Analysis A 2018.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 解析学A 2018\n",
    "\n",
    "黒木玄\n",
    "\n",
    "2018年7月30日(月)\n",
    "\n",
    "次のリンク先で綺麗に閲覧できる:\n",
    "\n",
    "* http://nbviewer.jupyter.org/github/genkuroki/Calculus/blob/master/Analysis%20A%202018.ipynb\n",
    "\n",
    "$\n",
    "\\newcommand\\eps{\\varepsilon}\n",
    "\\newcommand\\ds{\\displaystyle}\n",
    "\\newcommand\\Z{{\\mathbb Z}}\n",
    "\\newcommand\\R{{\\mathbb R}}\n",
    "\\newcommand\\C{{\\mathbb C}}\n",
    "\\newcommand\\QED{\\text{□}}\n",
    "\\newcommand\\root{\\sqrt}\n",
    "\\newcommand\\bra{\\langle}\n",
    "\\newcommand\\ket{\\rangle}\n",
    "\\newcommand\\d{\\partial}\n",
    "\\newcommand\\sech{\\operatorname{sech}}\n",
    "\\newcommand\\cosec{\\operatorname{cosec}}\n",
    "\\newcommand\\sign{\\operatorname{sign}}\n",
    "\\newcommand\\sinc{\\operatorname{sinc}}\n",
    "\\newcommand\\real{\\operatorname{Re}}\n",
    "\\newcommand\\imag{\\operatorname{Im}}\n",
    "\\newcommand\\Li{\\operatorname{Li}}\n",
    "\\newcommand\\np[1]{:\\!#1\\!:}\n",
    "\\newcommand\\PROD{\\mathop{\\coprod\\kern-1.35em\\prod}}\n",
    "$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "toc": true
   },
   "source": [
    "<h1>Table of Contents<span class=\"tocSkip\"></span></h1>\n",
    "<div class=\"toc\"><ul class=\"toc-item\"><li><span><a href=\"#[1]\" data-toc-modified-id=\"[1]-1\"><span class=\"toc-item-num\">1&nbsp;&nbsp;</span>[1]</a></span><ul class=\"toc-item\"><li><span><a href=\"#[1]-(1)\" data-toc-modified-id=\"[1]-(1)-1.1\"><span class=\"toc-item-num\">1.1&nbsp;&nbsp;</span>[1] (1)</a></span></li><li><span><a href=\"#[1]-(2)\" data-toc-modified-id=\"[1]-(2)-1.2\"><span class=\"toc-item-num\">1.2&nbsp;&nbsp;</span>[1] (2)</a></span></li></ul></li><li><span><a href=\"#[2]\" data-toc-modified-id=\"[2]-2\"><span class=\"toc-item-num\">2&nbsp;&nbsp;</span>[2]</a></span></li><li><span><a href=\"#[3]\" data-toc-modified-id=\"[3]-3\"><span class=\"toc-item-num\">3&nbsp;&nbsp;</span>[3]</a></span></li><li><span><a href=\"#[4]\" data-toc-modified-id=\"[4]-4\"><span class=\"toc-item-num\">4&nbsp;&nbsp;</span>[4]</a></span><ul class=\"toc-item\"><li><span><a href=\"#[4]-(1)\" data-toc-modified-id=\"[4]-(1)-4.1\"><span class=\"toc-item-num\">4.1&nbsp;&nbsp;</span>[4] (1)</a></span></li><li><span><a href=\"#[4]-(2)\" data-toc-modified-id=\"[4]-(2)-4.2\"><span class=\"toc-item-num\">4.2&nbsp;&nbsp;</span>[4] (2)</a></span></li><li><span><a href=\"#[4]-(3)\" data-toc-modified-id=\"[4]-(3)-4.3\"><span class=\"toc-item-num\">4.3&nbsp;&nbsp;</span>[4] (3)</a></span></li></ul></li><li><span><a href=\"#[5]\" data-toc-modified-id=\"[5]-5\"><span class=\"toc-item-num\">5&nbsp;&nbsp;</span>[5]</a></span><ul class=\"toc-item\"><li><span><a href=\"#[5]-(1)\" data-toc-modified-id=\"[5]-(1)-5.1\"><span class=\"toc-item-num\">5.1&nbsp;&nbsp;</span>[5] (1)</a></span></li><li><span><a href=\"#[5]-(2)\" data-toc-modified-id=\"[5]-(2)-5.2\"><span class=\"toc-item-num\">5.2&nbsp;&nbsp;</span>[5] (2)</a></span></li></ul></li><li><span><a href=\"#[6]\" data-toc-modified-id=\"[6]-6\"><span class=\"toc-item-num\">6&nbsp;&nbsp;</span>[6]</a></span><ul