5_Numerical_Differentiation_And_Integration.ipynb

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   "metadata": {},
   "source": [
    "# ★ Numerical Differentiaion and Integration ★"
   ]
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    "# Import modules\n",
    "import math\n",
    "import sympy as sym\n",
    "import numpy as np\n",
    "import scipy \n",
    "import matplotlib.pyplot as plt\n",
    "import plotly\n",
    "import plotly.plotly as ply\n",
    "import plotly.figure_factory as ply_ff\n",
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    "from IPython.display import display\n",
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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 5.1 Numerical Differentiation"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Two-point forward-difference formula\n",
    "\n",
    "## $f'(x) = \\frac{f(x+h) - f(x)}{h} - \\frac{h}{2}f''(c)$\n",
    "\n",
    "where $c$ is between $x$ and $x+h$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Use the two-point forward-difference formula with $h = 0.1$ to approximate the derivative of $f(x) = 1/x$ at $x = 2$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 53,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "approximate = -0.238095, real value = -0.250000, backward error = 0.011905\n"
     ]
    }
   ],
   "source": [
    "# Parameters\n",
    "x = 2\n",
    "h = 0.1\n",
    "\n",
    "# Symbolic computation\n",
    "sym_x = sym.Symbol('x')\n",
    "sym_deri_x1 = sym.diff(1 / sym_x, sym_x)\n",
    "sym_deri_x1_num = sym_deri_x1.subs(sym_x, x).evalf()\n",
    "\n",
    "# Approximation\n",
    "f = lambda x : 1 / x\n",
    "deri_x1 = (f(x + h) - f(x)) / h\n",
    "\n",
    "# Comparison\n",
    "print('approximate = %f, real value = %f, backward error = %f' %(deri_x1, sym_deri_x1_num, abs(deri_x1 - sym_deri_x1_num)) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Three-point centered-difference formula\n",
    "\n",
    "## $f'(x) =  \\frac{f(x+h) - f(x-h)}{2h} - \\frac{h^2}{6}f'''(c)$\n",
    "\n",
    "where $x-h < c < x+h$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Use the three-point centered-difference formula with $h = 0.1$ to approximate the derivative of $f(x) = 1 / x$ at $x = 2$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {},
   "outputs": [
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     "name": "stdout",
     "output_type": "stream",
     "text": [
      "approximate = -0.250627, real value = -0.250000, backward error = 0.000627\n"
     ]
    }
   ],
   "source": [
    "# Parameters\n",
    "x = 2\n",
    "h = 0.1\n",
    "f = lambda x : 1 / x\n",
    "\n",
    "# Symbolic computation\n",
    "sym_x = sym.Symbol('x')\n",
    "sym_deri_x1 = sym.diff(1 / sym_x, sym_x)\n",
    "sym_deri_x1_num = sym_deri_x1.subs(sym_x, x).evalf()\n",
    "\n",
    "# Approximation\n",
    "deri_x1 = (f(x + h) - f(x - h)) / (2 * h)\n",
    "\n",
    "# Comparison\n",
    "print('approximate = %f, real value = %f, backward error = %f' %(deri_x1, sym_deri_x1_num, abs(deri_x1 - sym_deri_x1_num)) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Three-point centered-difference formula for second derivative\n",
    "\n",
    "## $f''(x) = \\frac{f(x - h) - 2f(x) + f(x + h)}{h^2} - \\frac{h^2}{12}f^{(iv)}(c)$\n",
    "\n",
    "for some $c$ between $x - h$ and $x + h$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Rounding error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Approximate the derivative of $f(x) = e^x$ at $x = 0$"
   ]
  },
