{
"cells": [
{
"cell_type": "markdown",
"id": "36c8f10e-0361-4a4e-a24e-8fac5105c572",
"metadata": {},
"source": [
"定理 $P(q), q, z, x, y, X, A_k$が$P(q)=1-24\\,\\sum_{n=1}^{\\infty}\\frac{n\\,q^n}{1-q^n}$, $q=e^{-y}, z= {}_{2}F_{1}(\\frac12,\\frac12;1;x)$, $y=\\pi\\,\\frac{{}_{2}F_{1}(\\frac12,\\frac12;1;1-x)}{{}_{2}F_{1}(\\frac12,\\frac12;1;x)}$, $X=4\\,x\\,(1-x), A_k=\\frac{\\left(\\frac12\\right)_k^3}{k!^3}$として以下の式が成り立ちます。\n",
"$$P(q^2)=(1-2\\,x)\\sum_{k=0}^{\\infty}(3\\,k+1)\\,A_k\\,X^k$$\n",
"
\n",
"\n",
"証明の方針は、\n",
"$$P(q^2)=(1-2\\,x)\\,z^2 + 6\\,x\\,(1-x)\\,z\\,\\frac{dz}{dx}$$\n",
"に、\n",
"$$z^2={}_3F_2(\\frac12,\\frac12,\\frac12;1,1;X)=\\sum_{k=0}^{\\infty}A_k\\,X^k$$\n",
"及びそれを$x$で微分した、\n",
"$$2\\,z\\,\\frac{dz}{dx}=4\\,(1-2\\,x)\\sum_{k=0}^{\\infty}A_k\\,k\\,X^{k-1}$$\n",
"を代入して整理します。"
]
},
{
"cell_type": "markdown",
"id": "8e71a2e7-9e9e-4c6f-b278-e77602b2d925",
"metadata": {},
"source": [
"まず、$X,z$は$x$に依存しており、微分でその関係を使うのでこの依存関係を宣言します。"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "ddbc2375-ff37-4dd1-91b9-90e82abc3b86",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{0}$}\\left[ X\\left(x\\right) , z\\left(x\\right) \\right] \\]"
],
"text/plain": [
"(%o0) [X(x), z(x)]"
],
"text/x-maxima": [
"[X(x),z(x)]"
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"depends([X,z],x);"
]
},
{
"cell_type": "markdown",
"id": "41e1eee5-5f05-4614-a11e-3e8d06e8dd1c",
"metadata": {},
"source": [
"クローゼンの公式の特殊な場合を思い出します。それは${}_2F_1(\\frac12,\\frac12;1;x)^2={}_3F_2(\\frac12,\\frac12,\\frac12;1,1;4\\,x\\,(1-x))$というものでした。定理の設定から、左辺は$z^2$になります。また$X=4\\,x\\,(1-x)$から右辺は${}_3F_2(\\frac12,\\frac12,\\frac12;1,1;X)$になります。さらに一般超幾何関数の級数展開は${}_3F_2(a,b,c;d,e;z)=\\sum_{k=0}^{\\infty}\\frac{(a)_k\\,(b)_k\\,(c)_k\\,z^k}{(d)_k\\,(e)_k\\,k!}$でしたから$a=b=c=\\frac12, d=e=1$を代入すると、$(1)_k=k!$に注意して、\n",
"$${}_3F_2(\\frac12,\\frac12,\\frac12;1,1;X)=\\sum_{k=0}^{\\infty}\\frac{\\left(\\frac12\\right)_k^3}{k!^3}\\,X^k$$\n",
"となります。級数の係数は定理の前提で$A_k=\\frac{\\left(\\frac12\\right)_k^3}{k!^3}$でしたから以下の式$CL$が成り立つことがわかりました。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "fbc30e70-642c-4511-ae06-85f31117d864",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{1}$}z^2=\\sum_{k=0}^{\\infty }{X^{k}\\,A_{k}}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 2 \\ k\n",
"(%o1) z = > X A\n",
" / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"z^2 = 'sum(X^k*A[k],k,0,inf)"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL:z^2=sum(A[k]*X^k,k,0,inf);"
]
},
{
"cell_type": "markdown",
"id": "da418550-2376-4214-9541-4cb51cc97332",
"metadata": {},
"source": [
"両辺を$x$で微分します。得られた式を$CL1$とします。"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "6f05df08-5cd9-4893-9b0f-0a99b93a3659",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{2}$}2\\,z\\,\\left(\\frac{d}{d\\,x}\\,z\\right)=\\frac{d}{d\\,x}\\,X\\,\\sum_{k=0}^{\\infty }{X^{k-1}\\,k\\,A_{k}}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" dz dX \\ k - 1\n",
"(%o2) 2 z -- = -- > X k A\n",
" dx dx / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"2*z*'diff(z,x,1) = 'diff(X,x,1)*'sum(X^(k-1)*k*A[k],k,0,inf)"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL1:diff(CL,x);"
]
},
{
"cell_type": "markdown",
"id": "251deb3e-da9e-4758-90dd-981a041f3d87",
"metadata": {},
"source": [
"$X$の$x$による微分を計算して代入してみます。得られた式を$CL2$とします。"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "cd7a6b1b-4fbe-4856-8d44-657ca206cc86",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{3}$}2\\,z\\,\\left(\\frac{d}{d\\,x}\\,z\\right)=-4\\,\\left(\\sum_{k=0}^{\\infty }{X^{k-1}\\,k\\,A_{k}}\\right)\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" dz \\ k - 1\n",
