numpy_poly_bug.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Bug in numpy.poly"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import numpy as np\n",
      "print np.__version__"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "1.9.0\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Here we will compute the coefficients of the Wilkinson polynomial\n",
      "$$P(x) = \\prod_{i=1}^{20} (x-i)$$\n",
      "in two ways, both using the `np.poly` function.  First, we just give the vector of coefficients; second, we give a diagonal matrix with those coefficients on the diagonal.  The output of `np.poly` should be the same in both cases, but it's not:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "N = 20\n",
      "v = np.arange(1,N+1)\n",
      "c1 = np.poly(v)\n",
      "c2 = np.poly(np.diag(v))\n",
      "print c1-c2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "[  0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00\n",
        "   0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00\n",
        "   0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00\n",
        "   0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00\n",
        "  -1.02400000e+03   1.84467441e+19  -1.84467441e+19   0.00000000e+00\n",
        "   0.00000000e+00]\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "So at least one of the answers is wrong; which?"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "print c1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "[                   1                 -210                20615\n",
        "             -1256850             53327946          -1672280820\n",
        "          40171771630        -756111184500       11310276995381\n",
        "     -135585182899530     1307535010540395   -10142299865511450\n",
        "    63030812099294896  -311333643161390640  1206647803780373360\n",
        " -3599979517947607200  8037811822645051776  5575812828558562816\n",
        " -4642984320068847616 -8752948036761600000  2432902008176640000]\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Clearly `c1` is wrong, since the coefficients ought to have alternating signs."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "print c2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "[  1.00000000e+00  -2.10000000e+02   2.06150000e+04  -1.25685000e+06\n",
        "   5.33279460e+07  -1.67228082e+09   4.01717716e+10  -7.56111184e+11\n",
        "   1.13102770e+13  -1.35585183e+14   1.30753501e+15  -1.01422999e+16\n",
        "   6.30308121e+16  -3.11333643e+17   1.20664780e+18  -3.59997952e+18\n",
        "   8.03781182e+18  -1.28709312e+19   1.38037598e+19  -8.75294804e+18\n",
        "   2.43290201e+18]\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "For $N=19$, no such error occurs:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "N = 19\n",
      "v = np.arange(1,N+1)\n",
      "c1 = np.poly(v)\n",
      "c2 = np.poly(np.diag(v))\n",
      "print c1-c2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "[ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.\n",
        "  0.  0.]\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 5
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