{
"cells": [
{
"cell_type": "markdown",
"id": "7f80d5e8-bab3-461d-840b-cf4ecc53a6fd",
"metadata": {},
"source": [
"定理 $A_k=\\frac{\\left(\\frac12\\right)_k^3}{k!^3}$として\n",
"$$\\frac{16}{\\pi}=\\sum_{k=0}^{\\infty }{\\frac{\\left(42\\,k+5\\right)\\,A_{k}}{2^{6\\,k}}}$$\n",
"
\n",
"\n",
"証明の方針は以下の通りです。前回に得られた以下のアイゼンシュタイン級数の２つの式で$n=7$の場合を計算していきます。\n",
"$$ \\tag{${A1}$}P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right)=\\left(1-2\\,x_{n}\\right)\\,\\sum_{k=0}^{\\infty }{\\left(3\\,k+1\\right)\\,A_{k}\\,X_{n}^{k}}$$\n",
"$$ \\tag{${A2}$}\\frac{6\\,\\sqrt{n}}{\\pi}-P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{n}} }\\right)=n\\,P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right) $$\n",
"ただし以下の3つの事実を証明抜きで使うことにします。\n",
"$$\\tag{${A3}$}7\\,P\\left(e^ {- 2\\,\\sqrt{7}\\,\\pi }\\right)-P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{7}} }\\right)=\\frac{27\\,\\sqrt{7}\\,z_{7}^2}{8}$$\n",
"$$1-2\\,x_{7}=\\frac{3\\,\\sqrt{7}}{8}$$\n",
"$$X_7=\\frac{1}{2^6}$$\n",
"ここで$z_7^2=\\sum_{k=0}^{\\infty}A_k\\,X_7^k$です。また$X_n=4\\,x_n\\,(1-x_n)$でした。上記の値でこれが成立していることは簡単に確認できるので、証明抜きで使うのは実際は２つの事実ということになります。
\n",
"\n",
"\n",
"証明の流れは、まず$A2, A3$から$P\\left(e^ {- 2\\,\\pi\\,\\sqrt{n} }\\right)$を求めます。その結果と$X_7, x_7$を$A1$に代入して整理すると定理を証明することができます。\n",
"\n",
"\n",
"まず前回証明した２つの式を$A1, A2$として再掲します。"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "0b1a7e62-3d67-48c5-90eb-69a62afe7738",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{0}$}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",
"(%o0) 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": 1,
"metadata": {},
"output_type": "execute_result"
},
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{1}$}\\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",
"(%o1) --------- - 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": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A1:P(exp(-2*%pi*sqrt(n)))=(1-2*x[n])*sum((3*k+1)*A[k]*X[n]^k,k,0,inf);\n",
"A2:6*sqrt(n)/%pi-P(exp(-2*%pi/sqrt(n)))=n*P(exp(-2*%pi*sqrt(n)));"
]
},
{
"cell_type": "markdown",
"id": "45450682-9ad2-4682-a821-f04d0bccfd5c",
"metadata": {},
"source": [
"またラマヌジャンが導入した$f_n(q)$という関数を以下のように定義します。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "3a29bf0c-74a9-4815-a200-abb19473da06",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{2}$}f_{n}(q):=n\\,P\\left(q^{2\\,n}\\right)-P\\left(q^2\\right)\\]"
],
"text/plain": [
" 2 n 2\n",
"(%o2) f (q) := n P(q ) - P(q )\n",
" n"
],
"text/x-maxima": [
"f[n](q):=n*P(q^(2*n))-P(q^2)"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"f[n](q):=n*P(q^(2*n))-P(q^2);"
]
},
{
"cell_type": "markdown",
"id": "aeb24a0a-e879-4921-a492-9972e245ee6a",
"metadata": {},
"source": [
