Financial Maths, Part 1

For some of my students, interest calculations are troublesome: you can say that they quickly lose interest in interest.

If I still have your interest after that bad joke, I will continue. The two main types of interest are simple and compound interest. In simple interest, the principal (the amount initially invested) stays the same and interest is calculated on that amount at all times. In compound interest, the principal grows and the value upon which interest is calculated changes.

I have previously talked about percentages and how to take a percentage of a number. Please review that if you do not know how to take a percentage of a number.

As always, in any new topic, there are some definitions to know so that we understand each other. The following are the main definitions with the abbreviations for them that will be used in equations:

Principal (P): the amount invested or borrowed
Interest rate (r): a percentage to be applied to the principal. This can be a percentage (eg. 15%) or its decimal equivalent (0.15).
Interest (I): the dollar amount which results when the interest rate is applied to the principal
Time (t): the amount of time to be used in a problem
Period: the basic amount of time used by the interest rate. For example, 15% per annum (abbreviated p.a.) means that the period is 1 year.
Number of periods (n): The number of periods to be used in a given problem. Note that equations can be in terms of time (t) or number of periods (n).

Let’s start out with a simple interest situation. Suppose I invest $1000 at a simple interest rate of 3% p.a., that is 3% each year. Though I haven’t asked a question yet, let me identify the key items of this set up:

P = $1000
r = 3% or 0.03
period = 1 year

So my first question is: how much interest do I earn after 1 year? At the end of each year, if I keep that initial amount 0f $1000 in the investment, I will earn 3% of $1000 in interest. If you remember, to take a percentage “of” something, the “of” means to multiply. So after 1 year:

I = 3% × $1000 = (3/100) × 1000 or 0.03 × 1000 = $30

Note that in equations where you can put the interest rate in directly (the “3”), there will be a “/100” part in the equation. In equations where the decimal equivalent of the interest rate (0.03) is to be used, there will be no “/100” part. So the formulas to find the amount of interest (I) earned in 1 period are:

I = Pr/100, if you like to use the interest rate number directly (the “3”)

I = Pr, if you like to use the decimal equivalent of the interest rate (0.03)

This is why you may see different formulas in different books.

In my next post, I will ask some more questions about this investment and provide more formulas.

System of Equations, Part 4

Please read the previous posts on this topic if you do not understand this one.

To illustrate the power of matrices, consider the following system of equations:

w + x + y + z = 5
w + 2xy + 2z = 10
2w – 2xy + 3z = 11
2w + x + y – 3z = 0

This would take a while using the substitution or elimination methods. I will solve this using the matrix method. This system, in matrix form, is:

\[
\left[{\begin{array}{cccc}{1}&{1}&{1}&{1}\\{1}&{2}&{{-}{1}}&{2}\\{2}&{{-}{2}}&{{-}{1}}&{3}\\{2}&{1}&{1}&{{-}{3}}\end{array}}\right]\left[{\begin{array}{c}{w}\\{x}\\{y}\\{z}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{5}\\{10}\\{11}\\{0}\end{array}}\right]
\]

This system is in the form Ax = b and as shown in my last post, the solution to this is found by pre-multiplying both side by A-1. Now admittedly, finding A-1 manually for a 4 × 4 matrix would take some time. However, taking advantage of the internet (or a modern calculator), I find that

\[
{\mathbf{A}}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{59}\left[{\begin{array}{cccc}{1}&{4}&{12}&{15}\\{4}&{16}&{{-}{1}{1}}&{1}\\{39}&{{-}{2}{1}}&{{-}{4}}&{{-}{5}}\\{15}&{1}&{3}&{{-}{11}}\end{array}}\right]
\]

Here I factored out the 1/59 out of each element in the matrix to make it look nicer.

So from my last post, you know that the answer is found by pre-multiplying the b matrix:

Ax = bA-1Ax = A-1bIx = A-1bx = A-1b

So the solution is:

\[
{\mathbf{x}}\hspace{0.33em}{=}\hspace{0.33em}{\mathbf{A}}^{{-}{1}}{\mathbf{b}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{59}\left[{\begin{array}{cccc}{1}&{4}&{12}&{15}\\{4}&{16}&{{-}{1}{1}}&{1}\\{39}&{{-}{2}{1}}&{{-}{4}}&{{-}{5}}\\{15}&{1}&{3}&{{-}{11}}\end{array}}\right]\hspace{0.33em}\left[{\begin{array}{c}{5}\\{10}\\{11}\\{0}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{3}\\{1}\\{{-}{1}}\\{2}\end{array}}\right]
\]

So w = 3, x = 1, y = -1, and z = 2. Isn’t that fantastic!

