[2 points] Create a variable called x which is a list containing all even numbers less than 100 using the list(range(...)) syntax.
[1 point] Compute the sum of x without using a for loop
[4 points] Compute the sum of x with a for loop.
[1 point] Print the elements of x reversed using a slice
[2 point] Print the second half of x using the len(x) // 2 syntax
[4 points] Create a new empty list y. Using a for loop and the append keyword, make y contain the square of each element of x. So it should contain: [4, 16, ..., ].
In [1]:
#1.1
x = list(range(2,100,2))
print(x)
In [2]:
#1.2
print(sum(x))
In [3]:
#1.3
s = 0
for xi in x:
s += xi
print(s)
In [4]:
#1.4
print(x[::-1])
In [5]:
#1.5
print(x[len(x)//2:])
In [6]:
#1.6
y = []
for xi in x:
y.append(xi**2)
print(y)
[2 points] Create a variable called a which is an array containing all even numbers less than 100 using the numpy arange syntax.
[1 point] Compute the sum of a without using a for loop
[1 point] Print the elements of a reversed using a slice
[2 points] Print the minimum, maximum, and mean elements of a using numpy functions.
[1 point] Print the square of each element in a without using a for loop or lists. You should just have one line of numpy code.
In [7]:
#2.1
import numpy as np
a = np.arange(2,100,2)
print(a)
In [8]:
#2.2
print(np.sum(a))
In [9]:
#2.3
print(a[::-1])
In [10]:
#2.4
print(np.min(a), np.max(a), np.mean(a))
In [11]:
print(a**2)
For each problem below, use numpy to create x and y arrays which are plotted. Be sure to label your x-axis, y-axis, put the problem number as the title, use at least 500 points, make your figures be 4x3 inches, and add a legend if you have more than one line being plotted.
[6 points] Plot $y = x^2$ from $-1$ to $1$
[8 points] A hanging rope, wire, or chain follows a catenary curve ($y = a\textrm{cosh}\frac{x}{a} - a$), although many including Galileo mistakenly believed hanging ropes a parabolas. Compare the catenary and parabolic ($y = x^2$) curves over $-1$ to $1$ where $a = 0.63$.
[8 points] Compare the functions $\cos x$ and $\cosh x - 1$ from $-\pi/2$ to $\pi/2$
In [12]:
#3.1
import matplotlib.pyplot as plt
%matplotlib inline
x = np.linspace(-1, 1, 500)
y = x**2
plt.figure(figsize=(4,3))
plt.plot(x,y)
plt.xlabel('$x$')
plt.ylabel('$y$')
plt.title('Problem 3.1')
plt.show()
In [19]:
# 3.2
x = np.linspace(-1, 1, 500)
a = 0.62
y2 = a * np.cosh(x / a) - a
plt.figure(figsize=(4,3))
plt.plot(x,y, label='Parabolic')
plt.plot(x,y2, label='Catenary')
plt.legend()
plt.xlabel('$x$')
plt.ylabel('$y$')
plt.title('Problem 3.2')
plt.show()
In [20]:
#3.3
x = np.linspace(-np.pi/2, np.pi/2, 500)
y1 = np.cos(x)
y2 = np.cosh(x) - 1
plt.figure(figsize=(4,3))
plt.plot(x,y1, label='$\cos x$')
plt.plot(x,y2, label='$\cosh x$')
plt.legend()
plt.xlabel('$x$')
plt.ylabel('$y$')
plt.title('Problem 3.3')
plt.show()
Use this probability distribution:
$$ Q = \{\textrm{red},\textrm{ green}, \textrm{blue}\} $$$$ P(\textrm{red}) = 0.1,\, P(\textrm{green}) = 0.5,\, P(\textrm{blue}),\, = 0.4 $$[2 points] Create a dictionary called prob where the key is the colors as a string and the values are the probability.
[4 points] Starting with this fragment, show that your probability distribution is normalized.
for key,value in prob.items():
print(key, value)[4 points] Let's define a random variable $X$ that is 1 for red, 2 for green, and 0 for blue. Using your for loop from 4.2, use boolean statements to set a variable $x$ to what the value of $X$ should be using the key variable. Print out key, x, which should look like red, 1, green, 2...
[4 points] Compute the expected value of x using the for loop from 4.3
[4 points] Compute the variance by hand, showing the steps in Markdown
In [15]:
#4.1
prob = {'red': 0.1, 'green': 0.5, 'blue': 0.4}
In [16]:
#4.2
psum = 0
for key, value in prob.items():
psum += value
print(psum)
In [17]:
#4.3
for key,value in prob.items():
x = 0
if key == 'red':
x = 1
elif key == 'green':
x = 2
print(key, x)
In [18]:
#4.4
ev = 0
for key,value in prob.items():
x = 0
if key == 'red':
x = 1
elif key == 'green':
x = 2
ev += x * value
print(ev)