class=\"toc-item\"><li><span><a href=\"#[6]-(1)\" data-toc-modified-id=\"[6]-(1)-6.1\"><span class=\"toc-item-num\">6.1&nbsp;&nbsp;</span>[6] (1)</a></span></li><li><span><a href=\"#[6]-(2)\" data-toc-modified-id=\"[6]-(2)-6.2\"><span class=\"toc-item-num\">6.2&nbsp;&nbsp;</span>[6] (2)</a></span></li></ul></li><li><span><a href=\"#[7]\" data-toc-modified-id=\"[7]-7\"><span class=\"toc-item-num\">7&nbsp;&nbsp;</span>[7]</a></span><ul class=\"toc-item\"><li><span><a href=\"#[7]-(1)\" data-toc-modified-id=\"[7]-(1)-7.1\"><span class=\"toc-item-num\">7.1&nbsp;&nbsp;</span>[7] (1)</a></span></li><li><span><a href=\"#[7]-(2)\" data-toc-modified-id=\"[7]-(2)-7.2\"><span class=\"toc-item-num\">7.2&nbsp;&nbsp;</span>[7] (2)</a></span></li></ul></li><li><span><a href=\"#[8]\" data-toc-modified-id=\"[8]-8\"><span class=\"toc-item-num\">8&nbsp;&nbsp;</span>[8]</a></span></li></ul></div>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "using SymPy\n",
    "Beta(x,y) = gamma(x)*gamma(y)/gamma(x+y)\n",
    "# sympy[:init_printing](order=\"rev-lex\")\n",
    "# sympy[:init_printing](order=\"lex\") # default\n",
    "\n",
    "using Plots\n",
    "gr()\n",
    "ENV[\"PLOTS_TEST\"] = \"true\";"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[1\\]\n",
    "\n",
    "### \\[1\\] (1)\n",
    "\n",
    "$\\tan x$ のMaclaurin展開を5次の項まで求めよう. $f(x) = \\tan x$ とおくと, $f'=1+f^2$ なので $f''=2f+2f^3$, \n",
    "\n",
    "$$\n",
    "f'''=2+8f^2+6f^4, \\quad\n",
    "f^{(4)}=16f+40f^3+24f^5, \\quad\n",
    "f^{(5)}=16+136f^2+240f^4+120f^6\n",
    "$$\n",
    "\n",
    "となりので, \n",
    "\n",
    "$$\n",
    "f(0)=f''(0)=f^{(4)}(0)=0, \\quad\n",
    "f'(0)=1, \\quad\n",
    "\\frac{f'''(0)}{3!}=\\frac{2}{3!}=\\frac{1}{3}, \\quad\n",
    "\\frac{f^{(5)}(0)}{5!}=\\frac{16}{5!}=\\frac{2}{15}\n",
    "$$\n",
    "\n",
    "ゆえに, \n",
    "\n",
    "$$\n",
    "\\tan x = x + \\frac{x^3}{3} + \\frac{2x^5}{15} + O(x^6).\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$1 + \\tan^{2}{\\left (x \\right )}$$"
      ],
      "text/plain": [
       "       2   \n",
       "1 + tan (x)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$$2 \\tan{\\left (x \\right )} + 2 \\tan^{3}{\\left (x \\right )}$$"
      ],
      "text/plain": [
       "                3   \n",
       "2*tan(x) + 2*tan (x)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$$2 + 8 \\tan^{2}{\\left (x \\right )} + 6 \\tan^{4}{\\left (x \\right )}$$"
      ],
      "text/plain": [
       "         2           4   \n",
       "2 + 8*tan (x) + 6*tan (x)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$$16 \\tan{\\left (x \\right )} + 40 \\tan^{3}{\\left (x \\right )} + 24 \\tan^{5}{\\left (x \\right )}$$"
      ],
      "text/plain": [
       "                  3            5   \n",
       "16*tan(x) + 40*tan (x) + 24*tan (x)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$$16 + 136 \\tan^{2}{\\left (x \\right )} + 240 \\tan^{4}{\\left (x \\right )} + 120 \\tan^{6}{\\left (x \\right )}$$"
      ],
      "text/plain": [
       "            2             4             6   \n",
