4th

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"# 1 On the computation of $\\pi$" "# On the computation of $\\pi$"
] ]
}, },
{ {
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"## 1.1 Asking the maths library" "## Asking the maths library\n",
] "My computer tells me that $\\pi$ is *approximatively*"
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"My computer tells me that $\\pi$ is <i>approximatively</i>"
] ]
}, },
{ {
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"cell_type": "markdown", "cell_type": "markdown",
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"source": [ "source": [
"## 1.2 Buffon's needle" "## Buffon's needle\n",
] "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the <b>approximation</b>"
] ]
}, },
{ {
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"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"## 1.3 Using a surface fraction argument" "## 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:"
},
{
"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 $X ∼ U(0, 1)$ and $Y ∼ U(0, 1)$, then $P[X^2 + Y^2 ≤ 1] = π/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
] ]
}, },
{ {
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