"証明抜きで使う最初の事実は次のように表すことができます。この式を$A3$とします。"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "9e7edc5c-3549-4599-a041-755f4b645392",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{3}$}7\\,P\\left(e^ {- 2\\,\\sqrt{7}\\,\\pi }\\right)-P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{7}} }\\right)=\\frac{27\\,\\sqrt{7}\\,z_{7}^2}{8}\\]"
],
"text/plain": [
" 2 %pi 2\n",
" - ------- 27 sqrt(7) z\n",
" - 2 sqrt(7) %pi sqrt(7) 7\n",
"(%o3) 7 P(%e ) - P(%e ) = -------------\n",
" 8"
],
"text/x-maxima": [
"7*P(%e^-(2*sqrt(7)*%pi))-P(%e^-((2*%pi)/sqrt(7))) = (27*sqrt(7)*z[7]^2)/8"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A3:f[7](exp(-%pi/sqrt(7)))=27*sqrt(7)/8*z[7]^2;"
]
},
{
"cell_type": "markdown",
"id": "df937c9e-561c-4c4c-b30d-ef5852f0f618",
"metadata": {},
"source": [
"$A2$で$n=7$の場合を計算して結果を$A4$とします。"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "1153fffc-4475-4cfb-8f9f-ae86d9c1a68d",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{4}$}\\frac{6\\,\\sqrt{7}}{\\pi}-P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{7}} }\\right)=7\\,P\\left(e^ {- 2\\,\\sqrt{7}\\,\\pi }\\right)\\]"
],
"text/plain": [
" 2 %pi\n",
" - -------\n",
" 6 sqrt(7) sqrt(7) - 2 sqrt(7) %pi\n",
"(%o4) --------- - P(%e ) = 7 P(%e )\n",
" %pi"
],
"text/x-maxima": [
"(6*sqrt(7))/%pi-P(%e^-((2*%pi)/sqrt(7))) = 7*P(%e^-(2*sqrt(7)*%pi))"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A4:A2,n:7;"
]
},
{
"cell_type": "markdown",
"id": "b9e26580-990e-4e27-877c-2ab1f6bb62c1",
"metadata": {},
"source": [
"$P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{7}} }\\right), 7\\,P\\left(e^ {- 2\\,\\sqrt{7}\\,\\pi }\\right)$を変数と見立てて$A3, A4$を連立１次方程式として解きます。"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "b4bccc9a-2594-4303-a280-1a92a5422725",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{5}$}\\left[ \\left[ P\\left(e^ {- 2\\,\\sqrt{7}\\,\\pi }\\right)=\\frac{27\\,\\sqrt{7}\\,\\pi\\,z_{7}^2+48\\,\\sqrt{7}}{112\\,\\pi} , P\\left(e^ {- \\frac{2\\,\\pi}{\\sqrt{7}} }\\right)=-\\frac{27\\,\\sqrt{7}\\,\\pi\\,z_{7}^2-48\\,\\sqrt{7}}{16\\,\\pi} \\right] \\right] \\]"
],
"text/plain": [
" 2\n",
" 27 sqrt(7) %pi z + 48 sqrt(7)\n",
" - 2 sqrt(7) %pi 7\n",
"(%o5) [[P(%e ) = ------------------------------, \n",
" 112 %pi\n",
" 2 %pi 2\n",
" - ------- 27 sqrt(7) %pi z - 48 sqrt(7)\n",
" sqrt(7) 7\n",
" P(%e ) = - ------------------------------]]\n",
" 16 %pi"
],
"text/x-maxima": [
"[[P(%e^-(2*sqrt(7)*%pi)) = (27*sqrt(7)*%pi*z[7]^2+48*sqrt(7))/(112*%pi),\n",
" P(%e^-((2*%pi)/sqrt(7))) = -(27*sqrt(7)*%pi*z[7]^2-48*sqrt(7))/(16*%pi)]]"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"SOL:solve([A3,A4],[P(exp(-2*sqrt(7)*%pi)), P(exp(-2/sqrt(7)*%pi))]);"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "4bdffc59-4fd9-4041-97f4-75e9e6445910",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{6}$}P\\left(e^ {- 2\\,\\sqrt{7}\\,\\pi }\\right)=\\frac{27\\,\\sqrt{7}\\,\\pi\\,z_{7}^2+48\\,\\sqrt{7}}{112\\,\\pi}\\]"