       "16 + 136*tan (x) + 240*tan (x) + 120*tan (x)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "x = symbols(\"x\")\n",
    "f(x) = tan(x)\n",
    "sympy[:init_printing](order=\"rev-lex\")\n",
    "diff(f(x), x) |> display\n",
    "diff(f(x), x, 2) |> expand |> display\n",
    "diff(f(x), x, 3) |> expand |> display\n",
    "diff(f(x), x, 4) |> expand |> display\n",
    "diff(f(x), x, 5) |> expand |> display\n",
    "sympy[:init_printing](order=\"lex\") # default"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$x + \\frac{x^{3}}{3} + \\frac{2 x^{5}}{15} + \\frac{17 x^{7}}{315} + \\frac{62 x^{9}}{2835} + O\\left(x^{10}\\right)$$"
      ],
      "text/plain": [
       "     3      5       7       9         \n",
       "    x    2*x    17*x    62*x     / 10\\\n",
       "x + -- + ---- + ----- + ----- + O\\x  /\n",
       "    3     15     315     2835         "
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [1] (1)\n",
    "\n",
    "x = symbols(\"x\")\n",
    "series(tan(x), x, n=10)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### \\[1\\] (2)\n",
    "\n",
    "$x=0.5$ のとき,  \n",
    "\n",
    "$$\n",
    "x = 0.5, \\quad\n",
    "x + \\frac{x^3}{3} = 0.541666\\cdots, \\quad\n",
    "x + \\frac{x^3}{3} + \\frac{x^5}{5} = 0.548333\\cdots\n",
    "$$\n",
    "\n",
    "これより, $\\tan 0.5 = 0.54\\cdots$ であることがわかる."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "x = 0.5 = 0.5\n",
      "tan(x) = 0.5463024898437905\n",
      "x = 0.5\n",
      "x + x ^ 3 / 3 = 0.5416666666666666\n",
      "x + x ^ 3 / 3 + (2 * x ^ 5) / 15 = 0.5458333333333333\n",
      "x + x ^ 3 / 3 + (2 * x ^ 5) / 15 + (17 * x ^ 7) / 315 = 0.5462549603174602\n"
     ]
    }
   ],
   "source": [
    "# [1] (2)\n",
    "\n",
    "@show x = 0.5\n",
    "@show tan(x)\n",
    "@show x\n",
    "@show x+x^3/3\n",
    "@show x+x^3/3+2x^5/15\n",
    "@show x+x^3/3+2x^5/15+17*x^7/315;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[2\\]\n",
    "\n",
    "\n",
    "$|x|<1$ のとき, \n",
    "\n",
    "$$\n",
    "\\log\\frac{1+x}{1-x} = \n",
    "\\log(1+x)-\\log(1-x) =\n",
    "2\\sum_{k=0}^\\infty\\frac{x^{2k+1}}{2k+1} = \n",
    "2x+\\frac{2x^3}{3}+\\frac{2x^5}{5}+\\cdots.\n",
    "$$\n",
    "\n",
    "$x=1/3$ のとき, $\\ds\\frac{1+x}{1-x}=2$ であり, \n",
    "\n",
    "$$\n",
    "2x + \\frac{2x^3}{3} = 0.691\\cdots, \\quad\n",
    "\\frac{2x^5}{5} = \\frac{2}{5\\times 243} < \\frac{2}{1000} = 0.002\\cdots.\n",
    "$$\n",
    "\n",
    "これより $\\log 2 = 0.69\\cdots$ であることがわかる.\n",
    "\n",
    "**解説:** $\\log 2 \\approx 0.7$ は「70\\%ルール」として役にたっている. 年利 $r\\%$ の金利の複利計算で約何年後に2倍になるかは $70/r$ でわかる. なぜならば, $\\approx(1+r/100)^x = 2$ の解は $e^{rx/100}=2$ の解で近似され, その解 $\\ds x=\\frac{100\\log 2}{r}$ は $100\\log 2\\approx 70$ より $70/r$ で近似される. $\\QED$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$2 x + \\frac{2 x^{3}}{3} + \\frac{2 x^{5}}{5} + \\frac{2 x^{7}}{7} + \\frac{2 x^{9}}{9} + O\\left(x^{10}\\right)$$"
      ],
      "text/plain": [
       "         3      5      7      9         \n",
       "      2*x    2*x    2*x    2*x     / 10\\\n",
       "2*x + ---- + ---- + ---- + ---- + O\\x  /\n",
       "       3      5      7      9           "
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [2]\n",
