{
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  {
   "cell_type": "markdown",
   "id": "2429cc88",
   "metadata": {},
   "source": [
    "# 二次方程式を解く\n",
    "\n",
    "二次方程式\n",
    "\n",
    "$$\n",
    "ax^2+bx+c=0,\\qquad a\\ne0,\n",
    "$$\n",
    "\n",
    "の二つの解は、次の公式で与えられます。\n",
    "\n",
    "$$\n",
    "x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}.\n",
    "$$\n",
    "\n",
    "例ごとに解を個別に入力するのではなく、Egisonプログラムで多項式の係数を読み取り、\n",
    "判別式を組み立て、一つの型付き記号計算ソルバーを適用します。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f0c1634d",
   "metadata": {},
   "source": [
    "## 係数と判別式\n",
    "\n",
    "Egisonの coefficients 関数は、次数の昇順で係数を返します。したがって、\n",
    "パターン $[a_0,a_1,a_2]$ は $a_2x^2+a_1x+a_0$ を認識します。補助関数では、\n",
    "分母を払った判別式 $b^2-4ac$ をそのまま保ち、分母を明示的に $2a$ として\n",
    "まとめます。\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "23be8a3d",
   "metadata": {},
   "outputs": [],
   "source": [
    "declare symbol x, a, b, c: MathValue\n",
    "\n",
    "def solveQuadraticCoefficientsDemo\n",
    "  (a : MathValue)\n",
    "  (b : MathValue)\n",
    "  (c : MathValue)\n",
    "  : (MathValue, MathValue) :=\n",
    "  let discriminant := b ^ 2 - 4 * a * c\n",
    "   in ( ((- b) + sqrt discriminant) / (2 * a)\n",
    "      , ((- b) - sqrt discriminant) / (2 * a) )\n",
    "\n",
    "def solveQuadraticDemo\n",
    "  (f : MathValue)\n",
    "  (x : MathValue)\n",
    "  : (MathValue, MathValue) :=\n",
    "  match coefficients f x as list mathValue with\n",
    "    | [$a_0, $a_1, $a_2] ->\n",
    "      solveQuadraticCoefficientsDemo a_2 a_1 a_0\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "61c83725",
   "metadata": {},
   "source": [
    "## 円分多項式の例\n",
    "\n",
    "多項式 $x^2+x+1$ の判別式は $-3$ です。その解は、二つの1の原始3乗根です。\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "a97642ab",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "$(\\frac{1}{2} \\sqrt{3} i + \\frac{-1}{2}, \\frac{-1}{2} \\sqrt{3} i + \\frac{-1}{2})$"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "solveQuadraticDemo (x ^ 2 + x + 1) x\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c9217626",
   "metadata": {},
   "source": [
    "## 記号のまま得られる公式\n",
    "\n",
    "$a$、$b$、$c$ を記号のままにすると、特別な整形コードを使わなくても通常の\n",
    "判別式がそのまま現れます。\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "c72484c5",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "$(\\frac{-1}{2} a^{-1} b + \\frac{1}{2} \\sqrt{b^{2} - 4 a c} a^{-1}, \\frac{-1}{2} a^{-1} b + \\frac{-1}{2} \\sqrt{b^{2} - 4 a c} a^{-1})$"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "solveQuadraticDemo (a * x ^ 2 + b * x + c) x\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4ca393c5",
   "metadata": {},
   "source": [
    "## 便利な係数の取り方\n",
    "\n",
    "一次の係数を $2b$ と書けば、同値で簡潔な次の形が得られます。\n",
    "\n",
    "$$\n",
    "x=\\frac{-b\\pm\\sqrt{b^2-ac}}{a}.\n",
    "$$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "e43288c2",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "$(-a^{-1} b + \\sqrt{b^{2} - a c} a^{-1}, -a^{-1} b - \\sqrt{b^{2} - a c} a^{-1})$"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "solveQuadraticDemo (a * x ^ 2 + 2 * b * x + c) x\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d6bfd467",
   "metadata": {},
   "source": [
    "## まとめ\n",
    "\n",
    "パターンマッチと型付き補助関数により、計算を見通しのよい二段階に分けられます。\n",
    "まず多項式の係数を読み取り、次に判別式と、その平方根の正負両方を組み立てます。\n",
    "表示される解は事前計算した出力を埋め込んだものではなく、各入力多項式から\n",
    "実際に生成されたものです。\n"
   ]
  }
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