"My computer tells me that $\\pi$ is *approximatively*"
]
},
{
{
"cell_type": "code",
"cell_type": "code",
"execution_count": 1,
"execution_count": 1,
...
@@ -37,24 +18,6 @@
...
@@ -37,24 +18,6 @@
" print(pi)"
" print(pi)"
]
]
},
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"## Buffon's needle"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
]
},
{
{
"cell_type": "code",
"cell_type": "code",
"execution_count": 2,
"execution_count": 2,
...
@@ -80,24 +43,6 @@
...
@@ -80,24 +43,6 @@
" 2/(sum((x+np.sin(theta))>1)/N)"
" 2/(sum((x+np.sin(theta))>1)/N)"
]
]
},
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"## Using a surface fraction argument"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"A method that is easier to understand and does not make use of the $\\sin$ function is based on the fact that if $X\\sim U(0,1)$ 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:"
]
},
{
{
"cell_type": "code",
"cell_type": "code",
"execution_count": 3,
"execution_count": 3,
...
@@ -132,15 +77,6 @@
...
@@ -132,15 +77,6 @@
" ax.set_aspect('equal')"
" ax.set_aspect('equal')"
]
]
},
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"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:"
