{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "051adaa5-527d-4bd4-b704-994294a0af0a",
   "metadata": {},
   "source": [
    "コーシー積は以下の公式で求められます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "246c5f17-24ed-490e-affa-27cee0678944",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{0}$}\\left(\\sum_{n=0}^{\\infty }{A_{n}\\,x^{n}}\\right)\\,\\sum_{n=0}^{\\infty }{B_{n}\\,x^{n}}=\\sum_{k=0}^{\\infty }{\\left(\\sum_{j=0}^{k}{A_{j}\\,B_{k-j}}\\right)\\,x^{k}}\\]"
      ],
      "text/plain": [
       "             inf          inf           inf     k\n",
       "             ====         ====          ====   ====\n",
       "             \\         n  \\         n   \\      \\                 k\n",
       "(%o0)       ( >    A  x )  >    B  x  =  >    ( >    A  B     ) x\n",
       "             /      n     /      n      /      /      j  k - j\n",
       "             ====         ====          ====   ====\n",
       "             n = 0        n = 0         k = 0  j = 0"
      ],
      "text/x-maxima": [
       "('sum(A[n]*x^n,n,0,inf))*'sum(B[n]*x^n,n,0,inf)\n",
       "  = 'sum(('sum(A[j]*B[k-j],j,0,k))*x^k,k,0,inf)"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sum(A[n]*x^n,n,0,inf)*sum(B[n]*x^n,n,0,inf)=sum(sum(A[j]*B[k-j]*x^k,j,0,k),k,0,inf);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ba1b321f-23a7-4c09-83fe-3b1c7eaca70a",
   "metadata": {},
   "source": [
    "今回は冪級数の２乗ですから$A_n=B_n$として、$A_n$にポッホハマー記号を用いた定義式を与えます。そのために下記の定義式を思い出します。\n",
    "$${}_{2}F_1\\left(a,b;a+b+\\frac12;x\\right)=\\sum_{n=0}^{\\infty}{\\frac{(a)_n\\,(b)_n\\,x^n}{(a+b+\\frac12)_n\\,n!}}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "2613c790-e897-4351-9616-6a08dd34100c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{1}$}A_{n}:=\\frac{{\\it pochhammer}\\left(a , n\\right)\\,{\\it pochhammer}\\left(b , n\\right)}{{\\it pochhammer}\\left(a+b+\\frac{1}{2} , n\\right)\\,n!}\\]"
      ],
      "text/plain": [
       "                          pochhammer(a, n) pochhammer(b, n)\n",
       "(%o1)               A  := ---------------------------------\n",
       "                     n                          1\n",
       "                             pochhammer(a + b + -, n) n!\n",
       "                                                2"
      ],
      "text/x-maxima": [
       "A[n]:=(pochhammer(a,n)*pochhammer(b,n))/(pochhammer(a+b+1/2,n)*n!)"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A[n]:=pochhammer(a,n)*pochhammer(b,n)/(pochhammer(a+b+1/2,n)*n!);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "788ccaea-7867-4f8e-88e0-732a5db2ed74",
   "metadata": {},
   "source": [
    "$C_n$としてコーシー積の公式の右辺の$n$次の係数を計算します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "2b3ffbda-63d2-4b25-b142-8875c0c8fb3d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{2}$}C_{n}:={\\it sum}\\left(A_{j}\\,A_{n-j} , j , 0 , n\\right)\\]"
      ],
      "text/plain": [
       "(%o2)                    C  := sum(A  A     , j, 0, n)\n",
       "                          n         j  n - j"
      ],
      "text/x-maxima": [
       "C[n]:=sum(A[j]*A[n-j],j,0,n)"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "C[n]:=sum(A[j]*A[n-j],j,0,n);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "656ceb19-52fb-4aad-ae39-53fedac7f620",
   "metadata": {},
   "source": [
