ME2016-HW1

ME_2016/.ipynb_checkpoints/HW1-checkpoint.ipynb

@@ -0,0 +1,229 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "d9b87334-514c-41fa-aab5-ecef29c4c94d",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "import matplotlib.pyplot as plt\n",
+    "from sympy import *"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "6b45c0a5-2069-4b11-b208-e2fa3dcc1820",
+   "metadata": {},
+   "source": [
+    "**Dane Sabo**\n",
+    "\n",
+    "*September 18th, 2024*"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "0f1d7205-998f-4cbc-b3b3-df3afda5804c",
+   "metadata": {},
+   "source": [
+    "# Instructions\n",
+    "\n",
+    "Please do a written solution for problems 1 and 2. We will review them on Monday, Sept 16 in class prior to the assignment being due.\n",
+    "\n",
+    "Please upload a Jupyter Notebook for problems 3 and 4.\n",
+    "\n",
+    "Problems 1 and 2 are worth 10 points each, problems 3 and 4 are worth 15 points each.\n",
+    "\n",
+    "# Written Problems\n",
+    "## Problem 1\n",
+    "Please find the general solution of \n",
+    "$$\n",
+    "\\bf{\\dot{X}} = \n",
+    "\\begin{bmatrix}   \n",
+    "-1 & 5 & 2\\\\   \n",
+    "4 & -1 & -2\\\\\n",
+    "0 & 0 & 6\n",
+    "\\end{bmatrix}\n",
+    "\\bf{X}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "id": "fc73f387-de90-4c25-b24c-ec91ea774ce8",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "⎡d        ⎤                               \n",
+      "⎢──(x₁(t))⎥                               \n",
+      "⎢dt       ⎥                               \n",
+      "⎢         ⎥   ⎡-x₁(t) + 5⋅x₂(t) + 2⋅x₃(t)⎤\n",
+      "⎢d        ⎥   ⎢                          ⎥\n",
+      "⎢──(x₂(t))⎥ = ⎢4⋅x₁(t) - x₂(t) - 2⋅x₃(t) ⎥\n",
+      "⎢dt       ⎥   ⎢                          ⎥\n",
+      "⎢         ⎥   ⎣         6⋅x₃(t)          ⎦\n",
+      "⎢d        ⎥                               \n",
+      "⎢──(x₃(t))⎥                               \n",
+      "⎣dt       ⎦                               \n"
+     ]
+    }
+   ],
+   "source": [
+    "t, s = symbols('t, s')\n",
+    "x = Matrix([Function('x1')(t), Function('x2')(t), Function('x3')(t)])\n",
+    "x_dot = x.diff(t) \n",
+    "A = Matrix([[-1, 5, 2],[4, -1, -2], [0, 0, 6]])\n",
+    "\n",
+    "eq = Eq(x_dot,A*x)\n",
+    "print(pretty(eq))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9203ea18-e9e6-424b-bb2e-c12ef880f40b",
+   "metadata": {},
+   "source": [
+    "First we have to find the fundamental matrix $\\bf{\\Psi}(t) = e^{\\bf{A}t}$:\n",
+    "$$e^{\\bf{A}t} = \\mathcal{L}^{-1} \\{ (sI-\\bf{A})^{-1} \\} $$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "id": "bcfcdd55-34a5-4fb5-a91c-f1b540cef71f",
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "TypeError",
+     "evalue": "inverse_laplace_transform() missing 2 required positional arguments: 's' and 't'",
+     "output_type": "error",
+     "traceback": [
+      "\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
+      "\u001b[0;31mTypeError\u001b[0m                                 Traceback (most recent call last)",
+      "Cell \u001b[0;32mIn[10], line 1\u001b[0m\n\u001b[0;32m----> 1\u001b[0m psi_t \u001b[38;5;241m=\u001b[39m \u001b[43minverse_laplace_transform\u001b[49m\u001b[43m(\u001b[49m\u001b[43mMatPow\u001b[49m\u001b[43m(\u001b[49m\u001b[43ms\u001b[49m\u001b[38;5;241;43m*\u001b[39;49m\u001b[43meye\u001b[49m\u001b[43m(\u001b[49m\u001b[38;5;241;43m3\u001b[39;49m\u001b[43m)\u001b[49m\u001b[43m \u001b[49m\u001b[38;5;241;43m-\u001b[39;49m\u001b[43m \u001b[49m\u001b[43mA\u001b[49m\u001b[43m,\u001b[49m\u001b[38;5;241;43m-\u001b[39;49m\u001b[38;5;241;43m1\u001b[39;49m\u001b[43m)\u001b[49m\u001b[43m)\u001b[49m\n",
