"My computer tells me that $\\pi$ is *approximatively*"
]
},
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@@ -42,13 +42,13 @@
"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**"
"## 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": 5,
"execution_count": 2,
"metadata": {},
"outputs": [
{
...
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@@ -57,7 +57,7 @@
"3.128911138923655"
]
},
"execution_count": 5,
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
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@@ -75,8 +75,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"### Using a surface fraction argument\n",
"A method that is easier to understand and does not make use of the sin function is based on the fact that if $X \\sim U(0,1)$ and $Y \\sim U(0,1)$ then $P[X^2 + Y^2 \\le 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
"## Using a surface fraction argument\n",
"A method that is easier to understand and does not make use of the $\\sin$ function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then $P[X^2 + Y^2 \\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