    "実際に$C_0, C_1, C_2, C_3$を計算してみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "ba2c59ef-32f9-47d9-b4ea-edb5c6464538",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{3}$}\\left[ 1 , \\frac{2\\,a\\,b}{b+a+\\frac{1}{2}} , \\frac{a\\,\\left(a+1\\right)\\,b\\,\\left(b+1\\right)}{\\left(b+a+\\frac{1}{2}\\right)\\,\\left(b+a+\\frac{3}{2}\\right)}+\\frac{a^2\\,b^2}{\\left(b+a+\\frac{1}{2}\\right)^2} , \\frac{a\\,\\left(a+1\\right)\\,\\left(a+2\\right)\\,b\\,\\left(b+1\\right)\\,\\left(b+2\\right)}{3\\,\\left(b+a+\\frac{1}{2}\\right)\\,\\left(b+a+\\frac{3}{2}\\right)\\,\\left(b+a+\\frac{5}{2}\\right)}+\\frac{a^2\\,\\left(a+1\\right)\\,b^2\\,\\left(b+1\\right)}{\\left(b+a+\\frac{1}{2}\\right)^2\\,\\left(b+a+\\frac{3}{2}\\right)} \\right] \\]"
      ],
      "text/plain": [
       "                                                   2  2\n",
       "            2 a b      a (a + 1) b (b + 1)        a  b\n",
       "(%o3) [1, ---------, ----------------------- + ------------, \n",
       "                  1           1           3             1 2\n",
       "          b + a + -  (b + a + -) (b + a + -)   (b + a + -)\n",
       "                  2           2           2             2\n",
       "                                                        2          2\n",
       "               a (a + 1) (a + 2) b (b + 1) (b + 2)     a  (a + 1) b  (b + 1)\n",
       "              ------------------------------------- + ------------------------]\n",
       "                         1           3           5             1 2          3\n",
       "              3 (b + a + -) (b + a + -) (b + a + -)   (b + a + -)  (b + a + -)\n",
       "                         2           2           2             2            2"
      ],
      "text/x-maxima": [
       "[1,(2*a*b)/(b+a+1/2),\n",
       " (a*(a+1)*b*(b+1))/((b+a+1/2)*(b+a+3/2))+(a^2*b^2)/(b+a+1/2)^2,\n",
       " (a*(a+1)*(a+2)*b*(b+1)*(b+2))/(3*(b+a+1/2)*(b+a+3/2)*(b+a+5/2))\n",
       "  +(a^2*(a+1)*b^2*(b+1))/((b+a+1/2)^2*(b+a+3/2))]"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-UNIT-STEP in DEFUN\n",
      "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-POCHHAMMER in DEFUN\n"
     ]
    }
   ],
   "source": [
    "CC:[C[0],C[1],C[2],C[3]];"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "88bced85-319a-4671-9958-34a71f8766f9",
   "metadata": {},
   "source": [
    "同様に$${}_3F_2\\left(2\\,a,2\\,b,a+b;2\\,a+2\\,b,a+b+\\frac12;x\\right)=\\sum_{n=0}^{\\infty}{\\frac{(2\\,a)_n\\,(2\\,b)_n\\,(a+b)_n\\,x^n}{(2\\,a+2\\,b)_n\\,(a+b+\\frac12)_n\\,n!}}$$を思い出します。$D_n$をこの冪級数の$n$次の係数とします。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "cf62439e-672b-40c8-89bf-a08c69c0a55c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{4}$}D_{n}:=\\frac{\\left(\\left(2\\,a\\right)\\right)_{n}\\,\\left(\\left(2\\,b\\right)\\right)_{n}\\,\\left(\\left(a+b\\right)\\right)_{n}}{\\left(\\left(2\\,a+2\\,b\\right)\\right)_{n}\\,\\left(\\left(a+b+\\frac{1}{2}\\right)\\right)_{n}\\,n!}\\]"
      ],
      "text/plain": [
       "                            \"\"(2 a)  \"\"(2 b)  \"\"(a + b)\n",
       "                                   n        n          n\n",
       "(%o4)               D  := --------------------------------\n",
       "                     n                              1\n",
       "                          \"\"(2 a + 2 b)  \"\"(a + b + -)  n!\n",
       "                                       n            2 n"
      ],
      "text/x-maxima": [
       "D[n]:=(pochhammer(2*a,n)*pochhammer(2*b,n)*pochhammer(a+b,n))\n",
