{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Names in Python"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Question 1.** A famous fact in mathematics is that\n",
    "\n",
    "$$(1 + \\frac{1}{n})^{n}$$\n",
    "\n",
    "gets very close to the number $e$, which is roughly $2.718$, when $n$ is large.  Use this fact to compute an approximation to $e$, called `roughly_e`.  Your approximation should be correct to at least 5 decimal places."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.7182820533299275"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "roughly_e = (1 + (1/99999999999))**99999999999\n",
    "roughly_e"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Question 2.** In the previous question, you might have noticed that you had to type in the value of $n$ twice when you tried different values (once as `1/n` and once as `**n`).  You might have even accidentally used 2 different values.\n",
    "\n",
    "Redo the code in the cell below, giving your estimate of $e$ the name `also_roughly_e`.  Instead of writing the value of $n$ twice, give it the name `n` in your Python code, and then use that name in your expression for `also_roughly_e`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.7182820533299275"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = 99999999999\n",
    "also_roughly_e = (1+(1/n))**n\n",
    "also_roughly_e"
   ]
  }
 ],
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