{
"cells": [
{
"cell_type": "markdown",
"id": "f615c107-08cc-4972-9e63-39e4958f2ad4",
"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",
"
\n",
"
\n",
"**Chapter 5 Lemma 5.2.7**
\n",
"$$_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)=\\varphi\\left(q\\right)^2$$\n",
"が成り立つ。
\n",
"\n",
"この証明にはLemma 5.2.6で証明した下記の式F1を起点として使います。"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "b6b7d136-0edc-4e2e-beeb-4a5e60640a7f",
"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)=\\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",
"(%o0) 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": 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/phi(q^2)^2*hypergeometric([1/2,1/2],[1],1-phi(-q^2)^4/phi(q^2)^4);"
]
},
{
"cell_type": "markdown",
"id": "92a8b0bb-21ec-4372-b4ee-c422d2f13290",
"metadata": {},
"source": [
"この等式の$q$を$q^2$で置き換えてみます。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "06591b07-6e25-4533-b899-44c2021b4401",
"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(-q^2\\right)^4}{\\varphi\\left(q^2\\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^4\\right)^4}{\\varphi\\left(q^4\\right)^4}\\right)\\,\\varphi\\left(q^2\\right)^2}{\\varphi\\left(q^4\\right)^2}\\]"
],
"text/plain": [
" 4 2\n",
" 1 1 phi (- q )\n",
"(%o1) hypergeometric([-, -], [1], 1 - ----------) = \n",
" 2 2 4 2\n",
" phi (q )\n",
" 4 4\n",
" 1 1 phi (- q ) 2 2\n",
" hypergeometric([-, -], [1], 1 - ----------) phi (q )\n",
" 2 2 4 4\n",
" phi (q )\n",
" ----------------------------------------------------\n",
" 2 4\n",
" phi (q )"
],
"text/x-maxima": [
"hypergeometric([1/2,1/2],[1],1-phi(-q^2)^4/phi(q^2)^4)\n",
" = (hypergeometric([1/2,1/2],[1],1-phi(-q^4)^4/phi(q^4)^4)*phi(q^2)^2)\n",
" /phi(q^4)^2"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F2:subst(q^2,q,F1);"
]
},
{
"cell_type": "markdown",
"id": "da25d9cb-481a-45f4-90c2-3417ae0ea84c",
"metadata": {},
"source": [
"得られた式を使って元の式F1の右辺に現れる超幾何関数を書き換えます。"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "2a2bc211-e9d8-41df-9e7d-7aad182a9f3e",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{2}$}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^4\\right)^4}{\\varphi\\left(q^4\\right)^4}\\right)\\,\\varphi\\left(q\\right)^2}{\\varphi\\left(q^4\\right)^2}\\]"
],
"text/plain": [
" 4\n",
" 1 1 phi (- q)\n",
"(%o2) hypergeometric([-, -], [1], 1 - ---------) = \n",
" 2 2 4\n",
" phi (q)\n",
" 4 4\n",
" 1 1 phi (- q ) 2\n",
" hypergeometric([-, -], [1], 1 - ----------) phi (q)\n",
" 2 2 4 4\n",
" phi (q )\n",
" ---------------------------------------------------\n",
" 2 4\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^4)^4/phi(q^4)^4)*phi(q)^2)\n",
" /phi(q^4)^2"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F3:F1,F2;"
]
},
{
"cell_type": "markdown",
"id": "491c46a4-86a3-4364-9b71-eaae5ddf5c94",
"metadata": {},
"source": [
"元の等式F1の$q$を全て$q^4$で置き換えます。"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "1930bb82-a004-44a3-bf8c-cbc9810df5b5",
"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-\\frac{\\varphi\\left(-q^4\\right)^4}{\\varphi\\left(q^4\\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^8\\right)^4}{\\varphi\\left(q^8\\right)^4}\\right)\\,\\varphi\\left(q^4\\right)^2}{\\varphi\\left(q^8\\right)^2}\\]"
],
"text/plain": [
" 4 4\n",
" 1 1 phi (- q )\n",
"(%o3) hypergeometric([-, -], [1], 1 - ----------) = \n",
" 2 2 4 4\n",
" phi (q )\n",
" 4 8\n",
" 1 1 phi (- q ) 2 4\n",
" hypergeometric([-, -], [1], 1 - ----------) phi (q )\n",
" 2 2 4 8\n",
" phi (q )\n",
" ----------------------------------------------------\n",
" 2 8\n",
" phi (q )"
],
"text/x-maxima": [
