{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Gotchas"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "from sympy import *\n",
    "init_printing()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "For each exercise, fill in the function according to its docstring. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Symbols"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "What will be the output of the following code?\n",
    "\n",
    "    x = 3\n",
    "    y = symbols('y')\n",
    "    a = x + y\n",
    "    y = 5\n",
    "    print(a)\n",
    "\n",
    "Replace `???` in the below code with what you think the value of `a` will be.  Remember to define any Symbols you need!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def symbols_exercise():\n",
    "    \"\"\"\n",
    "    >>> def testfunc():\n",
    "    ...     x = 3\n",
    "    ...     y = symbols('y')\n",
    "    ...     a = x + y\n",
    "    ...     y = 5\n",
    "    ...     return a\n",
    "    >>> symbols_exercise() == testfunc()\n",
    "    True\n",
    "    \"\"\"\n",
    "    return ??? # Replace ??? with what you think the value of a is"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def testfunc():\n",
    "    x = 3\n",
    "    y = symbols('y')\n",
    "    a = x + y\n",
    "    y = 5\n",
    "    return a\n",
    "\n",
    "symbols_exercise() == testfunc()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Equality"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Write a function that takes two expressions as input, and returns a tuple of two booleans. The first if they are equal symbolically, and the second if they are equal mathematically."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def equality_exercise(a, b):\n",
    "    \"\"\"\n",
    "    Determine if a = b symbolically and mathematically.\n",
    "\n",
    "    Returns a tuple of two booleans. The first is True if a = b symbolically,\n",
    "    the second is True if a = b mathematically.  Note the second may be False\n",
    "    but the two still equal if SymPy is not powerful enough.\n",
    "\n",
    "    Examples\n",
    "    ========\n",
    "\n",
    "    >>> x = symbols('x')\n",
    "    >>> equality_exercise(x, 2)\n",
    "    (False, False)\n",
    "    >>> equality_exercise((x + 1)**2, x**2 + 2*x + 1)\n",
    "    (False, True)\n",
    "    >>> equality_exercise(2*x, 2*x)\n",
    "    (True, True)\n",
    "    \"\"\"\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "x = symbols('x')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "equality_exercise(x, 2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "equality_exercise((x + 1)**2, x**2 + 2*x + 1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "equality_exercise(2*x, 2*x)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## `^` and `/`"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Correct the following functions"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def operator_exercise1():\n",
    "    \"\"\"\n",
    "    >>> operator_exercise1()\n",
    "    x**2 + 2*x + 1/2\n",
    "    \"\"\"\n",
    "    x = symbols('x')\n",
    "    return x^2 + 2*x + 1/2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "operator_exercise1()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def operator_exercise2():\n",
    "    \"\"\"\n",
    "    >>> operator_exercise2()\n",
    "    (x**2/2 + 2*x + 3/4)**(3/2)\n",
    "    \"\"\"\n",
    "    x = symbols('x')\n",
    "    return (1/2*x^2 + 2*x + 3/4)^(3/2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "operator_exercise2()"
   ]
  }
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