"My computer tells me that $\\pi$ is *approximatively*"
]
},
{
...
...
@@ -33,7 +33,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1.2 Buffon’s needle\n",
"## Buffon’s needle\n",
"\n",
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
]
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@@ -67,10 +67,9 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1.3 Using a surface fraction argument\n",
"## Using a surface fraction argument\n",
"\n",
"A method that is easier to understand and does not make use of the sin function is based on the\n",
"fact that if *X ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
"A method that is easier to understand and does not make use of the sin function is based on the fact that if *X ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
]
},
{
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@@ -112,8 +111,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"It is then straightforward to obtain a (not really good) approximation to *π* by counting how\n",
"many times, on average, X<sup>2</sup> + Y<sup>2</sup>\n",
"It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, X<sup>2</sup> + Y<sup>2</sup>\n",
