{
"cells": [
{
"cell_type": "markdown",
"id": "40741790-0d96-452e-b5bc-5d70f0d38422",
"metadata": {},
"source": [
"### 超幾何関数に関するクローゼンの恒等式\n",
"\n",
"定理：$_{2}F_1\\left(a,b;a+b+\\frac12;x\\right)^2= {}_3F_2\\left(2\\,a,2\\,b,a+b;2\\,a+2\\,b,a+b+\\frac12;x\\right)$
\n",
"左辺はガウスの超幾何関数の2乗、右辺は一般超幾何関数です。\n",
"\n",
"証明の方針ですが、超幾何関数の恒等式の証明でよくある、左辺と右辺が同じ微分方程式を満たし、かつ必要な初期値が一致する、というやり方です。まず定理の右辺を処理します。一般超幾何関数$G(x)={}_3F_2(2\\,a,2\\,b,a+b;2\\,a+2\\,b,a+b+\\frac12;x)$が満たす微分方程式をもとに$F(x)^2=G(x)$となる$F(x)$が満たす微分方程式を求めます。といっても代入して整理するだけです。\n",
"\n",
"次にガウスの超幾何関数$F(x)={}_{2}F_1(,b;a+b+\\frac12;x)$が満たす微分方程式の適当な線型結合による新しい微分方程式を作ります。作り方から明らかに$F(x)$は新しい微分方程式も満たします。そしてこの新しい微分方程式が前の段落で求めた微分方程式と一致することを示します。といっても先に求めた微分方程式から後で求めた微分方程式を引き算して$0$になることを確認するだけです。\n",
"\n",
"今回の証明はLorenz Millaという数学者の2021年の論文[\"A DETAILED PROOF OF THE CHUDNOVSKY FORMULA WITH MEANS OF BASIC COMPLEX ANALYSIS\"](https://arxiv.org/pdf/1809.00533.pdf)のp24-29を参考にしました。"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "8d56c9d9-5cf9-4038-8247-4b5f48ae852d",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{0}$}\\mbox{ /Users/yasube/Programming2/maxima-5.44.0-install/share/maxima/5.44.0/share/to\\_poly\\_solve/to\\_poly\\_solve\\_extra.lisp }\\]"
],
"text/plain": [
"(%o0) /Users/yasube/Programming2/maxima-5.44.0-install/share/maxima/5.44.0/sha\\\n",
"re/to_poly_solve/to_poly_solve_extra.lisp"
],
"text/x-maxima": [
"\"/Users/yasube/Programming2/maxima-5.44.0-install/share/maxima/5.44.0/share/to_poly_solve/to_poly_solve_extra.lisp\""
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
},
{
"name": "stdout",
"output_type": "stream",
"text": [
"SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::$NONNEGINTEGERP in DEFUN\n",
"SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::$POLYNOMIALP in DEFUN\n",
"SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::POLYNOMIALP in DEFUN\n",
"SB-KERNEL:REDEFINITION-WITH-DEFMACRO: redefining MAXIMA::OPAPPLY in DEFMACRO\n",
"SB-KERNEL:REDEFINITION-WITH-DEFMACRO: redefining MAXIMA::OPCONS in DEFMACRO\n"
]
}
],
"source": [
"load(\"to_poly_solve_extra.lisp\");"
]
},
{
"cell_type": "markdown",
"id": "f39235d8-28ec-49ee-b255-74e73b5572fc",
"metadata": {},
"source": [
"一般超幾何関数$G(x)={}_3F_2\\left(a,b,c;d,e;x\\right)$が満たす微分方程式は以前の記事で求めました。それが次の微分方程式です。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "8b4f2a07-499b-46dc-a09b-5060e00aff4c",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{1}$}\\left(x-1\\right)\\,x^2\\,\\left(\\frac{d^3}{d\\,x^3}\\,G\\left(x\\right)\\right)+x\\,\\left(\\left(c+b+a+3\\right)\\,x-e-d-1\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,G\\left(x\\right)\\right)+\\left(\\left(b\\,c+a\\,c+c+a\\,b+b+a+1\\right)\\,x-d\\,e\\right)\\,\\left(\\frac{d}{d\\,x}\\,G\\left(x\\right)\\right)+a\\,b\\,c\\,G\\left(x\\right)=0\\]"
],
"text/plain": [
" 3 2\n",
