{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "B.C.Berndt: Number Theory in the Spirit of Ramanujanより\n",
    "\n",
    "Chapter 5 Corollary 5.1.10\n",
    "\n",
    "$x \\rightarrow\\,0^{+}$の時、\n",
    "$$ \\pi\\,F(\\frac{1}{2},\\frac{1}{2};1;1-x)\\sim -\\log(x)+C$$\n",
    "ただし$C$は定数である。\n",
    "\n",
    "\n",
    "まず$1-x$を$x$と置き直すとこの主張は、\n",
    "$x \\rightarrow\\,1^{-}$の時、\n",
    "$$ \\pi\\,F(\\frac{1}{2},\\frac{1}{2};1;x)\\sim -\\log(1-x)+C$$\n",
    "となります。\n",
    "\n",
    "また$K(x)=\\frac{\\pi}{2}\\,F(\\frac{1}{2},\\frac{1}{2};1;x^2)$だったので$K(\\sqrt{x})=\\frac{\\pi}{2}\\,F(\\frac{1}{2},\\frac{1}{2};1;x)$であることがわかります。まとめると、\n",
    "$$2\\,K(\\sqrt{x})\\sim -\\log(1-x)+C$$\n",
    "を示せば良いことになります。\n",
    "\n",
    "$x \\rightarrow\\,1^{-}$の時、$2\\,K(\\sqrt{x})+\\log(1-x) \\rightarrow\\,\\log 2$を示すことにします。\n",
    "\n",
    "楕円積分による$K(x)$の定義は以下の式でした。ここから証明を始めます。\n",
    "$$K(k)=\\int_{0}^{\\frac{\\pi}{2}}\\frac{d\\phi}{\\sqrt{1-k^2\\,\\sin^2\\phi}}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 221,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{255}$}K\\left(k\\right)=\\int_{0}^{\\frac{\\pi}{2}}{\\frac{1}{\\sqrt{1-k^2\\,\\sin ^2\\varphi}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                            %pi\n",
       "                            ---\n",
       "                             2\n",
       "                           /\n",
       "                           [              1\n",
       "(%o255)             K(k) = I    ---------------------- dphi\n",
       "                           ]              2    2\n",
       "                           /    sqrt(1 - k  sin (phi))\n",
       "                            0"
      ],
      "text/x-maxima": [
       "K(k) = 'integrate(1/sqrt(1-k^2*sin(phi)^2),phi,0,%pi/2)"
      ]
     },
     "execution_count": 221,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "DEFK:K(k)=integrate(1/sqrt(1-k^2*sin(phi)^2),phi,0,%pi/2);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "両辺を２倍し、$k=\\sqrt{x}$を代入します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 224,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{264}$}2\\,K\\left(\\sqrt{x}\\right)=2\\,\\int_{0}^{\\frac{\\pi}{2}}{\\frac{1}{\\sqrt{1-\\sin ^2\\varphi\\,x}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                                 %pi\n",
       "                                 ---\n",
       "                                  2\n",
       "                                /\n",
       "                                [              1\n",
       "(%o264)        2 K(sqrt(x)) = 2 I    --------------------- dphi\n",
       "                                ]                2\n",
       "                                /    sqrt(1 - sin (phi) x)\n",
       "                                 0"
      ],
      "text/x-maxima": [
       "2*K(sqrt(x)) = 2*'integrate(1/sqrt(1-sin(phi)^2*x),phi,0,%pi/2)"
      ]
     },
     "execution_count": 224,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "assume(0<x)$assume(x<1)$\n",
    "F1:2*DEFK,k=sqrt(x);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\sin\\phi=t$という変数変換を積分に施します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 338,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{378}$}2\\,K\\left(\\sqrt{x}\\right)=2\\,\\int_{0}^{1}{\\frac{1}{\\sqrt{1-t}\\,\\sqrt{t+1}\\,\\sqrt{1-t^2\\,x}}\\;dt}\\]"
      ],
      "text/plain": [
       "                           1\n",
       "                          /\n",
       "                          [                    1\n",
       "(%o378)  2 K(sqrt(x)) = 2 I  -------------------------------------- dt\n",
       "                          ]                                    2\n",