+      "\u001b[0;31mTypeError\u001b[0m: inverse_laplace_transform() missing 2 required positional arguments: 's' and 't'"
+     ]
+    }
+   ],
+   "source": [
+    "psi_t = inverse_laplace_transform(MatPow(s*eye(3) - A,-1))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "14643f79-07ea-4d87-b81e-39244b1a3fa7",
+   "metadata": {},
+   "source": [
+    "## Problem 2\n",
+    "Please find the general solution of \n",
+    "$$\n",
+    "\\bf{\\dot{X}} = \n",
+    "\\begin{bmatrix}   \n",
+    "-6 & 5 \\\\   \n",
+    "-5 & 4 \\\\\n",
+    "\\end{bmatrix}\n",
+    "\\bf{X}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "390b5ac8-c156-4606-8d69-1b86b80fcefd",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "markdown",
+   "id": "4c0f6f7a-d495-4f06-8d7c-a81f0f9ac2fc",
+   "metadata": {},
+   "source": [
+    "# Python Problems\n",
+    "## Problem 3\n",
+    "The Archimedes Spiral can be plotted by taking all the positive whole numbers (e.g.j = 0, 1, 2, 3, 4, 5, ...) and putting them into the format $n = (j,j)$ , and plotting them in polar coordinates where the first term, $n_1$, is the radius, and the second term, $n_2$, is the angle in radians.\n",
+    "### Part A\n",
+    "You need to plot the first 1000 terms in a scatter plot. In addition, we would like to only look at the top right quadrant! What you're going for is shown in Figure 1."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "5e4b28fb-f18f-44ae-887e-9088446f2e5d",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fde7a33e-6eb1-47b0-8867-85792164a17c",
+   "metadata": {},
+   "source": [
+    "### Part B\n",
+    "You need to plot the first 25 terms, looking at th eentire polar plot (all quadrants, and then, put a *smooth* line through it. What you're going for is shown in Figure 2.)\n",
+    "Hint: [This will be a useful reference](https://matplotlib.org/stable/gallery/pie_and_polar_charts/index.html)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "d03285bf-09a6-4b79-b6e2-aaf35c56f96f",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "markdown",
+   "id": "607410b5-74aa-44fd-ad88-4ed702dad2bc",
+   "metadata": {},
+   "source": [
+    "## Problem 4\n",
+    "Consider the following system:\n",
+    "$$\n",
+    "\\bf{\\dot{X}} = \n",
+    "\\begin{bmatrix}   \n",
+    "1 & 2 & 1\\\\   \n",
+    "3 & 1+x & 1\\\\\n",
+    "1 & 0 & 0\n",
+    "\\end{bmatrix}\n",
+    "\\bf{X}\n",
+    "$$\n",
+    "This linear differential equation system’s behavior is governed by its eigenvalues. In particular, the eigenvalues relate to stability and we may wish to see where they cross the 0 line (in terms of their real value). The constant x varies over the interval [−5, 5]. Using a Jupyter Notebook (local, or on Google Colab), Python, NumPy, and Matplotlib’s PyPlot, you should evaluate the eigenvalues for 50 evenly spaced values of x between −5 and 5, and produce a plot that visualizes the variation in the three eigenvalues as x varies. An example plot is shown in Figure 3 (for a different matrix!)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "368c0e47-f034-47e0-845c-f894ef601afc",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3 (ipykernel)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.12.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}

requirements.txt

@@ -49,6 +49,7 @@ MarkupSafe==2.1.5
 matplotlib==3.9.2
 matplotlib-inline==0.1.7
 mistune==3.0.2
+mpmath==1.3.0
 nbclient==0.10.0
 nbconvert==7.16.4
 nbformat==5.10.4
@@ -87,6 +88,7 @@ six==1.16.0
 sniffio==1.3.1
 soupsieve==2.6
 stack-data==0.6.3
+sympy==1.13.2
 terminado==0.18.1
 tinycss2==1.3.0
 tornado==6.4.1