       "  /(pochhammer(2*a+2*b,n)*pochhammer(a+b+1/2,n)*n!)"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "D[n]:=pochhammer(2*a,n)*pochhammer(2*b,n)*pochhammer(a+b,n)/(pochhammer(2*a+2*b,n)*pochhammer(a+b+1/2,n)*n!);"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5a4f4f3e-921f-4e91-a174-e854a5fcbadd",
   "metadata": {},
   "source": [
    "実際に$D_0, D_1, D_2, D_3$を計算してみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "0f6229da-9b45-435c-8d37-74efb082aa16",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{5}$}\\left[ 1 , \\frac{4\\,a\\,b\\,\\left(b+a\\right)}{\\left(b+a+\\frac{1}{2}\\right)\\,\\left(2\\,b+2\\,a\\right)} , \\frac{2\\,a\\,\\left(2\\,a+1\\right)\\,b\\,\\left(b+a\\right)\\,\\left(b+a+1\\right)\\,\\left(2\\,b+1\\right)}{\\left(b+a+\\frac{1}{2}\\right)\\,\\left(b+a+\\frac{3}{2}\\right)\\,\\left(2\\,b+2\\,a\\right)\\,\\left(2\\,b+2\\,a+1\\right)} , \\frac{2\\,a\\,\\left(2\\,a+1\\right)\\,\\left(2\\,a+2\\right)\\,b\\,\\left(b+a\\right)\\,\\left(b+a+1\\right)\\,\\left(b+a+2\\right)\\,\\left(2\\,b+1\\right)\\,\\left(2\\,b+2\\right)}{3\\,\\left(b+a+\\frac{1}{2}\\right)\\,\\left(b+a+\\frac{3}{2}\\right)\\,\\left(b+a+\\frac{5}{2}\\right)\\,\\left(2\\,b+2\\,a\\right)\\,\\left(2\\,b+2\\,a+1\\right)\\,\\left(2\\,b+2\\,a+2\\right)} \\right] \\]"
      ],
      "text/plain": [
       "               4 a b (b + a)\n",
       "(%o5) [1, -----------------------, \n",
       "                   1\n",
       "          (b + a + -) (2 b + 2 a)\n",
       "                   2\n",
       "   2 a (2 a + 1) b (b + a) (b + a + 1) (2 b + 1)\n",
       "---------------------------------------------------, \n",
       "         1           3\n",
       "(b + a + -) (b + a + -) (2 b + 2 a) (2 b + 2 a + 1)\n",
       "         2           2\n",
       "(2 a (2 a + 1) (2 a + 2) b (b + a) (b + a + 1) (b + a + 2) (2 b + 1) (2 b + 2))\n",
       "             1           3           5\n",
       "/(3 (b + a + -) (b + a + -) (b + a + -) (2 b + 2 a) (2 b + 2 a + 1)\n",
       "             2           2           2\n",
       " (2 b + 2 a + 2))]"
      ],
      "text/x-maxima": [
       "[1,(4*a*b*(b+a))/((b+a+1/2)*(2*b+2*a)),\n",
       " (2*a*(2*a+1)*b*(b+a)*(b+a+1)*(2*b+1))/((b+a+1/2)*(b+a+3/2)*(2*b+2*a)\n",
       "                                                 *(2*b+2*a+1)),\n",
       " (2*a*(2*a+1)*(2*a+2)*b*(b+a)*(b+a+1)*(b+a+2)*(2*b+1)*(2*b+2))\n",
       "  /(3*(b+a+1/2)*(b+a+3/2)*(b+a+5/2)*(2*b+2*a)*(2*b+2*a+1)*(2*b+2*a+2))]"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "DD:[D[0],D[1],D[2],D[3]];"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "aafcb298-691a-4500-ad05-17aa6d28b37e",
   "metadata": {},
   "source": [
    "$CC=[C_0, C_1, C_2, C_3]$から$DD=[D_0, D_1, D_2, D_3]$を引き算します。リストの引き算では独立に要素ごとに計算されます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "4d5626cf-cf7c-4b08-81d6-9ae77388fc5e",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{6}$}\\left[ 0 , 0 , 0 , 0 \\right] \\]"
      ],
      "text/plain": [
       "(%o6)                            [0, 0, 0, 0]"
      ],
      "text/x-maxima": [
       "[0,0,0,0]"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "CC-DD,ratsimp;"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b8900a54-e4d4-4103-979d-b04a7829ba8e",
   "metadata": {},
   "source": [
    "これで定理の両辺のマクローリン展開の$0\\sim 3$次の係数が等しいことがわかり、証明は終了しました。"
   ]
  }
 ],
 "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"
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 },
 "nbformat": 4,
 "nbformat_minor": 5
}