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-0.45, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 7, \"text\": \"$10^{-7}$\", \"showarrow\": false}, {\"x\": 0.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 7, \"text\": \"1.00000004943368\", \"showarrow\": false}, {\"x\": 1.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 7, \"text\": \"0.00000004943368\", \"showarrow\": false}, {\"x\": 2.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 7, \"text\": \"0.99999999947364\", \"showarrow\": false}, {\"x\": 3.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 7, \"text\": \"0.00000000052636\", \"showarrow\": false}, {\"x\": -0.45, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 8, \"text\": \"$10^{-8}$\", \"showarrow\": false}, {\"x\": 0.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 8, \"text\": \"0.99999999392253\", \"showarrow\": false}, {\"x\": 1.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 8, \"text\": \"0.00000000607747\", \"showarrow\": false}, {\"x\": 2.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 8, \"text\": \"0.99999999392253\", \"showarrow\": false}, {\"x\": 3.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 8, \"text\": \"0.00000000607747\", \"showarrow\": false}, {\"x\": -0.45, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 9, \"text\": \"$10^{-9}$\", \"showarrow\": false}, {\"x\": 0.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 9, \"text\": \"1.00000008274037\", \"showarrow\": false}, {\"x\": 1.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 9, \"text\": \"0.00000008274037\", \"showarrow\": false}, {\"x\": 2.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 9, \"text\": \"1.00000002722922\", \"showarrow\": false}, {\"x\": 3.55, \"align\": \"left\", \"xanchor\": \"left\", \"font\": {\"color\": \"#000000\"}, \"yref\": \"y1\", \"xref\": \"x1\", \"y\": 9, \"text\": \"0.00000002722922\", \"showarrow\": false}]}, {\"linkText\": \"Export to plot.ly\", \"showLink\": false})});</script>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# Parameters\n",
    "f = lambda x : math.exp(x)\n",
    "real_value = 1\n",
    "h_msg = \"$10^{-%d}$\"\n",
    "twp_deri_x1 = lambda x, h : ( f(x + h) - f(x) ) / h\n",
    "thp_deri_x1 = lambda x, h : ( f(x + h) - f(x - h) ) / (2 * h)\n",
    "\n",
    "data = [\n",
    "    [\"h\", \n",
    "     \"$f'(x) \\\\approx \\\\frac{e^{x+h} - e^x}{h}$\", \n",
    "     \"error\", \n",
    "     \"$f'(x) \\\\approx \\\\frac{e^{x+h} - e^{x-h}}{2h}$\", \n",
    "     \"error\"],\n",
    "]\n",
    "\n",
    "for i in range(1,10):\n",
    "    h = pow(10, -i)\n",
    "    twp_deri_x1_value = twp_deri_x1(0, h) \n",
    "    thp_deri_x1_value = thp_deri_x1(0, h)\n",
    "    row = [\"\", \"\", \"\", \"\", \"\"]\n",
    "    row[0] = h_msg %i\n",
    "    row[1] = '%.14f' %twp_deri_x1_value\n",
    "    row[2] = '%.14f' %abs(twp_deri_x1_value - real_value)\n",
    "    row[3] = '%.14f' %thp_deri_x1_value\n",
    "    row[4] = '%.14f' %abs(thp_deri_x1_value - real_value)\n",
    "    data.append(row)\n",
    "\n",
    "table = ply_ff.create_table(data)\n",
    "plotly.offline.iplot(table, show_link=False)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Extrapolation for order n formula\n",
    "\n",
    "## $ Q \\approx \\frac{2^nF(h/2) - F(h)}{2^n - 1} $"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$Derivative : e^{\\sin{\\left (x \\right )}} \\cos{\\left (x \\right )}$$"
      ],
      "text/plain": [
       "<IPython.core.display.Math object>"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sym.init_printing(use_latex=True)\n",
    "\n",
    "x = sym.Symbol('x')\n",
    "dx = sym.diff(sym.exp(sym.sin(x)), x)\n",
    "Math('Derivative : %s' %sym.latex(dx) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 5.2 Newton-Cotes Formulas For Numerical Integration"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Trapezoid Rule\n",
