"My computer tells me that $\\pi$ is approximatively"
"My computer tells me that $\\pi$ is *approximatively*"
]
},
{
...
...
@@ -57,14 +57,14 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"### 1.2 Buffon’s needle"
"## 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"
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
]
},
{
...
...
@@ -104,7 +104,7 @@
"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",
"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\"\n",
"on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
]
},
...
...
@@ -142,9 +142,16 @@
"ax.set_aspect('equal')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:"
