Lecture-5-Sympy.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Sympy - Symbolic algebra in Python"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "J.R. Johansson (robert@riken.jp) http://dml.riken.jp/~rob/\n",
      "\n",
      "The latest version of this [IPython notebook](http://ipython.org/notebook.html) lecture is available at [http://github.com/jrjohansson/scientific-python-lectures](http://github.com/jrjohansson/scientific-python-lectures).\n",
      "\n",
      "The other notebooks in this lecture series are indexed at [http://jrjohansson.github.com](http://jrjohansson.github.com)."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%pylab inline"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.kernel.zmq.pylab.backend_inline].\n",
        "For more information, type 'help(pylab)'.\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Introduction"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "There are two notable Computer Algebra Systems (CAS) for Python:\n",
      "\n",
      "* [SymPy](http://sympy.org/en/index.html) - A python module that can be used in any Python program, or in an IPython session, that provides powerful CAS features. \n",
      "* [Sage](http://www.sagemath.org/) - Sage is a full-featured and very powerful CAS enviroment that aims to provide an open source system that competes with Mathematica and Maple. Sage is not a regular Python module, but rather a CAS environment that uses Python as its programming language.\n",
      "\n",
      "Sage is in some aspects more powerful than SymPy, but both offer very comprehensive CAS functionality. The advantage of SymPy is that it is a regular Python module and integrates well with the IPython notebook. \n",
      "\n",
      "In this lecture we will therefore look at how to use SymPy with IPython notebooks. If you are interested in an open source CAS environment I also recommend to read more about Sage.\n",
      "\n",
      "To get started using SymPy in a Python program or notebook, import the module `sympy`:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy import *"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "To get nice-looking $\\LaTeX$ formatted output run:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "init_printing()\n",
      "\n",
      "# or with older versions of sympy/ipython, load the IPython extension\n",
      "#%load_ext sympy.interactive.ipythonprinting\n",
      "# or\n",
      "#%load_ext sympyprinting"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Symbolic variables"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "In SymPy we need to create symbols for the variables we want to work with. We can create a new symbol using the `Symbol` class:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x = Symbol('x')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 4
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "(pi + x)**2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(x + \\pi\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 5,
       "text": [
        "       2\n",
        "(x + \u03c0) "
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# alternative way of defining symbols\n",
      "a, b, c = symbols(\"a, b, c\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 6
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "type(a)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 7,
       "text": [
        "sympy.core.symbol.Symbol"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can add assumptions to symbols when we create them:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x = Symbol('x', real=True)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 8
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x.is_imaginary"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$False$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 9,
       "text": [
        "False"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x = Symbol('x', positive=True)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 10
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x > 0"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$True$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 11,
       "text": [
        "True"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Complex numbers"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The imaginary unit is denoted `I` in Sympy. "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "1+1*I"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1 + i$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 12,
       "text": [
        "1 + \u2148"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "I**2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$-1$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 13,
       "text": [
        "-1"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "(x * I + 1)**2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(i x + 1\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 14,
       "text": [
        "         2\n",
        "(\u2148\u22c5x + 1) "
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Rational numbers"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "There are three different numerical types in SymPy: `Real`, `Rational`, `Integer`: "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "r1 = Rational(4,5)\n",
      "r2 = Rational(5,4)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 15
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "r1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{4}{5}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 16,
       "text": [
        "4/5"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "r1+r2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{41}{20}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 17,
       "text": [
        "41\n",
        "\u2500\u2500\n",
        "20"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "r1/r2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{16}{25}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 18,
       "text": [
        "16\n",
        "\u2500\u2500\n",
        "25"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Numerical evaluation"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "SymPy uses a library for artitrary precision as numerical backend, and has predefined SymPy expressions for a number of mathematical constants, such as: `pi`, `e`, `oo` for infinity.\n",
      "\n",
      "To evaluate an expression numerically we can use the `evalf` function (or `N`). It takes an argument `n` which specifies the number of significant digits."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "pi.evalf(n=50)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$3.1415926535897932384626433832795028841971693993751$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 19,
       "text": [
        "3.1415926535897932384626433832795028841971693993751"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "y = (x + pi)**2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 20
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "N(y, 5) # same as evalf"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(x + 3.1416\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 21,
       "text": [
        "            2\n",
        "(x + 3.1416) "
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "When we numerically evaluate algebraic expressions we often want to substitute a symbol with a numerical value. In SymPy we do that using the `subs` function:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "y.subs(x, 1.5)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(1.5 + \\pi\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 22,
       "text": [
        "         2\n",
        "(1.5 + \u03c0) "
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "N(y.subs(x, 1.5))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$21.5443823618587$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 23,
       "text": [
        "21.5443823618587"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The `subs` function can of course also be used to substitute Symbols and expressions:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "y.subs(x, a+pi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(a + 2 \\pi\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 24,
       "text": [
        "         2\n",
        "(a + 2\u22c5\u03c0) "
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can also combine numerical evolution of expressions with NumPy arrays:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import numpy"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 25
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x_vec = numpy.arange(0, 10, 0.1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 26
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "y_vec = numpy.array([N(((x + pi)**2).subs(x, xx)) for xx in x_vec])"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 27
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "fig, ax = subplots()\n",
      "ax.plot(x_vec, y_vec);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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hKSmJ6upqvxYbiHbvhtmz4eKL4Sc/MX3155+HTz6BP/xBwS4i9vCpLVNRUUFF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\naFlJKJFyhJ8AAAAASUVORK5CYII=\n",
       "text": [
        "<matplotlib.figure.Figure at 0x7f54fafc4a90>"
       ]
      }
     ],
     "prompt_number": 28
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "However, this kind of numerical evolution can be very slow, and there is a much more efficient way to do it: Use the function `lambdify` to \"compile\" a Sympy expression into a function that is much more efficient to evaluate numerically:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f = lambdify([x], (x + pi)**2, 'numpy')  # the first argument is a list of variables that\n",
      "                                         # f will be a function of: in this case only x -> f(x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 29
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "y_vec = f(x_vec)  # now we can directly pass a numpy array and f(x) is efficiently evaluated"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 30
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The speedup when using \"lambdified\" functions instead of direct numerical evaluation can be significant, often several orders of magnitude. Even in this simple example we get a significant speed up:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%%timeit\n",
      "\n",
      "y_vec = numpy.array([N(((x + pi)**2).subs(x, xx)) for xx in x_vec])"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "10 loops, best of 3: 20.4 ms per loop\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%%timeit\n",
      "\n",
      "y_vec = f(x_vec)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "100000 loops, best of 3: 3.67 \u00b5s per loop\n"
       ]
      }
     ],
     "prompt_number": 32
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Algebraic manipulations"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "One of the main uses of an CAS is to perform algebraic manipulations of expressions. For example, we might want to expand a product, factor an expression, or simply an expression. The functions for doing these basic operations in SymPy are demonstrated in this section."
