Bisection.ipynb

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   "source": [
    "## Exercise for the course [Python for MATLAB users](http://sese.nu/python-for-matlab-users-ht15/), by Olivier Verdier"
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      "Using matplotlib backend: MacOSX\n",
      "Populating the interactive namespace from numpy and matplotlib\n"
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    "%pylab\n",
    "%matplotlib inline"
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    "A  continuous functions $f$ which changes its sign in an interval $[a,b]$, i.e. $f(a)f(b)< 0$, has at least one root in this interval. Such a root can be found by the *bisection method*. \n",
    "\n",
    "This method starts from the given interval. Then it investigates the sign changes in the subintervals $[a,\\frac{a+b}{2}]$ \n",
    "and $[\\frac{a+b}{2},b]$. If the sign changes in the first subinterval $ b$ is redefined to be \n",
    "$b:=\\frac{a+b}{2}$\n",
    "otherwise $a$ is redefined in the same manner to \n",
    "$a:=\\frac{a+b}{2}$,\n",
    "and the process is repeated until the $b-a$ is less than a given tolerance.\n"
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    "def bisect(f, a, b):\n",
    "    tol = 1e-9\n",
    "    for i in range(100):\n",
    "        if (b-a) < tol:\n",
    "            break\n",
    "        c = (a+b)/2\n",
    "        if f(a)*f(c) <= 0:\n",
    "            b = c\n",
    "        else:\n",
    "            a = c\n",
    "    return c"
   ]
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   "source": [
    "Implement the function `bisect` above until the following does not complain:"
   ]
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    "def lin(x):\n",
    "    return x\n",
    "assert allclose(bisect(sin, 3., 4.), pi)\n",
    "assert allclose(bisect(lin, -1,1), 0)"
   ]
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   "execution_count": null,
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