       "                          /  sqrt(1 - t) sqrt(t + 1) sqrt(1 - t  x)\n",
       "                           0"
      ],
      "text/x-maxima": [
       "2*K(sqrt(x)) = 2*'integrate(1/(sqrt(1-t)*sqrt(t+1)*sqrt(1-t^2*x)),t,0,1)"
      ]
     },
     "execution_count": 338,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F2:changevar(F1,sin(phi)=t,t,phi);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "一方、$\\log(1-x)$を積分で表します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 226,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{266}$}x\\,\\int_{0}^{1}{\\frac{1}{1-t\\,x}\\;dt}=-\\log \\left(1-x\\right)\\]"
      ],
      "text/plain": [
       "                           1\n",
       "                          /\n",
       "                          [     1\n",
       "(%o266)                 x I  ------- dt = - log(1 - x)\n",
       "                          ]  1 - t x\n",
       "                          /\n",
       "                           0"
      ],
      "text/x-maxima": [
       "x*'integrate(1/(1-t*x),t,0,1) = -log(1-x)"
      ]
     },
     "execution_count": 226,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F3:'integrate(x/(1-x*t),t,0,1)=integrate(x/(1-x*t),t,0,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "これで求める式の値を積分の和で表すことができました。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 230,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{270}$}\\log \\left(1-x\\right)+2\\,K\\left(\\sqrt{x}\\right)=2\\,\\int_{0}^{1}{\\frac{1}{\\sqrt{1-t}\\,\\sqrt{t+1}\\,\\sqrt{1-t^2\\,x}}\\;dt}-x\\,\\int_{0}^{1}{\\frac{1}{1-t\\,x}\\;dt}\\]"
      ],
      "text/plain": [
       "(%o270) log(1 - x) + 2 K(sqrt(x)) = 2\n",
       "                  1                                                1\n",
       "                 /                                                /\n",
       "                 [                    1                           [     1\n",
       "                 I  -------------------------------------- dt - x I  ------- dt\n",
       "                 ]                                    2           ]  1 - t x\n",
       "                 /  sqrt(1 - t) sqrt(t + 1) sqrt(1 - t  x)        /\n",
       "                  0                                                0"
      ],
      "text/x-maxima": [
       "log(1-x)+2*K(sqrt(x)) = 2*'integrate(1/(sqrt(1-t)*sqrt(t+1)*sqrt(1-t^2*x)),t,\n",
       "                                     0,1)\n",
       "                      -x*'integrate(1/(1-t*x),t,0,1)"
      ]
     },
     "execution_count": 230,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F4:lhs(F2)-rhs(F3)=rhs(F2)-lhs(F3);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "右辺の２つの積分は積分範囲も一致しているので、１つにまとめることが出来ます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 247,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{289}$}2\\,\\int_{0}^{1}{\\frac{1}{\\sqrt{1-t}\\,\\sqrt{t+1}\\,\\sqrt{1-t^2\\,x}}\\;dt}-x\\,\\int_{0}^{1}{\\frac{1}{1-t\\,x}\\;dt}=\\int_{0}^{1}{\\frac{2}{\\sqrt{1-t}\\,\\sqrt{t+1}\\,\\sqrt{1-t^2\\,x}}-\\frac{x}{1-t\\,x}\\;dt}\\]"
      ],
      "text/plain": [
       "           1                                                1\n",
       "          /                                                /\n",
       "          [                    1                           [     1\n",
       "(%o289) 2 I  -------------------------------------- dt - x I  ------- dt = \n",
       "          ]                                    2           ]  1 - t x\n",
       "          /  sqrt(1 - t) sqrt(t + 1) sqrt(1 - t  x)        /\n",
       "           0                                                0\n",
       "                        1\n",
       "                       /\n",
