{
"cells": [
{
"cell_type": "markdown",
"id": "b9aaf0b9-569b-4f24-8c1a-bece2b7032ba",
"metadata": {},
"source": [
"定理 $P(q)=1-24\\,\\sum_{n=1}^{\\infty}\\frac{n\\,q^n}{1-q^n},\\,a\\,b=\\pi^2$として、\n",
"$$6-a\\,P\\left(e^ {- 2\\,a }\\right)=b\\,P\\left(e^ {- 2\\,b }\\right)$$\n",
"が成り立つ。
\n",
"\n",
"\n",
"\n",
"証明はこの式の両辺の対数をとり$a$で微分して行きます。\n",
"$$a^{\\frac{1}{4}}\\,e^ {- \\frac{a}{12} }\\,\\prod_{n=1}^{\\infty }{\\left(1-e^ {- 2\\,a\\,n }\\right)}=b^{\\frac{1}{4}}\\,e^ {- \\frac{b}{12} }\\,\\prod_{n=1}^{\\infty }{\\left(1-e^ {- 2\\,b\\,n }\\right)}$$"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "c16e6ca9-34e9-40fd-a300-7e8e79f556ec",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{0}$}a^{\\frac{1}{4}}\\,e^ {- \\frac{a}{12} }\\,\\prod_{n=1}^{\\infty }{\\left(1-e^ {- 2\\,a\\,n }\\right)}=b^{\\frac{1}{4}}\\,e^ {- \\frac{b}{12} }\\,\\prod_{n=1}^{\\infty }{\\left(1-e^ {- 2\\,b\\,n }\\right)}\\]"
],
"text/plain": [
" inf inf\n",
" /===\\ /===\\\n",
" 1/4 - a/12 ! ! - 2 a n 1/4 - b/12 ! ! - 2 b n\n",
"(%o0) a %e ! ! (1 - %e ) = b %e ! ! (1 - %e )\n",
" ! ! ! !\n",
" n = 1 n = 1"
],
"text/x-maxima": [
"a^(1/4)*%e^-(a/12)*'product(1-%e^-(2*a*n),n,1,inf)\n",
" = b^(1/4)*%e^-(b/12)*'product(1-%e^-(2*b*n),n,1,inf)"
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A:a^(1/4)*exp(-a/12)*f(-exp(-2*a))=b^(1/4)*exp(-b/12)*f(-exp(-2*b)),f(q):=product(1-(-q)^n,n,1,inf);"
]
},
{
"cell_type": "markdown",
"id": "52a21136-d489-4487-b904-66adebc6cf85",
"metadata": {},
"source": [
"早速対数を取ります。その際に$logexpand$フラグを$all$に設定すると、総積も総和にしてくれます。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "41bec5b3-2011-452d-a405-7d520fb034d4",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{1}$}\\sum_{n=1}^{\\infty }{\\log \\left(1-e^ {- 2\\,a\\,n }\\right)}+\\frac{\\log a}{4}-\\frac{a}{12}=\\sum_{n=1}^{\\infty }{\\log \\left(1-e^ {- 2\\,b\\,n }\\right)}+\\frac{\\log b}{4}-\\frac{b}{12}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" \\ - 2 a n log(a) a\n",
"(%o1) > log(1 - %e ) + ------ - -- = \n",
" / 4 12\n",
" ====\n",
" n = 1\n",
" inf\n",
" ====\n",
" \\ - 2 b n log(b) b\n",
" > log(1 - %e ) + ------ - --\n",
" / 4 12\n",
" ====\n",
" n = 1"
],
"text/x-maxima": [
"'sum(log(1-%e^-(2*a*n)),n,1,inf)+log(a)/4-a/12\n",
" = 'sum(log(1-%e^-(2*b*n)),n,1,inf)+log(b)/4-b/12"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"B:log(A),logexpand:all;"
]
},
{
"cell_type": "markdown",
"id": "f2b12d21-1e04-48c1-b8a0-7c9d8f6b000f",
"metadata": {},
"source": [
"次に両辺を$a$で微分するのですが、その前に変数$b$が$a$に依存していることを宣言します。これによって右辺の$b$をそのままで$a$による微分を進めることが出来ます。"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "7e82aa4e-4296-4e75-8d0a-f4ed200a2545",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{3}$}2\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,a\\,n }}{1-e^ {- 2\\,a\\,n }}}+\\frac{1}{4\\,a}-\\frac{1}{12}=2\\,\\left(\\frac{d}{d\\,a}\\,b\\right)\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,b\\,n }}{1-e^ {- 2\\,b\\,n }}}+\\frac{\\frac{d}{d\\,a}\\,b}{4\\,b}-\\frac{\\frac{d}{d\\,a}\\,b}{12}\\]"
],
