Minor corrections

parent 7c3ac62f
......@@ -4,21 +4,15 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"# 1 On the computation of $\\pi$"
"# On the computation of $\\pi$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1.1 Asking the maths library"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"My computer tells me that $\\pi$ is approximatively"
"## Asking the maths library\n",
"My computer tells me that $\\pi$ is *approximatively*"
]
},
{
......@@ -43,14 +37,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1.2 Buffon's needle"
]
},
{
"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**"
"## Buffon's needle\n",
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
]
},
{
......@@ -82,14 +70,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1.3 Using a surface fraction argument"
]
},
{
"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 $X \\sim U(0,1)$ 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"
"## 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:"
]
},
{
......@@ -111,14 +93,17 @@
}
],
"source": [
"%matplotlib inline\n",
"%matplotlib inline \n",
"import matplotlib.pyplot as plt\n",
"\n",
"np.random.seed(seed=42)\n",
"N = 1000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n",
"y = np.random.uniform(size=N, low=0, high=1)\n",
"\n",
"accept = (x*x+y*y) <= 1\n",
"reject = np.logical_not(accept)\n",
"\n",
"fig, ax = plt.subplots(1)\n",
"ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n",
"ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
......@@ -129,7 +114,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
" It is then straightforward to obtain a (not really good) approximation to π 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:"
]
},
{
......
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