{
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"定理 次の２つの式が成り立つ。\n",
"$$ P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right)=\\left(1-2\\,x_{n}\\right)\\,\\sum_{k=0}^{\\infty }{\\left(3\\,k+1\\right)\\,A_{k}\\,X_{n}^{k}}$$\n",
"$$ \\frac{6\\,\\sqrt{n}}{\\pi}-P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{n}} }\\right)=n\\,P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right) $$\n",
"ただし$q, x_n, X_n, z_n$を次のように定義する。\n",
"$n$を適当な自然数として$q$を$q=e^{-\\pi\\,\\sqrt{n}}$と定義する。$q=exp(-\\pi\\,\\frac{{}_2F_1(\\frac12,\\frac12;1;1-x)}{{}_2F_1(\\frac12,\\frac12;1;x)})$だったから$\\sqrt{n}=\\frac{{}_2F_1(\\frac12,\\frac12;1;1-x)}{{}_2F_1(\\frac12,\\frac12;1;x)}$。そこでそうなる$x$を$x_n$と表記する。この$x_n$を使って$z_n, X_n$を次のように定義する。$z_n={}_2F_1(\\frac12,\\frac12;1;x_n), X_n=4\\,x_n\\,(1-x_n)$。
\n",
"
\n",
"\n",
"\n",
"系\n",
"$$1-24\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,\\pi\\,n }}{1-e^ {- 2\\,\\pi\\,n }}}=\\frac{3}{\\pi} $$\n",
"
\n",
"\n",
"\n",
"定理の証明は前回証明した次の２つの式変形して行います。とは言っても上記の$q=e^{-\\pi\\,\\sqrt{n}}$の定義を代入して整理するだけです。\n",
"$$ P(q^2)=(1-2\\,x)\\sum_{k=0}^{\\infty}(3\\,k+1)\\,A_k\\,X^k $$\n",
"$$ 6-a\\,P\\left(e^ {- 2\\,a }\\right)=b\\,P\\left(e^ {- 2\\,b }\\right) $$\n"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "28430f93-9c3a-4b23-933a-64fa7ee1ffc2",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{0}$}P\\left(q^2\\right)=\\left(\\sum_{k=0}^{\\infty }{X^{k}\\,\\left(3\\,k+1\\right)\\,A_{k}}\\right)\\,\\left(1-2\\,x\\right)\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 2 \\ k\n",
"(%o0) P(q ) = ( > X (3 k + 1) A ) (1 - 2 x)\n",
" / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"P(q^2) = ('sum(X^k*(3*k+1)*A[k],k,0,inf))*(1-2*x)"
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"P(q^2)=(1-2*x)*sum((3*k+1)*A[k]*X^k,k,0,inf);"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "a591b2ea-cb3b-4aa6-b8c3-ed4cbf1209e4",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{1}$}P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right)=\\left(1-2\\,x_{n}\\right)\\,\\sum_{k=0}^{\\infty }{\\left(3\\,k+1\\right)\\,A_{k}\\,X_{n}^{k}}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" - 2 %pi sqrt(n) \\ k\n",
"(%o1) P(%e ) = (1 - 2 x ) > (3 k + 1) A X\n",
" n / k n\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"P(%e^-(2*%pi*sqrt(n))) = (1-2*x[n])*'sum((3*k+1)*A[k]*X[n]^k,k,0,inf)"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%,q=exp(-%pi*sqrt(n)),x=x[n], X=X[n];"
]
},
{
"cell_type": "markdown",
"id": "125e3474-2291-42db-8866-efcbf9205fe7",
"metadata": {},
"source": [
"これで最初の式の証明は終了です。二番目の式の証明も簡明です。$a\\,b=\\pi^2$を満たす$a=\\frac{\\pi}{\\sqrt{n}}, b=\\pi\\,\\sqrt{n}$を次の式に代入して整理します。"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "67ad1b60-9aec-41c6-ba36-a65b94f8579a",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{2}$}6-a\\,P\\left(e^ {- 2\\,a }\\right)=b\\,P\\left(e^ {- 2\\,b }\\right)\\]"
],
"text/plain": [
" - 2 a - 2 b\n",
"(%o2) 6 - a P(%e ) = b P(%e )"
],
"text/x-maxima": [
