Restored previous version

module2/exo1/toy_notebook_en.ipynb

-# On the computation of $\pi$
-
-## Asking the maths library
-My computer tells me that $\pi$ is _approximatively_
-
-
-```python
-from math import *
-print(pi)
-```
-
-    3.141592653589793
-
-
-## Buffon’s needle
-Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
-
-
-```python
-import numpy as np
-np.random.seed(seed=42)
-N = 10000
-x = np.random.uniform(size=N, low=0, high=1)
-theta = np.random.uniform(size=N, low=0, high=pi/2)
-2/(sum((x+np.sin(theta))>1)/N)
-```
-
-
-
-
-    3.128911138923655
-
-
-
-## Using a surface fraction argument
-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 $P[X^2 + Y^2\le 1] = \pi/4$ (see ["Monte Carlo method"
-on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
-
-
-```python
-%matplotlib inline
-import matplotlib.pyplot as plt
-
-np.random.seed(seed=42)
-N = 1000
-x = np.random.uniform(size=N, low=0, high=1)
-y = np.random.uniform(size=N, low=0, high=1)
-accept = (x*x+y*y) <= 1
-reject = np.logical_not(accept)
-
-fig, ax = plt.subplots(1)
-ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)
-ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)
-ax.set_aspect('equal')
-```
-
-
-![png](output_6_0.png)
-
-
-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:
-
-
-```python
-4*np.mean(accept)
-```
-
-
-
-
-    3.112
-
-
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\ No newline at end of file