System of Equations, Part 3

So now that you know about matrices, we can use them to add a third way to solve a system of equations. You will need to read my previous 3 posts on matrices if you are unfamiliar with how to multiply matrices.

In the last post on System of Equations, I looked at the system:

2x + 3y = 51
3x + 2y = 49

And in my last post on Matrices, I showed you how a 2×2 matrix of numbers and a 2×1 matrix of unknowns can be multiplied together to get a 2×1 matrix that looks suspiciously like the left sides of a system of equations. This is , in fact, true. If I form a matrix using the coefficients on the left side of the above system, I get a matrix which I will call A:

\[
{\textbf{A}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right]
\]

Let me now define a matrix x (which is different from the single variable x which is in italics and not bold):

\[
{\textbf{x}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]
\]

Now I will define a matrix b:

\[
{\textbf{b}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{51}\\{49}\end{array}}\right]
\]

Now see what happens if I multiply A by x:

\[
{\textbf{A}}{\textbf{x}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right]\times\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{2}{x}{+}{3}{y}}\\{{3}{x}{+}{2}{y}}\end{array}}\right]
\]

The rows of this result look just like the left side of our system of equations. And b is the right side. So the matrix equivalent of the system is

\[
\begin{array}{c}
{{\textbf{A}}{\textbf{x}}\hspace{0.33em}{=}\hspace{0.33em}{\textbf{b}}}\\
{\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right]\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{51}\\{49}\end{array}}\right]}
\end{array}
\]

This is easy to form directly. You just form A as the matrix of coefficients (with the unknowns in the same order in each equation), x is the matrix of unknowns, and b is the matrix of the numbers on the right sides. So how do we solve this?

From my last post, I defined the inverse of a matrix A as A-1. This is the matrix that if I multiply A by its inverse, I get the identity matrix which is the equivalent of “1” in scalar maths.

The process of isolating (solving) for variables in a matrix equation is exactly the same as for scalar equations: you do the same thing to both sides with the goal of having the unknowns by themselves on one side. So if I pre-multiply (remember, order of multiplication in matrix maths is important) both sides of our matrix equation by A-1, the left side is the identity matrix times x which is equal to just x. The right side multiplies out to form the solution.

As I said before, finding A-1 is beyond the scope of this set of posts. I will just tell you what it is. However, many modern calculators will do this for you, and you can also use the internet and search for “matrix inverse calculator”. It turns out that A-1 is:

\[
\left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right]
\]

So taking the matrix equation and pre-multiply both sides by A-1 gives

A-1Ax = A-1bIx = A-1bx = A-1b

\[
\left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right]\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right]\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right]\left[{\begin{array}{c}{51}\\{49}\end{array}}\right]
\] \[
\Longrightarrow\hspace{0.33em}\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right]\left[{\begin{array}{c}{51}\\{49}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{9}\\{11}\end{array}}\right]
\]

Which is the same answer as before, x = 9 and y = 11.

This is a very powerful method for large systems of equations. Next time I will solve a system of 4 equations with 4 unknowns. For those of you who have done this manually, you will appreciate the ease matrix algebra provides.

Matrices, Part 3

In my last post, I multiplied a 2 × 2 matrix by another 2 × 2 matrix. Now let’s multiply a 2 × 2 matrix by a 2 × 1 matrix. If you were paying attention last time, this is possible because the inside dimensions are the same (2= 2) and the resulting matrix will be a 2 × 1 matrix, that is, the outside dimensions.

This is done exactly the same way as illustrated in my last post, only there is only 1 column in the second matrix to multiply with the first matrix:

\[
\left[{\begin{array}{cc}{1}&{{-}{2}}\\{3}&{{-}{4}}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{c}{5}\\{6}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{(}{5}\times{1}{)}{+}{(}{6}\times{(}{-}{2}{))}}\\{{(}{5}\times{3}{)}{+}{(}{6}\times{(}{-}{4}{))}}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{-}{7}}\\{{-}{9}}\end{array}}\right]
\]

Now let the second matrix be composed of variables. This does not change the method at all. It just means that the result is a matrix with algebraic expressions:

\[
\left[{\begin{array}{cc}{1}&{{-}{2}}\\{3}&{{-}{4}}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{(}{x}\times{1}{)}{+}{(}{y}\times{(}{-}{2}{))}}\\{{(}{x}\times{3}{)}{+}{(}{y}\times{(}{-}{4}{))}}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{x}{-}{2}{y}}\\{{3}{x}{-}{4}{y}}\end{array}}\right]
\]

Please keep this example in mind for my next post when I use matrices to solve a system of equations.

