{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# 解析学A 2018\n",
"\n",
"黒木玄\n",
"\n",
"2018年7月30日(月)\n",
"\n",
"次のリンク先で綺麗に閲覧できる:\n",
"\n",
"* http://nbviewer.jupyter.org/github/genkuroki/Calculus/blob/master/Analysis%20A%202018.ipynb\n",
"\n",
"$\n",
"\\newcommand\\eps{\\varepsilon}\n",
"\\newcommand\\ds{\\displaystyle}\n",
"\\newcommand\\Z{{\\mathbb Z}}\n",
"\\newcommand\\R{{\\mathbb R}}\n",
"\\newcommand\\C{{\\mathbb C}}\n",
"\\newcommand\\QED{\\text{□}}\n",
"\\newcommand\\root{\\sqrt}\n",
"\\newcommand\\bra{\\langle}\n",
"\\newcommand\\ket{\\rangle}\n",
"\\newcommand\\d{\\partial}\n",
"\\newcommand\\sech{\\operatorname{sech}}\n",
"\\newcommand\\cosec{\\operatorname{cosec}}\n",
"\\newcommand\\sign{\\operatorname{sign}}\n",
"\\newcommand\\sinc{\\operatorname{sinc}}\n",
"\\newcommand\\real{\\operatorname{Re}}\n",
"\\newcommand\\imag{\\operatorname{Im}}\n",
"\\newcommand\\Li{\\operatorname{Li}}\n",
"\\newcommand\\np[1]{:\\!#1\\!:}\n",
"\\newcommand\\PROD{\\mathop{\\coprod\\kern-1.35em\\prod}}\n",
"$"
]
},
{
"cell_type": "markdown",
"metadata": {
"toc": true
},
"source": [
"Table of Contents
\n",
""
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"using SymPy\n",
"Beta(x,y) = gamma(x)*gamma(y)/gamma(x+y)\n",
"# sympy[:init_printing](order=\"rev-lex\")\n",
"# sympy[:init_printing](order=\"lex\") # default\n",
"\n",
"using Plots\n",
"gr()\n",
"ENV[\"PLOTS_TEST\"] = \"true\";"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[1\\]\n",
"\n",
"### \\[1\\] (1)\n",
"\n",
"$\\tan x$ のMaclaurin展開を5次の項まで求めよう. $f(x) = \\tan x$ とおくと, $f'=1+f^2$ なので $f''=2f+2f^3$, \n",
"\n",
"$$\n",
"f'''=2+8f^2+6f^4, \\quad\n",
"f^{(4)}=16f+40f^3+24f^5, \\quad\n",
"f^{(5)}=16+136f^2+240f^4+120f^6\n",
"$$\n",
"\n",
"となりので, \n",
"\n",
"$$\n",
"f(0)=f''(0)=f^{(4)}(0)=0, \\quad\n",
"f'(0)=1, \\quad\n",
"\\frac{f'''(0)}{3!}=\\frac{2}{3!}=\\frac{1}{3}, \\quad\n",
"\\frac{f^{(5)}(0)}{5!}=\\frac{16}{5!}=\\frac{2}{15}\n",
"$$\n",
"\n",
"ゆえに, \n",
"\n",
"$$\n",
"\\tan x = x + \\frac{x^3}{3} + \\frac{2x^5}{15} + O(x^6).\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$1 + \\tan^{2}{\\left (x \\right )}$$"
],
"text/plain": [
" 2 \n",
"1 + tan (x)"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/latex": [
"$$2 \\tan{\\left (x \\right )} + 2 \\tan^{3}{\\left (x \\right )}$$"
],
"text/plain": [
" 3 \n",
"2*tan(x) + 2*tan (x)"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/latex": [
"$$2 + 8 \\tan^{2}{\\left (x \\right )} + 6 \\tan^{4}{\\left (x \\right )}$$"
],
"text/plain": [
" 2 4 \n",
"2 + 8*tan (x) + 6*tan (x)"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/latex": [
"$$16 \\tan{\\left (x \\right )} + 40 \\tan^{3}{\\left (x \\right )} + 24 \\tan^{5}{\\left (x \\right )}$$"
