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parent 0699f9b4
......@@ -4,7 +4,13 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"# On the computation of $\\pi$\n",
"# On the computation of $\\pi$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Asking the maths library\n",
"My computer tells me that $\\pi$ is *approximatively*"
]
......@@ -32,7 +38,7 @@
"metadata": {},
"source": [
"## Buffon's needle\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_"
]
},
{
......@@ -65,7 +71,7 @@
"metadata": {},
"source": [
"## Using a surface fraction argument\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 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:"
]
},
{
......@@ -87,17 +93,18 @@
}
],
"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",
"accept=(x*x+y*y)<=1\n",
"reject=np.logical_not(accept)\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",
"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",
"ax.set_aspect('equal')"
......
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