{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "4df199dd-2f12-46af-9375-c7911b9b93a7",
   "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。\n",
    "<br>\n",
    "<br>\n",
    "**Chapter 5 Lemma 5.2.6**<br>\n",
    "$$_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)=\\frac{\\varphi\\left(q\\right)^2}{\\varphi\\left(q^2\\right)^2}\\, _{2}F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-\\frac{\\varphi\\left(-q^2\\right)^4}{\\varphi\\left(q^2\\right)^4}\\right)$$\n",
    "が成り立つ。<br><br><br>\n",
    "\n",
    "この証明には以前に証明した結果をいくつか使うのでここに記載しておきます。<br>\n",
    "Lemma 5.2.2 における$x$の定義式：$\\frac{1-x}{1+x}=\\frac{\\varphi(-q)^2}{\\varphi(q)^2}$<br>\n",
    "と結論の式：$1-x^2=\\frac{\\varphi(-q^2)^4}{\\varphi(q^2)^4}$<br>\n",
    "テータ関数の恒等式：$\\varphi\\left(-q^2\\right)^2=\\varphi\\left(-q\\right)\\,\\varphi\\left(q\\right)$<br>\n",
    "Corollary 5.1.7 $F\\left(\\frac{1}{2},\\frac{1}{2};1;1-\\frac{\\left(x-1\\right)^2}{\\left(x+1\\right)^2}\\right)=(1+x)\\,F\\left(\\frac{1}{2},\\frac{1}{2};1;x^2\\right)$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1667ceaf-774c-498e-902d-db0e393d7c83",
   "metadata": {},
   "source": [
    "Lemma 5.2.2で$x$を$0 \\lt x \\lt 1$の範囲で定義しました。ここでは$0 \\lt x$を使うのでそれを宣言します。また上記の式を式変形で使えるようにMaximaのセッションの中で定義しておきます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "6b0a823b-5d96-4812-b661-d53df5696a4c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{0}$}\\left[ x>0 \\right] \\]"
      ],
      "text/plain": [
       "(%o0)                               [x > 0]"
      ],
      "text/x-maxima": [
       "[x > 0]"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{1}$}\\frac{1-x}{x+1}=\\frac{\\varphi\\left(-q\\right)^2}{\\varphi\\left(q\\right)^2}\\]"
      ],
      "text/plain": [
       "                                          2\n",
       "                               1 - x   phi (- q)\n",
       "(%o1)                          ----- = ---------\n",
       "                               x + 1       2\n",
       "                                        phi (q)"
      ],
      "text/x-maxima": [
       "(1-x)/(x+1) = phi(-q)^2/phi(q)^2"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{2}$}1-x^2=\\frac{\\varphi\\left(-q^2\\right)^4}{\\varphi\\left(q^2\\right)^4}\\]"
      ],
      "text/plain": [
       "                                          4    2\n",
       "                                   2   phi (- q )\n",
       "(%o2)                         1 - x  = ----------\n",
       "                                           4  2\n",
       "                                        phi (q )"
      ],
      "text/x-maxima": [
       "1-x^2 = phi(-q^2)^4/phi(q^2)^4"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{3}$}\\varphi\\left(-q^2\\right)^2=\\varphi\\left(-q\\right)\\,\\varphi\\left(q\\right)\\]"
      ],
      "text/plain": [
       "                            2    2\n",
       "(%o3)                    phi (- q ) = phi(- q) phi(q)"
      ],
      "text/x-maxima": [
       "phi(-q^2)^2 = phi(-q)*phi(q)"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{4}$}F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-\\frac{\\left(1-x\\right)^2}{\\left(x+1\\right)^2}\\right)=F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,x^2\\right)\\,\\left(x+1\\right)\\]"
      ],
      "text/plain": [
       "                                             2\n",
       "                      1  1            (1 - x)\n",
       "(%o4) hypergeometric([-, -], [1], 1 - --------) = \n",
       "                      2  2                   2\n",
       "                                      (x + 1)\n",
       "                                                        1  1         2\n",
       "                                        hypergeometric([-, -], [1], x ) (x + 1)\n",
       "                                                        2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-(1-x)^2/(x+1)^2)\n",
       "  = hypergeometric([1/2,1/2],[1],x^2)*(x+1)"
      ]
     },
     "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": [
    "assume(x>0);\n",
    "L522A:-(x-1)/(1+x)=phi(-q)^2/phi(q)^2;\n",
    "L522R:1-x^2=phi(-q^2)^4/phi(q^2)^4;\n",
    "F1332:phi(-q^2)^2=phi(-q)*phi(q);\n",
    "C517:hypergeometric([1/2,1/2],[1],1-(1-x)^2/(x+1)^2)=(1+x)*hypergeometric([1/2,1/2],[1],x^2);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "efc75509-5061-4e5e-a8a8-b4a511266709",
   "metadata": {},
   "source": [
    "最後に定義したC517から式変形を始めます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "4783885d-da2b-41d4-afab-7bc9297430a3",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{5}$}F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-\\frac{\\left(1-x\\right)^2}{\\left(x+1\\right)^2}\\right)=F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,x^2\\right)\\,\\left(x+1\\right)\\]"
      ],
      "text/plain": [
       "                                             2\n",
       "                      1  1            (1 - x)\n",
       "(%o5) hypergeometric([-, -], [1], 1 - --------) = \n",
       "                      2  2                   2\n",