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Expand and factor"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The first steps in an algebraic manipulation "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "(x+1)*(x+2)*(x+3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(x + 1\\right) \\left(x + 2\\right) \\left(x + 3\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 33,
       "text": [
        "(x + 1)\u22c5(x + 2)\u22c5(x + 3)"
       ]
      }
     ],
     "prompt_number": 33
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "expand((x+1)*(x+2)*(x+3))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$x^{3} + 6 x^{2} + 11 x + 6$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 34,
       "text": [
        " 3      2           \n",
        "x  + 6\u22c5x  + 11\u22c5x + 6"
       ]
      }
     ],
     "prompt_number": 34
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The `expand` function takes a number of keywords arguments which we can tell the functions what kind of expansions we want to have performed. For example, to expand trigonometric expressions, use the `trig=True` keyword argument:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "sin(a+b)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\sin{\\left (a + b \\right )}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 35,
       "text": [
        "sin(a + b)"
       ]
      }
     ],
     "prompt_number": 35
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "expand(sin(a+b), trig=True)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\sin{\\left (a \\right )} \\cos{\\left (b \\right )} + \\sin{\\left (b \\right )} \\cos{\\left (a \\right )}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 36,
       "text": [
        "sin(a)\u22c5cos(b) + sin(b)\u22c5cos(a)"
       ]
      }
     ],
     "prompt_number": 36
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "See `help(expand)` for a detailed explanation of the various types of expansions the `expand` functions can perform."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The opposite a product expansion is of course factoring. The factor an expression in SymPy use the `factor` function: "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "factor(x**3 + 6 * x**2 + 11*x + 6)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(x + 1\\right) \\left(x + 2\\right) \\left(x + 3\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 37,
       "text": [
        "(x + 1)\u22c5(x + 2)\u22c5(x + 3)"
       ]
      }
     ],
     "prompt_number": 37
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Simplify"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The `simplify` tries to simplify an expression into a nice looking expression, using various techniques. More specific alternatives to the `simplify` functions also exists: `trigsimp`, `powsimp`, `logcombine`, etc. \n",
      "\n",
      "The basic usages of these functions are as follows:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# simplify expands a product\n",
      "simplify((x+1)*(x+2)*(x+3))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(x + 1\\right) \\left(x + 2\\right) \\left(x + 3\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 38,
       "text": [
        "(x + 1)\u22c5(x + 2)\u22c5(x + 3)"
       ]
      }
     ],
     "prompt_number": 38
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# simplify uses trigonometric identities\n",
      "simplify(sin(a)**2 + cos(a)**2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 39,
       "text": [
        "1"
       ]
      }
     ],
     "prompt_number": 39
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "simplify(cos(x)/sin(x))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{1}{\\tan{\\left (x \\right )}}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 40,
       "text": [
        "  1   \n",
        "\u2500\u2500\u2500\u2500\u2500\u2500\n",
        "tan(x)"
       ]
      }
     ],
     "prompt_number": 40
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "apart and together"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "To manipulate symbolic expressions of fractions, we can use the `apart` and `together` functions:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f1 = 1/((a+1)*(a+2))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 41
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{1}{\\left(a + 1\\right) \\left(a + 2\\right)}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 42,
       "text": [
        "       1       \n",
        "\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n",
        "(a + 1)\u22c5(a + 2)"
       ]
      }
     ],
     "prompt_number": 42
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "apart(f1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$- \\frac{1}{a + 2} + \\frac{1}{a + 1}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 43,
       "text": [
        "    1       1  \n",
        "- \u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\n",
        "  a + 2   a + 1"
       ]
      }
     ],
     "prompt_number": 43
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f2 = 1/(a+2) + 1/(a+3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 44
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{1}{a + 3} + \\frac{1}{a + 2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 45,
       "text": [
        "  1       1  \n",
        "\u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\n",
        "a + 3   a + 2"
       ]
      }
     ],
     "prompt_number": 45
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "together(f2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{2 a + 5}{\\left(a + 2\\right) \\left(a + 3\\right)}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 46,
       "text": [
        "    2\u22c5a + 5    \n",
        "\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n",
        "(a + 2)\u22c5(a + 3)"
       ]
      }
     ],
     "prompt_number": 46
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Simplify usually combines fractions but does not factor: "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "simplify(f2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{2 a + 5}{\\left(a + 2\\right) \\left(a + 3\\right)}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 47,
       "text": [
        "    2\u22c5a + 5    \n",
        "\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n",
        "(a + 2)\u22c5(a + 3)"
       ]
      }
     ],
     "prompt_number": 47
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Calculus"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "In addition to algebraic manipulations, the other main use of CAS is to do calculus, like derivatives and integrals of algebraic expressions."
