"My computer tells me that $\\pi$ is *approximatively*"
"My computer tells me that $\\pi$ is *approximatively*"
]
]
},
},
...
@@ -43,13 +37,7 @@
...
@@ -43,13 +37,7 @@
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"cell_type": "markdown",
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"source": [
"### 1.2 Buffon’s needle"
"### 1.2 Buffon’s needle\n",
]
},
{
"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**"
]
]
},
},
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@@ -82,13 +70,7 @@
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@@ -82,13 +70,7 @@
"cell_type": "markdown",
"cell_type": "markdown",
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"source": [
"### 1.3 Using a surface fraction argument"
"### 1.3 Using a surface fraction argument\n",
]
},
{
"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 $X \\approx U(0, 1)$ and $Y \\approx U(0, 1)$, then $P[X^2 + Y^2 $\\le$ 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
"A method that is easier to understand and does not make use of the sin function is based on the fact that if $X \\approx U(0, 1)$ and $Y \\approx U(0, 1)$, then $P[X^2 + Y^2 $\\le$ 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