    "\n",
    "x = Sym(:x)\n",
    "series(log(1+x)-log(1-x), x, n=10)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(1 + x) / (1 - x) = 1.9999999999999998\n",
      "log(2) = 0.6931471805599453\n",
      "2x = 0.6666666666666666\n",
      "2x + (2 * x ^ 3) / 3 = 0.691358024691358\n",
      "2x + (2 * x ^ 3) / 3 + (2 * x ^ 5) / 5 = 0.6930041152263374\n",
      "2x + (2 * x ^ 3) / 3 + (2 * x ^ 5) / 5 + (2 * x ^ 7) / 7 = 0.6931347573322881\n"
     ]
    }
   ],
   "source": [
    "x = 1/3\n",
    "@show (1+x)/(1-x)\n",
    "@show log(2)\n",
    "@show 2x\n",
    "@show 2x + 2*x^3/3\n",
    "@show 2x + 2*x^3/3 + 2*x^5/5\n",
    "@show 2x + 2*x^3/3 + 2*x^5/5 + 2*x^7/7;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[3\\]\n",
    "\n",
    "$\\ds \\int_{-\\infty}^\\infty e^{-x^2/2}\\cos(px)\\,dx = \\sqrt{2\\pi}\\;e^{-p^2/2}$\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "&\n",
    "\\int_{-\\infty}^\\infty e^{-ax^2}x^{2k}\\,dx = \n",
    "\\left(-\\frac{\\d}{\\d a}\\right)^k\\int_{-\\infty}^\\infty e^{-ax^2}\\,dx =\n",
    "\\left(-\\frac{\\d}{\\d a}\\right)^k \\sqrt{\\pi}\\;a^{-1/2},\n",
    "\\\\ &\\qquad =\n",
    "\\sqrt{\\pi}\\;\\frac{1}{2}\\frac{3}{2}\\cdots\\frac{2k-1}{2}a^{-1/2-k} =\n",
    "\\sqrt{\\pi}\\;\\frac{(2k)!}{2^{2k}k!}a^{-1/2-k},\n",
    "\\\\ &\n",
    "\\int_{-\\infty}^\\infty e^{-x^2/2}x^{2k}\\,dx =\n",
    "\\sqrt{2\\pi}\\;\\frac{(2k)!}{2^k k!},\n",
    "\\\\ &\n",
    "\\int_{-\\infty}^\\infty e^{-x^2/2}\\cos(px)\\,dx =\n",
    "\\sum_{k=0}^\\infty (-1)^k\\frac{p^{2k}}{(2k)!}\\int_{-\\infty}^\\infty e^{-x^2/2}x^{2k}\\,dx \n",
    "\\\\ &\\qquad =\n",
    "\\sum_{k=0}^\\infty (-1)^k\\frac{p^{2k}}{(2k)!} \\sqrt{2\\pi}\\;\\frac{(2k)!}{2^k k!} =\n",
    "\\sqrt{2\\pi}\\sum_{k=0}^\\infty \\frac{(-p^2/2)^k}{k!} = \n",
    "\\sqrt{2\\pi}\\;e^{-p^2/2}.\n",
    "\\end{aligned}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\sqrt{2} \\sqrt{\\pi} e^{- \\frac{p^{2}}{2}}$$"
      ],
      "text/plain": [
       "                2 \n",
       "              -p  \n",
       "              ----\n",
       "  ___   ____   2  \n",
       "\\/ 2 *\\/ pi *e    "
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [3]\n",
    "\n",
    "p = symbols(\"p\", positive=true)\n",
    "x = symbols(\"x\", real=true)\n",
    "integrate(e^(-x^2/2)*cos(p*x), (x,-oo,oo))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[4\\]\n",
    "\n",
    "$$\n",
    "B(p,q) = \n",
    "\\int_0^1 x^{p-1}(1-x)^{q-1}\\,dx =\n",
    "\\int_0^\\infty \\frac{x^{p-1}}{(1+x)^{p+q}}\\,dx =\n",
    "2\\int_0^{\\pi/2}(\\cos x)^{2p-1}(\\sin x)^{2q-1}\\,dx\n",
    "$$\n",
    "\n",
    "および\n",
    "\n",
    "$$\n",
    "B(p,q) = \\frac{\\Gamma(p)\\Gamma(q}{\\Gamma(p+q)}, \\quad\n",
    "\\Gamma(s+1)=s\\Gamma(s),\\quad\n",
    "\\Gamma(1/2) = \\sqrt{\\pi}, \\quad\n",
    "\\Gamma(n+1) = n!\n",
    "$$\n",
    "\n",
    "を使う.\n",
    "\n",
    "### \\[4\\] (1)\n",
    "\n",
    "$$\n",
    "\\int_0^1 \\sqrt{\\frac{x^5}{1-x}}\\,dx = \n",
    "%\\int_0^1 x^{5/2}(1-x)^{-1/2}\\,dx =\n",
    "B(7/2,1/2) = \n",
    "\\frac{\\Gamma(7/2)\\Gamma(1/2)}{\\Gamma(4)} =\n",
