{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "2ce3b274-8119-4a14-a332-b98c5c168904",
   "metadata": {},
   "source": [
    "#### B.C.Berndt: Number Theory in the Spirit of Ramanujanより\n",
    "\n",
    "$-1\\lt q \\lt 1$の範囲でラマヌジャンのテータ関数の１つ$\\varphi(q)$を\n",
    "$$\\varphi(q)=\\sum_{n=-\\infty}^{\\infty}q^{n^2}$$\n",
    "と定義しました。また、\n",
    "$0\\lt x \\lt 1$の範囲で関数$F(x)$を\n",
    "$$F(x)=exp\\left(-\\pi\\,\\frac{{}_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-x\\right)}{{}_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;x\\right)}\\right)$$\n",
    "と定義しました\n",
    "<br>\n",
    "<br>\n",
    "**Chapter 5 Theorem 5.2.8**<br>\n",
    "$q=F(x)$として\n",
    "$$\\varphi\\left(q\\right)^2=_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;x\\right)$$\n",
    "が成り立つ。<br><br><br>\n",
    "\n",
    "この証明にはLemma 5.2.7の式を使うので再掲します。\n",
    "$$_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)=\\varphi\\left(q\\right)^2$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "60359794-eaef-490d-9481-31fda41a5380",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{0}$}F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)=\\varphi\\left(q\\right)^2\\]"
      ],
      "text/plain": [
       "                                                4\n",
       "                             1  1            phi (- q)       2\n",
       "(%o0)        hypergeometric([-, -], [1], 1 - ---------) = phi (q)\n",
       "                             2  2                4\n",
       "                                              phi (q)"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-phi(-q)^4/phi(q)^4) = phi(q)^2"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-HYPERGEOMETRIC in DEFUN\n"
     ]
    }
   ],
   "source": [
    "F1:hypergeometric([1/2,1/2],[1],1-phi(-q)^4/phi(q)^4)=phi(q)^2;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "0e937c17-8536-4287-bf2d-1822a7615aa0",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{1}$}F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-\\frac{\\varphi\\left(-F\\left(x\\right)\\right)^4}{\\varphi\\left(F\\left(x\\right)\\right)^4}\\right)=\\varphi\\left(F\\left(x\\right)\\right)^2\\]"
      ],
      "text/plain": [
       "                                             4\n",
       "                          1  1            phi (- F(x))       2\n",
       "(%o1)     hypergeometric([-, -], [1], 1 - ------------) = phi (F(x))\n",
       "                          2  2                4\n",
       "                                           phi (F(x))"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-phi(-F(x))^4/phi(F(x))^4) = phi(F(x))^2"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F2:F1,q=F(x);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "208b9432-42e6-44a4-a309-0a4b82f49a04",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{2}$}\\frac{\\varphi\\left(-F\\left(x\\right)\\right)^4}{\\varphi\\left(F\\left(x\\right)\\right)^4}=u\\]"
      ],
      "text/plain": [
       "                                  4\n",
       "                               phi (- F(x))\n",
       "(%o2)                          ------------ = u\n",
       "                                   4\n",
       "                                phi (F(x))"
      ],
      "text/x-maxima": [
       "phi(-F(x))^4/phi(F(x))^4 = u"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F3:phi(-q)^4/phi(q)^4=u,q=F(x);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "810a5859-883b-416a-8fea-f8e634ea1f79",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{3}$}F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-u\\right)=\\varphi\\left(F\\left(x\\right)\\right)^2\\]"
      ],
      "text/plain": [
       "                                1  1                    2\n",
       "(%o3)           hypergeometric([-, -], [1], 1 - u) = phi (F(x))\n",
       "                                2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-u) = phi(F(x))^2"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F4:F2,F3;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c319da83-36c4-44f9-a2b1-542ce813f879",
   "metadata": {},
   "source": [
    "Lemma 5.2.7の式から始めて上記F4の式を得ることができました。これはF4として覚えておくことにします。\n",
    "そして、今度は定理5.2.5を思い出し、式変形を行います。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "35ce604b-f9db-4eea-ac98-b9cc5e42a0e8",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{4}$}F\\left(1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)=q\\]"
      ],
      "text/plain": [
       "                                      4\n",
       "                                   phi (- q)\n",
       "(%o4)                        F(1 - ---------) = q\n",
       "                                       4\n",
       "                                    phi (q)"
      ],
      "text/x-maxima": [
       "F(1-phi(-q)^4/phi(q)^4) = q"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F5:F(1-phi(-q)^4/phi(q)^4)=q;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "7b94d95a-0af9-4ff0-b01b-cde52b88b95a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{5}$}F\\left(1-\\frac{\\varphi\\left(-F\\left(x\\right)\\right)^4}{\\varphi\\left(F\\left(x\\right)\\right)^4}\\right)=F\\left(x\\right)\\]"
      ],
      "text/plain": [
       "                                   4\n",
       "                                phi (- F(x))\n",
       "(%o5)                     F(1 - ------------) = F(x)\n",
       "                                    4\n",
       "                                 phi (F(x))"
      ],
      "text/x-maxima": [
