{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "超幾何関数に対するオイラー積分表示の証明の続きです。証明したいのは下記の(%o3)の式です。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{1}$}\\left[ z<1 \\right] \\]"
      ],
      "text/plain": [
       "(%o1)                               [z < 1]"
      ],
      "text/x-maxima": [
       "[z < 1]"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{2}$}\\left[ z>-1 \\right] \\]"
      ],
      "text/plain": [
       "(%o2)                              [z > - 1]"
      ],
      "text/x-maxima": [
       "[z > -1]"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{3}$}F\\left( \\left. \\begin{array}{c}a,\\;b\\\\c\\end{array} \\right |,z\\right)=\\frac{\\Gamma\\left(c\\right)\\,\\int_{0}^{1}{\\frac{\\left(1-t\\right)^{c-a-1}\\,t^{a-1}}{\\left(1-t\\,z\\right)^{b}}\\;dt}}{\\Gamma\\left(a\\right)\\,\\Gamma\\left(c-a\\right)}\\]"
      ],
      "text/plain": [
       "                                                 1\n",
       "                                                /         c - a - 1  a - 1\n",
       "                                                [  (1 - t)          t\n",
       "                                       gamma(c) I  ----------------------- dt\n",
       "                                                ]                 b\n",
       "                                                /        (1 - t z)\n",
       "                                                 0\n",
       "(%o3) hypergeometric([a, b], [c], z) = --------------------------------------\n",
       "                                               gamma(a) gamma(c - a)"
      ],
      "text/x-maxima": [
       "hypergeometric([a,b],[c],z) = (gamma(c)*'integrate(\n",
       "                                        ((1-t)^(c-a-1)*t^(a-1))/(1-t*z)^b,t,0,\n",
       "                                        1))\n",
       "                            /(gamma(a)*gamma(c-a))"
      ]
     },
     "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(z-1<0);\n",
    "assume(-1<z);\n",
    "F1:hypergeometric([a,b],[c],z)=gamma(c)/(gamma(c-a)*gamma(a))*'integrate(t^(a-1)*(1-t)^(c-a-1)*(1-t*z)^(-b),t,0,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "前回は、上記の右辺の積分について、下記の式まで証明しました。引き続き右辺の式変形を行います。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{4}$}\\int_{0}^{1}{\\frac{\\left(1-t\\right)^{c-a-1}\\,t^{a-1}}{\\left(1-t\\,z\\right)^{b}}\\;dt}=\\int_{0}^{1}{\\left(1-t\\right)^{c-a-1}\\,t^{a-1}\\,\\sum_{n=0}^{\\infty }{\\frac{\\left(b\\right)_{n}\\,t^{n}\\,z^{n}}{n!}}\\;dt}\\]"
      ],
      "text/plain": [
       "       1                               1\n",
       "      /         c - a - 1  a - 1      /\n",
       "      [  (1 - t)          t           [         c - a - 1  a - 1\n",
       "(%o4) I  ----------------------- dt = I  (1 - t)          t\n",
       "      ]                 b             ]\n",
       "      /        (1 - t z)              /\n",
       "       0                               0\n",
       "                                                            inf         n  n\n",
       "                                                            ====  (b)  t  z\n",
       "                                                            \\        n\n",
       "                                                             >    ---------- dt\n",
       "                                                            /         n!\n",
       "                                                            ====\n",
       "                                                            n = 0"
      ],
      "text/x-maxima": [
       "'integrate(((1-t)^(c-a-1)*t^(a-1))/(1-t*z)^b,t,0,1)\n",
       "  = 'integrate((1-t)^(c-a-1)*t^(a-1)\n",
       "                            *'sum((pochhammer(b,n)*t^n*z^n)/n!,n,0,inf),t,0,1)"
      ]
     },
     "execution_count": 2,
     "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": [