       "                                      (x + 1)\n",
       "                                                        1  1         2\n",
       "                                        hypergeometric([-, -], [1], x ) (x + 1)\n",
       "                                                        2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-(1-x)^2/(x+1)^2)\n",
       "  = hypergeometric([1/2,1/2],[1],x^2)*(x+1)"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F1:C517;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6f678e88-32db-4adf-a26d-37f5396bffae",
   "metadata": {},
   "source": [
    "次にL522Aをこの式変形で使える形にします。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "22700cd8-2b98-4f4a-bf57-19ad3edee848",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{6}$}1-\\frac{\\left(1-x\\right)^2}{\\left(x+1\\right)^2}=1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\]"
      ],
      "text/plain": [
       "                                    2          4\n",
       "                             (1 - x)        phi (- q)\n",
       "(%o6)                    1 - -------- = 1 - ---------\n",
       "                                    2           4\n",
       "                             (x + 1)         phi (q)"
      ],
      "text/x-maxima": [
       "1-(1-x)^2/(x+1)^2 = 1-phi(-q)^4/phi(q)^4"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F2:1-L522A^2;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "132f6342-035c-4a14-8293-9525fcc07447",
   "metadata": {},
   "source": [
    "左辺の第3引数にF2の等式を代入します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "85096bc2-4e61-4573-89a3-96cd1a5badab",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{7}$}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)=F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,x^2\\right)\\,\\left(x+1\\right)\\]"
      ],
      "text/plain": [
       "                                         4\n",
       "                      1  1            phi (- q)\n",
       "(%o7) hypergeometric([-, -], [1], 1 - ---------) = \n",
       "                      2  2                4\n",
       "                                       phi (q)\n",
       "                                                        1  1         2\n",
       "                                        hypergeometric([-, -], [1], x ) (x + 1)\n",
       "                                                        2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-phi(-q)^4/phi(q)^4)\n",
       "  = hypergeometric([1/2,1/2],[1],x^2)*(x+1)"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F3:F1,F2;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "bebb5464-4b14-4a48-b522-4689fcfbbdc1",
   "metadata": {},
   "source": [
    "右辺の第3引数($x^2$の部分)に1からL522Rの両辺を引いた等式を代入します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "ecef2322-ffa2-4322-bb75-5a1ab1eeb625",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{8}$}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)=F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-\\frac{\\varphi\\left(-q^2\\right)^4}{\\varphi\\left(q^2\\right)^4}\\right)\\,\\left(x+1\\right)\\]"
      ],
      "text/plain": [
       "                                         4\n",
       "                      1  1            phi (- q)\n",
       "(%o8) hypergeometric([-, -], [1], 1 - ---------) = \n",
       "                      2  2                4\n",
       "                                       phi (q)\n",
       "                                                               4    2\n",
       "                                            1  1            phi (- q )\n",
       "                            hypergeometric([-, -], [1], 1 - ----------) (x + 1)\n",
       "                                            2  2                4  2\n",
       "                                                             phi (q )"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-phi(-q)^4/phi(q)^4)\n",
       "  = hypergeometric([1/2,1/2],[1],1-phi(-q^2)^4/phi(q^2)^4)*(x+1)"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F4:F3,1-L522R;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2b440644-bf0d-49f0-8957-8ad6a81af8a0",
   "metadata": {},
   "source": [
    "この式の右辺に$(x+1)$が残っています。これを片付ければ所望の式が得られます。L522Aの分母と分子をひっくり返し、L522Rを辺辺にかけます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "bcc223da-ac93-45e8-aaac-1a317070b9c7",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{9}$}\\frac{\\left(x+1\\right)\\,\\left(1-x^2\\right)}{1-x}=\\frac{\\varphi\\left(q\\right)^2\\,\\varphi\\left(-q^2\\right)^4}{\\varphi\\left(-q\\right)^2\\,\\varphi\\left(q^2\\right)^4}\\]"
      ],
      "text/plain": [
       "                                   2       2       4    2\n",
       "                     (x + 1) (1 - x )   phi (q) phi (- q )\n",
       "(%o9)                ---------------- = ------------------\n",
       "                          1 - x            2         4  2\n",
       "                                        phi (- q) phi (q )"
      ],
      "text/x-maxima": [