       "F(1-phi(-F(x))^4/phi(F(x))^4) = F(x)"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F6:F5,q=F(x);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "a2e65fac-a72a-4260-82f3-e39828c0b044",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{6}$}F\\left(1-u\\right)=F\\left(x\\right)\\]"
      ],
      "text/plain": [
       "(%o6)                           F(1 - u) = F(x)"
      ],
      "text/x-maxima": [
       "F(1-u) = F(x)"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F7:F6,F3;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "401f45df-2b16-4448-bb0f-b70a2fa57016",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{7}$}F\\left(x\\right):=\\exp \\left(\\frac{-\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-x\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,x\\right)}\\right)\\]"
      ],
      "text/plain": [
       "                                               1  1\n",
       "                         - %pi hypergeometric([-, -], [1], 1 - x)\n",
       "                                               2  2\n",
       "(%o7)        F(x) := exp(----------------------------------------)\n",
       "                                              1  1\n",
       "                              hypergeometric([-, -], [1], x)\n",
       "                                              2  2"
      ],
      "text/x-maxima": [
       "F(x):=exp(\n",
       "  (-%pi*hypergeometric([1/2,1/2],[1],1-x))/hypergeometric([1/2,1/2],[1],x))"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F8:F(x):=exp(-%pi*hypergeometric([1/2,1/2],[1],1-x)/hypergeometric([1/2,1/2],[1],x));"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "d8042e45-ddbb-4612-9d27-4a81c64aa531",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{8}$}e^ {- \\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,u\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-u\\right)} }=e^ {- \\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-x\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,x\\right)} }\\]"
      ],
      "text/plain": [
       "          %pi hypergeometric([1/2, 1/2], [1], u)\n",
       "        - --------------------------------------\n",
       "          hypergeometric([1/2, 1/2], [1], 1 - u)\n",
       "(%o8) %e                                         = \n",
       "                                     %pi hypergeometric([1/2, 1/2], [1], 1 - x)\n",
       "                                   - ------------------------------------------\n",
       "                                         hypergeometric([1/2, 1/2], [1], x)\n",
       "                                 %e"
      ],
      "text/x-maxima": [
       "%e^-((%pi*hypergeometric([1/2,1/2],[1],u))/hypergeometric([1/2,1/2],[1],1-u))\n",
       "  = %e^-((%pi*hypergeometric([1/2,1/2],[1],1-x))\n",
       "       /hypergeometric([1/2,1/2],[1],x))"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F9:F7,F8;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "id": "10da56c9-d7f1-4d56-8209-f94e0906252a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{9}$}\\frac{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,u\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-u\\right)}=\\frac{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-x\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,x\\right)}\\]"
      ],
      "text/plain": [
       "                        1  1                               1  1\n",
       "        hypergeometric([-, -], [1], u)     hypergeometric([-, -], [1], 1 - x)\n",
       "                        2  2                               2  2\n",
       "(%o9) ---------------------------------- = ----------------------------------\n",
       "                      1  1                                   1  1\n",
       "      hypergeometric([-, -], [1], 1 - u)     hypergeometric([-, -], [1], x)\n",
       "                      2  2                                   2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],u)/hypergeometric([1/2,1/2],[1],1-u)\n",
       "  = hypergeometric([1/2,1/2],[1],1-x)/hypergeometric([1/2,1/2],[1],x)"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F10:log(F9)/(-%pi);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "42540469-6faf-46f3-9d7f-f70c9d84b03e",
   "metadata": {},
   "source": [
    "定理5.2.5の式から始めて式変形で上記F10の式が得られました。\n",
    "\n",
    "F4から、仮に任意の$x, 0\\lt x \\lt 1$について$$\\tag{A} _2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;x\\right)=_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-u\\right)$$が成り立てば定理5.2.8の式が得られます。そこで式$(A)$を証明することにします。その証明に上記のF10を使います。\n",
    "\n",
    "背理法を使います。ある$x=x_0$で$(A)$が成り立たないと仮定します。その時の$u$の値はF3を使えば計算できますが、ここでは$u_0$とします。$(A)$が成り立たないのですから、$x_0 \\neq 1-u_0$、従って$x_0 \\gt 1-u_0$か$x_0 \\lt 1-u_0$のどちらかです。仮に$x_0 \\lt 1-u_0$とすると、超幾何関数$_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;x\\right)$の$0\\lt x \\lt 1$での単調増加性（証明は別途必要ですが）から\n",
    "$$_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;x_0\\right) \\lt _2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-u_0\\right)$$\n",
    "この不等式の両辺はF10で$x=x_0$とした時の分母です。従ってF10が成り立つためには分子側で\n",
    "$$_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-x_0\\right) \\lt _2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;u_0\\right)$$\n",
    "が成り立たなければなりません。再びこの超幾何関数の単調増加性により、\n",
    "$$1-x_0 \\lt u_0$$\n",
    "がわかります。これは明らかに仮定した$x_0 \\lt 1-u_0$と矛盾します。<br>\n",
    "また$x_0 \\gt 1-u_0$を仮定しても同様の矛盾が生じます。これらから背理法の仮定は間違っていることが分かり、$(A)$はどんな$x$についても成り立つことがわかりました。\n",
    "\n",
    "F4と$(A)$より定理の式が成り立つことが分かりました。"
   ]
  }
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