"text/plain": [
" inf inf db db\n",
" ==== - 2 a n ==== - 2 b n -- --\n",
" \\ n %e 1 1 db \\ n %e da da\n",
"(%o3) 2 > ------------- + --- - -- = 2 -- > ------------- + --- - --\n",
" / - 2 a n 4 a 12 da / - 2 b n 4 b 12\n",
" ==== 1 - %e ==== 1 - %e\n",
" n = 1 n = 1"
],
"text/x-maxima": [
"2*'sum((n*%e^-(2*a*n))/(1-%e^-(2*a*n)),n,1,inf)+1/(4*a)-1/12\n",
" = 2*'diff(b,a,1)*'sum((n*%e^-(2*b*n))/(1-%e^-(2*b*n)),n,1,inf)\n",
" +'diff(b,a,1)/(4*b)-'diff(b,a,1)/12"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"depends(b,a)$\n",
"C:diff(B,a);"
]
},
{
"cell_type": "markdown",
"id": "be6c22b4-161a-4582-b4be-5ccd152c79c7",
"metadata": {},
"source": [
"微分の結果の右辺には$\\frac{d}{d\\,a}\\,b$が残っています。これを$b=\\frac{\\pi^2}{a}$より計算してその結果で置き換えてみます。"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "eac6bc09-2f0e-4324-96c4-e580c2ccb0c2",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{4}$}2\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,a\\,n }}{1-e^ {- 2\\,a\\,n }}}+\\frac{1}{4\\,a}-\\frac{1}{12}=-\\frac{2\\,\\pi^2\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,b\\,n }}{1-e^ {- 2\\,b\\,n }}}}{a^2}-\\frac{\\pi^2}{4\\,a^2\\,b}+\\frac{\\pi^2}{12\\,a^2}\\]"
],
"text/plain": [
" inf\n",
" ==== - 2 a n\n",
" \\ n %e 1 1\n",
"(%o4) 2 > ------------- + --- - -- = \n",
" / - 2 a n 4 a 12\n",
" ==== 1 - %e\n",
" n = 1\n",
" inf\n",
" ==== - 2 b n\n",
" 2 \\ n %e\n",
" 2 %pi > -------------\n",
" / - 2 b n\n",
" ==== 1 - %e 2 2\n",
" n = 1 %pi %pi\n",
" (- --------------------------) - ------ + -----\n",
" 2 2 2\n",
" a 4 a b 12 a"
],
"text/x-maxima": [
"2*'sum((n*%e^-(2*a*n))/(1-%e^-(2*a*n)),n,1,inf)+1/(4*a)-1/12\n",
" = (-(2*%pi^2*'sum((n*%e^-(2*b*n))/(1-%e^-(2*b*n)),n,1,inf))/a^2)\n",
" -%pi^2/(4*a^2*b)+%pi^2/(12*a^2)"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"D:subst(diff(%pi^2/a,a),diff(b,a),C);"
]
},
{
"cell_type": "markdown",
"id": "0cd00e28-c908-4f38-96d8-8b66a6bfb6da",
"metadata": {},
"source": [
"$\\pi^2$を全て$a\\,b$で置き換えます。"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "92d6932d-bdb5-40ad-93f5-4cbd9810f481",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{5}$}2\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,a\\,n }}{1-e^ {- 2\\,a\\,n }}}+\\frac{1}{4\\,a}-\\frac{1}{12}=-\\frac{2\\,b\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,b\\,n }}{1-e^ {- 2\\,b\\,n }}}}{a}+\\frac{b}{12\\,a}-\\frac{1}{4\\,a}\\]"
],
"text/plain": [
" inf\n",
" ==== - 2 a n\n",
" \\ n %e 1 1\n",
"(%o5) 2 > ------------- + --- - -- = \n",
" / - 2 a n 4 a 12\n",
" ==== 1 - %e\n",
" n = 1\n",
" inf\n",
" ==== - 2 b n\n",
" \\ n %e\n",
" 2 b > -------------\n",
" / - 2 b n\n",
" ==== 1 - %e\n",
" n = 1 b 1\n",
" (- -----------------------) + ---- - ---\n",
" a 12 a 4 a"
],
"text/x-maxima": [
"2*'sum((n*%e^-(2*a*n))/(1-%e^-(2*a*n)),n,1,inf)+1/(4*a)-1/12\n",
" = (-(2*b*'sum((n*%e^-(2*b*n))/(1-%e^-(2*b*n)),n,1,inf))/a)+b/(12*a)-1/(4*a)"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"E:D,%pi^2=a*b;"
]
},
{
"cell_type": "markdown",
"id": "4ed566f1-c8e9-41f6-a0de-53fdc5b5b7a6",