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Differentiation"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Differentiation is usually simple. Use the `diff` function. The first argument is the expression to take the derivative of, and the second argument is the symbol by which to take the derivative:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "y"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(x + \\pi\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 48,
       "text": [
        "       2\n",
        "(x + \u03c0) "
       ]
      }
     ],
     "prompt_number": 48
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "diff(y**2, x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$4 \\left(x + \\pi\\right)^{3}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 49,
       "text": [
        "         3\n",
        "4\u22c5(x + \u03c0) "
       ]
      }
     ],
     "prompt_number": 49
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "For higher order derivatives we can do:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "diff(y**2, x, x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$12 \\left(x + \\pi\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 50,
       "text": [
        "          2\n",
        "12\u22c5(x + \u03c0) "
       ]
      }
     ],
     "prompt_number": 50
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "diff(y**2, x, 2) # same as above"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$12 \\left(x + \\pi\\right)^{2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 51,
       "text": [
        "          2\n",
        "12\u22c5(x + \u03c0) "
       ]
      }
     ],
     "prompt_number": 51
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "To calculate the derivative of a multivariate expression, we can do:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "x, y, z = symbols(\"x,y,z\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 52
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f = sin(x*y) + cos(y*z)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 53
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$\\frac{d^3f}{dxdy^2}$"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "diff(f, x, 1, y, 2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$- x \\left(x y \\cos{\\left (x y \\right )} + 2 \\sin{\\left (x y \\right )}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 54,
       "text": [
        "-x\u22c5(x\u22c5y\u22c5cos(x\u22c5y) + 2\u22c5sin(x\u22c5y))"
       ]
      }
     ],
     "prompt_number": 54
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Integration"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Integration is done in a similar fashion:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\sin{\\left (x y \\right )} + \\cos{\\left (y z \\right )}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 55,
       "text": [
        "sin(x\u22c5y) + cos(y\u22c5z)"
       ]
      }
     ],
     "prompt_number": 55
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "integrate(f, x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$x \\cos{\\left (y z \\right )} + \\begin{cases} 0 & \\text{for}\\: y = 0 \\\\- \\frac{\\cos{\\left (x y \\right )}}{y} & \\text{otherwise} \\end{cases}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 56,
       "text": [
        "             \u239b\u23a7    0      for y = 0\u239e\n",
        "             \u239c\u23aa                    \u239f\n",
        "x\u22c5cos(y\u22c5z) + \u239c\u23a8-cos(x\u22c5y)           \u239f\n",
        "             \u239c\u23aa\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500  otherwise\u239f\n",
        "             \u239d\u23a9    y               \u23a0"
       ]
      }
     ],
     "prompt_number": 56
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "By providing limits for the integration variable we can evaluate definite integrals:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "integrate(f, (x, -1, 1))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$2 \\cos{\\left (y z \\right )}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 57,
       "text": [
        "2\u22c5cos(y\u22c5z)"
       ]
      }
     ],
     "prompt_number": 57
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "and also improper integrals"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "integrate(exp(-x**2), (x, -oo, oo))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\sqrt{\\pi}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 58,
       "text": [
        "  ___\n",
        "\u2572\u2571 \u03c0 "
       ]
      }
     ],
     "prompt_number": 58
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Remember, `oo` is the SymPy notation for inifinity."
