{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "e70bed39-e553-4e77-9e90-45acff2132cc",
   "metadata": {},
   "source": [
    "クローゼンの公式\n",
    "$$_{2}F_1\\left(a,b;a+b+\\frac12;x\\right)^2= {}_3F_2\\left(2\\,a,2\\,b,a+b;2\\,a+2\\,b,a+b+\\frac12;x\\right)$$\n",
    "２次変換公式\n",
    "$${}_2F_1\\left(a,b;\\frac{a+b+1}{2};z\\right)={}_2F_1\\left(\\frac{a}{2},\\frac{b}{2}; \\frac{a+b+1}{2}; 4\\,z\\,(1-z)\\right)$$\n",
    "から以下の式を導きます。\n",
    "$$_{2}F_1\\left(\\frac12,\\frac12;1;x\\right)^2= {}_3F_2\\left(\\frac12,\\frac12,\\frac12;1,1;4\\,x\\,(1-x)\\right)$$\n",
    "\n",
    "<br><br>まず、クローゼンの公式にCLという名前をつけます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "54dfbe14-8f65-4678-bd3b-eb08b13ecd3d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{0}$}F\\left( \\left. \\begin{array}{c}a,\\;b\\\\b+a+\\frac{1}{2}\\end{array} \\right |,x\\right)^2=F\\left( \\left. \\begin{array}{c}2\\,a,\\;2\\,b,\\;b+a\\\\b+a+\\frac{1}{2},\\;2\\,b+2\\,a\\end{array} \\right |,x\\right)\\]"
      ],
      "text/plain": [
       "                    2                  1\n",
       "(%o0) hypergeometric ([a, b], [b + a + -], x) = \n",
       "                                       2\n",
       "                                                              1\n",
       "                   hypergeometric([2 a, 2 b, b + a], [b + a + -, 2 b + 2 a], x)\n",
       "                                                              2"
      ],
      "text/x-maxima": [
       "hypergeometric([a,b],[b+a+1/2],x)^2 = hypergeometric([2*a,2*b,b+a],\n",
       "                                                     [b+a+1/2,2*b+2*a],x)"
      ]
     },
     "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": [
    "CL:hypergeometric([a,b],[a+b+1/2],x)^2=hypergeometric([2*a,2*b,a+b],[2*a+2*b,a+b+1/2],x);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "04e53774-2d2a-4bc6-a394-50f5c67518dd",
   "metadata": {},
   "source": [
    "この式で$a=b=\\frac{1}{4}, x=4\\,z\\,(1-z)$とし、その式にCL1と名前をつけます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "6dbbe7fd-941e-4391-8079-f6c35715bbb5",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{1}$}F\\left( \\left. \\begin{array}{c}\\frac{1}{4},\\;\\frac{1}{4}\\\\1\\end{array} \\right |,4\\,\\left(1-z\\right)\\,z\\right)^2=F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2},\\;\\frac{1}{2}\\\\1,\\;1\\end{array} \\right |,4\\,\\left(1-z\\right)\\,z\\right)\\]"
      ],
      "text/plain": [
       "                    2  1  1\n",
       "(%o1) hypergeometric ([-, -], [1], 4 (1 - z) z) = \n",
       "                       4  4\n",
       "                                                 1  1  1\n",
       "                                 hypergeometric([-, -, -], [1, 1], 4 (1 - z) z)\n",
       "                                                 2  2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/4,1/4],[1],4*(1-z)*z)^2\n",
       "  = hypergeometric([1/2,1/2,1/2],[1,1],4*(1-z)*z)"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "CL1:CL,a=1/4,b=1/4,x=4*z*(1-z);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "99d0c7ea-7815-4bfd-86b1-7891c2aedf6a",
   "metadata": {},
   "source": [
    "次に2次変換公式にQTという名前をつけます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "96ba7deb-aa1e-4b4d-a4f0-b79624d43d6e",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{2}$}F\\left( \\left. \\begin{array}{c}a,\\;b\\\\\\frac{b+a+1}{2}\\end{array} \\right |,z\\right)=F\\left( \\left. \\begin{array}{c}\\frac{a}{2},\\;\\frac{b}{2}\\\\\\frac{b+a+1}{2}\\end{array} \\right |,4\\,\\left(1-z\\right)\\,z\\right)\\]"
      ],
      "text/plain": [
       "                              b + a + 1\n",
       "(%o2) hypergeometric([a, b], [---------], z) = \n",
       "                                  2\n",