],
"text/plain": [
" 2\n",
" 27 sqrt(7) %pi z + 48 sqrt(7)\n",
" - 2 sqrt(7) %pi 7\n",
"(%o6) P(%e ) = ------------------------------\n",
" 112 %pi"
],
"text/x-maxima": [
"P(%e^-(2*sqrt(7)*%pi)) = (27*sqrt(7)*%pi*z[7]^2+48*sqrt(7))/(112*%pi)"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A5:SOL[1][1];"
]
},
{
"cell_type": "markdown",
"id": "4b992916-6bb5-402f-b43c-323169154c78",
"metadata": {},
"source": [
"$A1$で$n=7$とした式を計算し、上記の結果を代入します。"
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "11b0e65f-109a-4dc3-88b9-e689aec03746",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{7}$}P\\left(e^ {- 2\\,\\sqrt{7}\\,\\pi }\\right)=\\left(1-2\\,x_{7}\\right)\\,\\sum_{k=0}^{\\infty }{X_{7}^{k}\\,\\left(3\\,k+1\\right)\\,A_{k}}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" - 2 sqrt(7) %pi \\ k\n",
"(%o7) P(%e ) = (1 - 2 x ) > X (3 k + 1) A\n",
" 7 / 7 k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"P(%e^-(2*sqrt(7)*%pi)) = (1-2*x[7])*'sum(X[7]^k*(3*k+1)*A[k],k,0,inf)"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A6:A1,n:7;"
]
},
{
"cell_type": "code",
"execution_count": 27,
"id": "333f1a33-8891-4d2c-a591-112b4cdebc6a",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{27}$}\\frac{27\\,\\sqrt{7}\\,\\pi\\,z_{7}^2+48\\,\\sqrt{7}}{112\\,\\pi}=\\left(1-2\\,x_{7}\\right)\\,\\sum_{k=0}^{\\infty }{X_{7}^{k}\\,\\left(3\\,k+1\\right)\\,A_{k}}\\]"
],
"text/plain": [
" 2 inf\n",
" 27 sqrt(7) %pi z + 48 sqrt(7) ====\n",
" 7 \\ k\n",
"(%o27) ------------------------------ = (1 - 2 x ) > X (3 k + 1) A\n",
" 112 %pi 7 / 7 k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"(27*sqrt(7)*%pi*z[7]^2+48*sqrt(7))/(112*%pi)\n",
" = (1-2*x[7])*'sum(X[7]^k*(3*k+1)*A[k],k,0,inf)"
]
},
"execution_count": 27,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A7:A6,A5;"
]
},
{
"cell_type": "code",
"execution_count": 28,
"id": "9b31722a-c69b-4082-8cd2-977f9471e670",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{28}$}\\frac{27\\,z_{7}^2}{16\\,\\sqrt{7}}+\\frac{3}{\\sqrt{7}\\,\\pi}=\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}-2\\,x_{7}\\,\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}\\]"
],
"text/plain": [
" 2 inf\n",
" 27 z ====\n",
" 7 3 \\ k k\n",
"(%o28) ---------- + ----------- = > (3 X k A + X A )\n",
" 16 sqrt(7) sqrt(7) %pi / 7 k 7 k\n",
" ====\n",
" k = 0\n",
" inf\n",
" ====\n",
" \\ k k\n",
" - 2 x > (3 X k A + X A )\n",
" 7 / 7 k 7 k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"(27*z[7]^2)/(16*sqrt(7))+3/(sqrt(7)*%pi)\n",
" = 'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,inf)\n",
" -2*x[7]*'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,inf)"
]
},
"execution_count": 28,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%,expand;"
]
},
{
