add problem 25

problem0025.ipynb

@@ -0,0 +1,57 @@
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+    "# [1000-digit Fibonacci Number](https://projecteuler.net/problem=25)\n",
+    "\n",
+    "This problem can be solved with a scientific calculator!\n",
+    "\n",
+    "We're looking at [Fibonacci numbers](https://en.wikipedia.org/wiki/Fibonacci_sequence) again, after last seeing them in [problem 2](https://projecteuler.net/problem=2). Our goal is to find the first 1000 digit number in the sequence. Mathematically, we can state this is as the lowest $n$ such that\n",
+    "$$1+\\log{F_n} \\geq 1000$$\n",
+    "(Here, $\\log$ is the base 10 logarithm).\n",
+    "\n",
+    "To assist in finding $n$, we can employ Binet's formula:\n",
+    "$$F_n = \\frac{\\phi^n - (-\\phi)^{-n}}{\\sqrt{5}}$$\n",
+    "Here, $\\phi$ is the [golden ratio](https://en.wikipedia.org/wiki/Golden_ratio).\n",
+    "\n",
+    "Binet's formula is exact, but we can make our work a little easier by approximating.\n",
+    "$$F_n \\approx \\frac{\\phi^n}{\\sqrt{5}}$$\n",
+    "This approximation works since $(-\\phi)^{-n}$ approaches 0 as $n$ increases. This also means this approximation gets more accurate as $n$ increases, which is especially convenient since we're looking for a large $n$. Substituting this approximation into the above inequality, we have\n",
+    "$$1 + \\log{\\frac{\\phi^n}{\\sqrt{5}}} \\geq 1000$$\n",
+    "\n",
+    "Now with a little bit of algebra, we can solve for $n$.\n",
+    "$$n \\geq 999\\log_{\\phi}10 + \\log_{\\phi}\\sqrt{5}$$\n",
+    "If you're trying to solve this with a scientific calculator, you probably don't have a $\\log_{\\phi}$ button (*please let me know if you do*), but we can just use the [logarithmic change of base formula](https://en.wikipedia.org/wiki/List_of_logarithmic_identities). This ends up getting us\n",
+    "$$n \\geq 4781.85927\\ldots$$\n",
+    "Therefore, we want $n=4782$.\n",
+    "\n",
+    "## Relevant sequences\n",
+    "* Fibonacci numbers: [A000045](https://oeis.org/A000045)"
+   ]
+  }
+ ],
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