\n",
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"\n",
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"#Dekompozycja\n",
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"Q, R = np.linalg.qr(eigenvectors)\n",
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"\n",
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"#Normalizacja wektora własnego\n",
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"orthonormal_basis = Q\n",
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"\n",
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"print(\"Baza ortonormalnych wektorów własnych:\")\n",
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"print(orthonormal_basis)"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Zadanie 3.9"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Zadanie 9\n",
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"\n",
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"Oblicz metodą ''power iteration'' wartości własne macierzy\n",
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"\n",
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"$$\\left(\\begin{array}{rrr}\n",
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"1 & 2 & 3 \\\\\n",
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"4 & 5 & 6 \\\\\n",
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"7 & 8 & 9\n",
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"\\end{array}\\right)$$"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 146,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Dominujący wektor własny:\n",
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"[[0.231970]\n",
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" [0.525322]\n",
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" [0.818674]]\n",
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"Dominująca wartość własna (power iteration):\n",
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"16.116843969807043\n"
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]
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}
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],
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"source": [
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"import numpy as np\n",
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"\n",
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"A = np.array([[1,2,3],\n",
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" [4,5,6],\n",
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" [7,8,9]])\n",
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"\n",
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"S = np.array([[1], [1], [1]])\n",
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"\n",
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"def power_iteration(m, n, s):\n",
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" #Punkt startowy\n",
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" st = s\n",
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" eigv = 0\n",
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" for i in range(n):\n",
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" st = np.dot(A, st)\n",
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" eigv = np.max(st)\n",
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" st = st/eigv\n",
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" return eigv\n",
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"\n",
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"def power_iteration_vec(m, n, s):\n",
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" tolerance = 1e-6\n",
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" #Punkt startowy\n",
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" x = s\n",
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" for i in range(n): \n",
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" y = np.dot(m, x)\n",
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" #Normalizacja wektora\n",
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" x_new = y / np.linalg.norm(y)\n",
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" #Sprawdzenie warunku zbieżności\n",
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" if np.linalg.norm(x - x_new) < tolerance:\n",
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" break\n",
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" #Aktualizacja\n",
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" x = x_new\n",
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" return x\n",
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"\n",
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"y1 = power_iteration_vec(A, 1000, S)\n",
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"print(\"Dominujący wektor własny:\")\n",
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"print(y1)\n",
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"\n",
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"y2 = power_iteration(A, 1000, S)\n",
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"print(\"Dominująca wartość własna (power iteration):\")\n",
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"print(y2)\n",
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"\n"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.10.9"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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