" 2 d d\n",
"(%o1) (x - 1) x (--- (G(x))) + x ((c + b + a + 3) x - e - d - 1) (--- (G(x)))\n",
" 3 2\n",
" dx dx\n",
" d\n",
" + ((b c + a c + c + a b + b + a + 1) x - d e) (-- (G(x))) + a b c G(x) = 0\n",
" dx"
],
"text/x-maxima": [
"(x-1)*x^2*'diff(G(x),x,3)+x*((c+b+a+3)*x-e-d-1)*'diff(G(x),x,2)\n",
" +((b*c+a*c+c+a*b+b+a+1)*x-d*e)*'diff(G(x),x,1)\n",
" +a*b*c*G(x)\n",
" = 0"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF2:x^2*(x-1)*diff(G(x),x,3)\n",
"+x*(-(d+e+1)+(a+b+c+3)*x)*diff(G(x),x,2)\n",
"+(-d*e+(a+b+c+a*b+b*c+c*a+1)*x)*diff(G(x),x)\n",
"+a*b*c*G(x)=0;"
]
},
{
"cell_type": "markdown",
"id": "dba61c5d-6f48-4224-9601-47bfcca419e8",
"metadata": {},
"source": [
"${}_3F_2\\left(2\\,a,2\\,b,a+b;2\\,a+2\\,b,a+b+\\frac12;x\\right)$の満たす微分方程式は$a,b,c,d,e$を$2\\,a, 2\\,b, a+b, 2\\,a+2\\,b, a+b+\\frac12$に置き換えることで得られます。"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "719a9ada-ef51-4708-b590-3ace95ffa069",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{2}$}\\left(x-1\\right)\\,x^2\\,\\left(\\frac{d^3}{d\\,x^3}\\,G\\left(x\\right)\\right)+x\\,\\left(\\left(3\\,b+3\\,a+3\\right)\\,x-3\\,b-3\\,a-\\frac{3}{2}\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,G\\left(x\\right)\\right)+\\left(\\left(2\\,b\\,\\left(b+a\\right)+2\\,a\\,\\left(b+a\\right)+4\\,a\\,b+3\\,b+3\\,a+1\\right)\\,x-\\left(b+a+\\frac{1}{2}\\right)\\,\\left(2\\,b+2\\,a\\right)\\right)\\,\\left(\\frac{d}{d\\,x}\\,G\\left(x\\right)\\right)+4\\,a\\,b\\,\\left(b+a\\right)\\,G\\left(x\\right)=0\\]"
],
"text/plain": [
" 3\n",
" 2 d 3\n",
"(%o2) (x - 1) x (--- (G(x))) + x ((3 b + 3 a + 3) x - 3 b - 3 a - -)\n",
" 3 2\n",
" dx\n",
" 2\n",
" d\n",
" (--- (G(x))) + ((2 b (b + a) + 2 a (b + a) + 4 a b + 3 b + 3 a + 1) x\n",
" 2\n",
" dx\n",
" 1 d\n",
" - (b + a + -) (2 b + 2 a)) (-- (G(x))) + 4 a b (b + a) G(x) = 0\n",
" 2 dx"
],
"text/x-maxima": [
"(x-1)*x^2*'diff(G(x),x,3)+x*((3*b+3*a+3)*x-3*b-3*a-3/2)*'diff(G(x),x,2)\n",
" +((2*b*(b+a)+2*a*(b+a)+4*a*b+3*b+3*a+1)*x\n",
" -(b+a+1/2)*(2*b+2*a))\n",
" *'diff(G(x),x,1)+4*a*b*(b+a)*G(x)\n",
" = 0"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF21:subst_parallel([a=2*a,b=2*b,c=a+b,d=2*a+2*b,e=a+b+1/2],DF2);"
]
},
{
"cell_type": "markdown",
"id": "4b0641b5-713c-4b39-a5e6-58d1ca51b243",
"metadata": {},
"source": [
"自乗すると$G(x)$になる関数$F(x)$の満たす微分方程式は、DF21の$G(x)$に$F(x)^2$を代入して整理すれば得られます。これをDF22とします。ここまでで定理の右辺の処理は終わりです。"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "502553dd-9239-4bbd-afa4-d72263a9d3bb",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{3}$}\\left(2\\,x^3-2\\,x^2\\right)\\,F\\left(x\\right)\\,\\left(\\frac{d^3}{d\\,x^3}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,x^3-6\\,x^2\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,b+6\\,a+6\\right)\\,x^2+\\left(-6\\,b-6\\,a-3\\right)\\,x\\right)\\,F\\left(x\\right)\