],
"text/plain": [
" 3 5 \n",
"16*tan(x) + 40*tan (x) + 24*tan (x)"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/latex": [
"$$16 + 136 \\tan^{2}{\\left (x \\right )} + 240 \\tan^{4}{\\left (x \\right )} + 120 \\tan^{6}{\\left (x \\right )}$$"
],
"text/plain": [
" 2 4 6 \n",
"16 + 136*tan (x) + 240*tan (x) + 120*tan (x)"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"x = symbols(\"x\")\n",
"f(x) = tan(x)\n",
"sympy[:init_printing](order=\"rev-lex\")\n",
"diff(f(x), x) |> display\n",
"diff(f(x), x, 2) |> expand |> display\n",
"diff(f(x), x, 3) |> expand |> display\n",
"diff(f(x), x, 4) |> expand |> display\n",
"diff(f(x), x, 5) |> expand |> display\n",
"sympy[:init_printing](order=\"lex\") # default"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$x + \\frac{x^{3}}{3} + \\frac{2 x^{5}}{15} + \\frac{17 x^{7}}{315} + \\frac{62 x^{9}}{2835} + O\\left(x^{10}\\right)$$"
],
"text/plain": [
" 3 5 7 9 \n",
" x 2*x 17*x 62*x / 10\\\n",
"x + -- + ---- + ----- + ----- + O\\x /\n",
" 3 15 315 2835 "
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [1] (1)\n",
"\n",
"x = symbols(\"x\")\n",
"series(tan(x), x, n=10)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### \\[1\\] (2)\n",
"\n",
"$x=0.5$ のとき, \n",
"\n",
"$$\n",
"x = 0.5, \\quad\n",
"x + \\frac{x^3}{3} = 0.541666\\cdots, \\quad\n",
"x + \\frac{x^3}{3} + \\frac{x^5}{5} = 0.548333\\cdots\n",
"$$\n",
"\n",
"これより, $\\tan 0.5 = 0.54\\cdots$ であることがわかる."
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"x = 0.5 = 0.5\n",
"tan(x) = 0.5463024898437905\n",
"x = 0.5\n",
"x + x ^ 3 / 3 = 0.5416666666666666\n",
"x + x ^ 3 / 3 + (2 * x ^ 5) / 15 = 0.5458333333333333\n",
"x + x ^ 3 / 3 + (2 * x ^ 5) / 15 + (17 * x ^ 7) / 315 = 0.5462549603174602\n"
]
}
],
"source": [
"# [1] (2)\n",
"\n",
"@show x = 0.5\n",
"@show tan(x)\n",
"@show x\n",
"@show x+x^3/3\n",
"@show x+x^3/3+2x^5/15\n",
"@show x+x^3/3+2x^5/15+17*x^7/315;"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[2\\]\n",
"\n",
"\n",
"$|x|<1$ のとき, \n",
"\n",
"$$\n",
"\\log\\frac{1+x}{1-x} = \n",
"\\log(1+x)-\\log(1-x) =\n",
"2\\sum_{k=0}^\\infty\\frac{x^{2k+1}}{2k+1} = \n",
"2x+\\frac{2x^3}{3}+\\frac{2x^5}{5}+\\cdots.\n",
"$$\n",
"\n",
"$x=1/3$ のとき, $\\ds\\frac{1+x}{1-x}=2$ であり, \n",
"\n",
"$$\n",
"2x + \\frac{2x^3}{3} = 0.691\\cdots, \\quad\n",
"\\frac{2x^5}{5} = \\frac{2}{5\\times 243} < \\frac{2}{1000} = 0.002\\cdots.\n",
"$$\n",
"\n",
"これより $\\log 2 = 0.69\\cdots$ であることがわかる.\n",
"\n",
"**解説:** $\\log 2 \\approx 0.7$ は「70\\%ルール」として役にたっている. 年利 $r\\%$ の金利の複利計算で約何年後に2倍になるかは $70/r$ でわかる. なぜならば, $\\approx(1+r/100)^x = 2$ の解は $e^{rx/100}=2$ の解で近似され, その解 $\\ds x=\\frac{100\\log 2}{r}$ は $100\\log 2\\approx 70$ より $70/r$ で近似される. $\\QED$"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$2 x + \\frac{2 x^{3}}{3} + \\frac{2 x^{5}}{5} + \\frac{2 x^{7}}{7} + \\frac{2 x^{9}}{9} + O\\left(x^{10}\\right)$$"