"cell_type": "markdown",
"id": "50b599b8-6185-44ca-8f5a-89909693cfbb",
"metadata": {},
"source": [
"$z_7^2$にその級数展開を代入します。"
]
},
{
"cell_type": "code",
"execution_count": 29,
"id": "85d8d509-614a-45d4-bd05-c13e6129459b",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{29}$}\\frac{27\\,\\sum_{k=0}^{\\infty }{X_{7}^{k}\\,A_{k}}}{16\\,\\sqrt{7}}+\\frac{3}{\\sqrt{7}\\,\\pi}=\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}-2\\,x_{7}\\,\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" \\ k\n",
" 27 > X A\n",
" / 7 k inf\n",
" ==== ====\n",
" k = 0 3 \\ k k\n",
"(%o29) -------------- + ----------- = > (3 X k A + X A )\n",
" 16 sqrt(7) sqrt(7) %pi / 7 k 7 k\n",
" ====\n",
" k = 0\n",
" inf\n",
" ====\n",
" \\ k k\n",
" - 2 x > (3 X k A + X A )\n",
" 7 / 7 k 7 k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"(27*'sum(X[7]^k*A[k],k,0,inf))/(16*sqrt(7))+3/(sqrt(7)*%pi)\n",
" = 'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,inf)\n",
" -2*x[7]*'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,inf)"
]
},
"execution_count": 29,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%,z[7]^2=sum(A[k]*X[7]^k,k,0,inf);"
]
},
{
"cell_type": "code",
"execution_count": 30,
"id": "a0a921c4-5eac-4300-ac7b-a0568f406502",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{30}$}27\\,\\sum_{k=0}^{\\infty }{X_{7}^{k}\\,A_{k}}+\\frac{48}{\\pi}=16\\,\\sqrt{7}\\,\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}-32\\,\\sqrt{7}\\,x_{7}\\,\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}\\]"
],
"text/plain": [
" inf inf\n",
" ==== ====\n",
" \\ k 48 \\ k k\n",
"(%o30) 27 > X A + --- = 16 sqrt(7) > (3 X k A + X A )\n",
" / 7 k %pi / 7 k 7 k\n",
" ==== ====\n",
" k = 0 k = 0\n",
" inf\n",
" ====\n",
" \\ k k\n",
" - 32 sqrt(7) x > (3 X k A + X A )\n",
" 7 / 7 k 7 k\n",
" ====\n",
" k = 0"
],
"text/x-maxima": [
"27*'sum(X[7]^k*A[k],k,0,inf)+48/%pi = 16*sqrt(7)\n",
" *'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,\n",
" inf)\n",
" -32*sqrt(7)*x[7]\n",
" *'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,\n",
" inf)"
]
},
"execution_count": 30,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%*16*sqrt(7),expand;"
]
},
{
"cell_type": "code",
"execution_count": 31,
"id": "742e9365-6699-482c-8ef7-f22360c673d6",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{31}$}\\frac{48}{\\pi}=-32\\,\\sqrt{7}\\,x_{7}\\,\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}+16\\,\\sqrt{7}\\,\\sum_{k=0}^{\\infty }{\\left(3\\,X_{7}^{k}\\,k\\,A_{k}+X_{7}^{k}\\,A_{k}\\right)}-27\\,\\sum_{k=0}^{\\infty }{X_{7}^{k}\\,A_{k}}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 48 \\ k k\n",
"(%o31) --- = (- 32 sqrt(7) x > (3 X k A + X A ))\n",
" %pi 7 / 7 k 7 k\n",
" ====\n",
" k = 0\n",
" inf inf\n",
" ==== ====\n",
" \\ k k \\ k\n",