    "\n",
    "## $\\int_{x_0}^{x_1} f(x) dx = \\frac{h}{2}(y_0 + y_1) - \\frac{h^3}{12}f''(c)$\n",
    "\n",
    "where $h = x_1 - x_0$ and $c$ is between $x_0$ and $x_1$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Simpson's Rule\n",
    "\n",
    "## $\\int_{x_0}^{x_2} f(x) dx = \\frac{h}{3}(y_0 + 4y_1 + y_2) - \\frac{h^5}{90}f^{(iv)}(c)$\n",
    "\n",
    "where $h = x_2 - x_1 = x_1 - x_0$ and $c$ is between $x_0$ and $x_2$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Apply the Trapezoid Rule and Simpson's Rule to approximate $\\int_{1}^{2} \\ln(x) dx$ and find an upper bound for the error in your approximations"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Trapezoid rule : 0.346574 and upper bound error : 0.083333\n"
     ]
    }
   ],
   "source": [
    "# Apply Trapezoid Rule\n",
    "trapz = scipy.integrate.trapz([np.log(1), np.log(2)], [1, 2])\n",
    "\n",
    "# Evaluate the error term of Trapezoid Rule\n",
    "sym_x = sym.Symbol('x')\n",
    "expr = sym.diff(sym.log(sym_x), sym_x, 2)\n",
    "trapz_err = abs(expr.subs(sym_x, 1).evalf() / 12)\n",
    "\n",
    "# Print out results\n",
    "print('Trapezoid rule : %f and upper bound error : %f' %(trapz, trapz_err) )"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 59,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Simpson's rule : 0.385835 and upper bound error : 0.002083\n"
     ]
    }
   ],
   "source": [
    "# Apply Simpson's Rule\n",
    "area = scipy.integrate.simps([np.log(1), np.log(1.5), np.log(2)], [1, 1.5, 2])\n",
    "\n",
    "# Evaluate the error term\n",
    "sym_x = sym.Symbol('x')\n",
    "expr = sym.diff(sym.log(sym_x), sym_x, 4)\n",
    "simps_err = abs( pow(0.5, 5) / 90 * expr.subs(sym_x, 1).evalf() )\n",
    "\n",
    "# Print out results\n",
    "print('Simpson\\'s rule : %f and upper bound error : %f' %(area, simps_err) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Composite Trapezoid Rule\n",
    "\n",
    "## $\\int_{a}^{b} f(x) dx = \\frac{h}{2} \\left ( y_0 + y_m + 2\\sum_{i=1}^{m-1}y_i \\right ) - \\frac{(b-a)h^2}{12}f''(c)$\n",
    "\n",
    "where $h = (b - a) / m $ and $c$ is between $a$ and $b$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Composite Simpson's Rule\n",
    "\n",
    "## $ \\int_{a}^{b}f(x)dx = \\frac{h}{3}\\left [ y_0 + y_{2m} + 4\\sum_{i=1}^{m}y_{2i-1} + 2\\sum_{i=1}^{m - 1}y_{2i} \\right ] - \\frac{(b-a)h^4}{180}f^{(iv)}(c) $\n",
    "\n",
    "where $c$ is between $a$ and $b$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Carry out four-panel approximations of $\\int_{1}^{2} \\ln{x} dx$ using the composite Trapezoid Rule and composite Simpson's Rule"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Trapezoid Rule : 0.383700, error = 0.005208\n"
     ]
    }
   ],
   "source": [
    "# Apply composite Trapezoid Rule\n",
    "x = np.linspace(1, 2, 5)\n",
    "y = np.log(x)\n",
    "trapz = scipy.integrate.trapz(y, x)\n",
    "\n",
    "# Error term\n",
    "sym_x = sym.Symbol('x')\n",
    "expr = sym.diff(sym.log(sym_x), sym_x, 2)\n",
    "trapz_err = abs( (2 - 1) * pow(0.25, 2) / 12 * expr.subs(sym_x, 1).evalf() )\n",
    "\n",
    "print('Trapezoid Rule : %f, error = %f' %(trapz, trapz_err) )"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Simpson's Rule : 0.386292, error = 0.000008\n"
     ]
    }
   ],
   "source": [
    "# Apply composite Trapezoid Rule\n",
    "x = np.linspace(1, 2, 9)\n",
    "y = np.log(x)\n",
    "area = scipy.integrate.simps(y, x)\n",
    "\n",
    "# Error term\n",
    "sym_x = sym.Symbol('x')\n",
    "expr = sym.diff(sym.log(sym_x), sym_x, 4)\n",
    "simps_err = abs( (2 - 1) * pow(0.125, 4) / 180 * expr.subs(sym_x, 1).evalf() )\n",
    "\n",