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Sums and products"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can evaluate sums and products using the functions: 'Sum'"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "n = Symbol(\"n\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 59
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Sum(1/n**2, (n, 1, 10))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\sum_{n=1}^{10} n^{-2}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 60,
       "text": [
        "  10    \n",
        " ____   \n",
        " \u2572      \n",
        "  \u2572   1 \n",
        "   \u2572  \u2500\u2500\n",
        "   \u2571   2\n",
        "  \u2571   n \n",
        " \u2571      \n",
        " \u203e\u203e\u203e\u203e   \n",
        "n = 1   "
       ]
      }
     ],
     "prompt_number": 60
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Sum(1/n**2, (n,1, 10)).evalf()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1.54976773116654$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 61,
       "text": [
        "1.54976773116654"
       ]
      }
     ],
     "prompt_number": 61
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Sum(1/n**2, (n, 1, oo)).evalf()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1.64493406684823$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 62,
       "text": [
        "1.64493406684823"
       ]
      }
     ],
     "prompt_number": 62
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Products work much the same way:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Product(n, (n, 1, 10)) # 10!"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\prod_{n=1}^{10} n$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 63,
       "text": [
        "  10   \n",
        "\u252c\u2500\u2500\u2500\u252c  \n",
        "\u2502   \u2502 n\n",
        "\u2502   \u2502  \n",
        "n = 1  "
       ]
      }
     ],
     "prompt_number": 63
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Limits"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Limits can be evaluated using the `limit` function. For example, "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "limit(sin(x)/x, x, 0)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 64,
       "text": [
        "1"
       ]
      }
     ],
     "prompt_number": 64
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can use 'limit' to check the result of derivation using the `diff` function:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "f"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\sin{\\left (x y \\right )} + \\cos{\\left (y z \\right )}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 65,
       "text": [
        "sin(x\u22c5y) + cos(y\u22c5z)"
       ]
      }
     ],
     "prompt_number": 65
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "diff(f, x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$y \\cos{\\left (x y \\right )}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 66,
       "text": [
        "y\u22c5cos(x\u22c5y)"
       ]
      }
     ],
     "prompt_number": 66
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$\\displaystyle \\frac{\\mathrm{d}f(x,y)}{\\mathrm{d}x} = \\frac{f(x+h,y)-f(x,y)}{h}$"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "h = Symbol(\"h\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 67
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "limit((f.subs(x, x+h) - f)/h, h, 0)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$y \\cos{\\left (x y \\right )}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 68,
       "text": [
        "y\u22c5cos(x\u22c5y)"
       ]
      }
     ],
     "prompt_number": 68
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "OK!"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can change the direction from which we approach the limiting point using the `dir` keywork argument:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "limit(1/x, x, 0, dir=\"+\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\infty$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 69,
       "text": [
        "\u221e"
       ]
      }
     ],
     "prompt_number": 69
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "limit(1/x, x, 0, dir=\"-\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$-\\infty$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 70,
       "text": [
        "-\u221e"
       ]
      }
     ],
     "prompt_number": 70
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Series"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Series expansion is also one of the most useful features of a CAS. In SymPy we can perform a series expansion of an expression using the `series` function:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "series(exp(x), x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1 + x + \\frac{1}{2} x^{2} + \\frac{1}{6} x^{3} + \\frac{1}{24} x^{4} + \\frac{1}{120} x^{5} + \\mathcal{O}\\left(x^{6}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 71,
       "text": [
        "         2    3    4     5        \n",
        "        x    x    x     x     \u239b 6\u239e\n",
        "1 + x + \u2500\u2500 + \u2500\u2500 + \u2500\u2500 + \u2500\u2500\u2500 + O\u239dx \u23a0\n",
        "        2    6    24   120        "
       ]
      }
     ],
     "prompt_number": 71
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "By default it expands the expression around $x=0$, but we can expand around any value of $x$ by explicitly include a value in the function call:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "series(exp(x), x, 1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$e + e x + \\frac{1}{2} e x^{2} + \\frac{1}{6} e x^{3} + \\frac{1}{24} e x^{4} + \\frac{1}{120} e x^{5} + \\mathcal{O}\\left(x^{6}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 72,
       "text": [
        "             2      3      4      5        \n",
        "          \u212f\u22c5x    \u212f\u22c5x    \u212f\u22c5x    \u212f\u22c5x     \u239b 6\u239e\n",
        "\u212f + \u212f\u22c5x + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + O\u239dx \u23a0\n",
        "           2      6      24    120         "
       ]
      }
     ],
     "prompt_number": 72
