In order to successfully complete this assignment you need to participate both individually and in groups during class. If you attend class in-person then have one of the instructors check your notebook and sign you out before leaving class. If you are attending asyncronously, turn in your assignment using D2L no later than 11:59pm on the day of class. See links at the end of this document for access to the class timeline for your section.
Vector mathematics are used in physics all of the time. Consider the picture at the beginning of this notebook. While the space shuttle is attached to the Airplane they have the same velocity vector. This vector likely has three components; the velocity in the x direction, y direction and z direction. In fact this velocity is just a combination of multiple components such as thrust from the engines, wind speed and drag.
In this notebook we will be discussing basic vector mathematics and practicing our Python. We will be using the commands in this notebook all semester so make sure you have have some mastery before moving on.
Image from NASA February 1977
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For any vectors x
, y
, and z
of the same size/dimension, we have the following properties:
x + y = y + x
.(x + y) + z = x + (y + z)
. We can therefore write both as x + y + z
.x + 0 = 0 + x = x
. (This is an example where the size of the zero vector follows from the context, i.e., its size must be the same as the size of x
)x − x = 0
. Subtracting a vector from itself yields the zero vector. (Here too the size of 0 is the size of a.)For any vectors x
, y
, and scalars a
, b
, we have the following properties
ax = x * a
; This means that scalar-vector multiplication can be written in either order.(ab)x = a(bx)
a(x + y) = ax + ay
, (x+y)a = xa + ya
, and (a+b)x = ax + bx
.For those who are new to Python, there are many common mistakes happen in this course. Try to fix the following codes.
It means that the code does not make sense in Python. We would like to define a vector with four numbers.
✅ **DO THIS:** Fix the following code to creat three vectors with four numbers.
x = [1 2 3.4 4]
y = [1, 2, 3, 5]]
z = [[1, 2, 3, 6.3]
Although you may have been able to get rid of the error messages the answer to you problem may still not be correct. Throughout the semester we will be using a python program called answercheck
to allow you to check some of your answers. This program doesn't tell you the right answer but it is intended to be used as a way to get immediate feedback and accelerate learning.
✅ **DO THIS:** First we will need to download answercheck.py
to your current working directory. You only really need to do this once. However, if you delete this file by mistake sometime during the semester, you can come back to this notebook and download it again by running the following cell:
from urllib.request import urlretrieve
urlretrieve('https://raw.githubusercontent.com/colbrydi/jupytercheck/master/answercheck.py',
'answercheck.py');
✅ **DO THIS:** How just run the following command to see if you got x
, y
and z
correct when you fixed the code above.
from answercheck import checkanswer
checkanswer([x,y,z],'e80321644979a873b273aebbbcd0e450');
NOTE make sure you do not change the checkanswer
commands. The long string with numbers and letters is the secret code that encodes the true answer. This code is also called the HASH. Feel free to look at the answercheck.py
code and see if you can figure out how it works?
Numpy is a common way to represent vectors, and you are suggested to use numpy
unless otherwise specified. The benefit of numpy
is that it can perform the linear algebra operations listed in the previous section.
For example, the following code uses numpy.array
to define a vector of four elements.
import numpy as np
x_np = np.array([-1, 0, 2, 3.1])
x_np
In mathematics, 1-vector is considered as a scalar. But in Python, they are not the same.
x = 2.4
y = [2.4]
x == y
x == y[0]
We have a list of numpy arrays or a list of list. In this case, the vectors can have different dimensions.
✅ **DO THIS:** Modify the print statement using indexing to only print the value 3 from the list_of_vectors
defined below.
x_np = np.array([-1,0, 2 , 3.1])
y_np = np.array([1,-1,3])
z_np = np.array([0,1])
list_of_vectors = [x_np,y_np,z_np]
print(list_of_vectors)
The index of a vector runs from 0 to $n-1$ for a $n$-vector.