    "print('Simpson\\'s Rule : %f, error = %f' %(area, simps_err) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Midpoint Rule\n",
    "\n",
    "## $ \\int_{x_0}^{x_1} f(x)dx = hf(\\omega) + \\frac{h^3}{24}f''(c) $\n",
    "\n",
    "where $ h = (x_1 - x_0) $, $\\omega$ is the midpoint $ x_0 + h / 2 $, and $c$ is between $x_0$ and $x_1$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Composite Midpoint Rule\n",
    "\n",
    "## $ \\int_{a}^{b} f(x) dx = h \\sum_{i=1}^{m}f(\\omega_{i}) + \\frac{(b - a)h^2}{24} f''(c) $\n",
    "\n",
    "where $h = (b - a) / m$ and $c$ is between $a$ and $b$. The $\\omega_{i}$ are the midpoints of the $m$ equal subintervals of $[a,b]$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "Approximate $\\int_{0}^{1} \\frac{\\sin x}{x} dx$ by using the composite Midpoint Rule with $m = 10$ panels"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Composite Midpoint Rule : 0.94620858, error = 0.00009964\n"
     ]
    }
   ],
   "source": [
    "# Parameters\n",
    "m = 10\n",
    "h = (1 - 0) / m\n",
    "f = lambda x : np.sin(x) / x\n",
    "mids = np.arange(0 + h/2, 1, h)\n",
    "\n",
    "# Apply composite midpoint rule\n",
    "area = h * np.sum(f(mids))\n",
    "\n",
    "# Error term\n",
    "sym_x = sym.Symbol('x')\n",
    "expr = sym.diff(sym.sin(sym_x) / sym_x, sym_x, 2)\n",
    "mid_err = abs( (1 - 0) * pow(h, 2) / 24 * expr.subs(sym_x, 1).evalf() )\n",
    "\n",
    "# Print out\n",
    "print('Composite Midpoint Rule : %.8f, error = %.8f' %(area, mid_err) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 5.3 Romberg Integration"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 181,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def romberg(f, a, b, step):\n",
    "    R = np.zeros(step * step).reshape(step, step)\n",
    "    R[0][0] = (b - a) * (f(a) + f(b)) / 2\n",
    "    for j in range(1, step):\n",
    "        h = (b - a) / pow(2, j)\n",
    "        summ = 0\n",
    "        for i in range(1, pow(2, j - 1) + 1):\n",
    "            summ += h * f(a + (2 * i - 1) * h)\n",
    "        R[j][0] = 0.5 * R[j - 1][0] + summ\n",
    "        \n",
    "        for k in range(1, j + 1):\n",
    "            R[j][k] = ( pow(4, k) * R[j][k - 1] - R[j - 1][k - 1] ) / ( pow(4, k) - 1 )\n",
    "        \n",
    "    return R[step - 1][step - 1]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Apply Romberg Integration to approximate $\\int_{1}^{2} \\ln{x}dx$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 183,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Romberg Integration : 0.386294\n"
     ]
    }
   ],
   "source": [
    "f = lambda x : np.log(x)\n",
    "result = romberg(f, 1, 2, 4)\n",
    "print('Romberg Integration : %f' %(result) )"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 184,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Romberg integration of <function vectorize1.<locals>.vfunc at 0x7fa72aad8510> from [1, 2]\n",
      "\n",
      " Steps  StepSize   Results\n",
      "     1  1.000000  0.346574 \n",
      "     2  0.500000  0.376019  0.385835 \n",
      "     4  0.250000  0.383700  0.386260  0.386288 \n",
      "     8  0.125000  0.385644  0.386292  0.386294  0.386294 \n",
      "    16  0.062500  0.386132  0.386294  0.386294  0.386294  0.386294 \n",
      "    32  0.031250  0.386254  0.386294  0.386294  0.386294  0.386294  0.386294 \n",
      "\n",
      "The final result is 0.38629436112 after 33 function evaluations.\n",
      "Romberg Integration : 0.386294\n"
     ]
    }
   ],
   "source": [
    "f = lambda x : np.log(x)\n",
    "result = scipy.integrate.romberg(f, 1, 2, show=True)\n",