       "                                               a  b    b + a + 1\n",
       "                               hypergeometric([-, -], [---------], 4 (1 - z) z)\n",
       "                                               2  2        2"
      ],
      "text/x-maxima": [
       "hypergeometric([a,b],[(b+a+1)/2],z) = hypergeometric([a/2,b/2],[(b+a+1)/2],\n",
       "                                                     4*(1-z)*z)"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "QT:hypergeometric([a,b],[(a+b+1)/2],z)=hypergeometric([a/2,b/2],[(a+b+1)/2],4*z*(1-z));"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5681a6a5-709a-4143-a02f-7b052883ae03",
   "metadata": {},
   "source": [
    "この式で$a=b=\\frac12$としてみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "73678591-fee2-47c7-90a2-7b7ce46faff1",
   "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 |,z\\right)=F\\left( \\left. \\begin{array}{c}\\frac{1}{4},\\;\\frac{1}{4}\\\\1\\end{array} \\right |,4\\,\\left(1-z\\right)\\,z\\right)\\]"
      ],
      "text/plain": [
       "                      1  1                             1  1\n",
       "(%o3) hypergeometric([-, -], [1], z) = hypergeometric([-, -], [1], 4 (1 - z) z)\n",
       "                      2  2                             4  4"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],z) = hypergeometric([1/4,1/4],[1],4*(1-z)*z)"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "QT,a=1/2,b=1/2;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c7cb986b-6661-44e9-ba60-d383dea0c365",
   "metadata": {},
   "source": [
    "両辺を入れ替えます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "7e7de566-85dc-483b-982d-06f62fbe3e4a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{4}$}F\\left( \\left. \\begin{array}{c}\\frac{1}{4},\\;\\frac{1}{4}\\\\1\\end{array} \\right |,4\\,\\left(1-z\\right)\\,z\\right)=F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,z\\right)\\]"
      ],
      "text/plain": [
       "                      1  1                                       1  1\n",
       "(%o4) hypergeometric([-, -], [1], 4 (1 - z) z) = hypergeometric([-, -], [1], z)\n",
       "                      4  4                                       2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/4,1/4],[1],4*(1-z)*z) = hypergeometric([1/2,1/2],[1],z)"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "rhs(%)=lhs(%);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "43c9d939-8caa-4b23-a789-f903e185bb43",
   "metadata": {},
   "source": [
    "CL1に上記の式を代入します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "f260609b-48d8-428b-a841-0f01d14bb522",
   "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 |,z\\right)^2=F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2},\\;\\frac{1}{2}\\\\1,\\;1\\end{array} \\right |,4\\,\\left(1-z\\right)\\,z\\right)\\]"
      ],
      "text/plain": [
       "                    2  1  1\n",
       "(%o5) hypergeometric ([-, -], [1], z) = \n",
       "                       2  2\n",
       "                                                 1  1  1\n",
       "                                 hypergeometric([-, -, -], [1, 1], 4 (1 - z) z)\n",
       "                                                 2  2  2"
      ],
      "text/x-maxima": [
       "hypergeometric([1/2,1/2],[1],z)^2 = hypergeometric([1/2,1/2,1/2],[1,1],\n",
       "                                                   4*(1-z)*z)"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "CL1,%;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d1294b1f-305b-46fb-a6f8-45dedef49517",
   "metadata": {},
   "source": [
    "これで所望の式が得られました。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "82fda074-8b5d-4689-af2b-6fa8727a5602",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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