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "And we can explicitly define to which order the series expansion should be carried out:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "series(exp(x), x, 1, 10)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$e + e x + \\frac{1}{2} e x^{2} + \\frac{1}{6} e x^{3} + \\frac{1}{24} e x^{4} + \\frac{1}{120} e x^{5} + \\frac{1}{720} e x^{6} + \\frac{1}{5040} e x^{7} + \\frac{1}{40320} e x^{8} + \\frac{1}{362880} e x^{9} + \\mathcal{O}\\left(x^{10}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 73,
       "text": [
        "             2      3      4      5      6      7       8       9          \n",
        "          \u212f\u22c5x    \u212f\u22c5x    \u212f\u22c5x    \u212f\u22c5x    \u212f\u22c5x    \u212f\u22c5x     \u212f\u22c5x     \u212f\u22c5x      \u239b 10\u239e\n",
        "\u212f + \u212f\u22c5x + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\u2500 + O\u239dx  \u23a0\n",
        "           2      6      24    120    720    5040   40320   362880         "
       ]
      }
     ],
     "prompt_number": 73
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The series expansion includes the order of the approximation, which is very useful for keeping track of the order of validity when we do calculations with series expansions of different order:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "s1 = cos(x).series(x, 0, 5)\n",
      "s1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1 - \\frac{1}{2} x^{2} + \\frac{1}{24} x^{4} + \\mathcal{O}\\left(x^{5}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 74,
       "text": [
        "     2    4        \n",
        "    x    x     \u239b 5\u239e\n",
        "1 - \u2500\u2500 + \u2500\u2500 + O\u239dx \u23a0\n",
        "    2    24        "
       ]
      }
     ],
     "prompt_number": 74
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "s2 = sin(x).series(x, 0, 2)\n",
      "s2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$x + \\mathcal{O}\\left(x^{2}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 75,
       "text": [
        "     \u239b 2\u239e\n",
        "x + O\u239dx \u23a0"
       ]
      }
     ],
     "prompt_number": 75
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "expand(s1 * s2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$x + \\mathcal{O}\\left(x^{2}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 76,
       "text": [
        "     \u239b 2\u239e\n",
        "x + O\u239dx \u23a0"
       ]
      }
     ],
     "prompt_number": 76
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "If we want to get rid of the order information we can use the `removeO` method:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "expand(s1.removeO() * s2.removeO())"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{1}{24} x^{5} - \\frac{1}{2} x^{3} + x$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 77,
       "text": [
        " 5    3    \n",
        "x    x     \n",
        "\u2500\u2500 - \u2500\u2500 + x\n",
        "24   2     "
       ]
      }
     ],
     "prompt_number": 77
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "But note that this is not the correct expansion of $\\cos(x)\\sin(x)$ to $5$th order:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "(cos(x)*sin(x)).series(x, 0, 6)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$x - \\frac{2}{3} x^{3} + \\frac{2}{15} x^{5} + \\mathcal{O}\\left(x^{6}\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 78,
       "text": [
        "       3      5        \n",
        "    2\u22c5x    2\u22c5x     \u239b 6\u239e\n",
        "x - \u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500 + O\u239dx \u23a0\n",
        "     3      15         "
       ]
      }
     ],
     "prompt_number": 78
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Linear algebra"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Matrices"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Matrices are defined using the `Matrix` class:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "m11, m12, m21, m22 = symbols(\"m11, m12, m21, m22\")\n",
      "b1, b2 = symbols(\"b1, b2\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 79
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "A = Matrix([[m11, m12],[m21, m22]])\n",
      "A"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left[\\begin{smallmatrix}m_{11} & m_{12}\\\\m_{21} & m_{22}\\end{smallmatrix}\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 80,
       "text": [
        "\u23a1m\u2081\u2081  m\u2081\u2082\u23a4\n",
        "\u23a2        \u23a5\n",
        "\u23a3m\u2082\u2081  m\u2082\u2082\u23a6"
       ]
      }
     ],
     "prompt_number": 80
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "b = Matrix([[b1], [b2]])\n",
      "b"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left[\\begin{smallmatrix}b_{1}\\\\b_{2}\\end{smallmatrix}\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 81,
       "text": [
        "\u23a1b\u2081\u23a4\n",
        "\u23a2  \u23a5\n",
        "\u23a3b\u2082\u23a6"
       ]
      }
     ],
     "prompt_number": 81
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "With `Matrix` class instances we can do the usual matrix algebra operations:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "A**2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left[\\begin{smallmatrix}m_{11}^{2} + m_{12} m_{21} & m_{11} m_{12} + m_{12} m_{22}\\\\m_{11} m_{21} + m_{21} m_{22} & m_{12} m_{21} + m_{22}^{2}\\end{smallmatrix}\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 82,
       "text": [
        "\u23a1    2                               \u23a4\n",
        "\u23a2 m\u2081\u2081  + m\u2081\u2082\u22c5m\u2082\u2081    m\u2081\u2081\u22c5m\u2081\u2082 + m\u2081\u2082\u22c5m\u2082\u2082\u23a5\n",
        "\u23a2                                    \u23a5\n",
        "\u23a2                                 2  \u23a5\n",
        "\u23a3m\u2081\u2081\u22c5m\u2082\u2081 + m\u2082\u2081\u22c5m\u2082\u2082   m\u2081\u2082\u22c5m\u2082\u2081 + m\u2082\u2082   \u23a6"
       ]
      }
     ],
     "prompt_number": 82
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "A * b"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left[\\begin{smallmatrix}b_{1} m_{11} + b_{2} m_{12}\\\\b_{1} m_{21} + b_{2} m_{22}\\end{smallmatrix}\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 83,
       "text": [