    "\\frac{(5/2)(3/2)(1/2)\\sqrt{\\pi}\\;\\sqrt{\\pi}}{3!} =\n",
    "\\frac{5\\pi}{16}.\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\frac{5 \\pi}{16}$$"
      ],
      "text/plain": [
       "5*pi\n",
       "----\n",
       " 16 "
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [4] (1)\n",
    "\n",
    "x = symbols(\"x\", positive=true)\n",
    "integrate(x^(Sym(5)/2)/√(1-x), (x,0,1))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\frac{5 \\pi}{16}$$"
      ],
      "text/plain": [
       "5*pi\n",
       "----\n",
       " 16 "
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Beta(Sym(7)/2,Sym(1)/2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### \\[4\\] (2)\n",
    "\n",
    "$$\n",
    "\\int_0^\\infty \\frac{x^{5/2}}{(1+x)^5}\\,dx =\n",
    "B(7/2, 3/2) =\n",
    "\\frac{\\Gamma(7/2)\\Gamma(3/2)}{\\Gamma(5)} =\n",
    "\\frac{(5/2)(3/2)(1/2)\\sqrt{\\pi}\\;(1/2)\\sqrt{\\pi}}{4!} =\n",
    "\\frac{5\\pi}{128}.\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\frac{5 \\pi}{128}$$"
      ],
      "text/plain": [
       "5*pi\n",
       "----\n",
       "128 "
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [4] (2)\n",
    "\n",
    "x = symbols(\"x\", positive=true)\n",
    "integrate(x^(Sym(5)/2)/(1+x)^5, (x,0,oo))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\frac{5 \\pi}{128}$$"
      ],
      "text/plain": [
       "5*pi\n",
       "----\n",
       "128 "
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Beta(Sym(7)/2, Sym(3)/2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### \\[4\\] (3)\n",
    "\n",
    "$$\n",
    "2\\int_0^{\\pi/2} (\\cos x)^6 (\\sin x)^4\\,dx =\n",
    "B(7/2, 5/2) =\n",
    "\\frac{\\Gamma(7/2)\\Gamma(5/2)}{\\Gamma(6)} =\n",
    "\\frac{(5/2)(3/2)(1/2)\\sqrt{\\pi}\\;(3/2)(1/2)\\sqrt{\\pi}}{5!} =\n",
    "\\frac{3\\pi}{256.}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\frac{3 \\pi}{256}$$"
      ],
      "text/plain": [
       "3*pi\n",
       "----\n",
       "256 "
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [4] (3)\n",
    "\n",
    "2integrate(cos(x)^6*sin(x)^4, (x,0,PI/2))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\frac{3 \\pi}{256}$$"
      ],
      "text/plain": [
       "3*pi\n",
       "----\n",
       "256 "
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Beta(Sym(7)/2, Sym(5)/2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[5\\]\n",
    "\n",
    "### \\[5\\] (1) \n",
    "\n",
    "$x>0$, $f(x) =x\\log x$ のとき, $f'(x)=\\log x + 1$, $f''(x)=1/x>0$ なので, $f(x)$ は下に凸な函数である."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\log{\\left (x \\right )} + 1$$"
      ],
      "text/plain": [
       "log(x) + 1"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$$\\frac{1}{x}$$"
      ],
      "text/plain": [
       "1\n",
       "-\n",
       "x"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# [5] (1)\n",
    "\n",
    "f(x) = x * log(x)\n",
    "x = symbols(\"x\", positive=true)\n",
    "f1 = diff(f(x), x) |> display\n",
    "f2 = diff(f(x), x, x) |> display"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### \\[5\\] (2)\n",
    "\n",