The last skill I need to present is matrix division. When using matrices, you do not actually divide a matrix by another matrix. Rather, you multiply by the inverse of a matrix.

In scalar arithmetic, you can think of dividing a number say 4, by another number, say 2, as multiplying the 4 by the inverse (reciprocal) of 2:

\[
\frac{4}{2}\hspace{0.33em}{=}\hspace{0.33em}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}
\]

The same thing is done with matrices. However, finding the inverse of a matrix is a little involved and I will not cover that in this set of posts. Rather I will just give you the result when needed. However I will say a few things about the properties of matrix inverses.

In scalar arithmetic, multiplying a number by it’s reciprocal (inverse) equals 1:

\[
\frac{2}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}
\]

The same thing is true with matrices, only what is “1” in the matrix world?

The equivalent “1” for matrices is the Identity Matrix. This is a square (rows = columns) matrix with 1’s down its diagonal and 0’s everywhere else:

\[
\left[{\begin{array}{cc}{1}&{0}\\{0}&{1}\end{array}}\right]
\]

This is the identity matrix for a 2 × 2 matrix. The inverse of a matrix is that matrix where multiplying it by the original matrix, results in the identity matrix. The inverse of a matrix A, is denoted as A-1.

\[
{\mathbf{A}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathbf{A}}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{{-}{2}}&{1}\\{\frac{3}{2}}&{{-}\frac{1}{2}}\end{array}}\right]
\] \[
\begin{array}{l}
{{\mathbf{A}}\times{\mathbf{A}}^{{-}{1}}{=}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\times\left[{\begin{array}{cc}{{-}{2}}&{1}\\{\frac{3}{2}}&{{-}\frac{1}{2}}\end{array}}\right]{=}\left[{\begin{array}{cc}{{(}{-}{2}\times{1}{)}{+}{(}\frac{3}{2}\times{2}{)}}&{{(}{1}\times{1}{)}{+}{(}{-}\frac{1}{2}\times{2}{)}}\\{{(}{-}{2}\times{3}{)}{+}{(}\frac{3}{2}\times{4}{)}}&{{(}{1}\times{3}{)}{+}{(}{-}\frac{1}{2}\times{4}{)}}\end{array}}\right]}\\
{{=}\hspace{0.33em}\left[{\begin{array}{cc}{1}&{0}\\{0}&{1}\end{array}}\right]}
\end{array}
\]

And it turns out that when multiplying a matrix by its inverse, order does not matter: A × A-1 = A-1 × A.

In my next post, I will put all this talk about matrices in practice and use them to solve a system of equations.

Matrices, Part 2

In order to set up a system of equations using matrices, you need to understand how matrices multiply one another. Not all matrices can be multiplied together – they need to be compatible with one another. Not only that, unlike scalar (single number) arithmetic, multiplication does not commute, that is, the order of the multiplication will generally produce different results or one order may not even be possible. So what do I mean by compatible?

Let’s start with an example:

\[
\begin{array}{l}
{\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\times\left[{\begin{array}{cc}{5}&{6}\\{7}&{8}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{{(}{5}\times{1}{)}{+}{(}{7}\times{2}{)}}&{{(}{6}\times{1}{)}{+}{(}{8}\times{2}{)}}\\{{(}{5}\times{3}{)}{+}{(}{7}\times{4}{)}}&{{(}{6}\times{3}{)}{+}{(}{8}\times{4}{)}}\end{array}}\right]}\\
{{=}\hspace{0.33em}\left[{\begin{array}{cc}{19}&{22}\\{43}&{50}\end{array}}\right]}
\end{array}
\]

To multiply these two 2 × 2 matrices, you take the first column of the second matrix and lay it over the top of the first matrix:

\[
\begin{array}{l}
{\left[{\begin{array}{cc}{5}&{7}\end{array}}\right]}\\
{\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]}
\end{array}
\]

