improvements

Wrzodak_Koszarek_Zadania.ipynb

@@ -1,7 +1,6 @@
 {
  "cells": [
   {
-   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
@@ -9,7 +8,6 @@
    ]
   },
   {
-   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": []
@@ -32,31 +30,28 @@
     }
    ],
    "source": [
-    "import numpy as np\n",
-    "from sympy import symbols, Matrix\n",
-    "from numpy.linalg import eig\n",
+    "A=matrix(QQ,5,3,[2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 31])\n",
+    "print('A =')\n",
+    "print(A, '\\n')\n",
     "\n",
-    "A=np.matrix(QQ,5,3,[2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 31])\n",
-    "print(A.transpose()*A)\n",
-    "print((A.transpose()*A)^(-1))\n",
-    "mm=(A.transpose()*A)^(-1)\n",
-    "mm=(A.transpose()*A)^(-1)*A.transpose()\n",
-    "print(mm)\n",
+    "print('b =')\n",
+    "b=vector(QQ,[-1,0,1,0,1])\n",
+    "print(b, '\\n')\n",
     "\n",
-    "b1=np.vector([-1,0,1,0,1])\n",
-    "mm1=mm*b1\n",
-    "print(mm1)\n",
-    "print((b1-A*mm1))\n",
-    "b2=np.vector([1,1,1,1,1])\n",
-    "mm2=mm*b2\n",
-    "print(mm2)\n",
-    "print((b2-A*mm2))\n",
+    "print('A^T * A =')\n",
+    "print(A.transpose()*A, '\\n')\n",
+    "print('Macierz A^T*A jest kwadratowa, więc rozwiązanie istnieje\\n')\n",
     "\n",
-    "b2 in (m.transpose()).image()"
+    "u=(A.transpose()*A)^(-1)*A.transpose()*b\n",
+    "print('u = (A^T * A)^-1 * A^T * b =')\n",
+    "print(u, '\\n')\n",
+    "\n",
+    "\n",
+    "print('b - A * u = ')\n",
+    "print(b - A * u, '\\n')"
    ]
   },
   {
-   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
@@ -69,41 +64,22 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "zb1=[(1,1),(2,3),(4,5)]\n",
-    "zb2=[(1,1),(2,3),(3,4),(4,5),(5,7),(6,9)]\n",
-    "\n",
-    "m1=matrix(3,2,[1,exp(1.0),1,exp(2.0),1,exp(4.0)])\n",
-    "m2=matrix(6,2,[1,exp(1.0),1,exp(2.0),1,exp(3.0),1,exp(4.0),1,exp(5.0),1,exp(6.0)])\n",
+    "zbior=[(1,1),(2,3),(4,5)]\n",
+    "print('zbior punktów = ', zbior)\n",
+    "m=matrix(3,2,[1,exp(1.0),1,exp(2.0),1,exp(4.0)])\n",
     "\n",
     "a,b,t=var('a,b,t')\n",
     "\n",
-    "m1*vector([a,b])-vector([1,3,5])\n",
-    "m2*vector([a,b])-vector([1,3,4,5,7,9])\n",
+    "m*vector([a,b])-vector([1,3,5])\n",
     "\n",
+    "print('\\n (m^T * m)^-1 * m^T * vector =')\n",
+    "z = (m.transpose()*m)^(-1)*m.transpose()*vector([1,3,5])\n",
+    "print(z)\n",
     "\n",
-    "\n",
-    "M1=m1.transpose()*m1\n",
-    "M1.det()\n",
-    "\n",
-    "\n",
-    "\n",
-    "M2=m2.transpose()*m2\n",
-    "M2.det()\n",
-    "\n",
-    "\n",
-    "\n",
-    "M1^(-1)*m1.transpose()*vector([1,3,5])\n",
-    "\n",
-    "M2^(-1)*m2.transpose()*vector([1,3,4,5,7,9])\n",
-    "\n",
-    "\n",
-    "plot(1.64148598265947+ 0.0629860338045423*exp(t),(t,0,4))+sum([point(x) for x in zb1])\n",
-    "\n",
-    "plot(3.10041190358990+ 0.0163320609303546*exp(t),(t,0,6))+sum([point(x) for x in zb2])"
+    "plot(z[0] +z[1]*exp(t),(t,0,4))+sum([point(x) for x in zbior])"
    ]
   },
   {
-   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
@@ -132,7 +108,7 @@
  ],
  "metadata": {
   "kernelspec": {
-   "display_name": "base",
+   "display_name": "Python 3 (ipykernel)",
    "language": "python",
    "name": "python3"
   },
@@ -146,9 +122,8 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.10.9"
-  },
-  "orig_nbformat": 4
+   "version": "3.10.11"
+  }
  },
  "nbformat": 4,
  "nbformat_minor": 2