sqrt(3^pi) と sqrt(pi^pi)を評価する

pi81pi.ipynb

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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# $\\sqrt{3^\\pi}$と$\\sqrt{\\pi^\\pi}$を評価する\n",
    "\n",
    "ここでは$3.141 < \\pi < 3.142$，$1.098 < \\ln 3 < 1.099$とする．"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "import math\n",
    "PI_L = 3.141\n",
    "PI_H = 3.142\n",
    "LN3_L = 1.098\n",
    "LN3_H = 1.099"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\sqrt{3^\\pi} = 3^{\\pi/2}$を評価するために，$a^x$のマクローリン展開を考える\n",
    "$$\n",
    "a^x = 1 + x \\ln a + \\frac{1}{2} (c \\ln a)^2\n",
    "$$\n",
    "ただし，$c \\in [0, x]$である．\n",
    "$x^2$は$x>0$で単調増加関数のため，$0 <  \\frac{1}{2} (c \\ln a)^2 < \\frac{1}{2} (x \\ln a)^2$が成立する．\n",
    "また\n",
    "$$\n",
    "3^{3.141/2} = 3^{1.5705} < 3^{\\pi/2} < 3^{3.142/2} = 3^{1.571}\n",
    "$$\n",
    "である．\n",
    "まず，下限である$3^{1.5705}$を計算すると\n",
    "$$\n",
    "3^{1.5705} = 3^{1.5} \\times 3^{0.0705}\n",
    "$$\n",
    "である．\n",
    "これに，先ほどのマクローリン展開の下限を入れることで\n",
    "$$\n",
    "3^{1.5705} > 3^{1.5} \\times (1 + 0.0705 \\ln 3)\n",
    "$$\n",
    "である．$\\ln 3$の値にも下限値を代入すれば$3^{\\pi/2}$の下限値は以下の通りである．"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "5.59838138559593"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "3**1.5 * (1 + 0.0705 * LN3_L)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "上限値も同様に計算すると\n",
    "$$\n",
    "3^{1.571} < 3^{1.5} * (1 + 0.071 \\ln 3 + \\frac{1}{2} (0.071 \\ln 3)^2)\n",
    "$$\n",
    "であり，$\\ln 3$の値にも上限値を代入すれば$3^{\\pi/2}$の上限値は以下の通りである．"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "5.617421451649643"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "3**1.5 * (1 + 0.071 * LN3_H + 0.5 * (0.071 * LN3_H)**2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "従って，$5.598 < 3^{\\pi/2} < 5.618$であることがわかる．\n",
    "実際に$3^{\\pi/2}$の値は以下の通りであり，評価が妥当であることがわかる．"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "5.616429533092847"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "3**(math.pi/2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "次に，$\\pi^{\\pi/2}$を評価するために，$(1+x)^a$のマクローリン展開を考える．\n",
    "$$\n",
    "(1+x)^a = 1 + ax + \\frac{a(a-1)}{2} c^2\n",
    "$$\n",
    "ただし，$c \\in [0, x]$である．\n",
    "$x^2$は$x>0$で単調増加関数のため，$0 < \\frac{a(a-1)}{2} c^2 < \\frac{a(a-1)}{2} x^2$が成立する．\n",
    "これを用いて，$\\pi^{\\pi/2}$の下限値を評価するために，次のように変形する．\n",
    "$$\n",
    "\\pi^{\\pi/2} > 3.141^{3.141/2} = (3 + 0.141)^{1.5705} = 3^{1.5705}  \\left(1 + \\frac{0.141}{3} \\right)^{1.5705}\n",
    "$$\n",
    "$\\left(1 + \\frac{0.141}{3} \\right)^{1.5705}$の部分を評価すると\n",
    "$$\n",
    " \\left(1 + \\frac{0.141}{3} \\right)^{1.5705} > 1 + 1.5705 \\times \\frac{0.141}{3} \n",
    "$$\n",
    "であり，これを計算すると以下の通りである．"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1.0738135"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "1 + 1.5705 * 0.141 / 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "よって，$\\pi^{\\pi/2}$の下限値は，$ 3^{1.5705} > 5.598$を利用して以下の通りである．"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "6.006653999999999"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "5.598 * 1.073"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "上限値についても同様に評価する．\n",
    "$$\n",
    "\\pi^{\\pi/2} < 3.142^{3.142/2} = (3 + 0.142)^{1.571} = 3^{1.571}  \\left(1 + \\frac{0.142}{3} \\right)^{1.571}\n",
    "$$\n",
    "ただし，\n",
    "$$\n",
    " \\left(1 + \\frac{0.142}{3} \\right)^{1.571} < 1 + 1.571 \\times \\frac{0.142}{3} + \\frac{1.571 \\times 0.571}{2} \\left(\\frac{0.142}{3}\\right)^2\n",
    "$$\n",
    "でありこれを評価すると，"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1.0753655519291112"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "1 + 1.571 * (0.142/3) + (1.571 * 0.571)/2 * (0.142/3)**2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "結局，$\\pi^{\\pi/2}$の上限値は,$ 3^{1.571} > 5.618$を利用して"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "6.044968000000001"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "5.618 * 1.076"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "となり，$6.006 < \\pi^{\\pi/2} < 6.045$と評価できる．"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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