{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "第1種完全楕円積分\n", "$$K(k)=\\int_{0}^{\\frac{\\pi}{2}}{\\frac{1}{\\sqrt{1-k^2\\,\\sin ^2\\varphi}}\\;d\\varphi}$$\n", "に成り立つ次の関係式\n", "$$K(\\frac{2\\,\\sqrt{x}}{x+1})=(x+1)\\,K(x)$$\n", "を示します。" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{1}$}\\left[ x<1 \\right] \\]" ], "text/plain": [ "(%o1) [x < 1]" ], "text/x-maxima": [ "[x < 1]" ] }, "execution_count": 1, "metadata": {}, "output_type": "execute_result" }, { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{2}$}\\left[ x>-1 \\right] \\]" ], "text/plain": [ "(%o2) [x > - 1]" ], "text/x-maxima": [ "[x > -1]" ] }, "execution_count": 1, "metadata": {}, "output_type": "execute_result" } ], "source": [ "kill(all)$\n", "assume(x<1);\n", "assume(-1 ------------------------- + 1\n", " / 1\n", " ==== gamma(- - n) gamma(n + 1)\n", " n = 1 2" ], "text/x-maxima": [ "sqrt(%pi)*'sum(t^n/(gamma(1/2-n)*gamma(n+1)),n,1,inf)+1" ] }, "execution_count": 11, "metadata": {}, "output_type": "execute_result" } ], "source": [ "niceindices(makegamma(powerseries(1/sqrt(1+t),t,0))),niceindicespref:[n];" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "最初の項は$t=e^{-2\\,i\\,\\varphi}\\,x$なのでこの代入を行います。" ] }, { "cell_type": "code", "execution_count": 12, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{13}$}\\sqrt{\\pi}\\,\\sum_{n=1}^{\\infty }{\\frac{e^ {- 2\\,i\\,n\\,\\varphi }\\,x^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}+1\\]" ], "text/plain": [ " inf\n", " ==== - 2 %i n phi n\n", " \\ %e x\n", "(%o13) sqrt(%pi) > ------------------------- + 1\n", " / 1\n", " ==== gamma(- - n) gamma(n + 1)\n", " n = 1 2" ], "text/x-maxima": [ "sqrt(%pi)*'sum((%e^-(2*%i*n*phi)*x^n)/(gamma(1/2-n)*gamma(n+1)),n,1,inf)+1" ] }, "execution_count": 12, "metadata": {}, "output_type": "execute_result" } ], "source": [ "%,t:exp(-2*%i*phi)*x;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "第1項で$n=0$の場合1になることから第2項の1を総和の中に取り込みます。" ] }, { "cell_type": "code", "execution_count": 13, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{14}$}\\sqrt{\\pi}\\,\\sum_{n=0}^{\\infty }{\\frac{e^ {- 2\\,i\\,n\\,\\varphi }\\,x^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]" ], "text/plain": [ " inf\n", " ==== - 2 %i n phi n\n", " \\ %e x\n", "(%o14) sqrt(%pi) > -------------------------\n", " / 1\n", " ==== gamma(- - n) gamma(n + 1)\n", " n = 0 2" ], "text/x-maxima": [ "sqrt(%pi)*'sum((%e^-(2*%i*n*phi)*x^n)/(gamma(1/2-n)*gamma(n+1)),n,0,inf)" ] }, "execution_count": 13, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F8:substpart(0,part(%,1),2,3);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "総和のインデックス変数を$m$に変えて冪級数展開を求めます。" ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{15}$}\\sqrt{\\pi}\\,\\sum_{m=1}^{\\infty }{\\frac{t^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}+1\\]" ], "text/plain": [ " inf\n", " ==== m\n", " \\ t\n", "(%o15) sqrt(%pi) > ------------------------- + 1\n", " / 1\n", " ==== gamma(- - m) gamma(m + 1)\n", " m = 1 2" ], "text/x-maxima": [ "sqrt(%pi)*'sum(t^m/(gamma(1/2-m)*gamma(m+1)),m,1,inf)+1" ] }, "execution_count": 14, "metadata": {}, "output_type": "execute_result" } ], "source": [ "niceindices(makegamma(powerseries(1/sqrt(1+t),t,0))),niceindicespref:[m];" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "F7の被積分関数の第2項なので$t=e^{2\\,i\\,\\varphi}\\,x$の代入を行います。" ] }, { "cell_type": "code", "execution_count": 15, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{16}$}\\sqrt{\\pi}\\,\\sum_{m=1}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}+1\\]" ], "text/plain": [ " inf\n", " ==== 2 %i m phi m\n", " \\ %e x\n", "(%o16) sqrt(%pi) > ------------------------- + 1\n", " / 1\n", " ==== gamma(- - m) gamma(m + 1)\n", " m = 1 2" ], "text/x-maxima": [ "sqrt(%pi)*'sum((%e^(2*%i*m*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,1,inf)+1" ] }, "execution_count": 15, "metadata": {}, "output_type": "execute_result" } ], "source": [ "%,t:exp(2*%i*phi)*x;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "第2項の1を総和の$m=0$として取り込みます。" ] }, { "cell_type": "code", "execution_count": 16, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{17}$}\\sqrt{\\pi}\\,\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]" ], "text/plain": [ " inf\n", " ==== 2 %i m phi m\n", " \\ %e x\n", "(%o17) sqrt(%pi) > -------------------------\n", " / 1\n", " ==== gamma(- - m) gamma(m + 1)\n", " m = 0 2" ], "text/x-maxima": [ "sqrt(%pi)*'sum((%e^(2*%i*m*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,0,inf)" ] }, "execution_count": 16, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F9:substpart(0,part(%,1),2,3);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "従ってF7の被積分関数は次の式に等しくなります。" ] }, { "cell_type": "code", "execution_count": 17, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{18}$}\\pi\\,\\left(\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\right)\\,\\sum_{n=0}^{\\infty }{\\frac{e^ {- 2\\,i\\,n\\,\\varphi }\\,x^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]" ], "text/plain": [ " inf inf\n", " ==== 2 %i m phi m ==== - 2 %i n phi n\n", " \\ %e x \\ %e x\n", "(%o18) %pi ( > -------------------------) > -------------------------\n", " / 1 / 1\n", " ==== gamma(- - m) gamma(m + 1) ==== gamma(- - n) gamma(n + 1)\n", " m = 0 2 n = 0 2" ], "text/x-maxima": [ "%pi*('sum((%e^(2*%i*m*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,0,inf))\n", " *'sum((%e^-(2*%i*n*phi)*x^n)/(gamma(1/2-n)*gamma(n+1)),n,0,inf)" ] }, "execution_count": 17, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F8*F9;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "この式は2つの無限級数の積になっており、展開することで$0\\le n, m\\lt \\infty$の全ての組み合わせについて積を取ることになります。和の順番を変更し入れ子にすることで次の式と等しいことが分かります。" ] }, { "cell_type": "code", "execution_count": 18, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{19}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{n+m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]" ], "text/plain": [ " inf\n", " ==== 2 %i m phi - 2 %i n phi n + m\n", " \\ %e x\n", " > --------------------------------\n", " inf / 1\n", " ==== ==== gamma(- - m) gamma(m + 1)\n", " \\ m = 0 2\n", "(%o19) %pi > --------------------------------------\n", " / 1\n", " ==== gamma(- - n) gamma(n + 1)\n", " n = 0 2" ], "text/x-maxima": [ "%pi*'sum(('sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^(n+m))/(gamma(1/2-m)*gamma(m+1)),\n", " m,0,inf))\n", " /(gamma(1/2-n)*gamma(n+1)),n,0,inf)" ] }, "execution_count": 18, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F10:%pi*sum(sum(part(F8,2,1)*part(F9,2,1),m,0,inf),n,0,inf);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "2重総和の内側の総和を取り出します。" ] }, { "cell_type": "code", "execution_count": 19, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{20}$}\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{n+m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]" ], "text/plain": [ " inf\n", " ==== 2 %i m phi - 2 %i n phi n + m\n", " \\ %e x\n", "(%o20) > --------------------------------\n", " / 1\n", " ==== gamma(- - m) gamma(m + 1)\n", " m = 0 2" ], "text/x-maxima": [ "'sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^(n+m))/(gamma(1/2-m)*gamma(m+1)),m,0,inf)" ] }, "execution_count": 19, "metadata": {}, "output_type": "execute_result" } ], "source": [ "G1:part(F10,2,1,1);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "$x^n$は$m$と無関係なので総和の外にくくり出すことが来ます。" ] }, { "cell_type": "code", "execution_count": 20, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{21}$}x^{n}\\,\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]" ], "text/plain": [ " inf\n", " ==== 2 %i m phi - 2 %i n phi m\n", " n \\ %e x\n", "(%o21) x > ----------------------------\n", " / 1\n", " ==== gamma(- - m) gamma(m + 1)\n", " m = 0 2" ], "text/x-maxima": [ "x^n*'sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,0,inf)" ] }, "execution_count": 20, "metadata": {}, "output_type": "execute_result" } ], "source": [ "G2:x^n*substpart(part(G1,1)/x^n,G1,1);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "従ってF10は次の式と等しくなります。" ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{22}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{x^{n}\\,\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]" ], "text/plain": [ " inf\n", " ==== 2 %i m phi - 2 %i n phi m\n", " n \\ %e x\n", " x > ----------------------------\n", " inf / 1\n", " ==== ==== gamma(- - m) gamma(m + 1)\n", " \\ m = 0 2\n", "(%o22) %pi > -------------------------------------\n", " / 1\n", " ==== gamma(- - n) gamma(n + 1)\n", " n = 0 2" ], "text/x-maxima": [ "%pi*'sum((x^n*'sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),\n", " m,0,inf))\n", " /(gamma(1/2-n)*gamma(n+1)),n,0,inf)" ] }, "execution_count": 21, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F11:substpart(G2,F10,2,1,1);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "F7の被積分関数がこの2重総和と等しいことが分かりました。積分変数$\\varphi$でこの式を積分をすると、積分変数に関係する部分は以下の部分式だけです。" ] }, { "cell_type": "code", "execution_count": 22, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{23}$}e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\]" ], "text/plain": [ " 2 %i m phi - 2 %i n phi\n", "(%o23) %e" ], "text/x-maxima": [ "%e^(2*%i*m*phi-2*%i*n*phi)" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "G3:part(F11,2,1,1,2,1,1,1);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "$n,m$はどちらも0以上の整数であることに注意します。すると積分の値はm=nの場合にはその値に関わらず$\\pi$になります。" ] }, { "cell_type": "code", "execution_count": 23, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{24}$}\\pi\\]" ], "text/plain": [ "(%o24) %pi" ], "text/x-maxima": [ "%pi" ] }, "execution_count": 23, "metadata": {}, "output_type": "execute_result" } ], "source": [ "integrate(G3,phi,0,%pi),m=k,n=k;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "また$n\\neq m$の場合にはそれらの値に関わらず積分の値は0になります。例えば$m=4, n=9$で試しに計算してみます(これが成り立つのは積分範囲が$0\\le\\varphi\\le\\pi$の場合です。$0\\le\\varphi\\le\\frac{\\pi}{2}$では上手くいきません)。" ] }, { "cell_type": "code", "execution_count": 24, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{25}$}0\\]" ], "text/plain": [ "(%o25) 0" ], "text/x-maxima": [ "0" ] }, "execution_count": 24, "metadata": {}, "output_type": "execute_result" } ], "source": [ "integrate(G3,phi,0,%pi),m=4,n=9;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "これらのことからF11を$\\varphi$で積分すると内側の総和は$m=n$の場合だけ積分の値が$\\pi$として残り、それ以外の$m,n$の組み合わせは積分値が0になることから消えてしまいます。