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,b+6\\,a+6\\right)\\,x^2+\\left(-6\\,b-6\\,a-3\\right)\\,x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)^2+\\left(\\left(4\\,b^2+\\left(16\\,a+6\\right)\\,b+4\\,a^2+6\\,a+2\\right)\\,x-4\\,b^2+\\left(-8\\,a-2\\right)\\,b-4\\,a^2-2\\,a\\right)\\,F\\left(x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+\\left(4\\,a\\,b^2+4\\,a^2\\,b\\right)\\,F\\left(x\\right)^2=0\\]"
],
"text/plain": [
" 3\n",
" 3 2 d 3 2 d\n",
"(%o3) (2 x - 2 x ) F(x) (--- (F(x))) + ((6 x - 6 x ) (-- (F(x)))\n",
" 3 dx\n",
" dx\n",
" 2\n",
" 2 d\n",
" + ((6 b + 6 a + 6) x + ((- 6 b) - 6 a - 3) x) F(x)) (--- (F(x)))\n",
" 2\n",
" dx\n",
" 2 d 2\n",
" + ((6 b + 6 a + 6) x + ((- 6 b) - 6 a - 3) x) (-- (F(x)))\n",
" dx\n",
" 2 2 2 2\n",
" + ((4 b + (16 a + 6) b + 4 a + 6 a + 2) x - 4 b + ((- 8 a) - 2) b - 4 a\n",
" d 2 2 2\n",
" - 2 a) F(x) (-- (F(x))) + (4 a b + 4 a b) F (x) = 0\n",
" dx"
],
"text/x-maxima": [
"(2*x^3-2*x^2)*F(x)*'diff(F(x),x,3)+((6*x^3-6*x^2)*'diff(F(x),x,1)\n",
" +((6*b+6*a+6)*x^2+((-6*b)-6*a-3)*x)*F(x))\n",
" *'diff(F(x),x,2)\n",
" +((6*b+6*a+6)*x^2+((-6*b)-6*a-3)*x)\n",
" *('diff(F(x),x,1))^2\n",
" +((4*b^2+(16*a+6)*b+4*a^2+6*a+2)*x\n",
" -4*b^2+((-8*a)-2)*b-4*a^2-2*a)\n",
" *F(x)*'diff(F(x),x,1)\n",
" +(4*a*b^2+4*a^2*b)*F(x)^2\n",
" = 0"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF22:DF21,G(x):=F(x)^2,nouns,ratsimp;"
]
},
{
"cell_type": "markdown",
"id": "1adf6b69-d5d1-47a8-bf26-0c7f5efb580a",
"metadata": {},
"source": [
"次は定理の左辺の処理に取り掛かります。以前の記事でガウスの超幾何関数$F(x)={}_2F_1(a,b;c;x)$を満たす微分方程式を求めました。これをDF1とします。またDF1に$x$をかけたものをDF11とします。さらにDF11を$x$で微分したものをDF12とします。"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "168adfd0-b37e-4eef-a33c-e28015445149",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{4}$}\\left(x^2-x\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(b+a+1\\right)\\,x-c\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+a\\,b\\,F\\left(x\\right)=0\\]"
],
"text/plain": [
" 2\n",
" 2 d d\n",
"(%o4) (x - x) (--- (F(x))) + ((b + a + 1) x - c) (-- (F(x))) + a b F(x) = 0\n",
" 2 dx\n",
" dx"
],
"text/x-maxima": [
"(x^2-x)*'diff(F(x),x,2)+((b+a+1)*x-c)*'diff(F(x),x,1)+a*b*F(x) = 0"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF1:(x^2-x)*diff(F(x),x,2)+((a+b+1)*x-c)*diff(F(x),x)+a*b*F(x)=0;"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "791590e3-50a3-433e-9557-01f0fe320b2e",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{5}$}\\left(x^3-x^2\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(b+a+1\\right)\\,x^2-c\\,x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+a\\,b\\,x\\,F\\left(x\\right)=0\\]"
],
"text/plain": [
" 2\n",
" 3 2 d 2 d\n",
"(%o5) (x - x ) (--- (F(x))) + ((b + a + 1) x - c x) (-- (F(x)))\n",
" 2 dx\n",
" dx\n",
" + a b x F(x) = 0"
],
"text/x-maxima": [
