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    "# Compute root of Rosencrantz function\n",
    "\n",
    "**Randall Romero Aguilar, PhD**\n",
    "\n",
    "This demo is based on the original Matlab demo accompanying the  <a href=\"https://mitpress.mit.edu/books/applied-computational-economics-and-finance\">Computational Economics and Finance</a> 2001 textbook by Mario Miranda and Paul Fackler.\n",
    "\n",
    "Original (Matlab) CompEcon file: **demslv02.m**\n",
    "\n",
    "Running this file requires the Python version of CompEcon. This can be installed with pip by running\n",
    "\n",
    "    !pip install compecon --upgrade\n",
    "\n",
    "<i>Last updated: 2022-Sept-04</i>\n",
    "<hr>\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## About\n",
    "\n",
    "Compute root of \n",
    "\n",
    "\\begin{equation}\n",
    "f(x_1,x_2)= \\begin{bmatrix}200x_1(x_2-x_1^2) + 1-x_1 \\\\ 100(x_1^2-x_2)\\end{bmatrix}\n",
    "\\end{equation}\n",
    "\n",
    "using Newton and Broyden methods. Initial values generated randomly.  True root is $x_1=1, \\quad x_2=1$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import pandas as pd\n",
    "from compecon import NLP, tic, toc"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Set up the problem"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def f(x):\n",
    "    x1, x2 = x\n",
    "    fval = [200 * x1 * (x2 - x1 ** 2) + 1 - x1, 100 * (x1 ** 2 - x2)]\n",
    "    fjac = [[200 * (x2 - x1 ** 2) - 400 * x1 ** 2 - 1, 200 * x1],\n",
    "            [200 * x1, -100]]\n",
    "    return fval, fjac\n",
    "\n",
    "problem = NLP(f)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Randomly generate starting point"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "problem.x0 = np.random.randn(2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Compute root using Newton method"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "t0 = tic()\n",
    "x1 = problem.newton()\n",
    "t1 = 100 * toc(t0)\n",
    "n1 = problem.fnorm"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Compute root using Broyden method"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "t0 = tic()\n",
    "x2 = problem.broyden()\n",
    "t2 = 100 * toc(t0)\n",
    "n2 = problem.fnorm"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Print results"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Hundreds of seconds required to compute root of Rosencrantz function\n",
      "f(x1,x2)= [200*x1*(x2-x1^2)+1-x1;100*(x1^2-x2)] via Newton and Broyden\n",
      "methods, starting at x1 = -1.20 x2 = 0.26\n"
     ]
    },
    {
     "data": {
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       "\n",
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       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>Time</th>\n",
       "      <th>Norm of f</th>\n",
       "      <th>x1</th>\n",
       "      <th>x2</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Newton</th>\n",
       "      <td>7.397294</td>\n",
       "      <td>2.043809</td>\n",
       "      <td>-0.724236</td>\n",
       "      <td>0.522311</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Broyden</th>\n",
       "      <td>6.598163</td>\n",
       "      <td>2.320916</td>\n",
       "      <td>-0.617309</td>\n",
       "      <td>0.375372</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
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      "text/plain": [
       "             Time  Norm of f        x1        x2\n",
       "Newton   7.397294   2.043809 -0.724236  0.522311\n",
       "Broyden  6.598163   2.320916 -0.617309  0.375372"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "print('Hundreds of seconds required to compute root of Rosencrantz function')\n",
    "print('f(x1,x2)= [200*x1*(x2-x1^2)+1-x1;100*(x1^2-x2)] via Newton and Broyden')\n",
    "print('methods, starting at x1 = {:4.2f} x2 = {:4.2f}'.format(*problem.x0))\n",
    "\n",
    "pd.DataFrame({\n",
    "    'Time': [t1, t2],\n",
    "    'Norm of f': [n1, n2],\n",
    "    'x1': [x1[0], x2[0]],\n",
    "    'x2': [x1[1], x2[1]]},\n",
    "    index=['Newton', 'Broyden']\n",
    ")"
   ]
  }
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