Update toy_notebook_en.ipynb

module2/exo1/toy_notebook_en.ipynb

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+#  On the computation of $\pi$
+##  Asking the maths library
+My computer tells me that π is approximatively
 
+from math import *
+print(pi)
+
+## Buffon’s needle
+Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**
+
+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)
+
+## 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 $X ∼ U(0, 1)$ and $Y ∼ U(0, 1)$, then $P[X2 + Y2 ≤ 1] = π/4$ (see ["Monte Carlo method"on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
+
+%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)
+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')
+
+It is then straightforward to obtain a (not really good) approximation to  𝜋
+  by counting how many times, on average,  𝑋2+𝑌2
+  is smaller than 1:
+  
+  4*np.mean(accept)
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