Starting with the top row of the first matrix, Multiply the numbers in the same position together and add the result of each: (5 × 1) + (7 × 2) = 19. This result is the first row, first column number in the new matrix. Repeat this using the second row of the first matrix: (5 × 3) + (7 × 4) = 43. This is the first element of the second row of the new matrix. Now do the same with the second column of the second matrix:

\[
\begin{array}{l}
{\left[{\begin{array}{cc}{6}&{8}\end{array}}\right]}\\
{\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]}
\end{array}
\]

to get the second column of the new matrix. I will leave it as an exercise for you to confirm that if I reverse the order of the matrices, you will get a different result. That is,

\[
\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\times\left[{\begin{array}{cc}{5}&{6}\\{7}&{8}\end{array}}\right]\hspace{0.33em}\ne\hspace{0.33em}\left[{\begin{array}{cc}{5}&{6}\\{7}&{8}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]
\]

So this method works for any size matrices as long as they are compatible. From this example, you see that this works only if the second matrix has the same number of rows as the number of columns in the first matrix. This is easy to see if you put the dimensions together: (2 × 2) × (2 × 2). The inside numbers need to be the same if multiplication is possible (2 = 2). The outside numbers give the dimensions of the resulting matrix (2 × 2).

So you can multiply a 3 × 2 matrix by a 2 × 4 matrix to get a 3 × 4 matrix, but you cannot reverse the order because the inside dimensions will not be equal. It’s interesting that if you multiply a 1 × (anything) matrix by a (same anything) × 1 matrix, you will get a 1 × 1 matrix which is just a number (a scalar).

This multiplication works even if some or all of the elements of the matrices are variables. I will illustrate this in my next post.

Matrices, Part 1

In my two-part series on equation systems, I mentioned that there is a third method using matrices. Before I use this method, I have to explain what a matrix is. There is a whole new algebra surrounding matrices (the plural of matrix), so I will only explain what is needed to solve a system of equations.

All of the algebra we have used so far is called scalar algebra. That is, it is used on scalars. A scalar is just a single number like 3, -2.7, or x. It just has a value and you cannot get any more information from it like direction. In most engineering problems, 3-dimensional space is the playground and engineers are also interested in not only the speed of an object, but also the direction it is travelling in. Scalar algebra will not suffice. Enter matrix algebra. This is sometimes equated to linear algebra, but there are differences. I will continue to use matrix algebra as that is the most appropriate at this level.

A matrix is simply an array of numbers. Here are some examples:

\[
\left[{\begin{array}{cc}{{-}{1}}&{0}\\{3}&{2.4}\end{array}}\right]{,}\hspace{0.33em}\left[{\begin{array}{c}{0}\\{\sqrt{7}}\\{{-}{3}{.}{23}}\\{1}\end{array}}\right]{,}\hspace{0.33em}\left[{\begin{array}{ccc}{2}&{{-}{7}}&{0}\end{array}}\right]{,}\hspace{0.33em}\left[{\begin{array}{ccc}{2}&{0}&{3.8}\\{{-}{1}}&{0}&{0}\end{array}}\right]
\]

Of course, you cannot talk about a whole new field of maths without definitions. The first definition is dimension. The dimension of a matrix is its size. By convention, the dimension always indicates the number of rows first then the number of columns. So the first matrix in the examples above, is a 2 × 2 matrix because it has 2 rows and 2 columns. The second matrix is a 4 × 1 matrix. Matrices with 1 as one of the dimensions, can be called vectors as well. The third matrix is 1 × 3 and the last is 2 × 3.

It is very common to use bold typeface when using a single letter to represent a matrix so that the reader knows that they are talking about a matrix and not a scalar:

\[
{\mathbf{A}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{{-}{1}}&{0}\\{3}&{2.4}\end{array}}\right]
\]

So I can now talk about A and you know that A is a matrix.

So there are the same types of operations for matrices as there are for scalars: adding, subtracting, multiplying, and dividing. But there are rules associated with these matrix operations that are specific to matrices.

Since I am directing the use of matrices to solving a system of equations, I will only discuss the multiplying and dividing operations as they relate to matrices. This is where I will begin in my next post.

System of Equations, Part 2

So last time I solved a system of two equations using the substitution method where the information from one equation is inserted into the other equation. This is the method of choice if it is easy to solve for one of the unknowns. However, the example that I used also lends itself well to the other method: Elimination.