    "print('Romberg Integration : %f' %(result) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 5.4 Adaptive Quadrature"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 105,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "''' Use Trapezoid Rule '''\n",
    "\n",
    "def adaptive_quadrature(f, a, b, tol):\n",
    "    return adaptive_quadrature_recursively(f, a, b, tol, a, b, 0)\n",
    "    \n",
    "def adaptive_quadrature_recursively(f, a, b, tol, orig_a, orig_b, deep):\n",
    "    c = (a + b) / 2\n",
    "    S = lambda x, y : (y - x) * (f(x) + f(y)) / 2\n",
    "    if abs( S(a, b) - S(a, c) - S(c, b) ) < 3 * tol * (b - a) / (orig_b - orig_a) or deep > 20 :\n",
    "        return S(a, c) + S(c, b)\n",
    "    else:\n",
    "        return adaptive_quadrature_recursively(f, a, c, tol / 2, orig_a, orig_b, deep + 1) + adaptive_quadrature_recursively(f, c, b, tol / 2, orig_a, orig_b, deep + 1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 106,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "''' Use Simpon's Rule '''\n",
    "\n",
    "def adaptive_quadrature(f, a, b, tol):\n",
    "    return adaptive_quadrature_recursively(f, a, b, tol, a, b, 0)\n",
    "    \n",
    "def adaptive_quadrature_recursively(f, a, b, tol, orig_a, orig_b, deep):\n",
    "    c = (a + b) / 2\n",
    "    S = lambda x, y : (y - x) * ( f(x) + 4 * f((x + y) / 2) + f(y) ) / 6\n",
    "    if abs( S(a, b) - S(a, c) - S(c, b) ) < 15 * tol  or deep > 20 :\n",
    "        return S(a, c) + S(c, b)\n",
    "    else:\n",
    "        return adaptive_quadrature_recursively(f, a, c, tol / 2, orig_a, orig_b, deep + 1) + adaptive_quadrature_recursively(f, c, b, tol / 2, orig_a, orig_b, deep + 1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Use Adaptive Quadrature to approximate the integral $ \\int_{-1}^{1} (1 + \\sin{e^{3x}}) dx $"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 108,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2.50080911034\n"
     ]
    }
   ],
   "source": [
    "f = lambda x : 1 + np.sin(np.exp(3 * x))\n",
    "val = adaptive_quadrature(f, -1, 1, tol=1e-12)\n",
    "print(val)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 5.5 Gaussian Quadrature"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [],
   "source": [
    "poly = scipy.special.legendre(2)\n",
    "# Find roots of polynomials\n",
    "comp = scipy.linalg.companion(poly)\n",
    "roots = scipy.linalg.eig(comp)[0]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Approximate $\\int_{-1}^{1} e^{-\\frac{x^2}{2}}dx$ using Gaussian Quadrature"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.71124878401\n"
     ]
    }
   ],
   "source": [
    "f = lambda x : np.exp(-np.power(x, 2) / 2)\n",
    "quad = scipy.integrate.quadrature(f, -1, 1)\n",
    "print(quad[0])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.7120202452\n"
     ]
    }
   ],
   "source": [
    "# Parametes\n",
    "a = -1\n",
    "b = 1\n",
    "deg = 3\n",
    "f = lambda x : np.exp( -np.power(x, 2) / 2 )\n",
    "\n",
    "x, w = scipy.special.p_roots(deg) # Or use numpy.polynomial.legendre.leggauss\n",
    "quad = np.sum(w * f(x))\n",
    "    \n",
    "print(quad)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example\n",
    "\n",
    "Approximate the integral $\\int_{1}^{2} \\ln{x} dx$ using Gaussian Quadrature"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 66,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.38629449693871409"
      ]
     },
     "execution_count": 66,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# Parametes\n",
    "a = 1\n",
    "b = 2\n",
    "deg = 4\n",
    "f = lambda t : np.log( ((b - a) * t + b + a) / 2) * (b - a) / 2\n",
    "\n",
    "x, w = scipy.special.p_roots(deg)\n",
    "np.sum(w * f(x))"
   ]
  }
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