refer to relevant problems instead of re-implementing

notebooks/problem0065.ipynb

@@ -36,7 +36,7 @@
    "id": "396468b7",
    "metadata": {},
    "source": [
-    "If we wanted to compute the convergents ourselves, we could first make a generator for the partial denominators of the continued fraction of $e$."
+    "To compute the convergents ourselves, we'll first make a generator for the partial denominators of the continued fraction of $e$."
    ]
   },
   {
@@ -48,14 +48,9 @@
    "source": [
     "from itertools import count, chain\n",
     "\n",
-    "def partial_denominators_e(n):\n",
+    "def partial_denominators_e():\n",
     "    yield 2\n",
-    "    denominators = chain.from_iterable((1, 2 * k, 1) for k in count(1))\n",
-    "    for (i, b) in enumerate(denominators):\n",
-    "        if i >= n:\n",
-    "            break\n",
-    "            \n",
-    "        yield b"
+    "    yield from chain.from_iterable((1, 2 * k, 1) for k in count(1))"
    ]
   },
   {
@@ -63,7 +58,7 @@
    "id": "33288cd0",
    "metadata": {},
    "source": [
-    "Then write a function for computing a continued fraction from a sequence of partial denominators (outside of SageMath, you might want to use a [fraction type](https://docs.python.org/3/library/fractions.html))."
+    "Then we'll apply a simple algorithm for computing [convergents using the partial denominators](https://en.wikipedia.org/wiki/Simple_continued_fraction) (outside of SageMath, you might want to use a [fraction type](https://docs.python.org/3/library/fractions.html))."
    ]
   },
   {
@@ -73,13 +68,21 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "def cf(denominators):\n",
-    "    a = next(denominators)\n",
-    "    \n",
-    "    try:\n",
-    "        return a + 1 / cf(denominators)\n",
-    "    except StopIteration:\n",
-    "        return a"
+    "def convergents(partial_denoms):\n",
+    "    h, hprev = 1, 0\n",
+    "    k, kprev = 0, 1\n",
+    "    for b in partial_denoms:\n",
+    "        h, hprev = b * h + hprev, h\n",
+    "        k, kprev = b * k + kprev, k\n",
+    "        yield h/k"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "99b790f3",
+   "metadata": {},
+   "source": [
+    "Now just iterate until we reach the 100th convergent."
    ]
   },
   {
@@ -100,7 +103,11 @@
     }
    ],
    "source": [
-    "sum(cf(partial_denominators_e(99)).numerator().digits())"
+    "for (i, c) in enumerate(convergents(partial_denominators_e())):\n",
+    "    if i == 99:\n",
+    "        break\n",
+    "\n",
+    "sum(c.numerator().digits())"
    ]
   },
   {