Upload bisection.ipynb

bisection.ipynb

@@ -0,0 +1,151 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import mercury as mr\n",
+    "import numpy as np\n",
+    "import matplotlib.pyplot as plt\n",
+    "%matplotlib inline"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "name = mr.Text(value=\"Piotr\", label=\"What is your name?\")\n",
+    "print(f\"Hello {name.value}\")\n",
+    "# set application properites with App object\n",
+    "app = mr.App(show_code = True)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Consider one of the most basic problems:\n",
+    "\n",
+    "Finding the root of a function, i.e: $f(x) = 0$ .\n",
+    "\n",
+    "## Technique:\n",
+    "\n",
+    "We start with a boundary $[a,b]$ and then hope that there exists a point $p$ in that boundary where $f(p) = 0$. We half the end points of the boundary depending on if $f(a_i) \\text{ or } f(b_i)$ is negative or positive, until we reach the point $p$\n",
+    "\n",
+    "## Example\n",
+    "Suppose we want to look for the zero for $f(x)= \\sin(x) + 0.5$ between $[-1,2]$. "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 42,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "26\n"
+     ]
+    },
+    {
+     "data": {
+      "text/plain": [
+       "-0.523598775267601"
+      ]
+     },
+     "execution_count": 42,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "def f(x): \n",
+    "    return np.sin(x)+0.5\n",
+    "\n",
+    "def validate_interval(f,x0,x1):\n",
+    "    return f(x0)*f(x1) < 0\n",
+    "\n",
+    "\n",
+    "def bisection(f, interval, n, tol):\n",
+    "    x0, x1 = interval[0], interval[1] #extract interval \n",
+    "    if not validate_interval(f, x0, x1): #check interval can be solved for roots\n",
+    "        return \"Not valid interval\"\n",
+    "\n",
+    "    counter = 1\n",
+    "    while True:\n",
+    "        p = x0 + ((x1-x0)/2)\n",
+    "        y = f(p)\n",
+    "        if -tol < y < tol:\n",
+    "            print(counter)\n",
+    "            return p\n",
+    "        if validate_interval(f,x0,p):\n",
+    "            x1 = p\n",
+    "        else:\n",
+    "            x0 = p\n",
+    "        counter += 1\n",
+    "\n",
+    "\n",
+    "\n",
+    "bisection(f,[-1,2], 50, 0.000000001) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def graph(x):\n",
+    "    #need function\n",
+    "    #need tangents\n",
+    "    #root\n",
+    "    fig, ax = plt.subplots(figsize=(10, 10))\n",
+    "    y=f(x)\n",
+    "    plt.plot(x,y)\n",
+    "   \n",
+    "    plt.ylabel('some numbers')\n",
+    "    plt.axis('tight')\n",
+    "    plt.grid(True)\n",
+    "    plt.show()\n",
+    "    ax.spines['top'].set_visible(False)\n",
+    "    ax.spines['right'].set_visible(False)\n",
+    "\n",
+    "graph(np.arange(-np.pi,np.pi,np.pi/32))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "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": "3.10.13"
+  },
+  "orig_nbformat": 4
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}