],
"text/plain": [
" 3 5 7 9 \n",
" 2*x 2*x 2*x 2*x / 10\\\n",
"2*x + ---- + ---- + ---- + ---- + O\\x /\n",
" 3 5 7 9 "
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [2]\n",
"\n",
"x = Sym(:x)\n",
"series(log(1+x)-log(1-x), x, n=10)"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(1 + x) / (1 - x) = 1.9999999999999998\n",
"log(2) = 0.6931471805599453\n",
"2x = 0.6666666666666666\n",
"2x + (2 * x ^ 3) / 3 = 0.691358024691358\n",
"2x + (2 * x ^ 3) / 3 + (2 * x ^ 5) / 5 = 0.6930041152263374\n",
"2x + (2 * x ^ 3) / 3 + (2 * x ^ 5) / 5 + (2 * x ^ 7) / 7 = 0.6931347573322881\n"
]
}
],
"source": [
"x = 1/3\n",
"@show (1+x)/(1-x)\n",
"@show log(2)\n",
"@show 2x\n",
"@show 2x + 2*x^3/3\n",
"@show 2x + 2*x^3/3 + 2*x^5/5\n",
"@show 2x + 2*x^3/3 + 2*x^5/5 + 2*x^7/7;"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[3\\]\n",
"\n",
"$\\ds \\int_{-\\infty}^\\infty e^{-x^2/2}\\cos(px)\\,dx = \\sqrt{2\\pi}\\;e^{-p^2/2}$\n",
"\n",
"$$\n",
"\\begin{aligned}\n",
"&\n",
"\\int_{-\\infty}^\\infty e^{-ax^2}x^{2k}\\,dx = \n",
"\\left(-\\frac{\\d}{\\d a}\\right)^k\\int_{-\\infty}^\\infty e^{-ax^2}\\,dx =\n",
"\\left(-\\frac{\\d}{\\d a}\\right)^k \\sqrt{\\pi}\\;a^{-1/2},\n",
"\\\\ &\\qquad =\n",
"\\sqrt{\\pi}\\;\\frac{1}{2}\\frac{3}{2}\\cdots\\frac{2k-1}{2}a^{-1/2-k} =\n",
"\\sqrt{\\pi}\\;\\frac{(2k)!}{2^{2k}k!}a^{-1/2-k},\n",
"\\\\ &\n",
"\\int_{-\\infty}^\\infty e^{-x^2/2}x^{2k}\\,dx =\n",
"\\sqrt{2\\pi}\\;\\frac{(2k)!}{2^k k!},\n",
"\\\\ &\n",
"\\int_{-\\infty}^\\infty e^{-x^2/2}\\cos(px)\\,dx =\n",
"\\sum_{k=0}^\\infty (-1)^k\\frac{p^{2k}}{(2k)!}\\int_{-\\infty}^\\infty e^{-x^2/2}x^{2k}\\,dx \n",
"\\\\ &\\qquad =\n",
"\\sum_{k=0}^\\infty (-1)^k\\frac{p^{2k}}{(2k)!} \\sqrt{2\\pi}\\;\\frac{(2k)!}{2^k k!} =\n",
"\\sqrt{2\\pi}\\sum_{k=0}^\\infty \\frac{(-p^2/2)^k}{k!} = \n",
"\\sqrt{2\\pi}\\;e^{-p^2/2}.\n",
"\\end{aligned}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\sqrt{2} \\sqrt{\\pi} e^{- \\frac{p^{2}}{2}}$$"
],
"text/plain": [
" 2 \n",
" -p \n",
" ----\n",
" ___ ____ 2 \n",
"\\/ 2 *\\/ pi *e "
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [3]\n",
"\n",
"p = symbols(\"p\", positive=true)\n",
"x = symbols(\"x\", real=true)\n",
"integrate(e^(-x^2/2)*cos(p*x), (x,-oo,oo))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[4\\]\n",
"\n",
"$$\n",
"B(p,q) = \n",
"\\int_0^1 x^{p-1}(1-x)^{q-1}\\,dx =\n",
"\\int_0^\\infty \\frac{x^{p-1}}{(1+x)^{p+q}}\\,dx =\n",
"2\\int_0^{\\pi/2}(\\cos x)^{2p-1}(\\sin x)^{2q-1}\\,dx\n",
"$$\n",
"\n",
"および\n",
"\n",
"$$\n",