       "                       [                     2                         x\n",
       "                       I  (-------------------------------------- - -------) dt\n",
       "                       ]                                     2      1 - t x\n",
       "                       /   sqrt(1 - t) sqrt(t + 1) sqrt(1 - t  x)\n",
       "                        0"
      ],
      "text/x-maxima": [
       "2*'integrate(1/(sqrt(1-t)*sqrt(t+1)*sqrt(1-t^2*x)),t,0,1)\n",
       " -x*'integrate(1/(1-t*x),t,0,1)\n",
       "  = 'integrate(2/(sqrt(1-t)*sqrt(t+1)*sqrt(1-t^2*x))-x/(1-t*x),t,0,1)"
      ]
     },
     "execution_count": 247,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "assume(sqrt(x)<1)$\n",
    "F5:rhs(F4)=integrate(2*part(F4,2,1,2,1)-x*part(F4,2,2,1,2,1),t,0,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "右辺の積分は$x=1$で連続なので（連続と本に書いてありました）$x=1$を代入して計算を続けてみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 250,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{292}$}\\int_{0}^{1}{\\frac{2}{\\sqrt{1-t}\\,\\sqrt{t+1}\\,\\sqrt{1-t^2}}-\\frac{1}{1-t}\\;dt}\\]"
      ],
      "text/plain": [
       "              1\n",
       "             /\n",
       "             [                    2                       1\n",
       "(%o292)      I  (------------------------------------ - -----) dt\n",
       "             ]                                     2    1 - t\n",
       "             /   sqrt(1 - t) sqrt(t + 1) sqrt(1 - t )\n",
       "              0"
      ],
      "text/x-maxima": [
       "'integrate(2/(sqrt(1-t)*sqrt(t+1)*sqrt(1-t^2))-1/(1-t),t,0,1)"
      ]
     },
     "execution_count": 250,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F6:rhs(F5),x=1;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "計算を進めてみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 343,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{379}$}\\int_{0}^{1}{\\frac{1}{t+1}\\;dt}\\]"
      ],
      "text/plain": [
       "                                   1\n",
       "                                  /\n",
       "                                  [    1\n",
       "(%o379)                           I  ----- dt\n",
       "                                  ]  t + 1\n",
       "                                  /\n",
       "                                   0"
      ],
      "text/x-maxima": [
       "'integrate(1/(t+1),t,0,1)"
      ]
     },
     "execution_count": 343,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "radcan(F6);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 344,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{380}$}\\log 2\\]"
      ],
      "text/plain": [
       "(%o380)                             log(2)"
      ],
      "text/x-maxima": [
       "log(2)"
      ]
     },
     "execution_count": 344,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "%,nouns;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "ここまでで証明は終了です。２つの関数の差が定数に収束するのですからその漸近的振る舞いは一致します。\n",
    "\n",
    "超幾何関数$\\pi\\,F(\\frac{1}{2},\\frac{1}{2};1;x)$（あるいは楕円積分$2\\,K(\\sqrt{x}) $）と対数関数$-\\log{x}$の漸近性を視覚で確認してみたくなりますよね。まず$\\log{2}$の大きさを確認しておきましょう。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 345,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{381}$}0.6931471805599453\\]"
      ],
      "text/plain": [
       "(%o381)                       0.6931471805599453"
      ],
      "text/x-maxima": [
       "0.6931471805599453"
      ]
     },