"hypergeometric([1/2,1/2],[1],1-phi(-q^4)^4/phi(q^4)^4)\n",
" = (hypergeometric([1/2,1/2],[1],1-phi(-q^8)^4/phi(q^8)^4)*phi(q^4)^2)\n",
" /phi(q^8)^2"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F4:subst(q^4,q,F1);"
]
},
{
"cell_type": "markdown",
"id": "b5beb6c7-6016-46e6-b596-846729c3210c",
"metadata": {},
"source": [
"得られた式F4を使って等式F3の右辺の超幾何関数を書き換えます。"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "5c3f6946-a5c5-43f0-80ff-d377ea510500",
"metadata": {},
"outputs": [
{
"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{\\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^8\\right)^4}{\\varphi\\left(q^8\\right)^4}\\right)\\,\\varphi\\left(q\\right)^2}{\\varphi\\left(q^8\\right)^2}\\]"
],
"text/plain": [
" 4\n",
" 1 1 phi (- q)\n",
"(%o4) hypergeometric([-, -], [1], 1 - ---------) = \n",
" 2 2 4\n",
" phi (q)\n",
" 4 8\n",
" 1 1 phi (- q ) 2\n",
" hypergeometric([-, -], [1], 1 - ----------) phi (q)\n",
" 2 2 4 8\n",
" phi (q )\n",
" ---------------------------------------------------\n",
" 2 8\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^8)^4/phi(q^8)^4)*phi(q)^2)\n",
" /phi(q^8)^2"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F5:F3,F4;"
]
},
{
"cell_type": "markdown",
"id": "3c7d1d55-742a-41ac-a15e-664154a70f4e",
"metadata": {},
"source": [
"この操作を繰り返すことにより$m$を自然数として（正確には数学的帰納法で）次の式は明らかです。"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "70a4a929-20e3-4196-82cd-b2ad092bca86",
"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{\\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^{m}}\\right)^4}{\\varphi\\left(q^{2^{m}}\\right)^4}\\right)\\,\\varphi\\left(q\\right)^2}{\\varphi\\left(q^{2^{m}}\\right)^2}\\]"
],
"text/plain": [
" 4\n",
" 1 1 phi (- q)\n",
"(%o5) hypergeometric([-, -], [1], 1 - ---------) = \n",
" 2 2 4\n",
" phi (q)\n",
" m\n",
" 4 2\n",
" 1 1 phi (- q ) 2\n",
" hypergeometric([-, -], [1], 1 - -----------) phi (q)\n",
" 2 2 m\n",
" 4 2\n",
" phi (q )\n",
" ----------------------------------------------------\n",
" m\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^m)^4/phi(q^2^m)^4)*phi(q)^2)\n",
" /phi(q^2^m)^2"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F6:subst(2^m,8,F5);"
]
},
{
"cell_type": "markdown",
"id": "21cbfd5e-3987-4ead-9e9f-01a2e5b46202",
"metadata": {},
"source": [
"$m\\rightarrow \\infty$とすると$q^{2^m}\\rightarrow 0$となり、$\\varphi(\\pm{q^{2^m}})\\rightarrow \\varphi(0)=1$となります。これと、"
]
},
{
"cell_type": "code",
"execution_count": 10,
"id": "d7b5269c-d936-46c8-923e-b05915252dbc",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{9}$}{\\it hypergeometric}\\left(\\left[ \\frac{1}{2} , \\frac{1}{2} \\right] , \\left[ 1 \\right] , 0\\right)=1\\]"
],
"text/plain": [
" 1 1\n",
"(%o9) hypergeometric([-, -], [1], 0) = 1\n",
" 2 2"
],
"text/x-maxima": [
"'hypergeometric([1/2,1/2],[1],0) = 1"
]
},
"execution_count": 10,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"'hypergeometric([1/2,1/2],[1],0)=hypergeometric([1/2,1/2],[1],0);"
]
},
{
"cell_type": "markdown",
"id": "4af078c5-006f-4e0f-a4f5-5e905028322d",
"metadata": {},
"source": [
"および超幾何関数とテータ関数の必要な点での連続性により、"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "1d898bcb-4e1e-4423-b0b7-62edeb6b4bd6",
"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)=\\varphi\\left(q\\right)^2\\]"
],
"text/plain": [
" 4\n",
" 1 1 phi (- q) 2\n",
"(%o7) 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": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"lhs(F6)=phi(q)^2;"
]
},
{
"cell_type": "markdown",
"id": "c4d045a3-3157-40b4-83fc-bde6d3d5c62a",
"metadata": {},
"source": [
"が得られます。"
]
}
],
"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
}