"6-a*P(%e^-(2*a)) = b*P(%e^-(2*b))"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A1:6-a*P(exp(-2*a))=b*P(exp(-2*b));"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "4aed9887-1258-4f2c-9869-266031dee05b",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{3}$}6-\\frac{\\pi\\,P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{n}} }\\right)}{\\sqrt{n}}=\\pi\\,\\sqrt{n}\\,P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right)\\]"
],
"text/plain": [
" 2 %pi\n",
" - -------\n",
" sqrt(n)\n",
" %pi P(%e ) - 2 %pi sqrt(n)\n",
"(%o3) 6 - ------------------ = %pi sqrt(n) P(%e )\n",
" sqrt(n)"
],
"text/x-maxima": [
"6-(%pi*P(%e^-((2*%pi)/sqrt(n))))/sqrt(n) = %pi*sqrt(n)*P(%e^-(2*%pi*sqrt(n)))"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A2:A1,a=%pi/sqrt(n),b=%pi*sqrt(n);"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "afa83cc9-1068-4694-9ae0-117cca359397",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{4}$}\\frac{6\\,\\sqrt{n}}{\\pi}-P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{n}} }\\right)=n\\,P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right)\\]"
],
"text/plain": [
" 2 %pi\n",
" - -------\n",
" 6 sqrt(n) sqrt(n) - 2 %pi sqrt(n)\n",
"(%o4) --------- - P(%e ) = n P(%e )\n",
" %pi"
],
"text/x-maxima": [
"(6*sqrt(n))/%pi-P(%e^-((2*%pi)/sqrt(n))) = n*P(%e^-(2*%pi*sqrt(n)))"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A3:A2*sqrt(n)/%pi,expand;"
]
},
{
"cell_type": "markdown",
"id": "dfd26ea4-87bb-415b-9716-563ceca0abe4",
"metadata": {},
"source": [
"これで証明は終了です。
\n",
"\n",
"系の証明はこの式で$n=1$とすることで自明に得られます。"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "7fcc108c-dc29-45c9-a39a-372f64d3fb54",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{5}$}\\frac{6}{\\pi}-P\\left(e^ {- 2\\,\\pi }\\right)=P\\left(e^ {- 2\\,\\pi }\\right)\\]"
],
"text/plain": [
" 6 - 2 %pi - 2 %pi\n",
"(%o5) --- - P(%e ) = P(%e )\n",
" %pi"
],
"text/x-maxima": [
"6/%pi-P(%e^-(2*%pi)) = P(%e^-(2*%pi))"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%,n=1;"
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "aeb18e37-36b1-465e-849f-cf72da879d98",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{6}$}\\left[ P\\left(e^ {- 2\\,\\pi }\\right)=\\frac{3}{\\pi} \\right] \\]"
],
"text/plain": [
" - 2 %pi 3\n",
"(%o6) [P(%e ) = ---]\n",
" %pi"
],
"text/x-maxima": [
"[P(%e^-(2*%pi)) = 3/%pi]"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"solve(%,rhs(%));"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "1deac3dc-d0ac-44d8-820e-845680abbf1b",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{7}$}\\left[ 1-24\\,\\sum_{n=1}^{\\infty }{\\frac{n\\,e^ {- 2\\,\\pi\\,n }}{1-e^ {- 2\\,\\pi\\,n }}}=\\frac{3}{\\pi} \\right] \\]"
],
"text/plain": [
" inf\n",
" ==== - 2 %pi n\n",
" \\ n %e 3\n",
"(%o7) [1 - 24 > --------------- = ---]\n",
" / - 2 %pi n %pi\n",
" ==== 1 - %e\n",
" n = 1"
],
"text/x-maxima": [
"[1-24*'sum((n*%e^-(2*%pi*n))/(1-%e^-(2*%pi*n)),n,1,inf) = 3/%pi]"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%,P(q):=1-24*sum(n*q^n/(1-q^n),n,1,inf);"
]
}
],
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