✅ **DO THIS:** The following code tries to get the third element of x_np
, which is the number 2.0
. Fix this code to provide the correct answer.
print(x_np(3))
✅ **DO THIS:** Replace only the third element of x_np
with the number 20.0
such that the new values of x_np
is [-1, 0, 20., 3.1]
# Replace the third element using 20.0, then the resulting element is
#####Start your code here #####
#####End of your code here#####
print(x_np)
from answercheck import checkanswer
checkanswer(x_np,'993d5cbc6ddeb10776ed48159780a5d3');
There is a special index -1, which represents the last element in an array. There are several ways to get more than one consecutive elements.
✅ **DO THIS:** you are given a vector (x_np
) of $n$ elements, define a new vector (d
) of size $n-1$ such that $d_i = x_{i+1}-x_i$ for $i=1,\dots,n-1$. Hint try doing this without writing your own loop. You should be able to use simple numpy
indexing as described above.
x_np = np.array([1,8,3,2,1,9,7])
## Put your answer to the above question here.
from answercheck import checkanswer
checkanswer(d,'14205415f0ed56e608d0a87e7253fa70');
Take a look at the following code.
x_np
y_np = x_np
y_np
x_np
is also changedThis looks weired and may not make sense for those uses other languages such as MATLAB.
The reason for this is that we are not creating a copy of x_np
and name it as y_np
. What we did is that we give a new name y_np
to the same array x_np
. Therefore, if one is changed, and the other one is also changed, because they refer to the same array.
x_np = np.array([-1, 0, 2, 3.1])
y_np = x_np
y_np[2] = 20.0
x_np
✅ **DO THIS:** There is a method named copy
that can be used to create a new array. You can search how it works and fix the code below. If this is done correctly the x_np
vector should stay the same and the y_np
you now be [-1 0 2 3.1]
.
## modify the following code to copy the x_np instead of just giving it a new name
x_np = np.array([-1, 0, 2, 3.1])
y_np = x_np
y_np[2] = 20.0
print(x_np)
from answercheck import checkanswer
checkanswer(x_np,'0ba269d18692155ba252e6addedf37ad');
from answercheck import checkanswer
checkanswer(y_np,'993d5cbc6ddeb10776ed48159780a5d3');
The relational operator (==
, <
, >
, !=
, etc.) can be used to check whether the vectors are same or not. However, they will act differently if the code is comparing numpy.array
objects or a list
. In numpy
, In numpy
relational operators checks the equality for each element in the array
. For list
, relational operators check all elements.
x = [-1, 0, 2, 3.1]
y = x.copy()
y[2] = 20.2
x_np = np.array(x)
y_np = np.array(y)
x == y
np.array(x_np) == np.array(y_np)
✅ **DO THIS:** Create a zero vector (called zero_np
) with the same dimension as vector x_np
. Create a ones vector (called ones+np
) also with the same dimension as vector x_np
.
x_np = np.array([-1, 0, 2, 3.1])
### Define zero_np and ones_np here
from answercheck import checkanswer
checkanswer([zero_np, ones_np],'7f874c2e446786655ff96f0bbae8a0c6');
random_np = np.random.random(2)
print(random_np)
In this section, you will understand why we use numpy for linear algebra opeartions. If x and y are numpy arrays of the same size, we can have x + y and x-y for their addition and subtraction, respectively.
x_np = np.array([1,2,3])
y_np = np.array([100,200,300])
v_sum = x_np + y_np
v_diff = x_np + y_np
print (f'Sum of vectors: {v_sum}')
print (f'Difference of vectors: {v_diff}')
For comparison, we also put the addition of two lists below. Recall from the pre-class assignment, we have to define a function to add two lists for linear algebra.