     "execution_count": 345,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "%,numer;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "超幾何関数$\\pi\\,F(\\frac{1}{2},\\frac{1}{2};1;x)$、楕円積分$2\\,K(\\sqrt{x}) $）、対数関数$-\\log{x}$を準備して、それら及び差分をグラフに書いてみます。ただし超幾何関数の浮動小数点計算には制約があるのか、$x=1$に近すぎると計算できません。そこで$1$の少し手前($1-\\epsilon$)までのグラフを書きます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 262,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{302}$}F_{1}\\left(x\\right):=\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,x\\right)\\]"
      ],
      "text/plain": [
       "                                               1  1\n",
       "(%o302)           F1(x) := %pi hypergeometric([-, -], [1], x)\n",
       "                                               2  2"
      ],
      "text/x-maxima": [
       "F1(x):=%pi*hypergeometric([1/2,1/2],[1],x)"
      ]
     },
     "execution_count": 262,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{303}$}F_{2}\\left(x\\right):=-\\log \\left(1-x\\right)\\]"
      ],
      "text/plain": [
       "(%o303)                      F2(x) := - log(1 - x)"
      ],
      "text/x-maxima": [
       "F2(x):=-log(1-x)"
      ]
     },
     "execution_count": 262,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{304}$}F_{3}\\left(x\\right):=2\\,{\\it elliptic\\_kc}\\left(x\\right)\\]"
      ],
      "text/plain": [
       "(%o304)                    F3(x) := 2 elliptic_kc(x)"
      ],
      "text/x-maxima": [
       "F3(x):=2*elliptic_kc(x)"
      ]
     },
     "execution_count": 262,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{305}$}0.013\\]"
      ],
      "text/plain": [
       "(%o305)                              0.013"
      ],
      "text/x-maxima": [
       "0.013"
      ]
     },
     "execution_count": 262,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F1(x):=%pi*hypergeometric([1/2,1/2],[1],x);\n",
    "F2(x):=-log(1-x);\n",
    "F3(x):=2*elliptic_kc(x);\n",
    "epsilon:0.013;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 358,
   "metadata": {},
   "outputs": [
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      "text/plain": [
       "/var/folders/3n/jp9c5wkw8xjgmy006s6t_pt80000gn/T/maxplot.svg"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{394}$}\\left[ \\mbox{ /var/folders/3n/jp9c5wkw8xjgmy006s6t\\_pt80000gn/T/maxout65155.gnuplot } , \\mbox{ /var/folders/3n/jp9c5wkw8xjgmy006s6t\\_pt80000gn/T/maxplot.svg } \\right] \\]"
      ],
      "text/plain": [
       "(%o394) [/var/folders/3n/jp9c5wkw8xjgmy006s6t_pt80000gn/T/maxout65155.gnuplot, \n",
       "                  /var/folders/3n/jp9c5wkw8xjgmy006s6t_pt80000gn/T/maxplot.svg]"
      ],
      "text/x-maxima": [
       "[\"/var/folders/3n/jp9c5wkw8xjgmy006s6t_pt80000gn/T/maxout65155.gnuplot\",\n",
       " \"/var/folders/3n/jp9c5wkw8xjgmy006s6t_pt80000gn/T/maxplot.svg\"]"
      ]
     },
     "execution_count": 358,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "plot2d([F1(x),F2(x),F3(x),F3(x)-F2(x)],[x,0.0,1.0-epsilon]);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "超幾何関数と楕円積分は完全に重なっており、計算はうまく行っているようです。$x\\rightarrow 1$で楕円積分と対数関数はどちらも無限大に向かって発散していますが、その差分は少しづつ減少しながら$2.7$位の値に収束しているようにみます。まさに定理の通りです、、、？？？\n",
    "\n",
    "証明の結果としてその差は$\\log{2}=0.6931\\dots$にいくのではなかったでしょうか。何か変ですね。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 105,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{126}$}0.6931471805599453\\]"
      ],
      "text/plain": [
       "(%o126)                       0.6931471805599453"
      ],
      "text/x-maxima": [
       "0.6931471805599453"
      ]
     },
     "execution_count": 105,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "log(2.0);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 334,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{377}$}2.789328410773694\\]"
      ],
      "text/plain": [
       "(%o377)                        2.789328410773694"
      ],
      "text/x-maxima": [