The elimination method, like the substitution method, uses the two equations to generate one equation with one unknown which can be solved. The system I worked on last time was:

x + y = 108
xy = 38

The elimination method is simply adding the two equations together with the goal of eliminating one of the unknowns. So let’s add these two equations:

x + y = 108
xy = 38
2x = 146

Notice that I now have an equation with one unknown. Dividing both sides of this new equation by 2 gives:

2x/2 = 146/2 ⟹ x = 73

the same answer as before. Now use this value for x in one of the original equations. Let’s use the first one:

x + y = 108 ⟹ y = 108 – x = 108 – 73 = 35

So we get (thankfully) the same solution as before. But this example is rather contrived in that the y‘s were conveniently of opposite signs in the given equations. So let’s consider the following system:

2x + 3y = 51
3x + 2y = 49

Adding these two equations together will just give us another equation with two unknowns. But just like we do with single variable equations, we can modify one or both of these. Let me take the first equation, and multiply it by – 2. Why I am doing this will soon be revealed:

-2(2x + 3y) = 51(-2) ⟹ -4x -6y = -102

I will now multiply the second equation by 3:

3(3x + 2y) = 49(3) ⟹ 9x + 6y = 147

Now let’s repeat the system, replacing the equations with the new ones:

-4x -6y = -102
9x + 6y = 147

Notice that if I now add these equations, the y variable will disappear:

-4x -6y = -102
9x + 6y = 147
5x = 45 ⟹ x = 45/5 = 9

I will now substitute this partial solution into the original second equation:

3x + 2y = 49 ⟹ 3(9) + 2y = 49 ⟹ 27 + 2y = 49 ⟹ 2y = 49 – 27
⟹ 2y = 22 ⟹ y = 11

So x = 9 and y = 11 solves both of these equations.

Both methods, substitution and elimination, can either be used to solve a system of equations, but one method may be less work than the other.

System of Equations, Part 1

The equations that I have solved so far, have been equations with one unknown (variable). Suppose we have two unknowns in a problem and wish to solve for both. There is a rule in maths that you have to have the same number of equations as the number of unknowns if you are to solve for all unknowns. These equations need to be independent. what do I mean by that?

Suppose I need to find two numbers that add up to 108. In equation form, I want to find x and y such that x + y = 108. So I need another piece of information (equation) if I need a specific solution. Suppose you say “I can get another equation by just multiplying both sides of the given equation by 2: 2x + 2y = 216.” The problem here is that this second equation depends on the first equation. This is what I meant by “independent”. The second equation must be completely new information.

So let’s say that these two numbers must also have a difference of 38 : xy = 38. This is a completely new (independent) requirement. So we now have the two equations:

x + y = 108
xy = 38

So what two numbers satisfy these two requirements? There are two main methods for solving this: Substitution and Elimination. Actually, there is a third method which uses matrices, but I will explain what a matrix is and how to use it in a subsequent post. Let’s first talk about the substitution method.

The substitution method is basically inserting information from one of the equations into the other equation. Let’s solve the second equation for x. That is, use algebra to get x on one side and everything else on the other side. If I add y to both sides of the second equation, I get

xy + y = 38 + yx = 38 + y

Now put this definition of x into the first equation. That is, replace x in the first equation with what x is equal to from the second equation:

x + y = 108 ⟹ 38 + y + y = 108 ⟹ 38 + 2y =108

Notice that this new equation, which combines the information of the two equations into one, is a single equation with a single unknown. This can now be solved with the techniques we have seen before:

38 + 2y =108 ⟹ 2y = 108 – 38 = 70 ⟹ y = 70/2 = 35

So y is 35. You can now use this solution into any of the equations involving x and y, then solve for x. Let’s use x = 38 + y:

x = 38 + y = 38 + 35 = 73

So the two numbers are 35 and 73. They both add up to 108 and subtract to equal 38.

I’ll start the next post doing another example.

Newton’s Clock, Part 4

So here is the last of this series explaining the expressions on Newton’s clock:

We are now up to 8. Let’s look at

\[
\mathop{\prod}\limits_{{k}{=}{0}}\limits^{1}{{(}{2}{k}{+}{2}{)}}
\]

This is another excellent example of how concise the words in maths can be. The symbol “𝚷” is the capital version of 𝜋 which corresponds to the english “P”. The “P” here stands for “product” which is the result of multiplying two or more numbers. The expression on the clock means : “Take the expression 2k + 2 and successively replace the k with the number at the bottom of the 𝚷 symbol (in this case “0”), evaluate the expression to get a number, and increment k by 1 and repeat until you reach the number at the top of the 𝚷 symbol (in this case “1”). Then multiply all these numbers together.”