    "F2:'integrate(t^(a-1)*(1-t)^(c-a-1)*(1-t*z)^(-b),t,0,1)='integrate(t^(a-1)*(1-t)^(c-a-1)*sum(pochhammer(b,n)/n!*(z*t)^n,n,0,inf),t,0,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "総和記号の外側の積の項を内側に移し、積分と総和を入れ替えます（この入れ替えには収束の議論が別途必要ですがそこは省略です）。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{5}$}\\int_{0}^{1}{\\sum_{n=0}^{\\infty }{\\frac{\\left(b\\right)_{n}\\,\\left(1-t\\right)^{c-a-1}\\,t^{n+a-1}\\,z^{n}}{n!}}\\;dt}\\]"
      ],
      "text/plain": [
       "                 1 inf               c - a - 1  n + a - 1  n\n",
       "                /  ====  (b)  (1 - t)          t          z\n",
       "                [  \\        n\n",
       "(%o5)           I   >    ----------------------------------- dt\n",
       "                ]  /                     n!\n",
       "                /  ====\n",
       "                 0 n = 0"
      ],
      "text/x-maxima": [
       "'integrate('sum((pochhammer(b,n)*(1-t)^(c-a-1)*t^(n+a-1)*z^n)/n!,n,0,inf),t,0,\n",
       "           1)"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F3:intosum(rhs(F2));"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{6}$}\\sum_{n=0}^{\\infty }{\\frac{\\left(b\\right)_{n}\\,\\left(\\int_{0}^{1}{\\left(1-t\\right)^{c-a-1}\\,t^{n+a-1}\\;dt}\\right)\\,z^{n}}{n!}}\\]"
      ],
      "text/plain": [
       "                            1\n",
       "                           /\n",
       "                           [         c - a - 1  n + a - 1      n\n",
       "                     (b)  (I  (1 - t)          t          dt) z\n",
       "               inf      n  ]\n",
       "               ====        /\n",
       "               \\            0\n",
       "(%o6)           >    -------------------------------------------\n",
       "               /                         n!\n",
       "               ====\n",
       "               n = 0"
      ],
      "text/x-maxima": [
       "'sum((pochhammer(b,n)*('integrate((1-t)^(c-a-1)*t^(n+a-1),t,0,1))*z^n)/n!,n,0,\n",
       "     inf)"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F4:sum('integrate(part(F3,1,1),t,0,1),n,0,inf);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この積分はベータ関数になっていることがわかります。実際、ベータ関数の定義は\n",
    "$$\\beta (x,y)= \\int_0^1 {(1-t)^{x-1}\\,t^{y-1}}$$\n",
    "です。この積分をベータ関数に書き直せば、ベータ関数に関する恒等式を使えるようになります。そこでこの積分をベータ関数に書き直すルールを定義しましょう。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{7}$}\\mathbf{done}\\]"
      ],
      "text/plain": [
       "(%o7)                                done"
      ],
      "text/x-maxima": [
       "done"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{8}$}{\\it int\\_to\\_beta}:\\int_{0}^{1}{\\left(1-t\\right)^{{\\it exp}_{1}-1}\\,t^{{\\it exp}_{2}-1}\\;dt}\\rightarrow \\beta\\left({\\it exp}_{1} , {\\it exp}_{2}\\right)\\]"
      ],
      "text/plain": [
       "                      1\n",
       "                     /\n",
       "                     [         exp1 - 1  exp2 - 1\n",
       "(%o8)  int_to_beta : I  (1 - t)         t         dt -> beta(exp1, exp2)\n",
       "                     ]\n",
       "                     /\n",
       "                      0"
      ],
      "text/x-maxima": [
       "int_to_beta:marrow('integrate((1-t)^(exp1-1)*t^(exp2-1),t,0,1),\n",
       "                   beta(exp1,exp2))"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "matchdeclare([exp1,exp2],true);\n",
    "defrule(int_to_beta,'integrate((1-t)^(exp1-1)*t^(exp2-1),t,0,1), beta(exp1,exp2));"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "%o6にこのルールを適用することで次式を得ます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{9}$}\\sum_{n=0}^{\\infty }{\\frac{\\left(b\\right)_{n}\\,\\beta\\left(c-a , n+a\\right)\\,z^{n}}{n!}}\\]"
      ],
      "text/plain": [
       "                       inf                            n\n",
