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   "cell_type": "markdown",
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   "source": [
    "# Newton method for square root"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Consider the function\n",
    "$$\n",
    "f(x) = x^2 - a\n",
    "$$\n",
    "The roots are $\\pm \\sqrt{a}$. The Newton method is\n",
    "$$\n",
    "x_{n+1} = \\frac{1}{2} \\left( x_n + \\frac{a}{x_n} \\right)\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [],
   "source": [
    "from math import sqrt\n",
    "a = 2.0\n",
    "r = sqrt(a)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Define the function and its derivative."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [],
   "source": [
    "def f(x):\n",
    "    return x**2 - a\n",
    "\n",
    "def df(x):\n",
    "    return 2.0*x"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now implement Newton method as a function."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [],
   "source": [
    "def newton(f,df,x0,M=100,eps=1.0e-15):\n",
    "    x = x0\n",
    "    for i in range(M):\n",
    "        dx = - f(x)/df(x)\n",
    "        x  = x + dx\n",
    "        print('%3d %20.14f %20.14e %20.14e'%(i,x,x-r,abs(f(x))))\n",
    "        if abs(dx) < eps * abs(x):\n",
    "            return x\n",
    "    print('No convergence, current root = %e' % x)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We call Newton method with a complex initial guess for the root."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "  0     1.50000000000000 8.57864376269049e-02 2.50000000000000e-01\n",
      "  1     1.41666666666667 2.45310429357160e-03 6.94444444444464e-03\n",
      "  2     1.41421568627451 2.12390141474117e-06 6.00730488287127e-06\n",
      "  3     1.41421356237469 1.59472435257157e-12 4.51061410444709e-12\n",
      "  4     1.41421356237310 0.00000000000000e+00 4.44089209850063e-16\n",
      "  5     1.41421356237309 -2.22044604925031e-16 4.44089209850063e-16\n"
     ]
    }
   ],
   "source": [
    "x0 = 2.0\n",
    "x = newton(f,df,x0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "From the last column, we observe that in each iteration, the number of correct decimal places doubles."
   ]
  }
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