You can see why maths expressions are much more concise than English. So to evaluate this expression, we first replace the k with 0, then work out 2(0) + 2. This equals 2. Now increment the k by 1 to get 1, then work out 2(1) + 2. This equals 4. Since k is now at the number at the top of 𝚷, we are done increasing k. Now multiply these numbers together. 2 × 4 = 8 which is the correct number at this position on the clock.

Most of you now know what

\[
\sqrt{81}
\]

means. It means “what number multiplied by itself equals 81?” The answer, of course is 9 as 9 × 9 = 81.

The next hour is

\[
{\log}_{2}1024
\]

The basics of this expression have already been explained for position “2” on the clock. This expression is asking the question “what does the exponent of 2 have to be so that 2x = 1024?”. Hopefully, the answer is “10” and it is because 210 = 1024.

Now let’s look at B16. Remember when I explained position “7” on the clock: 01112? That was a number in the base 2 system of counting. Another common base used with computers is the base 16 counting system. We are familiar with the base 10 counting system that has 10 symbols used to count with: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The base 16 system needs 16 symbols. So what is used after 9 is reached? Well, we resort to the letters of the alphabet. The numbers up to 9 in base 16 correspond to the same numbers in base 10. The next number in base 16 is “A” which corresponds to 10 in base 10 and the next number is “B” which is 11 in base 10. So B16 = 11.

Finally, the last expression:

\[
\mathop{\sum}\limits_{{i}{=}{1}}\limits^{3}{{(}{3}{i}{-}{2}{)}}
\]

The Σ symbol is the Greek capital “sigma” and corresponds to the english “S”. This letter stands for “Sum” which is the addition of two or more numbers. This expression is just like the one in position “8” on the clock except that you add the resulting numbers together instead of multiplying them. So starting at i = 1, 3(1) – 2 = 1, 3(2) -2 = 4, 3(3) – 2 = 7, and we are done as i now equals the numbers on top of the Σ. So now add these numbers together: 1 + 4 + 7 = 12. It is now high noon and that completes Newton’s clock.

Newton’s Clock, Part 3

So now I’m up to “4” on Newton’s clock:

So the expression

\[
{\left({2\sin\frac{\mathit{\pi}}{2}}\right)}^{2}
\]

uses the sine function which has been talked about many posts before. Only this time, it is using radian measure of angles instead of degrees. If your calculator is in degree mode, you can substitute 90° in place of 𝜋/2 to get the same answer. The sine of 𝜋/2 radians or 90° is 1. So in the brackets we have 2 × 1 = 2. 2² = 4, hence its position on the clock.

Now let’s look at

\[
\sqrt[3]{125}
\]

This is the cube root of 125. This expression is asking the question: “What number multiplied 3 times equals 125?”. The answer to that is 5 because 5 × 5 × 5 = 125. So once again, the clock does not lie.

Now let’s look at 3! This is pronounced “3 factorial”. The factorial of a number is that number successively multiplied by a number which is 1 less. So 5! = 5 × 4 × 3 × 2 × 1 = 120. So 3! = 3 × 2 × 1 = 6. Factorials are used a lot in probability. I have touched on this before but perhaps there is another future post here.

Now let’s look at 01112. We are very familiar with decimal system way of counting. This system is a base 10 system because we use 10 distinct digits (symbols) to count: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. When we run out of digits, like when we count up to 9, we add another place holder to the right of the number and put the starting digit 0 there: 10. And then successively increase it’s digits until we get to 9 again. then we increase the left digit by 1 and start over again: 20, 21, … . There are other number systems based on numbers other than ten.

Computers are composed of switches based on two states, on or off. We mathematically say that off is 0 and on is 1. Computers essentially count with just o’s and 1’s: a base 2 system. Counting in base 2 is done exactly as we do in base 10, we just have fewer digits to work with.

So we if we start counting we get 0, 1, but we’ve ran out of digits so we add a place holder to the right and start again: 0, 1, 10, 11. Ran out of digits again so add another place holder and start over: 0, 1, 10, 11, 100, 101, 110, 111. If you are keeping track, 111 in base 2 is equal to 7 in base 10. It is a convention to subscript a number with its base when dealing with other base systems, so 01112 means 7 in base 10. The leading 0 doesn’t add to the value but in computer maths, base 2 numbers are typically written 4 digit places at a time.