"B(p,q) = \\frac{\\Gamma(p)\\Gamma(q}{\\Gamma(p+q)}, \\quad\n",
"\\Gamma(s+1)=s\\Gamma(s),\\quad\n",
"\\Gamma(1/2) = \\sqrt{\\pi}, \\quad\n",
"\\Gamma(n+1) = n!\n",
"$$\n",
"\n",
"を使う.\n",
"\n",
"### \\[4\\] (1)\n",
"\n",
"$$\n",
"\\int_0^1 \\sqrt{\\frac{x^5}{1-x}}\\,dx = \n",
"%\\int_0^1 x^{5/2}(1-x)^{-1/2}\\,dx =\n",
"B(7/2,1/2) = \n",
"\\frac{\\Gamma(7/2)\\Gamma(1/2)}{\\Gamma(4)} =\n",
"\\frac{(5/2)(3/2)(1/2)\\sqrt{\\pi}\\;\\sqrt{\\pi}}{3!} =\n",
"\\frac{5\\pi}{16}.\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\frac{5 \\pi}{16}$$"
],
"text/plain": [
"5*pi\n",
"----\n",
" 16 "
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [4] (1)\n",
"\n",
"x = symbols(\"x\", positive=true)\n",
"integrate(x^(Sym(5)/2)/√(1-x), (x,0,1))"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\frac{5 \\pi}{16}$$"
],
"text/plain": [
"5*pi\n",
"----\n",
" 16 "
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Beta(Sym(7)/2,Sym(1)/2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### \\[4\\] (2)\n",
"\n",
"$$\n",
"\\int_0^\\infty \\frac{x^{5/2}}{(1+x)^5}\\,dx =\n",
"B(7/2, 3/2) =\n",
"\\frac{\\Gamma(7/2)\\Gamma(3/2)}{\\Gamma(5)} =\n",
"\\frac{(5/2)(3/2)(1/2)\\sqrt{\\pi}\\;(1/2)\\sqrt{\\pi}}{4!} =\n",
"\\frac{5\\pi}{128}.\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\frac{5 \\pi}{128}$$"
],
"text/plain": [
"5*pi\n",
"----\n",
"128 "
]
},
"execution_count": 10,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [4] (2)\n",
"\n",
"x = symbols(\"x\", positive=true)\n",
"integrate(x^(Sym(5)/2)/(1+x)^5, (x,0,oo))"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\frac{5 \\pi}{128}$$"
],
"text/plain": [
"5*pi\n",
"----\n",
"128 "
]
},
"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Beta(Sym(7)/2, Sym(3)/2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### \\[4\\] (3)\n",
"\n",
"$$\n",
"2\\int_0^{\\pi/2} (\\cos x)^6 (\\sin x)^4\\,dx =\n",
"B(7/2, 5/2) =\n",
"\\frac{\\Gamma(7/2)\\Gamma(5/2)}{\\Gamma(6)} =\n",
"\\frac{(5/2)(3/2)(1/2)\\sqrt{\\pi}\\;(3/2)(1/2)\\sqrt{\\pi}}{5!} =\n",
"\\frac{3\\pi}{256.}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\frac{3 \\pi}{256}$$"
],
"text/plain": [
"3*pi\n",
"----\n",
"256 "
]
},
"execution_count": 12,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [4] (3)\n",
"\n",
"2integrate(cos(x)^6*sin(x)^4, (x,0,PI/2))"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\frac{3 \\pi}{256}$$"
],
"text/plain": [
"3*pi\n",
"----\n",
"256 "
]
},
"execution_count": 13,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Beta(Sym(7)/2, Sym(5)/2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[5\\]\n",
"\n",
"### \\[5\\] (1) \n",
"\n",
"$x>0$, $f(x) =x\\log x$ のとき, $f'(x)=\\log x + 1$, $f''(x)=1/x>0$ なので, $f(x)$ は下に凸な函数である."
