"My computer tells me that *$\\pi$* is *approximatively*"
"## Asking the maths library\n",
"My computer tells me that $\\pi$ is *approximatively*"
]
},
{
...
...
@@ -57,14 +51,8 @@
"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 *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__"
]
},
{
...
...
@@ -96,16 +84,8 @@
"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\n",
"fact that if $X ∼ U(0, 1)$ and $Y ∼ U(0, 1)$, then $P[$X^2$ + $Y^2$ ≤ 1] = $\\pi$/4$ (see [\"Monte Carlo method\"\n",
"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:"
