minor changes

parent a645391b
...@@ -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*"
] ]
}, },
...@@ -33,13 +31,7 @@ ...@@ -33,13 +31,7 @@
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"## Buffon's needle" "## 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**"
] ]
}, },
...@@ -73,7 +65,6 @@ ...@@ -73,7 +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 thesin function is based on thefact that if $X \\sim U(0,1)$ and $Y \\sim U(0,1)$, then $P[X^2 + Y^2 \\leq 1] = \\frac{\\pi}{4}$ (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 thesin function is based on thefact that if $X \\sim U(0,1)$ and $Y \\sim U(0,1)$, then $P[X^2 + Y^2 \\leq 1] = \\frac{\\pi}{4}$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
] ]
}, },
...@@ -116,7 +107,7 @@ ...@@ -116,7 +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, $X^2+Y^2$ is smaller than 1:" "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:"
] ]
}, },
{ {
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment