exponentielle_regression_beispiel.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "%matplotlib inline\n",
    "import numpy as np\n",
    "import matplotlib.pyplot as plt"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Wenn das Radium perfekt einem exponentiellen Zerfall genügen würde, dann würde es folgendem Gesetz gehorchen:\n",
    "$$m = d \\cdot e^{kt}$$\n",
    "Wenn wir beide Seiten logarithmieren, dann erhalten wir:\n",
    "$$ log(m) = log(d) + kt$$\n",
    "Da wir aber sowohl Messfehler als auch Modellfehler berücksichtigen müssen, wird diese Gleichung niemals exakt erfüllt sein. Stattdessen suchen wir nun bestmögliche $k$ und $d$ finden. In unserem Fall verwenden wir als Maß für \"bestmöglich\" die Summe der Abstandsquadrate, die wir minimieren wollen.\n",
    "\n",
    "$$V(k,d) = \\sum_{i=1}^{n} [log(m_i) - (log(d) + kt_i)]^2$$\n",
    "\n",
    "Mit etwas anderer Notation: $\\hat{m}_i := log(m_i)$ und $\\hat{d} := log(d_i)$ können wir die Formel etwas einfacher hinschreiben.\n",
    "$$V(k,\\hat{d}) = \\sum_{i=1}^{n} [\\hat{m}_i - (\\hat{d} + kt_i)]^2$$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Die Funktion $V$, die von $k$ und $\\hat{d}$ abhängt, wird minimiert indem wir die Nullstellen ihrer Ableitung finden.\n",
    "Die Ableitung von $V$ an einer Stelle $(k, \\hat{d})$ ist folgender $1 \\times 2$ Vektor:\n",
    "\n",
    "$$dV = (\\partial_k V, \\partial_\\hat{d} V) = \\left(\\sum_{i=1}^n 2\\cdot[\\hat{m}_i - (\\hat{d} + kt_i)]\\cdot(-t_i),  \\sum_{i=1}^n 2\\cdot[\\hat{m}_i - (\\hat{d} + kt_i)]\\cdot(-1)\\right)$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Wir setzen diesen Vektor also gleich $(0,0)$:\n",
    "\n",
    "$$dV = 2 \\cdot \\left(\\sum_{i=1}^n [\\hat{m}_i - (\\hat{d} + kt_i)]\\cdot(-t_i),  \\sum_{i=1}^n [\\hat{m}_i - (\\hat{d} + kt_i)]\\cdot(-1)\\right) = (0,0)$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Die zwei Gleichungen die sich aus den Komponenten ergeben kann man nun wie folgt umformen um schlussendlich ein linear Gleichungssystem zu erhalten.\n",
    "\n",
    "#### Erste Gleichung:\n",
    "\n",
    "\n",
    "$$\\sum_{i=1}^n (\\hat{m}_i - (\\hat{d} + kt_i))(-t_i) = 0$$\n",
    "$$\\sum_{i=1}^n -t_i\\hat{m}_i + \\sum_{i=1}^n t_i\\hat{d} + \\sum_{i=1}^n k t_i^2 = 0$$\n",
    "$$\\hat{d} \\sum_{i=1}^n t_i + k \\sum_{i=1}^n t_i^2 = \\sum_{i=1}^n t_i \\hat{m}_i$$\n",
    "\n",
    "#### Zweite Gleichung:\n",
    "\n",
    "$$ \\sum_{i=1}^n -(\\hat{m}_i - (\\hat{d} + kt_i)) = 0$$\n",
    "$$ -\\sum_{i=1}^n \\hat{m}_i + \\sum_{i=1}^n \\hat{d} + k\\sum_{i=1}^n t_i = 0$$\n",
    "$$ \\hat{d} \\sum_{i=1}^n (1) + k\\sum_{i=1}^n t_i = \\sum_{i=1}^n \\hat{m}_i$$\n",
    "\n",
    "\n",
    "#### Beide Gleichungen in Matrix-Darstellung:\n",
    "\n",
    "$$ \\begin{pmatrix}\n",
    "  \\sum_{i=1}^n t_i & \\sum_{i=1}^n t_i^2 \\\\\n",
    "  \\sum_{i=1}^n (1) & \\sum_{i=1}^n t_i \\\\ \n",
    "  \\end{pmatrix}\n",
    "  \\cdot\n",
    "  \\begin{pmatrix}\n",
    "  \\hat{d} \\\\\n",
    "  k\n",
    "  \\end{pmatrix}\n",
    " = \n",
    "   \\begin{pmatrix}\n",
    "  \\sum_{i=1}^n t_i \\hat{m}_i \\\\\n",
    "  \\sum_{i=1}^n \\hat{m}_i\n",
    "  \\end{pmatrix}\n",
    "$$\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Nun können wir unsere echten Daten, die $t_i$ und $\\hat{m}_i$ in die Gleichung einsetzen und erhalten folgende Lösung:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "t =  [ 0.322  1.184  2.321  3.165  5.43 ]\n",
      "m =  [ 2.133  1.811  1.457  1.242  0.808]\n",
      "mhat =  [ 0.75752944  0.59387918  0.37637953  0.21672298 -0.21319322]\n"
     ]
    }
   ],
   "source": [
    "t = np.asarray([0.322, 1.184, 2.321, 3.165, 5.430])\n",
    "m = np.asarray([2.133, 1.811, 1.457, 1.242, 0.808])\n",
    "mhat = np.log(m)\n",
    "print \"t = \", t\n",
    "print \"m = \", m\n",
    "print \"mhat = \", mhat"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "A = \n",
      "[[ 12.422     46.394706]\n",
      " [  5.        12.422   ]]\n"
     ]
    }
   ],
   "source": [
    "A = np.matrix([[np.sum(t), np.sum(t**2)], [len(t), np.sum(t)]])\n",
    "print \"A = \"\n",
    "print A"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "b = \n",
      "[[ 1.34894337]\n",
      " [ 1.73131791]]\n"
     ]
    }
   ],
   "source": [
    "b = np.matrix([[np.sum(t*mhat)],[np.sum(mhat)]])\n",
    "print \"b = \"\n",
    "print b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "dhat, k =  [ 0.81845631] [-0.19006308]\n",
      "d, k =  [ 2.26699758] [-0.19006308]\n"
     ]
    }
   ],
   "source": [
    "x = np.linalg.solve(A,b)\n",
    "dhat, k = np.asarray(x)\n",
    "print \"dhat, k = \", dhat, k\n",
    "\n",
    "d, k = np.exp(dhat), k\n",
    "print \"d, k = \", d, k"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[<matplotlib.lines.Line2D at 0x7f8230477b50>]"
      ]
     },
     "execution_count": 46,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
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+kQ9EOh1LxO/p7ofiF1JTYdQo2Lu2DJVSvuRvlV6i7cdt6bGgB8fTjjsdT8T1VOTiVamp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\nIuJyKnIREZdTkYuIuJyKXETE5f4/QixMsyj/Wy0AAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f8230477510>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.plot(t, m,'x')\n",
    "t_fine = np.linspace(0, 6, 100)\n",
    "plt.plot(t_fine, d*np.exp(k*t_fine))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
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   },
   "outputs": [],
   "source": []
  }
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