       "((x+1)*(1-x^2))/(1-x) = (phi(q)^2*phi(-q^2)^4)/(phi(-q)^2*phi(q^2)^4)"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "1/L522A*L522R;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b300e9bf-e9d5-46a4-abb5-7172f9f53456",
   "metadata": {},
   "source": [
    "左辺を整理します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "927a6d98-5d11-4221-aad5-2334c6f9dd31",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{10}$}\\left(x+1\\right)^2=\\frac{\\varphi\\left(q\\right)^2\\,\\varphi\\left(-q^2\\right)^4}{\\varphi\\left(-q\\right)^2\\,\\varphi\\left(q^2\\right)^4}\\]"
      ],
      "text/plain": [
       "                                       2       4    2\n",
       "                                2   phi (q) phi (- q )\n",
       "(%o10)                   (x + 1)  = ------------------\n",
       "                                       2         4  2\n",
       "                                    phi (- q) phi (q )"
      ],
      "text/x-maxima": [
       "(x+1)^2 = (phi(q)^2*phi(-q^2)^4)/(phi(-q)^2*phi(q^2)^4)"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "factor(%);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "daa31e25-6b18-4448-acdf-6e8b5f6c032a",
   "metadata": {},
   "source": [
    "F1332の二乗を辺辺に掛けます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "64cc69b7-57c4-4e5d-a1fc-e423134f3242",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{11}$}\\left(x+1\\right)^2=\\frac{\\varphi\\left(q\\right)^4}{\\varphi\\left(q^2\\right)^4}\\]"
      ],
      "text/plain": [
       "                                            4\n",
       "                                     2   phi (q)\n",
       "(%o11)                        (x + 1)  = --------\n",
       "                                            4  2\n",
       "                                         phi (q )"
      ],
      "text/x-maxima": [
       "(x+1)^2 = phi(q)^4/phi(q^2)^4"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "%,F1332^2;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dc05cd8b-0330-4e89-9a9c-a7212b599e43",
   "metadata": {},
   "source": [
    "両辺の平方根を取ります。$0\\lt x$を宣言してあるので絶対値記号は付きません。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "ff4124cb-9279-436e-9251-4d09a7791022",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{12}$}x+1=\\frac{\\varphi\\left(q\\right)^2}{\\varphi\\left(q^2\\right)^2}\\]"
      ],
      "text/plain": [
       "                                          2\n",
       "                                       phi (q)\n",
       "(%o12)                         x + 1 = --------\n",
       "                                          2  2\n",
       "                                       phi (q )"
      ],
      "text/x-maxima": [
       "x+1 = phi(q)^2/phi(q^2)^2"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sqrt(%);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c622ace7-5e85-4650-b8c6-96195608285c",
   "metadata": {},
   "source": [
    "得られた式を用いてF4の右辺に残る$x+1$を書き換えます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "id": "c09d43ef-917b-47e4-a378-bdffea154895",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{13}$}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)=\\frac{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-\\frac{\\varphi\\left(-q^2\\right)^4}{\\varphi\\left(q^2\\right)^4}\\right)\\,\\varphi\\left(q\\right)^2}{\\varphi\\left(q^2\\right)^2}\\]"
      ],
      "text/plain": [
       "                                          4\n",
       "                       1  1            phi (- q)\n",
       "(%o13) hypergeometric([-, -], [1], 1 - ---------) = \n",
       "                       2  2                4\n",
       "                                        phi (q)\n",
       "                                                               4    2\n",
       "                                            1  1            phi (- q )     2\n",
       "                            hypergeometric([-, -], [1], 1 - ----------) phi (q)\n",
       "                                            2  2                4  2\n",
       "                                                             phi (q )\n",
       "                            ---------------------------------------------------\n",
       "                                                    2  2\n",
       "                                                 phi (q )"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],1-phi(-q)^4/phi(q)^4)\n",
       "  = (hypergeometric([1/2,1/2],[1],1-phi(-q^2)^4/phi(q^2)^4)*phi(q)^2)\n",
       "  /phi(q^2)^2"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F4,%;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "61150b09-03ed-4d2a-a999-77167ec2f357",
   "metadata": {},
   "source": [
    "これで証明が終了しました。"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Maxima",
   "language": "maxima",
   "name": "maxima"
  },
  "language_info": {
   "codemirror_mode": "maxima",
   "file_extension": ".mac",
   "mimetype": "text/x-maxima",
   "name": "maxima",
   "pygments_lexer": "maxima",
   "version": "5.44.0"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 5
}