Uploaded 1st IPythonNotebook of simple algorithms

Different Ways of Calculating Pi.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.141592653589793"
+      ]
+     },
+     "execution_count": 22,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "import math\n",
+    "math.pi"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Implementing the methods for calculating pi from the WikiHow article on the subject   \n",
+    "https://www.wikihow.com/Calculate-Pi#:~:text=The%20circumference%20of%20a%20circle%20is%20found%20with%20the%20formula,result%20should%20be%20roughly%203.14.\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Gregory-Leibniz Series**\n",
+    "\n",
+    "This series takes many calculations to reach even a few accurate decimal places of pi. The calculation is done by alternating adding and subtracing calculations of 4 divided by an odd integer, starting with 0 add 4 divided by 1. According to the WikiHow article, it takes 500,000 calculations to obtain 5 accurate digits of pi. "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.14159264958921"
+      ]
+     },
+     "execution_count": 9,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "def GregoryLeibnizSeries(endInteger):\n",
+    "    pi = 0\n",
+    "    current_operator = \"+\"\n",
+    "    for i in range(1, endInteger, 2):\n",
+    "        current_calculation = 4/i\n",
+    "        if current_operator == \"+\":\n",
+    "            pi += current_calculation\n",
+    "            current_operator = \"-\"\n",
+    "        elif current_operator == \"-\":\n",
+    "            pi -= current_calculation\n",
+    "            current_operator = \"+\"\n",
+    "    return(pi)\n",
+    "\n",
+    "GregoryLeibnizSeries(500000000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Nilakantha Series**\n",
+    "\n",
+    "This series converges faster than the *Gregory-Leibniz Series*, but is a mildly more complex calculation. Again, alternating addition and subtraction, starting with addition, and dividing 4 by a value. This time, the value 4 is divided by is three numbers multiplied together, starting with an even number and the next two numbers are that even number plus 1 and plus 2. So, the first two calculations are starting with 3, add 4 divided by (2 × 3 × 4), and then subtract 4 divided by (4 × 5 × 6). "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.141592653589787"
+      ]
+     },
+     "execution_count": 12,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "def NilakanthaSeries(endInteger):\n",
+    "    pi = 3\n",
+    "    current_operator = \"+\"\n",
+    "    for i in range(2, endInteger, 2):\n",
+    "        current_calculation = 4/(i*(i+1)*(i+2))\n",
+    "        if current_operator == \"+\":\n",
+    "            pi += current_calculation\n",
+    "            current_operator = \"-\"\n",
+    "        elif current_operator == \"-\":\n",
+    "            pi -= current_calculation\n",
+    "            current_operator = \"+\"\n",
+    "    return(pi)\n",
+    "\n",
+    "NilakanthaSeries(500000000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Pi using sine**  \n",
+    "\n",
+    "This one is easier, pick a big number, multiple that big number by the sine of 180 / big number, sine needs to be in degrees. In the python code below you can see the sin calculation is converting degrees to radians with `math.radians` because trig functions in the `math` library calculate with radians. "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.141592653589793"
+      ]
+     },
+     "execution_count": 21,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "def LimitsCalculation(BigNumber):\n",
+    "    current_calculation = BigNumber * math.sin(math.radians(180/BigNumber))\n",
+    "    return(current_calculation)\n",
+    "\n",
+    "LimitsCalculation(900000000000)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
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+   "nbconvert_exporter": "python",
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+}