        "\u23a1b\u2081\u22c5m\u2081\u2081 + b\u2082\u22c5m\u2081\u2082\u23a4\n",
        "\u23a2               \u23a5\n",
        "\u23a3b\u2081\u22c5m\u2082\u2081 + b\u2082\u22c5m\u2082\u2082\u23a6"
       ]
      }
     ],
     "prompt_number": 83
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "And calculate determinants and inverses, and the like:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "A.det()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$m_{11} m_{22} - m_{12} m_{21}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 84,
       "text": [
        "m\u2081\u2081\u22c5m\u2082\u2082 - m\u2081\u2082\u22c5m\u2082\u2081"
       ]
      }
     ],
     "prompt_number": 84
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "A.inv()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left[\\begin{smallmatrix}\\frac{1}{m_{11}} + \\frac{m_{12} m_{21}}{m_{11}^{2} \\left(m_{22} - \\frac{m_{12} m_{21}}{m_{11}}\\right)} & - \\frac{m_{12}}{m_{11} \\left(m_{22} - \\frac{m_{12} m_{21}}{m_{11}}\\right)}\\\\- \\frac{m_{21}}{m_{11} \\left(m_{22} - \\frac{m_{12} m_{21}}{m_{11}}\\right)} & \\frac{1}{m_{22} - \\frac{m_{12} m_{21}}{m_{11}}}\\end{smallmatrix}\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 85,
       "text": [
        "\u23a1 1          m\u2081\u2082\u22c5m\u2082\u2081                 -m\u2081\u2082       \u23a4\n",
        "\u23a2\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500  \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u23a5\n",
        "\u23a2m\u2081\u2081      2 \u239b      m\u2081\u2082\u22c5m\u2082\u2081\u239e      \u239b      m\u2081\u2082\u22c5m\u2082\u2081\u239e\u23a5\n",
        "\u23a2      m\u2081\u2081 \u22c5\u239cm\u2082\u2082 - \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u239f  m\u2081\u2081\u22c5\u239cm\u2082\u2082 - \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u239f\u23a5\n",
        "\u23a2           \u239d        m\u2081\u2081  \u23a0      \u239d        m\u2081\u2081  \u23a0\u23a5\n",
        "\u23a2                                               \u23a5\n",
        "\u23a2           -m\u2082\u2081                      1         \u23a5\n",
        "\u23a2   \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500         \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500   \u23a5\n",
        "\u23a2       \u239b      m\u2081\u2082\u22c5m\u2082\u2081\u239e               m\u2081\u2082\u22c5m\u2082\u2081   \u23a5\n",
        "\u23a2   m\u2081\u2081\u22c5\u239cm\u2082\u2082 - \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u239f         m\u2082\u2082 - \u2500\u2500\u2500\u2500\u2500\u2500\u2500   \u23a5\n",
        "\u23a3       \u239d        m\u2081\u2081  \u23a0                 m\u2081\u2081     \u23a6"
       ]
      }
     ],
     "prompt_number": 85
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Solving equations"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "For solving equations and systems of equations we can use the `solve` function:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "solve(x**2 - 1, x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\begin{bmatrix}-1, & 1\\end{bmatrix}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 86,
       "text": [
        "[-1, 1]"
       ]
      }
     ],
     "prompt_number": 86
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "solve(x**4 - x**2 - 1, x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\begin{bmatrix}- i \\sqrt{- \\frac{1}{2} + \\frac{1}{2} \\sqrt{5}}, & i \\sqrt{- \\frac{1}{2} + \\frac{1}{2} \\sqrt{5}}, & - \\sqrt{\\frac{1}{2} + \\frac{1}{2} \\sqrt{5}}, & \\sqrt{\\frac{1}{2} + \\frac{1}{2} \\sqrt{5}}\\end{bmatrix}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 87,
       "text": [
        "\u23a1        _____________         _____________        ___________       ________\n",
        "\u23a2       \u2571         ___         \u2571         ___        \u2571       ___       \u2571       _\n",
        "\u23a2      \u2571    1   \u2572\u2571 5         \u2571    1   \u2572\u2571 5        \u2571  1   \u2572\u2571 5       \u2571  1   \u2572\u2571 \n",
        "\u23a2-\u2148\u22c5  \u2571   - \u2500 + \u2500\u2500\u2500\u2500\u2500 , \u2148\u22c5  \u2571   - \u2500 + \u2500\u2500\u2500\u2500\u2500 , -  \u2571   \u2500 + \u2500\u2500\u2500\u2500\u2500 ,   \u2571   \u2500 + \u2500\u2500\u2500\n",
        "\u23a3   \u2572\u2571      2     2       \u2572\u2571      2     2      \u2572\u2571    2     2     \u2572\u2571    2     2\n",
        "\n",
        "___\u23a4\n",
        "__ \u23a5\n",
        "5  \u23a5\n",
        "\u2500\u2500 \u23a5\n",
        "   \u23a6"
       ]
      }
     ],
     "prompt_number": 87
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "System of equations:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "solve([x + y - 1, x - y - 1], [x,y])"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\begin{Bmatrix}x : 1, & y : 0\\end{Bmatrix}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 88,
       "text": [
        "{x: 1, y: 0}"
       ]
      }
     ],
     "prompt_number": 88
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "In terms of other symbolic expressions:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "solve([x + y - a, x - y - c], [x,y])"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\begin{Bmatrix}x : \\frac{1}{2} a + \\frac{1}{2} c, & y : \\frac{1}{2} a - \\frac{1}{2} c\\end{Bmatrix}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 89,
       "text": [
        "\u23a7   a   c     a   c\u23ab\n",
        "\u23a8x: \u2500 + \u2500, y: \u2500 - \u2500\u23ac\n",
        "\u23a9   2   2     2   2\u23ad"
       ]
      }
     ],
     "prompt_number": 89
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Quantum mechanics: noncommuting variables"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "How about non-commuting symbols? In quantum mechanics we need to work with noncommuting operators, and SymPy has a nice support for noncommuting symbols and even a subpackage for quantum mechanics related calculations!"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy.physics.quantum import *"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "States"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can define symbol states, kets and bras:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Ket('psi')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$${\\left|\\psi\\right\\rangle }$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 91,
