{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Names in Python" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**Question 1.** A famous fact in mathematics is that\n", "\n", "$$(1 + \\frac{1}{n})^{n}$$\n", "\n", "gets very close to the number $e$, which is roughly $2.718$, when $n$ is large. Use this fact to compute an approximation to $e$, called `roughly_e`. Your approximation should be correct to at least 5 decimal places." ] }, { "cell_type": "code", "execution_count": 7, "metadata": {}, "outputs": [], "source": [ "roughly_e = ...\n", "roughly_e" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**Question 2.** In the previous question, you might have noticed that you had to type in the value of $n$ twice when you tried different values (once as `1/n` and once as `**n`). You might have even accidentally used 2 different values.\n", "\n", "Redo the code in the cell below, giving your estimate of $e$ the name `also_roughly_e`. Instead of writing the value of $n$ twice, give it the name `n` in your Python code, and then use that name in your expression for `also_roughly_e`." ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [], "source": [ "n = ...\n", "also_roughly_e = ...\n", "also_roughly_e" ] } ], "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", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.8.3" } }, "nbformat": 4, "nbformat_minor": 1 }