" + 16 sqrt(7) > (3 X k A + X A ) - 27 > X A\n",
" / 7 k 7 k / 7 k\n",
" ==== ====\n",
" k = 0 k = 0"
],
"text/x-maxima": [
"48/%pi = (-32*sqrt(7)*x[7]*'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,inf))\n",
" +16*sqrt(7)*'sum(3*X[7]^k*k*A[k]+X[7]^k*A[k],k,0,inf)\n",
" -27*'sum(X[7]^k*A[k],k,0,inf)"
]
},
"execution_count": 31,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A8:%-first(lhs(%));"
]
},
{
"cell_type": "markdown",
"id": "ef6b85ec-ee57-46ca-8d35-7c6ebfdea531",
"metadata": {},
"source": [
"級数の項を全て右辺に集めて整理すると目的の式の片鱗が見えてきます。"
]
},
{
"cell_type": "code",
"execution_count": 32,
"id": "e6628a1f-2792-4a95-9755-a2624c092fb3",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{32}$}\\frac{48}{\\pi}=-\\sum_{k=0}^{\\infty }{X_{7}^{k}\\,\\left(\\left(96\\,\\sqrt{7}\\,x_{7}-48\\,\\sqrt{7}\\right)\\,k+32\\,\\sqrt{7}\\,x_{7}-16\\,\\sqrt{7}+27\\right)\\,A_{k}}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 48 \\ k\n",
"(%o32) --- = - > X ((96 sqrt(7) x - 48 sqrt(7)) k + 32 sqrt(7) x\n",
" %pi / 7 7 7\n",
" ====\n",
" k = 0\n",
" - 16 sqrt(7) + 27) A\n",
" k"
],
"text/x-maxima": [
"48/%pi = -'sum(X[7]^k*((96*sqrt(7)*x[7]-48*sqrt(7))*k\n",
" +32*sqrt(7)*x[7]-16*sqrt(7)+27)*A[k],k,0,inf)"
]
},
"execution_count": 32,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A9:factor(sumcontract(intosum(A8)));"
]
},
{
"cell_type": "markdown",
"id": "6327ade6-1787-4ec9-801d-e90ca2a89f53",
"metadata": {},
"source": [
"$X_7=\\frac{1}{2^6}$を代入します。"
]
},
{
"cell_type": "code",
"execution_count": 35,
"id": "aa302a9a-cc19-4222-8734-532c1ca7a2ee",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{34}$}\\frac{48}{\\pi}=-\\sum_{k=0}^{\\infty }{\\frac{\\left(\\left(96\\,\\sqrt{7}\\,x_{7}-48\\,\\sqrt{7}\\right)\\,k+32\\,\\sqrt{7}\\,x_{7}-16\\,\\sqrt{7}+27\\right)\\,A_{k}}{64^{k}}}\\]"
],
"text/plain": [
" 48\n",
"(%o34) --- = \n",
" %pi\n",
" inf\n",
" ==== ((96 sqrt(7) x - 48 sqrt(7)) k + 32 sqrt(7) x - 16 sqrt(7) + 27) A\n",
" \\ 7 7 k\n",
" - > ---------------------------------------------------------------------\n",
" / k\n",
" ==== 64\n",
" k = 0"
],
"text/x-maxima": [
"48/%pi = -'sum((((96*sqrt(7)*x[7]-48*sqrt(7))*k+32*sqrt(7)*x[7]-16*sqrt(7)+27)\n",
" *A[k])\n",
" /64^k,k,0,inf)"
]
},
"execution_count": 35,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A10:A9, X[7]=1/2^6;"
]
},
{
"cell_type": "markdown",
"id": "e02cd64e-2971-4044-a0e2-c7873665849b",
"metadata": {},
"source": [
"事実その2の式を$x_7$について解きます。"
]
},
{
"cell_type": "code",
"execution_count": 16,
"id": "fecdcb33-9da8-47bf-a7bc-647eace130df",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{16}$}1-2\\,x_{7}=\\frac{3\\,\\sqrt{7}}{8}\\]"
],
"text/plain": [
" 3 sqrt(7)\n",
"(%o16) 1 - 2 x = ---------\n",
" 7 8"
],
"text/x-maxima": [
"1-2*x[7] = (3*sqrt(7))/8"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"1-2*x[7]=3*sqrt(7)/8;"