"(%o3) 2 z -- = - 4 ( > X k A ) (2 x - 1)\n",
" dx / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"2*z*'diff(z,x,1) = -4*('sum(X^(k-1)*k*A[k],k,0,inf))*(2*x-1)"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL2:subst(ev(diff(X,x),X:4*x*(1-x)),diff(X,x),CL1),factor;"
]
},
{
"cell_type": "markdown",
"id": "729ac8fc-f9fe-4711-9dd3-9a26843c6a1f",
"metadata": {},
"source": [
"定理5.4.9で証明した以下の式を思い出しましょう。"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "73f086c1-9457-4bb9-b979-f8401f3a87ff",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{4}$}P\\left(q^2\\right)=6\\,\\left(1-x\\right)\\,x\\,z\\,\\left(\\frac{d}{d\\,x}\\,z\\right)+\\left(1-2\\,x\\right)\\,z^2\\]"
],
"text/plain": [
" 2 dz 2\n",
"(%o4) P(q ) = 6 (1 - x) x z -- + (1 - 2 x) z\n",
" dx"
],
"text/x-maxima": [
"P(q^2) = 6*(1-x)*x*z*'diff(z,x,1)+(1-2*x)*z^2"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"TH549:P(q^2)=(1-2*x)*z^2+6*x*(1-x)*z*diff(z,x);"
]
},
{
"cell_type": "markdown",
"id": "105e7930-34de-4a5c-91bf-5efe3d8b2e06",
"metadata": {},
"source": [
"$CL2$の両辺を$2\\,z$で割れば$\\frac{d}{dx}z$が得られます。これをTH549に代入した結果を$CL3$とします。"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "5062beff-7b5a-4d19-bad3-3b8a0d56a554",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{5}$}P\\left(q^2\\right)=\\left(1-2\\,x\\right)\\,z^2-12\\,\\left(\\sum_{k=0}^{\\infty }{X^{k-1}\\,k\\,A_{k}}\\right)\\,\\left(1-x\\right)\\,x\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 2 2 \\ k - 1\n",
"(%o5) P(q ) = (1 - 2 x) z - 12 ( > X k A ) (1 - x) x (2 x - 1)\n",
" / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"P(q^2) = (1-2*x)*z^2-12*('sum(X^(k-1)*k*A[k],k,0,inf))*(1-x)*x*(2*x-1)"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL3:TH549,CL2/(2*z);"
]
},
{
"cell_type": "markdown",
"id": "93928322-eaf7-44ee-ad64-4845869db2ea",
"metadata": {},
"source": [
"$CL3$の以下の部分式に注目します。"
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "e35e058e-92d5-4a9e-91c3-09fccf8c6786",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{6}$}12\\,\\left(\\sum_{k=0}^{\\infty }{X^{k-1}\\,k\\,A_{k}}\\right)\\,\\left(1-x\\right)\\,x\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" \\ k - 1\n",
"(%o6) 12 ( > X k A ) (1 - x) x (2 x - 1)\n",
" / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"12*('sum(X^(k-1)*k*A[k],k,0,inf))*(1-x)*x*(2*x-1)"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL31:part(CL3,2,2,1);"
]
},
{
"cell_type": "markdown",
"id": "28761ed9-89a6-4ce4-9396-681f8ed51773",
"metadata": {},
"source": [
"$12\\,(1-x)\\,x\\,(2\\,x-1)=3\\,(2\\,x-1)\\,4\\,(1-x)\\,x=3\\,(2\\,x-1)\\,X$であることから以下のように整理できます。"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "5aef189d-fa72-478b-8b83-03c10dc3146a",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{7}$}3\\,X\\,\\left(\\sum_{k=0}^{\\infty }{X^{k-1}\\,k\\,A_{k}}\\right)\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" \\ k - 1\n",
"(%o7) 3 X ( > X k A ) (2 x - 1)\n",
" / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"3*X*('sum(X^(k-1)*k*A[k],k,0,inf))*(2*x-1)"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL32:CL31*X/(4*x*(1-x));"
]
},
{
"cell_type": "markdown",
"id": "10439ef7-e884-44ba-bf56-5df390cc61b9",
"metadata": {},
"source": [
"$X$は$k$とは関係ないので、そのままintosum()を使って総和記号の中に持ち込むことが出来ます。"
]
},
{
"cell_type": "code",
"execution_count": 9,
"id": "816ee088-85a6-4803-857e-a6f2d9969e8a",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{8}$}\\sum_{k=0}^{\\infty }{3\\,X^{k}\\,k\\,A_{k}\\,\\left(2\\,x-1\\right)}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" \\ k\n",
"(%o8) > 3 X k A (2 x - 1)\n",
" / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"'sum(3*X^k*k*A[k]*(2*x-1),k,0,inf)"
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL33:intosum(CL32);"
]
},
{
"cell_type": "markdown",
"id": "0aec954f-e8b9-40f5-952e-63b71fad02ef",
"metadata": {},
"source": [
"従って$CL3$を以下のように書き直すことが出来ます。これを$CL4$とします。"
]
},
{
"cell_type": "code",
"execution_count": 10,
"id": "a40d591f-3ebe-4e22-9200-7afcae06a491",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{9}$}P\\left(q^2\\right)=\\left(1-2\\,x\\right)\\,z^2-3\\,\\left(\\sum_{k=0}^{\\infty }{X^{k}\\,k\\,A_{k}}\\right)\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 2 2 \\ k\n",
"(%o9) P(q ) = (1 - 2 x) z - 3 ( > X k A ) (2 x - 1)\n",
" / k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"P(q^2) = (1-2*x)*z^2-3*('sum(X^k*k*A[k],k,0,inf))*(2*x-1)"
]
},
"execution_count": 10,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL4:substpart(CL33,CL3,2,2,1);"
]
},
{
"cell_type": "markdown",
"id": "24669fbd-7005-4fb7-a26a-5100114a042f",
"metadata": {},
"source": [
"右辺第一項の中の$z^2$は$CL$を使って書き換えることが出来ます。"
]
},
{
"cell_type": "code",
"execution_count": 11,
"id": "dfa5e1f4-33ef-45af-a4e1-06564a25e5a4",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{10}$}P\\left(q^2\\right)=\\left(\\sum_{k=0}^{\\infty }{X^{k}\\,A_{k}}\\right)\\,\\left(1-2\\,x\\right)-3\\,\\left(\\sum_{k=0}^{\\infty }{X^{k}\\,k\\,A_{k}}\\right)\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf inf\n",
" ==== ====\n",
" 2 \\ k \\ k\n",
"(%o10) P(q ) = ( > X A ) (1 - 2 x) - 3 ( > X k A ) (2 x - 1)\n",
" / k / k\n",
" ==== ====\n",
" k = 0 k = 0"
],
"text/x-maxima": [
"P(q^2) = ('sum(X^k*A[k],k,0,inf))*(1-2*x)-3*('sum(X^k*k*A[k],k,0,inf))*(2*x-1)"
]
},
"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL5:CL4,CL;"
]
},
{
"cell_type": "markdown",
"id": "bcb32767-ba7e-444c-924e-989e99d7a73b",
"metadata": {},
"source": [
"後は因数分解して、"
]
},
{
"cell_type": "code",
"execution_count": 12,
"id": "9e695a46-ccf8-4b4e-9eb4-ed73d63d5dba",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{11}$}P\\left(q^2\\right)=-\\left(3\\,\\sum_{k=0}^{\\infty }{X^{k}\\,k\\,A_{k}}+\\sum_{k=0}^{\\infty }{X^{k}\\,A_{k}}\\right)\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf inf\n",
" ==== ====\n",
" 2 \\ k \\ k\n",
"(%o11) P(q ) = - (3 > X k A + > X A ) (2 x - 1)\n",
" / k / k\n",
" ==== ====\n",
" k = 0 k = 0"
],
"text/x-maxima": [
"P(q^2) = -(3*'sum(X^k*k*A[k],k,0,inf)+'sum(X^k*A[k],k,0,inf))*(2*x-1)"
]
},
"execution_count": 12,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL6:CL5,factor;"
]
},
{
"cell_type": "markdown",
"id": "9a65fedd-63cf-4ed8-864a-f1cfb3b0bbac",
"metadata": {},
"source": [
"総和記号を１つにまとめれば証明したかった式を得られました。"
]
},
{
"cell_type": "code",
"execution_count": 13,
"id": "5e506559-6f08-45d8-965c-c2076d1cd5d5",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{12}$}P\\left(q^2\\right)=-\\left(\\sum_{k=0}^{\\infty }{\\left(3\\,X^{k}\\,k\\,A_{k}+X^{k}\\,A_{k}\\right)}\\right)\\,\\left(2\\,x-1\\right)\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 2 \\ k k\n",
"(%o12) P(q ) = - ( > (3 X k A + X A )) (2 x - 1)\n",
" / k k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"P(q^2) = -('sum(3*X^k*k*A[k]+X^k*A[k],k,0,inf))*(2*x-1)"
]
},
"execution_count": 13,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"CL7:substpart(sumcontract(intosum(part(CL6,2,1,1))),CL6,2,1,1);"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "5f5a3c2f-721e-4ec4-99d6-5918eea8f41d",
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
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},
"language_info": {
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