

In [1]:

    
import numpy as np
import matplotlib.pyplot as plt
from sympy import *
%matplotlib inline

Exercise 1

We proceed building the alogrithm for testing the accuracy of the numerical derivative



In [2]:

    
def f(x):
    return np.exp(np.sin(x))

def df(x):
    return f(x) * np.cos(x)

def absolute_err(f, df, h):
    g = (f(h) - f(0)) / h
    return np.abs(df(0) - g)

hs = 10. ** -np.arange(15)
epsilons = np.empty(15)
for i, h in enumerate(hs):
    epsilons[i] = absolute_err(f, df, h)

a)



In [3]:

    
plt.plot(hs, epsilons, 'o')
plt.yscale('log')
plt.xscale('log')
plt.xlabel(r'h')
plt.ylabel(r'$\epsilon(h)$')
plt.grid(linestyle='dotted')

We can see that until $h = 10^7$ the trend is that the absolute error diminishes, but after that it goes back up. This is due to the fact that when we compute $f(h) - f(0)$ we are using an ill-conditioned operation. In fact these two values are really close to each other.

Exercise 2

a.

We can easily see that when $\|x\| \ll 1$ we have that both $\frac{1 - x }{x + 1}$ and $\frac{1}{2 x + 1}$ are almost equal to 1, so the subtraction is ill-conditined.



In [4]:

    
x_1 = symbols('x_1')
fun1 = 1 / (1 + 2*x_1) - (1 - x_1) / (1 + x_1)
fun1









    Out[4]:





-(-x_1 + 1)/(x_1 + 1) + 1/(2*x_1 + 1)

We can modify the previous expression for being well conditioned around 0. This is well conditoned.



In [5]:

    
fun2 = simplify(fun1)
fun2









    Out[5]:





2*x_1**2/(2*x_1**2 + 3*x_1 + 1)

A comparison between the two ways of computing this value. we can clearly see that if we are far from 1 the methods are nearly identical, but the closer you get to 0 the two methods diverge



In [6]:

    
def f1(x):
    return 1 / (1 + 2*x) - (1 - x) / (1 + x)

def f2(x):
    return 2*x**2/(2*x**2 + 3*x + 1)

for i, h in enumerate(hs):
    epsilons[i] = (f1(h) - f2(h)) / h
plt.plot(hs, epsilons, 'o')
plt.yscale('log')
plt.xscale('log')
plt.xlabel(r'h')
plt.ylabel(r'$\epsilon(h)$')
plt.grid(linestyle='dotted')

$ \sqrt{x + \frac{1}{x}} - \sqrt{x - \frac{1}{x}} = \sqrt{x + \frac{1}{x}} - \sqrt{x - \frac{1}{x}} \frac{\sqrt{x + \frac{1}{x}} + \sqrt{x - \frac{1}{x}} }{\sqrt{x + \frac{1}{x}} + \sqrt{x - \frac{1}{x}}} = \frac{2x}{\sqrt{x + \frac{1}{x}} + \sqrt{x - \frac{1}{x}}}$

Exercise 3.

a.

If we assume we posess a 6 faced dice, we have at each throw three possible outcome. So we have to take all the combination of 6 numbers repeating 3 times. It is intuitive that our $\Omega$ will be composed by $6^3$ samples, and will be of type: $(1, 1, 1), (1, 1, 2), (1, 1, 3), ... (6, 6, 5), (6, 6, 6)$



In [7]:

    
import itertools
x = [1, 2, 3, 4, 5, 6]
omega = set([p for p in itertools.product(x, repeat=3)])
print(r'Omega has', len(omega), 'elements and they are:')
print(omega)









    



Omega has 216 elements and they are:

Concerning the $\sigma$-algebra we need to state that there does not exist only a $\sigma$-algebra for a given $\Omega$, but it this case a reasonable choice would be the powerset of $\Omega$.

b.

In case of fairness of dice we will have the discrete uniform distribution. And for computing the value of $\rho(\omega)$ we just need to compute the inverse of our sample space $\rho(\omega) = \frac{1}{6^3}$



In [8]:

    
1/(6**3)









    Out[8]:





0.004629629629629629

c.

If we want to determine the set $A$ we can take in consideration its complementary $A^c = \{\text{Not even one throw is 6}\}$. This event is analogous the sample space of a 5-faced dice. So it's dimension will be $5^3$. For computing the size of $A$ we can simply compute $6^3 - 5^3$ and for the event its self we just need to $\Omega \setminus A^c = A$



In [9]:

    
print('Size of A^c:', 5**3)
print('Size of A:   ', 6 ** 3 - 5 ** 3)









    



Size of A^c: 125
Size of A:    91



In [10]:

    
36 + 5 *6 + 5 * 5









    Out[10]:





91



In [11]:

    
x = [1, 2, 3, 4, 5]
A_c =  set([p for p in itertools.product(x, repeat=3)])
print('A^c has ', len(A_c), 'elements.\nA^c =', A_c)
print('A has ', len(omega - A_c), 'elements.\nA^c =', omega - A_c)



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