       "                       ====  (b)  beta(c - a, n + a) z\n",
       "                       \\        n\n",
       "(%o9)                   >    --------------------------\n",
       "                       /                 n!\n",
       "                       ====\n",
       "                       n = 0"
      ],
      "text/x-maxima": [
       "'sum((pochhammer(b,n)*beta(c-a,n+a)*z^n)/n!,n,0,inf)"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F5:apply1(F4,int_to_beta);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この式にmakegamma関数を適用するとポッホハマー記号、ベータ関数、階乗が全てガンマ関数で表示されます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{10}$}\\frac{\\Gamma\\left(c-a\\right)\\,\\sum_{n=0}^{\\infty }{\\frac{\\Gamma\\left(n+a\\right)\\,\\Gamma\\left(n+b\\right)\\,z^{n}}{\\Gamma\\left(n+1\\right)\\,\\Gamma\\left(n+c\\right)}}}{\\Gamma\\left(b\\right)}\\]"
      ],
      "text/plain": [
       "                             inf\n",
       "                             ====                             n\n",
       "                             \\     gamma(n + a) gamma(n + b) z\n",
       "                gamma(c - a)  >    ----------------------------\n",
       "                             /      gamma(n + 1) gamma(n + c)\n",
       "                             ====\n",
       "                             n = 0\n",
       "(%o10)          -----------------------------------------------\n",
       "                                   gamma(b)"
      ],
      "text/x-maxima": [
       "(gamma(c-a)*'sum((gamma(n+a)*gamma(n+b)*z^n)/(gamma(n+1)*gamma(n+c)),n,0,inf))\n",
       " /gamma(b)"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F6:makegamma(F5);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{11}$}\\sum_{n=0}^{\\infty }{\\frac{\\Gamma\\left(c-a\\right)\\,\\Gamma\\left(n+a\\right)\\,\\Gamma\\left(n+b\\right)\\,z^{n}}{\\Gamma\\left(b\\right)\\,\\Gamma\\left(n+1\\right)\\,\\Gamma\\left(n+c\\right)}}\\]"
      ],
      "text/plain": [
       "                inf\n",
       "                ====                                          n\n",
       "                \\     gamma(c - a) gamma(n + a) gamma(n + b) z\n",
       "(%o11)           >    -----------------------------------------\n",
       "                /        gamma(b) gamma(n + 1) gamma(n + c)\n",
       "                ====\n",
       "                n = 0"
      ],
      "text/x-maxima": [
       "'sum((gamma(c-a)*gamma(n+a)*gamma(n+b)*z^n)/(gamma(b)*gamma(n+1)*gamma(n+c)),\n",
       "     n,0,inf)"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F7:intosum(F6);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "あと一息です。(%o3)を振り返ってみると、右辺の積分が(%o11)と等しいことがわかりました。左辺を定義の級数の形で書き、右辺は積分のところを上記の総和で置き換えると、証明するべき式は次式であること、あるいはそれを少し変形した(%o13)であることがわかります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{12}$}\\sum_{n=0}^{\\infty }{\\frac{\\left(a\\right)_{n}\\,\\left(b\\right)_{n}\\,z^{n}}{\\left(c\\right)_{n}\\,n!}}=\\frac{\\Gamma\\left(c\\right)\\,\\sum_{n=0}^{\\infty }{\\frac{\\Gamma\\left(c-a\\right)\\,\\Gamma\\left(n+a\\right)\\,\\Gamma\\left(n+b\\right)\\,z^{n}}{\\Gamma\\left(b\\right)\\,\\Gamma\\left(n+1\\right)\\,\\Gamma\\left(n+c\\right)}}}{\\Gamma\\left(a\\right)\\,\\Gamma\\left(c-a\\right)}\\]"
      ],
      "text/plain": [
       "       inf              n\n",
       "       ====  (a)  (b)  z\n",
       "       \\        n    n\n",
       "(%o12)  >    ------------ = \n",
       "       /       (c)  n!\n",
       "       ====       n\n",
       "       n = 0\n",
       "                                inf\n",
       "                                ====                                          n\n",
       "                                \\     gamma(c - a) gamma(n + a) gamma(n + b) z\n",
       "                       gamma(c)  >    -----------------------------------------\n",
       "                                /        gamma(b) gamma(n + 1) gamma(n + c)\n",
