try 3 I think I have it now

parent 767aec72
{
"cells": [
{
"cell_type": "code",
"execution_count": null,
"cell_type": "markdown",
"metadata": {},
"outputs": [],
"source": [
"# On the computation of $\\pi$\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"cell_type": "markdown",
"metadata": {},
"outputs": [],
"source": [
"## Asking the maths library\n",
"My computer tells me that $\\pi$ is approximately"
]
},
{
"cell_type": "markdown",
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from math import *\n",
"print(pi)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##Buffon's needle\n",
"Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"import numpy as np\n",
"np.random.seed(seed=42)\n",
"N = 10000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n",
"theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
"2/(sum(x+np.sin(theta))>1/N)"
]
}
],
"metadata": {
......
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