{
"cells": [
{
"cell_type": "markdown",
"id": "174ca2ed-b340-4188-899e-0e322699ad82",
"metadata": {},
"source": [
"#### B.C.Berndt: Number Theory in the Spirit of Ramanujanより\n",
"\n",
"$0\\lt x \\lt 1$の範囲で関数$F(x)$を\n",
"$$F(x)=exp\\left(-\\pi\\,\\frac{{}_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;1-x\\right)}{{}_2 F_1\\left(\\frac{1}{2},\\frac{1}{2};1;x\\right)}\\right)$$\n",
"と定義しました(超幾何関数もFですが、引数の形が異なるので区別できると思います）。\n",
"
\n",
"
\n",
"**Chapter 5 Corollary 5.2.4**
\n",
"$m$を$0$以上の整数として、$n=2^m$であるとき、次式が成り立つ：
\n",
"$$F\\left(1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)^{n}=F\\left(1-\\frac{\\varphi\\left(-q^{n}\\right)^4}{\\varphi\\left(q^{n}\\right)^4}\\right)$$\n",
"\n",
"今回はこの補題を証明していきます。
\n",
"証明の方針としてまず今回の補題を少し一般化して、$F(a)=F(b)^n$ならば$F(1-a)^n=F(1-b)$が成り立つことを示します。これと前回証明したCorollary 5.2.3 $F\\left(\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)=F\\left(\\frac{\\varphi\\left(-q^{n}\\right)^4}{\\varphi\\left(q^{n}\\right)^4}\\right)^{n}$ を使えば証明は終了します。"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "fa4f9e20-93d8-4e03-8f5a-e613aeefbf3d",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{0}$}F\\left(x\\right):=\\exp \\left(\\frac{-\\pi\\,{\\it hypergeometric}\\left(\\left[ \\frac{1}{2} , \\frac{1}{2} \\right] , \\left[ 1 \\right] , 1-x\\right)}{{\\it hypergeometric}\\left(\\left[ \\frac{1}{2} , \\frac{1}{2} \\right] , \\left[ 1 \\right] , x\\right)}\\right)\\]"
],
"text/plain": [
" 1 1\n",
" - %pi hypergeometric([-, -], [1], 1 - x)\n",
" 2 2\n",
"(%o0) F(x) := exp(----------------------------------------)\n",
" 1 1\n",
" hypergeometric([-, -], [1], x)\n",
" 2 2"
],
"text/x-maxima": [
"F(x):=exp(\n",
" (-%pi*hypergeometric([1/2,1/2],[1],1-x))/hypergeometric([1/2,1/2],[1],x))"
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F(x):=exp(-%pi*hypergeometric([1/2,1/2],[1],1-x)/hypergeometric([1/2,1/2],[1],x));"
]
},
{
"cell_type": "markdown",
"id": "d7ef84c8-8498-49ee-b74e-86d2c928c654",
"metadata": {},
"source": [
"一般化したステートメントの仮定の部分の式からスタートして、結論の式を等式変形で導きます。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "ef505a56-d481-4eb1-8805-eee6f3a43068",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{1}$}F\\left(a\\right)=F\\left(b\\right)^{n}\\]"
],
"text/plain": [
" n\n",
"(%o1) F(a) = F (b)"
],
"text/x-maxima": [
"'F(a) = 'F(b)^n"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"'F(a)='F(b)^n;"
]
},
{
"cell_type": "markdown",
"id": "bad4bf40-1619-4a17-98ba-b9545fbff805",
"metadata": {},
"source": [
"定義に基づいて展開します。"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "6f204e35-b7ef-4cd3-b474-cc17d8beb71c",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{2}$}e^ {- \\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-a\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,a\\right)} }=e^ {- \\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-b\\right)\\,n}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,b\\right)} }\\]"
],
"text/plain": [
" %pi hypergeometric([1/2, 1/2], [1], 1 - a)\n",
" - ------------------------------------------\n",
" hypergeometric([1/2, 1/2], [1], a)\n",
"(%o2) %e = \n",
" %pi hypergeometric([1/2, 1/2], [1], 1 - b) n\n",
" - --------------------------------------------\n",
" hypergeometric([1/2, 1/2], [1], b)\n",
" %e"
],
"text/x-maxima": [
"%e^-((%pi*hypergeometric([1/2,1/2],[1],1-a))/hypergeometric([1/2,1/2],[1],a))\n",
" = %e^-((%pi*hypergeometric([1/2,1/2],[1],1-b)*n)\n",
" /hypergeometric([1/2,1/2],[1],b))"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
},
{
"name": "stdout",
"output_type": "stream",
"text": [
"SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-HYPERGEOMETRIC in DEFUN\n"
]
}
],
"source": [
"%,nouns;"
]
},
{
"cell_type": "markdown",
"id": "6fbbc36d-cce8-4ed5-8722-c39aea63a36c",
"metadata": {},
"source": [
"両辺の対数をとり、逆数をとり、さらに両辺に$n$をかけます。"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "78b27a8d-e870-4147-ba63-c8976078bd4e",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{3}$}-\\frac{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,a\\right)\\,n}{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-a\\right)}=-\\frac{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,b\\right)}{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-b\\right)}\\]"
],
"text/plain": [
" 1 1\n",
" hypergeometric([-, -], [1], a) n\n",
" 2 2\n",
"(%o3) - -------------------------------------- = \n",
" 1 1\n",