"(x^3-x^2)*'diff(F(x),x,2)+((b+a+1)*x^2-c*x)*'diff(F(x),x,1)+a*b*x*F(x) = 0"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF11:DF1*x,ratsimp;"
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "e6f47f1f-7e7e-4f7e-a1de-036786733b6f",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{6}$}\\left(x^3-x^2\\right)\\,\\left(\\frac{d^3}{d\\,x^3}\\,F\\left(x\\right)\\right)+\\left(\\left(b+a+4\\right)\\,x^2+\\left(-c-2\\right)\\,x\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(\\left(a+2\\right)\\,b+2\\,a+2\\right)\\,x-c\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+a\\,b\\,F\\left(x\\right)=0\\]"
],
"text/plain": [
" 3 2\n",
" 3 2 d 2 d\n",
"(%o6) (x - x ) (--- (F(x))) + ((b + a + 4) x + ((- c) - 2) x) (--- (F(x)))\n",
" 3 2\n",
" dx dx\n",
" d\n",
" + (((a + 2) b + 2 a + 2) x - c) (-- (F(x))) + a b F(x) = 0\n",
" dx"
],
"text/x-maxima": [
"(x^3-x^2)*'diff(F(x),x,3)+((b+a+4)*x^2+((-c)-2)*x)*'diff(F(x),x,2)\n",
" +(((a+2)*b+2*a+2)*x-c)*'diff(F(x),x,1)+a*b*F(x)\n",
" = 0"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF12:diff(DF11,x),ratsimp;"
]
},
{
"cell_type": "markdown",
"id": "26bdd2ce-d939-4387-9c60-afcfbb938955",
"metadata": {},
"source": [
"次のステップが若干謎なのですが、上記のDF1, DF11, DF12を素材として$(2\\,a+2\\,b-1)\\,2\\,F(x)\\,DF1 + 6\\,\\frac{d\\,F(x)}{d\\,x}\\,DF11 + 2\\,F(x)\\,DF12$という微分方程式を構成してみます。ちなみにこのような作り方ですから元の$F(x)={}_{2}F_1(a,b;c;x)$がこの微分方程式を満たすことは明らかです。"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "a334382c-cc3b-468b-bfef-1765889724c5",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{7}$}2\\,F\\left(x\\right)\\,\\left(\\left(x^3-x^2\\right)\\,\\left(\\frac{d^3}{d\\,x^3}\\,F\\left(x\\right)\\right)+\\left(\\left(b+a+4\\right)\\,x^2+\\left(-c-2\\right)\\,x\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(\\left(a+2\\right)\\,b+2\\,a+2\\right)\\,x-c\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+a\\,b\\,F\\left(x\\right)\\right)+6\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)\\,\\left(\\left(x^3-x^2\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(b+a+1\\right)\\,x^2-c\\,x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+a\\,b\\,x\\,F\\left(x\\right)\\right)+2\\,\\left(2\\,b+2\\,a-1\\right)\\,F\\left(x\\right)\\,\\left(\\left(x^2-x\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(b+a+1\\right)\\,x-c\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+a\\,b\\,F\\left(x\\right)\\right)=0\\]"
],
"text/plain": [
" 3\n",
" 3 2 d 2\n",
"(%o7) 2 F(x) ((x - x ) (--- (F(x))) + ((b + a + 4) x + ((- c) - 2) x)\n",
" 3\n",
" dx\n",
" 2\n",
" d d\n",
" (--- (F(x))) + (((a + 2) b + 2 a + 2) x - c) (-- (F(x))) + a b F(x))\n",
" 2 dx\n",
" dx\n",
" 2\n",
" d 3 2 d 2 d\n",
" + 6 (-- (F(x))) ((x - x ) (--- (F(x))) + ((b + a + 1) x - c x) (-- (F(x)))\n",
" dx 2 dx\n",
" dx\n",
" + a b x F(x)) + 2 (2 b + 2 a - 1) F(x)\n",
" 2\n",
