"My computer tells me that $\\pi$ is approximatively"
"My computer tells me that $\\pi$ is *approximatively*"
]
]
},
},
{
{
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@@ -57,14 +57,14 @@
...
@@ -57,14 +57,14 @@
"cell_type": "markdown",
"cell_type": "markdown",
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"source": [
"### 1.2 Buffon’s needle"
"## Buffon’s needle"
]
]
},
},
{
{
"cell_type": "markdown",
"cell_type": "markdown",
"metadata": {},
"metadata": {},
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"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__"
]
]
},
},
{
{
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@@ -104,7 +104,7 @@
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@@ -104,7 +104,7 @@
"metadata": {},
"metadata": {},
"source": [
"source": [
"A method that is easier to understand and does not make use of the sin function is based on the\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 $P[X^2 + Y^2 ≤ 1] = \\pi/4$ (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:"
"on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
]
]
},
},
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@@ -142,9 +142,16 @@
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@@ -142,9 +142,16 @@
"ax.set_aspect('equal')"
"ax.set_aspect('equal')"
]
]
},
},
{
"cell_type": "markdown",
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"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:"