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\log{\\left (x \\right )} + 1$$"
],
"text/plain": [
"log(x) + 1"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/latex": [
"$$\\frac{1}{x}$$"
],
"text/plain": [
"1\n",
"-\n",
"x"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"# [5] (1)\n",
"\n",
"f(x) = x * log(x)\n",
"x = symbols(\"x\", positive=true)\n",
"f1 = diff(f(x), x) |> display\n",
"f2 = diff(f(x), x, x) |> display"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### \\[5\\] (2)\n",
"\n",
"$\\displaystyle E[g(x)] = \\sum_{i=1}^n p_i g(q_i/p_i)$ は期待値汎函数なので, Jensenの不等式より, 下に凸な函数 $g(x)$ に対して $E[g(x)]\\leqq g(E[x])$ なので, それを $f(x)$ に適用すると,\n",
"\n",
"$$\n",
"\\begin{aligned}\n",
"\\sum_{i=1}^n q_i\\log\\frac{q_i}{p_i} =\n",
"\\sum_{i=1}^n p_i\\frac{q_i}{p_i}\\log\\frac{q_i}{p_i} =\n",
"E[f(x)] \\geqq\n",
"f(E[x]) = \n",
"f\\left(\\sum_{i=1}^n p_i\\frac{q_i}{p_i}\\right) =\n",
"f(1) = 0.\n",
"\\end{aligned}\n",
"$$\n",
"\n",
"**解説:** この結果は**Gibbsの(情報)不等式**と呼ばれている. $\\QED$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[6\\]\n",
"\n",
"### \\[6\\] (1)\n",
"\n",
"$\\ds f_n(y) =e^{-\\sqrt{n}\\;y}\\left(1+\\frac{y}{\\sqrt{n}}\\right)^n$ とおくと, $|y/\\sqrt{n}|<1$ のとき,\n",
"\n",
"$$\n",
"\\begin{aligned}\n",
"\\log f_n(y) &= -\\sqrt{n}\\;y + n\\log\\left(1+\\frac{y}{\\sqrt{n}}\\right) =\n",
"-\\sqrt{n};y + n\\left(\\frac{y}{\\sqrt{n}} - \\frac{y^2}{2n} + \\frac{y^3}{3n\\sqrt{n}} - \\frac{y^4}{4n^2} + \\cdots\\right)\n",
"\\\\ &= -\n",
"\\frac{y^2}{2} + \\frac{y^3}{3\\sqrt{n}} - \\frac{y^4}{4n} +\\cdots \\to -\\frac{y^2}{2}\n",
"\\qquad(n\\to\\infty).\n",
"\\end{aligned}\n",
"$$\n",
"\n",
"ゆえに $n\\to\\infty$ のとき, $f_n(y)\\to e^{-y^2/2}$."