       "                                ====\n",
       "                                n = 0\n",
       "                       --------------------------------------------------------\n",
       "                                        gamma(a) gamma(c - a)"
      ],
      "text/x-maxima": [
       "'sum((pochhammer(a,n)*pochhammer(b,n)*z^n)/(pochhammer(c,n)*n!),n,0,inf)\n",
       "  = (gamma(c)*'sum((gamma(c-a)*gamma(n+a)*gamma(n+b)*z^n)\n",
       "                    /(gamma(b)*gamma(n+1)*gamma(n+c)),n,0,inf))\n",
       "  /(gamma(a)*gamma(c-a))"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F8:sum(pochhammer(a,n)*pochhammer(b,n)/pochhammer(c,n)*z^n/n!,n,0,inf)=gamma(c)/gamma(a)/gamma(c-a)*F7;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{13}$}\\sum_{n=0}^{\\infty }{\\frac{\\left(a\\right)_{n}\\,\\left(b\\right)_{n}\\,z^{n}}{\\left(c\\right)_{n}\\,n!}}=\\sum_{n=0}^{\\infty }{\\frac{\\Gamma\\left(c\\right)\\,\\Gamma\\left(n+a\\right)\\,\\Gamma\\left(n+b\\right)\\,z^{n}}{\\Gamma\\left(a\\right)\\,\\Gamma\\left(b\\right)\\,\\Gamma\\left(n+1\\right)\\,\\Gamma\\left(n+c\\right)}}\\]"
      ],
      "text/plain": [
       "       inf              n   inf\n",
       "       ====  (a)  (b)  z    ====                                         n\n",
       "       \\        n    n      \\        gamma(c) gamma(n + a) gamma(n + b) z\n",
       "(%o13)  >    ------------ =  >    -------------------------------------------\n",
       "       /       (c)  n!      /     gamma(a) gamma(b) gamma(n + 1) gamma(n + c)\n",
       "       ====       n         ====\n",
       "       n = 0                n = 0"
      ],
      "text/x-maxima": [
       "'sum((pochhammer(a,n)*pochhammer(b,n)*z^n)/(pochhammer(c,n)*n!),n,0,inf)\n",
       "  = 'sum((gamma(c)*gamma(n+a)*gamma(n+b)*z^n)\n",
       "          /(gamma(a)*gamma(b)*gamma(n+1)*gamma(n+c)),n,0,inf)"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F9:lhs(F8)=intosum(rhs(F8));"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "収束条件を満たす任意の$z$について両辺が等しいためには、両辺の総和の各項の係数が全て等しいことが十分条件です。そこで係数の差を計算してみましょう。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{14}$}\\frac{\\left(a\\right)_{n}\\,\\left(b\\right)_{n}}{\\left(c\\right)_{n}\\,n!}-\\frac{\\Gamma\\left(c\\right)\\,\\Gamma\\left(n+a\\right)\\,\\Gamma\\left(n+b\\right)}{\\Gamma\\left(a\\right)\\,\\Gamma\\left(b\\right)\\,\\Gamma\\left(n+1\\right)\\,\\Gamma\\left(n+c\\right)}\\]"
      ],
      "text/plain": [
       "            (a)  (b)\n",
       "               n    n       gamma(c) gamma(n + a) gamma(n + b)\n",
       "(%o14)      --------- - -------------------------------------------\n",
       "             (c)  n!    gamma(a) gamma(b) gamma(n + 1) gamma(n + c)\n",
       "                n"
      ],
      "text/x-maxima": [
       "(pochhammer(a,n)*pochhammer(b,n))/(pochhammer(c,n)*n!)\n",
       " -(gamma(c)*gamma(n+a)*gamma(n+b))/(gamma(a)*gamma(b)*gamma(n+1)*gamma(n+c))"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F10:part(F9,1,1)/z^n-part(F9,2,1)/z^n;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この式のポッホハマー記号、階乗を全てガンマ関数に書き直して簡約してみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{15}$}0\\]"
      ],
      "text/plain": [
       "(%o15)                                 0"
      ],
      "text/x-maxima": [
       "0"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "makegamma(F10);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "というわけで(%o13)、従って\n",
    "$$F\\left( \\left. \\begin{array}{c}a,\\;b\\\\c\\end{array} \\right |,z\\right)=\\frac{\\Gamma\\left(c\\right)\\,\\int_{0}^{1}{\\frac{\\left(1-t\\right)^{c-a-1}\\,t^{a-1}}{\\left(1-t\\,z\\right)^{b}}\\;dt}}{\\Gamma\\left(a\\right)\\,\\Gamma\\left(c-a\\right)}$$\n",
    "が証明できました！"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
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