]
},
{
"cell_type": "code",
"execution_count": 17,
"id": "c7107d3d-66ff-4a78-9857-46ed7569c20c",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{17}$}\\left[ x_{7}=-\\frac{3\\,\\sqrt{7}-8}{16} \\right] \\]"
],
"text/plain": [
" 3 sqrt(7) - 8\n",
"(%o17) [x = - -------------]\n",
" 7 16"
],
"text/x-maxima": [
"[x[7] = -(3*sqrt(7)-8)/16]"
]
},
"execution_count": 17,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"SOL2:solve(%,x[7]);"
]
},
{
"cell_type": "markdown",
"id": "e050fcc2-d15a-43db-9d1a-55fe6fb1f110",
"metadata": {},
"source": [
"結果を$A10$に代入します。"
]
},
{
"cell_type": "code",
"execution_count": 18,
"id": "f6b132c9-943f-4011-a954-2e73fc10d1e8",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{18}$}\\frac{48}{\\pi}=-\\sum_{k=0}^{\\infty }{\\frac{\\left(\\left(-6\\,\\sqrt{7}\\,\\left(3\\,\\sqrt{7}-8\\right)-48\\,\\sqrt{7}\\right)\\,k-2\\,\\sqrt{7}\\,\\left(3\\,\\sqrt{7}-8\\right)-16\\,\\sqrt{7}+27\\right)\\,A_{k}}{64^{k}}}\\]"
],
"text/plain": [
" inf\n",
" ====\n",
" 48 \\\n",
"(%o18) --- = - > ((((- 6 sqrt(7) (3 sqrt(7) - 8)) - 48 sqrt(7)) k\n",
" %pi /\n",
" ====\n",
" k = 0\n",
" k\n",
" - 2 sqrt(7) (3 sqrt(7) - 8) - 16 sqrt(7) + 27) A )/64\n",
" k"
],
"text/x-maxima": [
"48/%pi = -'sum(((((-6*sqrt(7)*(3*sqrt(7)-8))-48*sqrt(7))*k\n",
" -2*sqrt(7)*(3*sqrt(7)-8)-16*sqrt(7)+27)\n",
" *A[k])\n",
" /64^k,k,0,inf)"
]
},
"execution_count": 18,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A11:A10,SOL2;"
]
},
{
"cell_type": "code",
"execution_count": 19,
"id": "141050d6-b3b6-4486-a845-147bbc2c46e1",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{19}$}\\frac{48}{\\pi}=\\sum_{k=0}^{\\infty }{\\frac{\\left(126\\,k+15\\right)\\,A_{k}}{2^{6\\,k}}}\\]"
],
"text/plain": [
" inf\n",
" ==== (126 k + 15) A\n",
" 48 \\ k\n",
"(%o19) --- = > ---------------\n",
" %pi / 6 k\n",
" ==== 2\n",
" k = 0"
],
"text/x-maxima": [
"48/%pi = 'sum(((126*k+15)*A[k])/2^(6*k),k,0,inf)"
]
},
"execution_count": 19,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A12:radcan(A11);"
]
},
{
"cell_type": "markdown",
"id": "32526f18-c336-4703-95d7-1da53e1f8233",
"metadata": {},
"source": [
"整理して両辺を$3$で割れば証明は終了します。"
]
},
{
"cell_type": "code",
"execution_count": 20,
"id": "8477f513-01af-4abf-8b5e-7be525fcdceb",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{20}$}\\frac{16}{\\pi}=\\sum_{k=0}^{\\infty }{\\frac{\\left(42\\,k+5\\right)\\,A_{k}}{2^{6\\,k}}}\\]"
],
"text/plain": [
" inf\n",
" ==== (42 k + 5) A\n",
" 16 \\ k\n",
"(%o20) --- = > -------------\n",
" %pi / 6 k\n",
" ==== 2\n",
" k = 0"
],
"text/x-maxima": [
"16/%pi = 'sum(((42*k+5)*A[k])/2^(6*k),k,0,inf)"
]
},
"execution_count": 20,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A13:ratsimp(intosum(A12/3));"
]
}
],
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