    "$\\displaystyle E[g(x)] = \\sum_{i=1}^n p_i g(q_i/p_i)$ は期待値汎函数なので, Jensenの不等式より, 下に凸な函数 $g(x)$ に対して $E[g(x)]\\leqq g(E[x])$ なので, それを $f(x)$ に適用すると,\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\sum_{i=1}^n q_i\\log\\frac{q_i}{p_i} =\n",
    "\\sum_{i=1}^n p_i\\frac{q_i}{p_i}\\log\\frac{q_i}{p_i} =\n",
    "E[f(x)] \\geqq\n",
    "f(E[x]) = \n",
    "f\\left(\\sum_{i=1}^n p_i\\frac{q_i}{p_i}\\right) =\n",
    "f(1) = 0.\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "**解説:** この結果は**Gibbsの(情報)不等式**と呼ばれている. $\\QED$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[6\\]\n",
    "\n",
    "### \\[6\\] (1)\n",
    "\n",
    "$\\ds f_n(y) =e^{-\\sqrt{n}\\;y}\\left(1+\\frac{y}{\\sqrt{n}}\\right)^n$ とおくと,  $|y/\\sqrt{n}|<1$ のとき,\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\log f_n(y) &= -\\sqrt{n}\\;y + n\\log\\left(1+\\frac{y}{\\sqrt{n}}\\right) =\n",
    "-\\sqrt{n};y + n\\left(\\frac{y}{\\sqrt{n}} - \\frac{y^2}{2n} + \\frac{y^3}{3n\\sqrt{n}} - \\frac{y^4}{4n^2} + \\cdots\\right)\n",
    "\\\\ &= -\n",
    "\\frac{y^2}{2} + \\frac{y^3}{3\\sqrt{n}} - \\frac{y^4}{4n}  +\\cdots \\to -\\frac{y^2}{2}\n",
    "\\qquad(n\\to\\infty).\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "ゆえに $n\\to\\infty$ のとき, $f_n(y)\\to e^{-y^2/2}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$e^{- \\frac{y^{2}}{2}}$$"
      ],
      "text/plain": [
       "   2 \n",
       " -y  \n",
       " ----\n",
       "  2  \n",
       "e    "
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [6] (1)\n",
    "\n",
    "y = symbols(\"y\", positive=true)\n",
    "n = symbols(\"n\", positive=true)\n",
    "limit(e^(-√n*y)*(1+y/√n)^n, n=>oo)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\frac{y^{5}}{5 n^{\\frac{3}{2}}} + \\frac{y^{3}}{3 \\sqrt{n}} - \\frac{y^{4}}{4 n} - \\frac{y^{2}}{2} + O\\left(y^{6}\\right)$$"
      ],
      "text/plain": [
       "   5         3       4    2        \n",
       "  y         y       y    y     / 6\\\n",
       "------ + ------- - --- - -- + O\\y /\n",
       "   3/2       ___   4*n   2         \n",
       "5*n      3*\\/ n                    "
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$$- \\frac{y^{2}}{2}$$"
      ],
      "text/plain": [
       "  2 \n",
       "-y  \n",
       "----\n",
       " 2  "
      ]
     },
     "execution_count": 16,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "y = symbols(\"y\", real=true)\n",
    "n = symbols(\"n\", positive=true)\n",
    "series(log(e^(-√n*y)*(1+y/√n)^n), y) |> display\n",
    "limit(series(log(e^(-√n*y)*(1+y/√n)^n), y), n=>oo)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### \\[6\\] (2)\n",
    "\n",
    "$x = n + \\sqrt{n}\\;y = n(1+y/\\sqrt{n})$ とおくと, $n\\to\\infty$ のとき,\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\frac{1}{n^n e^{-n}\\sqrt{n}}\\int_0^\\infty e^{-x}x^n\\,dx &=\n",
    "\\frac{1}{n^n e^{-n}\\sqrt{n}}\\int_{-\\sqrt{n}}^\\infty e^{-n-\\sqrt{n}\\;y}n^n\\left(1+\\frac{y}{\\sqrt{n}}\\right)^n\\sqrt{n}\\,dy\n",
    "\\\\ &=\n",
    "\\int_{-\\sqrt{n}}^\\infty e^{-\\sqrt{n}\\;y}\\left(1+\\frac{y}{\\sqrt{n}}\\right)^n\\,dy\n",
    "\\to\\int_{-\\infty}^\\infty e^{-y^2/2}\\,dy = \\sqrt{2\\pi}.\n",