" 2 d d\n",
" ((x - x) (--- (F(x))) + ((b + a + 1) x - c) (-- (F(x))) + a b F(x)) = 0\n",
" 2 dx\n",
" dx"
],
"text/x-maxima": [
"2*F(x)\n",
" *((x^3-x^2)*'diff(F(x),x,3)+((b+a+4)*x^2+((-c)-2)*x)*'diff(F(x),x,2)\n",
" +(((a+2)*b+2*a+2)*x-c)*'diff(F(x),x,1)+a*b*F(x))\n",
" +6*'diff(F(x),x,1)\n",
" *((x^3-x^2)*'diff(F(x),x,2)+((b+a+1)*x^2-c*x)*'diff(F(x),x,1)+a*b*x*F(x))\n",
" +2*(2*b+2*a-1)*F(x)\n",
" *((x^2-x)*'diff(F(x),x,2)+((b+a+1)*x-c)*'diff(F(x),x,1)+a*b*F(x))\n",
" = 0"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF13:(2*a+2*b-1)*2*F(x)*DF1+6*diff(F(x),x)*DF11+2*F(x)*DF12;"
]
},
{
"cell_type": "markdown",
"id": "7929a4fb-69ab-4dec-8235-27bf75a0fdce",
"metadata": {},
"source": [
"若干整理します。"
]
},
{
"cell_type": "code",
"execution_count": 9,
"id": "65c44ac4-8e59-422c-a6f2-015dfbf5fe9e",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{8}$}\\left(2\\,x^3-2\\,x^2\\right)\\,F\\left(x\\right)\\,\\left(\\frac{d^3}{d\\,x^3}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,x^3-6\\,x^2\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,b+6\\,a+6\\right)\\,x^2+\\left(-2\\,c-4\\,b-4\\,a-2\\right)\\,x\\right)\\,F\\left(x\\right)\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,b+6\\,a+6\\right)\\,x^2-6\\,c\\,x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)^2+\\left(\\left(4\\,b^2+\\left(16\\,a+6\\right)\\,b+4\\,a^2+6\\,a+2\\right)\\,x+\\left(-4\\,b-4\\,a\\right)\\,c\\right)\\,F\\left(x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+\\left(4\\,a\\,b^2+4\\,a^2\\,b\\right)\\,F\\left(x\\right)^2=0\\]"
],
"text/plain": [
" 3\n",
" 3 2 d 3 2 d\n",
"(%o8) (2 x - 2 x ) F(x) (--- (F(x))) + ((6 x - 6 x ) (-- (F(x)))\n",
" 3 dx\n",
" dx\n",
" 2\n",
" 2 d\n",
" + ((6 b + 6 a + 6) x + ((- 2 c) - 4 b - 4 a - 2) x) F(x)) (--- (F(x)))\n",
" 2\n",
" dx\n",
" 2 d 2\n",
" + ((6 b + 6 a + 6) x - 6 c x) (-- (F(x)))\n",
" dx\n",
" 2 2\n",
" + ((4 b + (16 a + 6) b + 4 a + 6 a + 2) x + ((- 4 b) - 4 a) c) F(x)\n",
" d 2 2 2\n",
" (-- (F(x))) + (4 a b + 4 a b) F (x) = 0\n",
" dx"
],
"text/x-maxima": [
"(2*x^3-2*x^2)*F(x)*'diff(F(x),x,3)+((6*x^3-6*x^2)*'diff(F(x),x,1)\n",
" +((6*b+6*a+6)*x^2+((-2*c)-4*b-4*a-2)*x)\n",
" *F(x))\n",
" *'diff(F(x),x,2)\n",
" +((6*b+6*a+6)*x^2-6*c*x)*('diff(F(x),x,1))^2\n",
" +((4*b^2+(16*a+6)*b+4*a^2+6*a+2)*x\n",
" +((-4*b)-4*a)*c)\n",
" *F(x)*'diff(F(x),x,1)\n",
" +(4*a*b^2+4*a^2*b)*F(x)^2\n",
" = 0"
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF14:ratsimp(DF13);"
]
},
{
"cell_type": "markdown",
"id": "94cacaa9-c629-47d1-93e8-4d4db88e3923",
"metadata": {},
"source": [
"少し特別な$c=a+b+\\frac12$の場合、$F(x)={}_{2}F_1(a,b;a+b+\\frac12;x)$は次の微分方程式DF15を満たします。"
]
},
{
"cell_type": "code",
"execution_count": 10,
