floating_point_normalize_example.ipynb

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   "source": [
    "We might like to think that normalizing a vector by its sum would be idempotent.\n",
    "But due to floating point arithmetic, this isn't necessarily true."
   ]
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      "text/plain": [
       "array([1., 1., 1., 1., 1., 1.])"
      ]
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     "execution_count": 1,
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   "source": [
    "import numpy as np\n",
    "\n",
    "x = np.ones(6)\n",
    "x"
   ]
  },
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     "text": [
      "0.9999999999999999\n",
      "1.0000000000000002\n",
      "0.9999999999999999\n",
      "1.0000000000000002\n",
      "0.9999999999999999\n",
      "1.0000000000000002\n",
      "0.9999999999999999\n",
      "1.0000000000000002\n",
      "0.9999999999999999\n",
      "1.0000000000000002\n"
     ]
    }
   ],
   "source": [
    "for i in range(10):\n",
    "    x = x/np.sum(x)\n",
    "    print(sum(x))"
   ]
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   "cell_type": "markdown",
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   "source": [
    "You get different answers depending on how you sum up the elements of `x`."
   ]
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    {
     "data": {
      "text/plain": [
       "0.16666666666666669"
      ]
     },
     "execution_count": 3,
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   "source": [
    "a = x[0]\n",
    "a"
   ]
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       "1.0000000000000002"
      ]
     },
     "execution_count": 4,
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   "source": [
    "a+a+a+a+a+a"
   ]
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      "text/plain": [
       "1.0"
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   "source": [
    "(a+a)+(a+a)+(a+a)"
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     "data": {
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       "0.16666666666666666"
      ]
     },
     "execution_count": 6,
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   "source": [
    "x = x/np.sum(x)\n",
    "a = x[0]\n",
    "a"
   ]
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   "execution_count": 7,
   "id": "a506821f-81e9-4048-9023-4fb91e893eee",
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    {
     "data": {
      "text/plain": [
       "0.9999999999999999"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
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   "source": [
    "a+a+a+a+a+a"
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       "1.0"
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    "(a+a)+(a+a)+(a+a)"
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