✅ **DO THIS:** Modify the following code to properly add and subtract the two lists. HINT it is perfectly okay NOT to write your own function try you should be able to cast the lists as arrays:
x = [1,2,3]
y = [100,200,300]
v_sum = x + y
v_diff = x + y
print (f'Sum of vectors: {v_sum}')
print (f'Difference of vectors: {v_diff}')
A scalar-vector addition means that the scalar (or a 1-vector) is added to all elements of the vector.
✅ **DO THIS:** Add a scalar 20.20 to all elements of the following vector x_np
and store teh result back into x_np
x_np = np.array([1.0,2.0,3.0])
from answercheck import checkanswer
checkanswer(x_np,'2f8cbcce405fa12b8608422ff28544bb');
When a
is a scalar and x
is numpy
array. We can express the scalar-vector multiplication as a*x
or x*a
.
We can also do scalar-vector division for x/a
or a/x
. (note that x/a
and a/x
are different)
✅ **DO THIS:** Divide all elements of the following vector x_np
by 20.20
and put it into y_np
x_np = np.array([1,2,3])
#####Start your code here #####
y_np =
#####End of your code here#####
print(y_np)
from answercheck import checkanswer
checkanswer(y_np,'90c1b8639f9d350af1d971d89209a0c6');
As stated above relational operations on numpy
arrays are performed element-wise. Examples we mentioned before are
==
operator +
and subtraction -
Note for this to work the two vectors have to be the same dimensions.
If they are not have the same dimension, such as a scalar and a vector, we can think about expanding the scalar to have the same dimension as the vector and perform the operations. For example.
✅ **DO THIS:** Assume that you invested three assets with initial values stored in p_initial
, and after one week, their values are stored in p_final
. Then what are the asset return ratio (r
) for these three assets (i.e. price change over the initial value).
p_initial = np.array([22.15, 89.32, 56.77])
p_final = np.array([23.05, 87.32, 53.13])
from answercheck import checkanswer
checkanswer(r,'0e231e6cfbef65cf178208cf377af85c');
We have two vectors $x$ and $y$ we can get the linear combination of these two vectors as $ax + by$ where $a$ and $b$ are scalar coefficients.
In the following example, we are given two vectors (x_np
and y_np
), and two scalars (alpha
and beta
), we obtain the linear combination alpha*x_np + beta*y_np
.
x_np = np.array([1,2])
y_np = np.array([3,4])
alpha = 0.5
beta = -0.8
c = alpha*x_np + beta*y_np
print(c)
We can also define a function lincomb
to performn the linear combination.
✅ **DO THIS:** Finish the following code for lincomb and compare the results we just get.
def lincomb(coef, vectors):
n = len(vectors[0]) # get the dimension of the vectors. note they have to be of the same dimension
comb = np.zeros(n) # initial the value with all zeros.
### Add code here to calculate the linear combination of the input vecotrs and the coefficients.
return comb
from answercheck import checkanswer
combination = lincomb([alpha, beta], [x_np,y_np])
checkanswer(combination,'8bab7329c94f3e3cda423add411685c2');
We can also test the functions ourselves by using values for which we know the answer. For example, the following tests are multiplying and adding by zero we know what these answers should be and can check them.
combination = lincomb([0, 0], [x_np,y_np])
combination == np.zeros(combination.shape)
combination = lincomb([2, 2], [combination,np.zeros(combination.shape)])
combination == 2*combination
If you want to check that all values in a numpy.array
are the same you could convert it to a list or there is a method called alltrue
which checks if everything is true. It is a good idea to use this method if vectors get big.
combination = lincomb([2, 2], [combination,np.zeros(combination.shape)])
np.alltrue(combination == 2*combination)
In your first Homework you are going to convert data into vectors and practice your understanding of vector math. Lets go over this a little bit so everyone knows what we need to do. Lets go over the homework together in class to make sure we all know what is going on.
If you attend class in-person then have one of the instructors check your notebook and sign you out before leaving class. If you are attending remote, turn in your assignment using D2L.
Writen by Drs. Ming Yan and Dirk Colbry, Michigan State University
This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.