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parent 504aaa49
...@@ -5,9 +5,7 @@ ...@@ -5,9 +5,7 @@
"metadata": {}, "metadata": {},
"source": [ "source": [
"# On the computation of $\\pi$\n", "# On the computation of $\\pi$\n",
"\n",
"## Asking the maths library\n", "## Asking the maths library\n",
"\n",
"My computer tells me that $\\pi$ is *approximatively*" "My computer tells me that $\\pi$ is *approximatively*"
] ]
}, },
...@@ -34,7 +32,6 @@ ...@@ -34,7 +32,6 @@
"metadata": {}, "metadata": {},
"source": [ "source": [
"## Buffon’s needle\n", "## Buffon’s needle\n",
"\n",
"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**"
] ]
}, },
...@@ -68,7 +65,6 @@ ...@@ -68,7 +65,6 @@
"metadata": {}, "metadata": {},
"source": [ "source": [
"## Using a surface fraction argument\n", "## Using a surface fraction argument\n",
"\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 ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/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 ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
] ]
}, },
...@@ -111,8 +107,7 @@ ...@@ -111,8 +107,7 @@
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, X<sup>2</sup> + Y<sup>2</sup>\n", "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, X<sup>2</sup> + Y<sup>2</sup> is smaller than 1:"
"is smaller than 1:"
] ]
}, },
{ {
......
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