       "text": [
        "\u2758\u03c8\u27e9"
       ]
      }
     ],
     "prompt_number": 91
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Bra('psi')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$${\\left\\langle \\psi\\right|}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 92,
       "text": [
        "\u27e8\u03c8\u2758"
       ]
      }
     ],
     "prompt_number": 92
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "u = Ket('0')\n",
      "d = Ket('1')\n",
      "\n",
      "a, b = symbols('alpha beta', complex=True)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 93
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "phi = a * u + sqrt(1-abs(a)**2) * d; phi"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\alpha {\\left|0\\right\\rangle } + \\sqrt{- \\left\\lvert{\\alpha}\\right\\rvert^{2} + 1} {\\left|1\\right\\rangle }$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 94,
       "text": [
        "           ____________    \n",
        "          \u2571      2         \n",
        "\u03b1\u22c5\u27580\u27e9 + \u2572\u2571  - \u2502\u03b1\u2502  + 1 \u22c5\u27581\u27e9"
       ]
      }
     ],
     "prompt_number": 94
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Dagger(phi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\overline{\\alpha} {\\left\\langle 0\\right|} + \\overline{\\sqrt{- \\left\\lvert{\\alpha}\\right\\rvert^{2} + 1}} {\\left\\langle 1\\right|}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 95,
       "text": [
        "        _______________    \n",
        "           ____________    \n",
        "_         \u2571      2         \n",
        "\u03b1\u22c5\u27e80\u2758 + \u2572\u2571  - \u2502\u03b1\u2502  + 1 \u22c5\u27e81\u2758"
       ]
      }
     ],
     "prompt_number": 95
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Dagger(phi) * d"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left(\\overline{\\alpha} {\\left\\langle 0\\right|} + \\overline{\\sqrt{- \\left\\lvert{\\alpha}\\right\\rvert^{2} + 1}} {\\left\\langle 1\\right|}\\right) {\\left|1\\right\\rangle }$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 96,
       "text": [
        "\u239b        _______________    \u239e    \n",
        "\u239c           ____________    \u239f    \n",
        "\u239c_         \u2571      2         \u239f    \n",
        "\u239d\u03b1\u22c5\u27e80\u2758 + \u2572\u2571  - \u2502\u03b1\u2502  + 1 \u22c5\u27e81\u2758\u23a0\u22c5\u27581\u27e9"
       ]
      }
     ],
     "prompt_number": 96
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Use `qapply` to distribute a mutiplication:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "qapply(Dagger(phi) * d)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\overline{\\alpha} \\left\\langle 0 \\right. {\\left|1\\right\\rangle } + \\overline{\\sqrt{- \\left\\lvert{\\alpha}\\right\\rvert^{2} + 1}} \\left\\langle 1 \\right. {\\left|1\\right\\rangle }$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 97,
       "text": [
        "          _______________      \n",
        "             ____________      \n",
        "_           \u2571      2           \n",
        "\u03b1\u22c5\u27e80\u27581\u27e9 + \u2572\u2571  - \u2502\u03b1\u2502  + 1 \u22c5\u27e81\u27581\u27e9"
       ]
      }
     ],
     "prompt_number": 97
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "qapply(Dagger(phi) * u)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\overline{\\alpha} \\left\\langle 0 \\right. {\\left|0\\right\\rangle } + \\overline{\\sqrt{- \\left\\lvert{\\alpha}\\right\\rvert^{2} + 1}} \\left\\langle 1 \\right. {\\left|0\\right\\rangle }$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 98,
       "text": [
        "          _______________      \n",
        "             ____________      \n",
        "_           \u2571      2           \n",
        "\u03b1\u22c5\u27e80\u27580\u27e9 + \u2572\u2571  - \u2502\u03b1\u2502  + 1 \u22c5\u27e81\u27580\u27e9"
       ]
      }
     ],
     "prompt_number": 98
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Operators"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "A = Operator('A')\n",
      "B = Operator('B')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Check if they are commuting!"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "A * B == B * A"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$False$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 100,
       "text": [
        "False"
       ]
      }
     ],
     "prompt_number": 100
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "expand((A+B)**3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$A B A + A \\left(B\\right)^{2} + \\left(A\\right)^{2} B + \\left(A\\right)^{3} + B A B + B \\left(A\\right)^{2} + \\left(B\\right)^{2} A + \\left(B\\right)^{3}$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 101,
       "text": [
        "           2    2      3              2    2      3\n",
        "A\u22c5B\u22c5A + A\u22c5B  + A \u22c5B + A  + B\u22c5A\u22c5B + B\u22c5A  + B \u22c5A + B "
       ]
      }
     ],
     "prompt_number": 101
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "c = Commutator(A,B)\n",
      "c"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\left[A,B\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 102,
       "text": [
        "[A,B]"
       ]
      }
     ],
     "prompt_number": 102
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can use the `doit` method to evaluate the commutator:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "c.doit()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$A B - B A$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 103,
       "text": [
        "A\u22c5B - B\u22c5A"
       ]
      }
     ],
     "prompt_number": 103
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "We can mix quantum operators with C-numbers:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "c = Commutator(a * A, b * B)\n",
      "c"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\alpha \\beta \\left[A,B\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 104,
       "text": [
        "\u03b1\u22c5\u03b2\u22c5[A,B]"
       ]
      }
     ],
     "prompt_number": 104