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$e^{- \\frac{y^{2}}{2}}$$"
],
"text/plain": [
" 2 \n",
" -y \n",
" ----\n",
" 2 \n",
"e "
]
},
"execution_count": 15,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [6] (1)\n",
"\n",
"y = symbols(\"y\", positive=true)\n",
"n = symbols(\"n\", positive=true)\n",
"limit(e^(-√n*y)*(1+y/√n)^n, n=>oo)"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$\\frac{y^{5}}{5 n^{\\frac{3}{2}}} + \\frac{y^{3}}{3 \\sqrt{n}} - \\frac{y^{4}}{4 n} - \\frac{y^{2}}{2} + O\\left(y^{6}\\right)$$"
],
"text/plain": [
" 5 3 4 2 \n",
" y y y y / 6\\\n",
"------ + ------- - --- - -- + O\\y /\n",
" 3/2 ___ 4*n 2 \n",
"5*n 3*\\/ n "
]
},
"metadata": {},
"output_type": "display_data"
},
{
"data": {
"text/latex": [
"$$- \\frac{y^{2}}{2}$$"
],
"text/plain": [
" 2 \n",
"-y \n",
"----\n",
" 2 "
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"y = symbols(\"y\", real=true)\n",
"n = symbols(\"n\", positive=true)\n",
"series(log(e^(-√n*y)*(1+y/√n)^n), y) |> display\n",
"limit(series(log(e^(-√n*y)*(1+y/√n)^n), y), n=>oo)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### \\[6\\] (2)\n",
"\n",
"$x = n + \\sqrt{n}\\;y = n(1+y/\\sqrt{n})$ とおくと, $n\\to\\infty$ のとき,\n",
"\n",
"$$\n",
"\\begin{aligned}\n",
"\\frac{1}{n^n e^{-n}\\sqrt{n}}\\int_0^\\infty e^{-x}x^n\\,dx &=\n",
"\\frac{1}{n^n e^{-n}\\sqrt{n}}\\int_{-\\sqrt{n}}^\\infty e^{-n-\\sqrt{n}\\;y}n^n\\left(1+\\frac{y}{\\sqrt{n}}\\right)^n\\sqrt{n}\\,dy\n",
"\\\\ &=\n",
"\\int_{-\\sqrt{n}}^\\infty e^{-\\sqrt{n}\\;y}\\left(1+\\frac{y}{\\sqrt{n}}\\right)^n\\,dy\n",
"\\to\\int_{-\\infty}^\\infty e^{-y^2/2}\\,dy = \\sqrt{2\\pi}.\n",
"\\end{aligned}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[7\\]\n",
"\n",
"### \\[7\\] (1)\n",
"\n",
"$\\displaystyle \\log n^{1/n}=\\frac{1}{n}\\log n\\to 0$ ($n\\to\\infty$) なので $n^{1/n}\\to 1$ である."
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$1$$"
],
"text/plain": [
"1"
]
},
"execution_count": 17,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [7] (1)\n",
"\n",
"n = symbols(\"n\", positive=true)\n",
"limit(n^(1/n), n=>oo)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### \\[7\\] (2)\n",
"\n",
"Stirlingの公式と(1)より, $n\\to\\infty$ のとき\n",
"\n",
"$$\n",
"\\begin{aligned}\n",
"&\n",
"\\binom{n(a+b)}{na} =\n",
"\\frac\n",
"{(n(a+b))^{n(a+b)}e^{-n(a+b)}\\sqrt{2\\pi n(a+b)}}\n",
"{(na)^{na}(nb)^{nb}e^{-na-nb}\\sqrt{2\\pi na\\;2\\pi nb}} =\n",
"\\left(\n",
"\\frac\n",
"{(a+b)^{a+b}}\n",
"{a^a b^b}\n",
"\\right)^n\n",
"\\sqrt{\\frac{a+b}{2\\pi nab}}\n",
"\\end{aligned}\n",
"$$\n",
"\n",
"なので,\n",
"\n",
"$$\n",
"\\lim_{n\\to\\infty}\\binom{n(a+b)}{na}^{1/n} =\n",
"\\frac{(a+b)^{a+b}}{a^a b^b}.\n",
"$$\n",
"\n",
"**解説:** この結果の対数はエントロピーの話と関係がある. $\\QED$"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {},
"outputs": [
{
"data": {
"text/latex": [
"$$a^{- a} b^{- b} \\left(a + b\\right)^{a + b}$$"
],
"text/plain": [
" -a -b a + b\n",
"a *b *(a + b) "
]
},
"execution_count": 18,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# [7] (2)\n",
"\n",
"n, a, b = symbols(\"n a b\", positive=true)\n",
"simplify(limit((gamma(n*(a+b)+1)/(gamma(n*a+1)*gamma(n*b+1)))^(1/n), n=>oo))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## \\[8\\]\n",
"\n",
"二項展開によって, $0