"id": "d863b3e5-d8b5-41ff-9808-59d9fb57137e",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{9}$}\\left(2\\,x^3-2\\,x^2\\right)\\,F\\left(x\\right)\\,\\left(\\frac{d^3}{d\\,x^3}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,x^3-6\\,x^2\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,b+6\\,a+6\\right)\\,x^2+\\left(-6\\,b-6\\,a-3\\right)\\,x\\right)\\,F\\left(x\\right)\\right)\\,\\left(\\frac{d^2}{d\\,x^2}\\,F\\left(x\\right)\\right)+\\left(\\left(6\\,b+6\\,a+6\\right)\\,x^2+\\left(-6\\,b-6\\,a-3\\right)\\,x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)^2+\\left(\\left(4\\,b^2+\\left(16\\,a+6\\right)\\,b+4\\,a^2+6\\,a+2\\right)\\,x-4\\,b^2+\\left(-8\\,a-2\\right)\\,b-4\\,a^2-2\\,a\\right)\\,F\\left(x\\right)\\,\\left(\\frac{d}{d\\,x}\\,F\\left(x\\right)\\right)+\\left(4\\,a\\,b^2+4\\,a^2\\,b\\right)\\,F\\left(x\\right)^2=0\\]"
],
"text/plain": [
" 3\n",
" 3 2 d 3 2 d\n",
"(%o9) (2 x - 2 x ) F(x) (--- (F(x))) + ((6 x - 6 x ) (-- (F(x)))\n",
" 3 dx\n",
" dx\n",
" 2\n",
" 2 d\n",
" + ((6 b + 6 a + 6) x + ((- 6 b) - 6 a - 3) x) F(x)) (--- (F(x)))\n",
" 2\n",
" dx\n",
" 2 d 2\n",
" + ((6 b + 6 a + 6) x + ((- 6 b) - 6 a - 3) x) (-- (F(x)))\n",
" dx\n",
" 2 2 2 2\n",
" + ((4 b + (16 a + 6) b + 4 a + 6 a + 2) x - 4 b + ((- 8 a) - 2) b - 4 a\n",
" d 2 2 2\n",
" - 2 a) F(x) (-- (F(x))) + (4 a b + 4 a b) F (x) = 0\n",
" dx"
],
"text/x-maxima": [
"(2*x^3-2*x^2)*F(x)*'diff(F(x),x,3)+((6*x^3-6*x^2)*'diff(F(x),x,1)\n",
" +((6*b+6*a+6)*x^2+((-6*b)-6*a-3)*x)*F(x))\n",
" *'diff(F(x),x,2)\n",
" +((6*b+6*a+6)*x^2+((-6*b)-6*a-3)*x)\n",
" *('diff(F(x),x,1))^2\n",
" +((4*b^2+(16*a+6)*b+4*a^2+6*a+2)*x\n",
" -4*b^2+((-8*a)-2)*b-4*a^2-2*a)\n",
" *F(x)*'diff(F(x),x,1)\n",
" +(4*a*b^2+4*a^2*b)*F(x)^2\n",
" = 0"
]
},
"execution_count": 10,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"DF15:DF14,c=a+b+1/2,ratsimp;"
]
},
{
"cell_type": "code",
"execution_count": 11,
"id": "a406adbc-a3dd-4133-8179-d8510b3ca5d9",
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"\\[\\tag{${\\it \\%o}_{10}$}0\\]"
],
"text/plain": [
"(%o10) 0"
],
"text/x-maxima": [
"0"
]
},
"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"lhs(DF22)-lhs(DF15),expand;"
]
},
{
"cell_type": "markdown",
"id": "bf11c724-b1af-4a46-b50d-7e4b3cc7350a",
"metadata": {},
"source": [
"なんと素晴らしいことにDF22の左辺とDF15の左辺は全く等しかったのです。すなわち$F(x)={}_{2}F_1(a,b;a+b+\\frac12;x)$はDF22も満たすことがわかりました。これで$_{2}F_1\\left(a,b;a+b+\\frac12;x\\right)^2$と${}_3F_2\\left(2\\,a,2\\,b,a+b;2\\,a+2\\,b,a+b+\\frac12;x\\right)$は同じ微分方程式を満たすことがわかりました。
\n",
"\n",
"$F(x)=_{2}F_1\\left(a,b;a+b+\\frac12;x\\right)$と$G(x)={}_3F_2\\left(2\\,a,2\\,b,a+b;2\\,a+2\\,b,a+b+\\frac12;x\\right)$で$F(x)^2$と$G(x)$の初期値が十分に一致すること、つまり$F(0)^2=G(0), F'(0)^2=G'(0), F''(0)^2=G''(0), F'''(0)^2=G'''(0)$を示すことは次回にしましょう。"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "43676d9f-494e-4418-b0aa-bac88363331c",
"metadata": {},
"outputs": [],
"source": []
}
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