"My computer tells me that $\\pi$ is <i>approximatively</i>"
"## Asking the maths library\n",
"My computer tells me that $\\pi$ is *approximatively*"
]
},
{
...
...
@@ -43,14 +37,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1.2 Buffon's needle"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the <b>approximation</b>"
"## Buffon's needle\n",
"Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
]
},
{
...
...
@@ -82,14 +70,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1.3 Using a surface fraction argument"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"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 $P[X^2 + Y^2 ≤ 1] = π/4$ (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:"