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "To expand the commutator, use the `expand` method with the `commutator=True` keyword argument:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "c = Commutator(A+B, A*B)\n",
      "c.expand(commutator=True)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$- \\left[A,B\\right] B + A \\left[A,B\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 105,
       "text": [
        "-[A,B]\u22c5B + A\u22c5[A,B]"
       ]
      }
     ],
     "prompt_number": 105
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Dagger(Commutator(A, B))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$- \\left[A^{\\dagger},B^{\\dagger}\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 106,
       "text": [
        " \u23a1 \u2020  \u2020\u23a4\n",
        "-\u23a3A ,B \u23a6"
       ]
      }
     ],
     "prompt_number": 106
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "ac = AntiCommutator(A,B)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 107
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "ac.doit()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$A B + B A$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 108,
       "text": [
        "A\u22c5B + B\u22c5A"
       ]
      }
     ],
     "prompt_number": 108
    },
    {
     "cell_type": "heading",
     "level": 4,
     "metadata": {},
     "source": [
      "Example: Quadrature commutator"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Let's look at the commutator of the electromagnetic field quadatures $x$ and $p$. We can write the quadrature operators in terms of the creation and annihilation operators as:\n",
      "\n",
      "$\\displaystyle x = (a + a^\\dagger)/\\sqrt{2}$\n",
      "\n",
      "$\\displaystyle p = -i(a - a^\\dagger)/\\sqrt{2}$\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "X = (A + Dagger(A))/sqrt(2)\n",
      "X"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$\\frac{1}{2} \\sqrt{2} \\left(A^{\\dagger} + A\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 109,
       "text": [
        "  ___ \u239b \u2020    \u239e\n",
        "\u2572\u2571 2 \u22c5\u239dA  + A\u23a0\n",
        "\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n",
        "      2       "
       ]
      }
     ],
     "prompt_number": 109
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "P = -I * (A - Dagger(A))/sqrt(2)\n",
      "P"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$- \\frac{1}{2} \\sqrt{2} i \\left(- A^{\\dagger} + A\\right)$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 110,
       "text": [
        "   ___   \u239b   \u2020    \u239e\n",
        "-\u2572\u2571 2 \u22c5\u2148\u22c5\u239d- A  + A\u23a0\n",
        "\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n",
        "         2         "
       ]
      }
     ],
     "prompt_number": 110
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Let's expand the commutator $[x,p]$"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Commutator(X, P).expand(commutator=True).expand(commutator=True)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$- i \\left[A^{\\dagger},A\\right]$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 111,
       "text": [
        "   \u23a1 \u2020  \u23a4\n",
        "-\u2148\u22c5\u23a3A ,A\u23a6"
       ]
      }
     ],
     "prompt_number": 111
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Here we see directly that the well known commutation relation for the quadratures\n",
      "\n",
      "$[x,p]=i$\n",
      "\n",
      "is a directly related to\n",
      "\n",
      "$[A, A^\\dagger]=1$ \n",
      "\n",
      "(which SymPy does not know about, and does not simplify)."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "For more details on the quantum module in SymPy, see:\n",
      "\n",
      "* http://docs.sympy.org/0.7.2/modules/physics/quantum/index.html\n",
      "* http://nbviewer.ipython.org/urls/raw.github.com/ipython/ipython/master/docs/examples/notebooks/sympy_quantum_computing.ipynb"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Further reading"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "* http://sympy.org/en/index.html - The SymPy projects web page.\n",
      "* https://github.com/sympy/sympy - The source code of SymPy.\n",
      "* http://live.sympy.org - Online version of SymPy for testing and demonstrations."
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Versions"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%reload_ext version_information\n",
      "\n",
      "%version_information numpy, sympy"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "html": [
        "<table><tr><th>Software</th><th>Version</th></tr><tr><td>Python</td><td>3.3.2+ (default, Feb 28 2014, 00:52:16) [GCC 4.8.1]</td></tr><tr><td>IPython</td><td>2.2.0</td></tr><tr><td>OS</td><td>posix [linux]</td></tr><tr><td>numpy</td><td>1.8.2</td></tr><tr><td>sympy</td><td>0.7.5</td></tr><tr><td colspan='2'>Tue Aug 26 22:57:37 2014 JST</td></tr></table>"
       ],
       "json": [
        "{ \"Software versions\" : [{ \"module\" : \"Python\", \"version\" : \"3.3.2+ (default, Feb 28 2014, 00:52:16) [GCC 4.8.1]\" }, { \"module\" : \"IPython\", \"version\" : \"2.2.0\" }, { \"module\" : \"OS\", \"version\" : \"posix [linux]\" }, { \"module\" : \"numpy\", \"version\" : \"1.8.2\" }, { \"module\" : \"sympy\", \"version\" : \"0.7.5\" } ] }"
       ],
       "latex": [
        "\\begin{table}\n",
        "\\begin{tabular}{|l|l|}\\hline\n",
        "{\\bf Software} & {\\bf Version} \\\\ \\hline\\hline\n",
        "Python & 3.3.2+ (default, Feb 28 2014, 00:52:16) [GCC 4.8.1] \\\\ \\hline\n",
        "IPython & 2.2.0 \\\\ \\hline\n",
        "OS & posix [linux] \\\\ \\hline\n",
        "numpy & 1.8.2 \\\\ \\hline\n",
        "sympy & 0.7.5 \\\\ \\hline\n",
        "\\hline \\multicolumn{2}{|l|}{Tue Aug 26 22:57:37 2014 JST} \\\\ \\hline\n",
        "\\end{tabular}\n",
        "\\end{table}\n"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 7,
       "text": [
        "Software versions\n",
        "Python 3.3.2+ (default, Feb 28 2014, 00:52:16) [GCC 4.8.1]\n",
        "IPython 2.2.0\n",
        "OS posix [linux]\n",
        "numpy 1.8.2\n",
        "sympy 0.7.5\n",
        "\n",
        "Tue Aug 26 22:57:37 2014 JST"
       ]
      }
     ],
     "prompt_number": 7
    }
   ],
   "metadata": {}
  }
 ]
}