" %pi hypergeometric([-, -], [1], 1 - a)\n",
" 2 2\n",
" 1 1\n",
" hypergeometric([-, -], [1], b)\n",
" 2 2\n",
" - --------------------------------------\n",
" 1 1\n",
" %pi hypergeometric([-, -], [1], 1 - b)\n",
" 2 2"
],
"text/x-maxima": [
"-(hypergeometric([1/2,1/2],[1],a)*n)/(%pi*hypergeometric([1/2,1/2],[1],1-a))\n",
" = -hypergeometric([1/2,1/2],[1],b)/(%pi*hypergeometric([1/2,1/2],[1],1-b))"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"n/log(%);"
]
},
{
"cell_type": "markdown",
"id": "f920a90c-3abf-4b3f-be63-deadbfe433fa",
"metadata": {},
"source": [
"両辺に$\\pi^2$をかけてから、指数関数を適用します。"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "9a4253b6-2777-479a-9f6e-1e5ef67a1046",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{4}$}e^{-\\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,a\\right)\\,n}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-a\\right)}=-\\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,b\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-b\\right)}}\\]"
],
"text/plain": [
" 1 1\n",
" %pi hypergeometric([-, -], [1], a) n\n",
" 2 2\n",
"(%o4) expt(%e, - ------------------------------------ = \n",
" 1 1\n",
" hypergeometric([-, -], [1], 1 - a)\n",
" 2 2\n",
" 1 1\n",
" %pi hypergeometric([-, -], [1], b)\n",
" 2 2\n",
" - ----------------------------------)\n",
" 1 1\n",
" hypergeometric([-, -], [1], 1 - b)\n",
" 2 2"
],
"text/x-maxima": [
"%e^(-(%pi*hypergeometric([1/2,1/2],[1],a)*n)/hypergeometric([1/2,1/2],[1],1-a)\n",
" = -(%pi*hypergeometric([1/2,1/2],[1],b))\n",
" /hypergeometric([1/2,1/2],[1],1-b))"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"exp(%*%pi^2);"
]
},
{
"cell_type": "markdown",
"id": "2487da77-1cbd-4758-9cc4-3d62604fc30f",
"metadata": {},
"source": [
"目標とする式の左辺、右辺を表示すると一致していることがわかります。"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "10f1a969-6764-4a0d-99f4-16440be9bb03",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{5}$}\\left[ e^ {- \\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,a\\right)\\,n}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-a\\right)} } , e^ {- \\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,b\\right)}{F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,1-b\\right)} } \\right] \\]"
],
"text/plain": [
" %pi hypergeometric([1/2, 1/2], [1], a) n\n",
" - ----------------------------------------\n",
" hypergeometric([1/2, 1/2], [1], 1 - a)\n",
"(%o5) [%e , \n",
" %pi hypergeometric([1/2, 1/2], [1], b)\n",
" - --------------------------------------\n",
" hypergeometric([1/2, 1/2], [1], 1 - b)\n",
" %e ]"
],
"text/x-maxima": [
"[%e^-((%pi*hypergeometric([1/2,1/2],[1],a)*n)\n",
" /hypergeometric([1/2,1/2],[1],1-a)),\n",
" %e^-((%pi*hypergeometric([1/2,1/2],[1],b))\n",
" /hypergeometric([1/2,1/2],[1],1-b))]"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"[F(1-a)^n, F(1-b)];"
]
},
{
"cell_type": "markdown",
"id": "8064545f-1170-492d-8ff2-3c117ecba01e",
"metadata": {},
"source": [
"という訳で仮定から出発して結論の式を導くことが出来ました。"
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "732fd392-61f7-440b-9e27-a5bac77415d9",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{6}$}F\\left(1-a\\right)^{n}=F\\left(1-b\\right)\\]"
],
"text/plain": [
" n\n",
"(%o6) F (1 - a) = F(1 - b)"
],
"text/x-maxima": [
"'F(1-a)^n = 'F(1-b)"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"'F(1-a)^n='F(1-b);"
]
},
{
"cell_type": "markdown",
"id": "9aec020f-a2e5-4bba-813d-74226f5ca9e5",
"metadata": {},
"source": [
"$F\\left(a\\right)=F\\left(b\\right)^{n}$ならば$F\\left(1-a\\right)^{n}=F\\left(1-b\\right)$であることが示ました。
\n",
"最後のステップです。$a=\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}, b=\\frac{\\varphi\\left(-q^{n}\\right)^4}{\\varphi\\left(q^{n}\\right)^4}$と置きます。すると仮定の部分は$F\\left(\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)=F\\left(\\frac{\\varphi\\left(-q^{n}\\right)^4}{\\varphi\\left(q^{n}\\right)^4}\\right)^{n}$となります。これはCorollary 5.2.3そのものですから成立することがわかります。\n",
"\n",
"また結論の部分は$F\\left(1-\\frac{\\varphi\\left(-q\\right)^4}{\\varphi\\left(q\\right)^4}\\right)^{n}=F\\left(1-\\frac{\\varphi\\left(-q^{n}\\right)^4}{\\varphi\\left(q^{n}\\right)^4}\\right)$となります。これは今回証明したい式ですから、これで証明は終わりました。"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Maxima",
"language": "maxima",
"name": "maxima"
},
"language_info": {
"codemirror_mode": "maxima",
"file_extension": ".mac",
"mimetype": "text/x-maxima",
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