従ってF11の2重総和の内側の総和は次の式と等しいことが分かります。" ] }, { "cell_type": "code", "execution_count": 25, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{26}$}\\pi\\,x^{n}\\,\\sum_{m=n}^{n}{\\frac{x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]" ], "text/plain": [ " n\n", " ==== m\n", " n \\ x\n", "(%o26) %pi x > -------------------------\n", " / 1\n", " ==== gamma(- - m) gamma(m + 1)\n", " m = n 2" ], "text/x-maxima": [ "%pi*x^n*'sum(x^m/(gamma(1/2-m)*gamma(m+1)),m,n,n)" ] }, "execution_count": 25, "metadata": {}, "output_type": "execute_result" } ], "source": [ "G4:substpart(%pi,substpart(n,substpart(n,G2,2,3),2,4),2,1,1,1);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "この総和は自明に簡約することが出来ます。" ] }, { "cell_type": "code", "execution_count": 26, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{27}$}\\frac{\\pi\\,x^{2\\,n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}\\]" ], "text/plain": [ " 2 n\n", " %pi x\n", "(%o27) -------------------------\n", " 1\n", " gamma(- - n) gamma(n + 1)\n", " 2" ], "text/x-maxima": [ "(%pi*x^(2*n))/(gamma(1/2-n)*gamma(n+1))" ] }, "execution_count": 26, "metadata": {}, "output_type": "execute_result" } ], "source": [ "G5:G4,simpsum:true;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "上記のG5の式で内側の総和を置き換えることでF7の積分は以下の式に変形できます。" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{28}$}\\pi^2\\,\\sum_{n=0}^{\\infty }{\\frac{x^{2\\,n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)^2\\,\\Gamma\\left(n+1\\right)^2}}\\]" ], "text/plain": [ " inf\n", " ==== 2 n\n", " 2 \\ x\n", "(%o28) %pi > ---------------------------\n", " / 2 1 2\n", " ==== gamma (- - n) gamma (n + 1)\n", " n = 0 2" ], "text/x-maxima": [ "%pi^2*'sum(x^(2*n)/(gamma(1/2-n)^2*gamma(n+1)^2),n,0,inf)" ] }, "execution_count": 27, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F12:substpart(G5,F11,2,1,1);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "以下はこの式を超幾何関数の形に変形していきます。\n", "まずポッホハマー記号$\\left(\\frac{1}{2}\\right)_n$を思い出します。" ] }, { "cell_type": "code", "execution_count": 28, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{29}$}\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}\\]" ], "text/plain": [ " 1\n", "(%o29) (-)\n", " 2 n" ], "text/x-maxima": [ "pochhammer(1/2,n)" ] }, "execution_count": 28, "metadata": {}, "output_type": "execute_result" }, { "name": "stdout", "output_type": "stream", "text": [ "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-UNIT-STEP in DEFUN\n", "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-POCHHAMMER in DEFUN\n" ] } ], "source": [ "pochhammer(1/2,n);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "$\\left(\\frac{1}{2}\\right)_n$を単純にガンマ関数で表示すると以下のようになりますが、これではちょっと上手くいきません。" ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{30}$}\\frac{\\Gamma\\left(n+\\frac{1}{2}\\right)}{\\sqrt{\\pi}}\\]" ], "text/plain": [ " 1\n", " gamma(n + -)\n", " 2\n", "(%o30) ------------\n", " sqrt(%pi)" ], "text/x-maxima": [ "gamma(n+1/2)/sqrt(%pi)" ] }, "execution_count": 29, "metadata": {}, "output_type": "execute_result" } ], "source": [ "makegamma(%);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "ポッホハマー記号について次の等式が成り立ちます。" ] }, { "cell_type": "code", "execution_count": 30, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{31}$}\\left(\\left(1-x\\right)\\right)_{n}=\\frac{\\left(-1\\right)^{n}\\,\\Gamma\\left(x\\right)}{\\Gamma\\left(x-n\\right)}\\]" ], "text/plain": [ " n\n", " (- 1) gamma(x)\n", "(%o31) (1 - x) = ---------------\n", " n gamma(x - n)" ], "text/x-maxima": [ "pochhammer(1-x,n) = ((-1)^n*gamma(x))/gamma(x-n)" ] }, "execution_count": 30, "metadata": {}, "output_type": "execute_result" } ], "source": [ "PoId:pochhammer(1-x,n)=gamma(x)*(-1)^n/gamma(x-n);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "この等式で$x=\\frac{1}{2}$を代入すると次の式を得ます。" ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{32}$}\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}=\\frac{\\sqrt{\\pi}\\,\\left(-1\\right)^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)}\\]" ], "text/plain": [ " n\n", " 1 sqrt(%pi) (- 1)\n", "(%o32) (-) = ----------------\n", " 2 n 1\n", " gamma(- - n)\n", " 2" ], "text/x-maxima": [ "pochhammer(1/2,n) = (sqrt(%pi)*(-1)^n)/gamma(1/2-n)" ] }, "execution_count": 31, "metadata": {}, "output_type": "execute_result" } ], "source": [ "PoId,x:1/2;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "$\\Gamma\\left(\\frac{1}{2}-n\\right)$を作り出すことが出来ました。この式をこのガンマ関数について解いてみます。" ] }, { "cell_type": "code", "execution_count": 32, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{33}$}\\left[ \\Gamma\\left(\\frac{1}{2}-n\\right)=\\frac{\\sqrt{\\pi}\\,\\left(-1\\right)^{n}}{\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}} \\right] \\]" ], "text/plain": [ " n\n", " 1 sqrt(%pi) (- 1)\n", "(%o33) [gamma(- - n) = ----------------]\n", " 2 1\n", " (-)\n", " 2 n" ], "text/x-maxima": [ "[gamma(1/2-n) = (sqrt(%pi)*(-1)^n)/pochhammer(1/2,n)]" ] }, "execution_count": 32, "metadata": {}, "output_type": "execute_result" } ], "source": [ "PoId2:solve(%,gamma(1/2-n));" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "この式をF13に代入します。" ] }, { "cell_type": "code", "execution_count": 33, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{34}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}^2\\,x^{2\\,n}}{\\Gamma\\left(n+1\\right)^2}}\\]" ], "text/plain": [ " 2\n", " 1 2 n\n", " inf (-) x\n", " ==== 2 n\n", " \\\n", "(%o34) %pi > -------------\n", " / 2\n", " ==== gamma (n + 1)\n", " n = 0" ], "text/x-maxima": [ "%pi*'sum((pochhammer(1/2,n)^2*x^(2*n))/gamma(n+1)^2,n,0,inf)" ] }, "execution_count": 33, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F13:F12,PoId2;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "ガンマ関数を階乗に直してみます。" ] }, { "cell_type": "code", "execution_count": 34, "metadata": {}, "outputs": [ { "data": { "text/latex": [ "\\[\\tag{${\\it \\%o}_{35}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}^2\\,x^{2\\,n}}{n!^2}}\\]" ], "text/plain": [ " 2\n", " 1 2 n\n", " inf (-) x\n", " ==== 2 n\n", " \\\n", "(%o35) %pi > ----------\n", " / 2\n", " ==== n!\n", " n = 0" ], "text/x-maxima": [ "%pi*'sum((pochhammer(1/2,n)^2*x^(2*n))/n!^2,n,0,inf)" ] }, "execution_count": 34, "metadata": {}, "output_type": "execute_result" } ], "source": [ "makefact(F13);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "F7の積分を変形して上記の式が得られました。この式を超幾何関数の定義と見比べれば\n", "$$\\pi\\,F\\left(\\frac{1}{2},\\frac{1}{2};1,x^2\\right)$$\n", "と等しいことが分かります。\n", " \n", "忘れずに$(x+1)$を掛けて2で割り\n", "$$(x+1)\\,\\frac{\\pi}{2}\\,F\\left(\\frac{1}{2},\\frac{1}{2};1,x^2\\right)$$\n", "を得ます。\n", "\n", "既に証明した\n", "$$\\int_{0}^{\\frac{\\pi}{2}}{\\frac{1}{\\sqrt{1-k^2\\,\\sin ^2\\vartheta}}\\;d\\vartheta}=\\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,k^2\\right)}{2}$$\n", "と合わせると、\n", "$$(x+1)\\,K(x)$$\n", "となります。というわけで、\n", "$$K(\\frac{2\\,\\sqrt{x}}{x+1})=(x+1)\\,K(x)$$\n", "が証明できました。" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Maxima", "language": "maxima", "name": "maxima" }, "language_info": { "codemirror_mode": "maxima", "file_extension": ".mac", "mimetype": "text/x-maxima", "name": "maxima", "pygments_lexer": "maxima", "version": "5.45.1" } }, "nbformat": 4, "nbformat_minor": 4 }