73 lines
1.8 KiB
Text
73 lines
1.8 KiB
Text
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Names in Python"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**Question 1.** A famous fact in mathematics is that\n",
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"\n",
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"$$(1 + \\frac{1}{n})^{n}$$\n",
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"\n",
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"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."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"metadata": {},
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"outputs": [],
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"source": [
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"roughly_e = ...\n",
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"roughly_e"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**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",
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"\n",
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"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`."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"n = ...\n",
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"also_roughly_e = ...\n",
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"also_roughly_e"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.3"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 1
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}
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