БДЗ 2, задача 14

BHW2.ipynb

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    "$f(x) = \\sin \\frac{\\pi}{2} x^2 \\quad [a, b] = [0, \\frac{3}{2}]$\n",
    "\n",
    "$f'(x) = \\pi x \\cos \\frac{\\pi}{2} x^2$\n",
    "\n",
    "$f'(x) = 0$\n",
    "\n",
    "$x = 0, x^2 = 2k + 1, k \\in \\mathbb{Z}$\n",
    "\n",
    "На $[0, \\frac{3}{2}]$ $x = 0, 1$\n",
    "\n",
    "На $(0, 1)$ $f'(x) > 0 \\rightarrow f(x) \\uparrow$ на $[0, 1]$\n",
    "\n",
    "На $(1, \\frac{3}{2})$ $f'(x) < 0 \\rightarrow f(x) \\downarrow$ на $[1, \\frac{3}{2}]$\n",
    "\n",
    "Тогда если $\\tau_n: x_0 = 0, x_i = \\frac{3}{2n} i$, то:\n",
    "\n",
    "если $x_i < x_{i + 1} \\leq 1$, то $\\sup_{[x_i, x_{i + 1}]}(f(x)) = f(x_{i + 1}), \\quad \\inf_{[x_i, x_{i + 1}]}(f(x)) = f(x_i)$\n",
    "\n",
    "если $1 \\leq x_i < x_{i + 1}$, то $\\sup_{[x_i, x_{i + 1}]}(f(x)) = f(x_{i}), \\quad \\inf_{[x_i, x_{i + 1}]}(f(x)) = f(x_{i + 1})$\n",
    "\n",
    "если $x_i < 1 < x_{i + 1}$, то $\\sup_{[x_i, x_{i + 1}]}(f(x)) = f(1), \\quad \\inf_{[x_i, x_{i + 1}]}(f(x)) = \\min(f(x_i), f(x_{i + 1}))$\n",
    "\n",
    "$\\overline S_{\\tau_n} = \\sum_{i = 0}^{n - 1} \\sup_{[x_i, x_{i + 1}]}(f(x)) \\frac{3}{2n}$\n",
    "\n",
    "$\\underline S_{\\tau_n} = \\sum_{i = 0}^{n - 1} \\inf_{[x_i, x_{i + 1}]}(f(x)) \\frac{3}{2n}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import pandas as pd"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "def f(x):\n",
    "    return np.sin(np.pi * x * x / 2.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "def get(n):\n",
    "    upper = 0.\n",
    "    lower = 0.\n",
    "    d = 3. / (2. * n)\n",
    "    \n",
    "    for i in range(n):\n",
    "        x_i = d * i\n",
    "        x_ii = d * (i + 1)\n",
    "        if x_i < x_ii <= 1:\n",
    "            sup = f(x_ii)\n",
    "            inf = f(x_i)\n",
    "        elif 1 <= x_i < x_ii:\n",
    "            sup = f(x_i)\n",
    "            inf = f(x_ii)\n",
    "        else:\n",
    "            sup = f(1)\n",
    "            inf = min(f(x_i), f(x_ii))\n",
    "        upper += sup * d\n",
    "        lower += inf * d\n",
    "    return lower, upper"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
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       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>lower</th>\n",
       "      <th>upper</th>\n",
       "      <th>delta</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>n</th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>10</th>\n",
       "      <td>0.512500</td>\n",
       "      <td>0.867962</td>\n",
       "      <td>0.355462</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>100</th>\n",
       "      <td>0.679555</td>\n",
       "      <td>0.715293</td>\n",
       "      <td>0.035738</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1000</th>\n",
       "      <td>0.695717</td>\n",
       "      <td>0.699291</td>\n",
       "      <td>0.003574</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>10000</th>\n",
       "      <td>0.697326</td>\n",
       "      <td>0.697684</td>\n",
       "      <td>0.000357</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>100000</th>\n",
       "      <td>0.697487</td>\n",
       "      <td>0.697523</td>\n",
       "      <td>0.000036</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
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      "text/plain": [
       "           lower     upper     delta\n",
       "n                                   \n",
       "10      0.512500  0.867962  0.355462\n",
       "100     0.679555  0.715293  0.035738\n",
       "1000    0.695717  0.699291  0.003574\n",
       "10000   0.697326  0.697684  0.000357\n",
       "100000  0.697487  0.697523  0.000036"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "interesting = [10, 100, 1000, 10000, 100000]\n",
    "\n",
    "results = []\n",
    "\n",
    "for n in interesting:\n",
    "    lower, upper = get(n)\n",
    "    results.append([lower, upper, upper - lower])\n",
    "\n",
    "df = pd.DataFrame(index=interesting, columns=['lower', 'upper', 'delta'], data=results)\n",
    "df.index.name = 'n'\n",
    "df"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\overline S_{\\tau_n} - \\underline S_{\\tau_n} < 1\\mathrm{e}{-4}$ для $n = 100000$\n",
    "\n",
    "Значение интеграла: $0.697487$"
   ]
  }
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