       "2.789328410773694"
      ]
     },
     "execution_count": 334,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F3(1.0-epsilon)-F2(1.0-epsilon);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この２つの数の差を埋めることが出来ずに１週間くらい悩みました。上記式変形がうまく行ったのですぐにもブログ記事を書きたかったのですが、この差について何が起こっているのか説明できないと気持ち悪いですし、記事も中途半端です。\n",
    "\n",
    "hypergeometric( )やelliptic\\_kc( )とlog( )を独立に計算してはダメでその差を直接計算する必要があると考えました。そのための道具はその差を１つの積分に表した次の式です。\n",
    "$$\\int_{0}^{1}{\\frac{2}{\\sqrt{1-t}\\,\\sqrt{t+1}\\,\\sqrt{1-t^2\\,x}}-\\frac{x}{1-t\\,x}\\;dt}$$\n",
    "この積分を$x$の値を変えながら数値計算してみることにしました。数値積分にはromberg()という組み込み関数を使えます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 128,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{155}$}\\mbox{ /usr/local/Cellar/maxima/5.45.1\\_1/share/maxima/5.45.1/share/numeric/romberg.lisp }\\]"
      ],
      "text/plain": [
       "(%o155) /usr/local/Cellar/maxima/5.45.1_1/share/maxima/5.45.1/share/numeric/ro\\\n",
       "mberg.lisp"
      ],
      "text/x-maxima": [
       "\"/usr/local/Cellar/maxima/5.45.1_1/share/maxima/5.45.1/share/numeric/romberg.lisp\""
      ]
     },
     "execution_count": 128,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::ROMBERG-SUBR in DEFUN\n",
      "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::$ROMBERG in DEFUN\n"
     ]
    }
   ],
   "source": [
    "load(romberg);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "積分範囲の上の値を$1.0$にすると数値積分がうまく計算できないため、ここでは$0.99$までとしました。\n",
    "そのため計算結果は小さめに出るはずです。\n",
    "それ以外は式そのままです。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 323,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{366}$}g\\left(k\\right):={\\it romberg}\\left({\\it ev}\\left(\\frac{2}{\\sqrt{1-t^2}\\,\\sqrt{1-x\\,t^2}}-\\frac{x}{1-x\\,t} , x:k\\right) , t , 0 , 0.99\\right)\\]"
      ],
      "text/plain": [
       "                                        2                   x\n",
       "(%o366) g(k) := romberg(ev(--------------------------- - -------, x : k), t, \n",
       "                                     2              2    1 - x t\n",
       "                           sqrt(1 - t ) sqrt(1 - x t )\n",
       "                                                                       0, 0.99)"
      ],
      "text/x-maxima": [
       "g(k):=romberg(ev(2/(sqrt(1-t^2)*sqrt(1-x*t^2))-x/(1-x*t),x:k),t,0,0.99)"
      ]
     },
     "execution_count": 323,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "g(k):=romberg(ev(2/(sqrt(1-t^2)*sqrt(1-x*t^2))-x/(1-x*t),x:k),t,0,0.99);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "g()の引数を$0.8$から$0.99999999$まで下記のように変化させて計算してみました。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 348,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\(2.319341662583405\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\(2.069439436893744\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\(1.176056896939411\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\(0.7594185658227455\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\(0.6956354529343399\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\(0.688889226547345\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\(0.6882161811091833\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "\\(0.6881508251625317\\)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "for a in [0.8,0.9,0.99,0.999,0.9999,0.99999,0.999999,0.9999999] do print(g(a))$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$0.9$から$0.99999$の間で急激に小さくなることがわかります。このような動きを最初のグラフでは表現できていませんでした。この積分が$x=1$の直前でこのような急激な値の変動を起こして$\\log{2}$に行くことがあのグラフの原因であったわけです。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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