solution.ipynb

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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Solutions to mock class test\n",
    "\n",
    "# Question 1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
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   "outputs": [
    {
     "data": {
      "text/plain": [
       "6050"
      ]
     },
     "execution_count": 1,
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   "source": [
    "nbrs = [] \n",
    "i = 0\n",
    "while len(nbrs) < 10:  \n",
    "# Using a while loop to keep track of how many numbers are in the list\n",
    "    i += 1  # Increment the number I am testing\n",
    "    if i % 10 == 0 and i % 11 == 0:  # Test for divisibility\n",
    "        nbrs.append(i)  # Put the number in the list\n",
    "sum(nbrs)  # Sum the number"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Question 2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "import math  # Typo\n",
    "def hyp(a, b):  # Need two arguments\n",
    "    return math.sqrt(a ** 2 + b ** 2)  # Fix the brack and missing exponent on a"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "95.0"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "hyp(76, 57)  # Test the required triangle"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Question 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Part a"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def p(k):\n",
    "    \"\"\"Return the kth term of the summation\"\"\"\n",
    "    return ((2 ** k) * (math.factorial(k)) ** 2) / (math.factorial(2 * k + 1))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Part b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def sum_n(n):\n",
    "    \"\"\"Return the sum of the first nth terms of the summation\"\"\"\n",
    "    return sum(p(k) for k in range(n + 1))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Part c"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0 2.0\n",
      "1 2.6666666666666665\n",
      "2 2.933333333333333\n",
      "3 3.0476190476190474\n",
      "4 3.098412698412698\n"
     ]
    }
   ],
   "source": [
    "def approx_pi(n):\n",
    "    \"\"\"Return 2 times the first n terms of the summation\"\"\"\n",
    "    return 2 * sum_n(n)\n",
    "\n",
    "# Using a for loop get the required numbers\n",
    "for n in range(5):\n",
    "    print(n, approx_pi(n))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Part d"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "with open(\"pi.csv\", \"w\") as outfile:\n",
    "    for n in range(50):\n",
    "        outfile.write(str(approx_pi(n)) + \"\\n\")"
   ]
  }
 ],
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