"metadata": {},
"source": [
"両辺に$12\\,a$をかけて整理します。"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "17da1dd0-012a-44de-a929-43c6b3d7c149",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{6}$}24\\,a\\,\\sum_{n=1}^{\\infty }{\\frac{n}{e^{2\\,a\\,n}-1}}-a+3=-24\\,b\\,\\sum_{n=1}^{\\infty }{\\frac{n}{e^{2\\,b\\,n}-1}}+b-3\\]"
],
"text/plain": [
" inf inf\n",
" ==== ====\n",
" \\ n \\ n\n",
"(%o6) 24 a > ----------- - a + 3 = (- 24 b > -----------) + b - 3\n",
" / 2 a n / 2 b n\n",
" ==== %e - 1 ==== %e - 1\n",
" n = 1 n = 1"
],
"text/x-maxima": [
"24*a*'sum(n/(%e^(2*a*n)-1),n,1,inf)-a+3 = (-24*b\n",
" *'sum(n/(%e^(2*b*n)-1),n,1,inf))\n",
" +b-3"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F:E*12*a,expand;"
]
},
{
"cell_type": "markdown",
"id": "a3a4ed36-c123-4179-8570-5274ef2c2be8",
"metadata": {},
"source": [
"両辺に$3$を足します。"
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "cb540e97-bce5-4afc-8ae0-6ec4dbcfeceb",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{7}$}24\\,a\\,\\sum_{n=1}^{\\infty }{\\frac{n}{e^{2\\,a\\,n}-1}}-a+6=b-24\\,b\\,\\sum_{n=1}^{\\infty }{\\frac{n}{e^{2\\,b\\,n}-1}}\\]"
],
"text/plain": [
" inf inf\n",
" ==== ====\n",
" \\ n \\ n\n",
"(%o7) 24 a > ----------- - a + 6 = b - 24 b > -----------\n",
" / 2 a n / 2 b n\n",
" ==== %e - 1 ==== %e - 1\n",
" n = 1 n = 1"
],
"text/x-maxima": [
"24*a*'sum(n/(%e^(2*a*n)-1),n,1,inf)-a+6 = b\n",
" -24*b*'sum(n/(%e^(2*b*n)-1),n,1,inf)"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"F+3;"
]
},
{
"cell_type": "markdown",
"id": "c4eaad0a-1095-4900-9829-fd9d9dac1600",
"metadata": {},
"source": [
"これで証明終了なのですが分かりますか？念の為、証明すべき式の$P(q)$をその定義で置き換えた式を下記に示します。上の式と全く同じですね。"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "5e1120b6-3383-432f-bb6b-4b1550b08107",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{8}$}24\\,a\\,\\sum_{n=1}^{\\infty }{\\frac{n}{e^{2\\,a\\,n}-1}}-a+6=b-24\\,b\\,\\sum_{n=1}^{\\infty }{\\frac{n}{e^{2\\,b\\,n}-1}}\\]"
],
"text/plain": [
" inf inf\n",
" ==== ====\n",
" \\ n \\ n\n",
"(%o8) 24 a > ----------- - a + 6 = b - 24 b > -----------\n",
" / 2 a n / 2 b n\n",
" ==== %e - 1 ==== %e - 1\n",
" n = 1 n = 1"
],
"text/x-maxima": [
"24*a*'sum(n/(%e^(2*a*n)-1),n,1,inf)-a+6 = b\n",
" -24*b*'sum(n/(%e^(2*b*n)-1),n,1,inf)"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"-a*P(exp(-2*a))+6=b*P(exp(-2*b)),P(q):=1-24*sum(n*q^n/(1-q^n),n,1,inf),expand;"
]
},
{
"cell_type": "markdown",
"id": "996a2e25-72b9-41c4-b387-28c178f17080",
"metadata": {},
"source": [
"というわけで以下の式が成り立つことがわかりました。"
]
},
{
"cell_type": "code",
"execution_count": 9,
"id": "ebf0a02c-2807-4455-af5b-585e365f2eef",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{9}$}6-a\\,P\\left(e^ {- 2\\,a }\\right)=b\\,P\\left(e^ {- 2\\,b }\\right)\\]"
],
"text/plain": [
" - 2 a - 2 b\n",
"(%o9) 6 - a P(%e ) = b P(%e )"
],
"text/x-maxima": [
"6-a*P(%e^-(2*a)) = b*P(%e^-(2*b))"
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"-a*P(exp(-2*a))+6=b*P(exp(-2*b));"
]
}
],
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