    "\\end{aligned}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[7\\]\n",
    "\n",
    "### \\[7\\] (1)\n",
    "\n",
    "$\\displaystyle \\log n^{1/n}=\\frac{1}{n}\\log n\\to 0$ ($n\\to\\infty$) なので $n^{1/n}\\to 1$ である."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$1$$"
      ],
      "text/plain": [
       "1"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [7] (1)\n",
    "\n",
    "n = symbols(\"n\", positive=true)\n",
    "limit(n^(1/n), n=>oo)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### \\[7\\] (2)\n",
    "\n",
    "Stirlingの公式と(1)より, $n\\to\\infty$ のとき\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "&\n",
    "\\binom{n(a+b)}{na} =\n",
    "\\frac\n",
    "{(n(a+b))^{n(a+b)}e^{-n(a+b)}\\sqrt{2\\pi n(a+b)}}\n",
    "{(na)^{na}(nb)^{nb}e^{-na-nb}\\sqrt{2\\pi na\\;2\\pi nb}} =\n",
    "\\left(\n",
    "\\frac\n",
    "{(a+b)^{a+b}}\n",
    "{a^a b^b}\n",
    "\\right)^n\n",
    "\\sqrt{\\frac{a+b}{2\\pi nab}}\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "なので,\n",
    "\n",
    "$$\n",
    "\\lim_{n\\to\\infty}\\binom{n(a+b)}{na}^{1/n} =\n",
    "\\frac{(a+b)^{a+b}}{a^a b^b}.\n",
    "$$\n",
    "\n",
    "**解説:** この結果の対数はエントロピーの話と関係がある. $\\QED$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$a^{- a} b^{- b} \\left(a + b\\right)^{a + b}$$"
      ],
      "text/plain": [
       " -a  -b        a + b\n",
       "a  *b  *(a + b)     "
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [7] (2)\n",
    "\n",
    "n, a, b = symbols(\"n a b\", positive=true)\n",
    "simplify(limit((gamma(n*(a+b)+1)/(gamma(n*a+1)*gamma(n*b+1)))^(1/n), n=>oo))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## \\[8\\]\n",
    "\n",
    "二項展開によって, $0<x<1$ のとき,\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "&\n",
    "(-1)^n\\binom{-1/2}{n} = \\frac{(1/2)(3/2)\\cdots((2n+1)/2)}{n!} =\n",
    "\\frac{1\\cdot3\\cdots(2n-1)}{2^n n!} =\n",
    "\\frac{(2n)!}{2^{2n} (n!)^2} =\n",
    "\\binom{2n}{n}2^{-2n},\n",
    "\\\\ &\n",
    "(1-x)^{-1/2} = \\sum_{n=0}^\\infty \\binom{-1/2}{n}(-x)^n =\n",
    "\\sum_{n=0}^\\infty \\binom{2n}{n}2^{-2n}x^n\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "なので, $\\Gamma(1/2)=\\sqrt{\\pi}$, $\\Gamma(1)=1$ を使うと, \n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\pi &= \\frac{\\Gamma(1/2)\\Gamma(1/2)}{\\Gamma(1)} = B(1/2,1/2) = \\int_0^1 x^{-1/2}(1-x)^{-1/2}\\,dx \n",
    "\\\\ &=\n",
    "\\sum_{n=0}^\\infty \\binom{2n}{n}2^{-2n}\\int_0^1 x^{n-1/2}\\,dx =\n",
    "\\sum_{n=0}^\\infty \\binom{2n}{n}\\frac{2^{-2n}}{n+1/2}.\n",
    "\\end{aligned}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\pi$$"
      ],
      "text/plain": [
       "pi"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# [8]\n",
    "\n",
    "n = symbols(\"n\", integer=true)\n",
    "doit(Sum(binomial(2n,n)*2^(-2n)/(n+Sym(1)/2), (n,0,oo)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\pi$$"
      ],
      "text/plain": [
       "pi"
      ]
     },
     "execution_count": 20,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Beta(Sym(1)/2, Sym(1)/2) # B(1/2,1/2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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