# Category: Teaching Ideas

## When Imaginary gets Real

- Post author By Amir
- Post date
- No Comments on When Imaginary gets Real

One of the many interesting aspects of imaginary numbers is that we can use them to find out “real” facts (facts about real numbers).

Perhaps the most used examples are the derivation of the trigonometric identities for \(sin 2\theta\) and \(cos 2\theta\)*.

This post offers something more exciting and less-known: the radical forms of \(cos \frac{\pi}{12}\) and \(sin \frac{\pi}{12}\).

Consider \(\frac{\frac{1}{2}+\frac{\sqrt{3}}{2}i}{\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}\)

**Algebraically**, simplify it and write it in the form of \(a+bi\).

\[\frac{\sqrt6+\sqrt2}{4}+\frac{\sqrt6-\sqrt2}{4} i\]

We know the answer. But as a wise man once said, mathematics is all about seeing the same thing from different angles (or something like this). So let us, calculate the same thing in some other way.

We write \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and \(\frac {\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) in the form \(r e^{i\alpha}\). And then evaluate the original fraction , that is going to be:

\[\frac{e^{i \frac{\pi}{3}}}{e^{i \frac{\pi}{4}}}\]

**the same things, represented in two different forms.**Thus, their real parts should be equal with each other, and also their imaginary part should be equal with each other. So, we have:

\[cos\frac{\pi}{12} = \frac{\sqrt6+\sqrt2}{4}\]

\[sin\frac{\pi}{12} = \frac{\sqrt6-\sqrt2}{4}\]

Design a problem to find \(cos \frac{\pi}{8}\) and \(sin \frac{\pi}{8}\) in the radical form.

* We know that (a post will be written for this):

\[(cos \alpha + i sin \alpha)^2=cos (2\alpha) + i sin(2\alpha)\]

Remember what the wise man once said and find \((cos\alpha+i sin\alpha)^{2}\) by actually finding the power. Then we have:

\[(cos \alpha + i sin \alpha)^2=(cos^2\alpha -sin^2\alpha)+ i (2 sin\alpha cos\alpha)\] The right sides of the above two equality are the same represented in two different forms (both are equal to the same left side). So the real parts of the two should be equal, and the imaginary parts of the two should be also equal.

\[cos 2\alpha=cos^2 \alpha -sin^2 \alpha\]\[sin 2\alpha=2 sin \alpha cos\alpha\]

**Design** a problem to find \(cos 3\alpha\) and \(sin 3\alpha\) .

- Tags Calculus, Complex numbers

Where is \(i^{127}\)?

If you have read i Cycle, the Sum of Powers, you have already experienced the powers of i, and their hidden cycle. This is a much basic problem and in fact, it is a prerequisite for understanding i Cycle, the Sum of Powers and everything else about complex numbers (okay, it is a bit exaggeration to say so, but you got the point, it is very important).

**What is the problem here? **

Notice that I could ask the “same” problem just using a different power. For example, I could ask: Where is \(i^{39}\)? You see, the problem here is not about finding this or that power of \(i\), it is about finding the pattern governing these powers.

The fundamental power in this case is two, as it is the power used in the definition of \(i\): \[i^2=-1\].

Furthermore, So if our question was Where is \(i^2\), we knew the answer is C. From, this starting point, we can see \(i^3=i^2*i=-i\) is the point D, \(i^4=i^2*i^2=1\) is on the point A, \(i^5=i^4*i=i\) is on the point B, and \(i^6=i^4*i^2=-1\) is on the point C, that happens to be the point we started our 90 degrees anticlockwise journey.

It doesn’t matter where we start our rotation. But, maybe, the point A is the most natural starting point. Starting from A, the powers are:

\[i^0, i^1, i^2, i^3, i^4, i^5, …\]

They look different from each other, but we know that there are only four fundamentally different number, as the above sequence is in fact the following sequence:

\[1, i, -1, -i, 1, i, -1, -i, …\]

So, to find the answer to our original question, we only need to put aside the fours that goes into 127 and see what remains.

Starting from the point A, has another interesting advantage. It gives meaning the the very definition of \(i\). Think of \(i\) as \(i \times 1\). You might say, okay so what, isn’t it obvious that \(i \times 1=i\). Yes, but, notice that the left side is the process of multiplication and the right side is the result of the process. In this case, the process has an important interpretation, that is rotating 90 degrees anticlockwise. That is to say, when we multiply the point A (that happens to be \((1,0)\)) by \(i\) it goes to the point B, and likewise, when we multiply the point B by \(i\), it goes to the point C. In other words, \(i\times i\times 1=-1\).

Voila: \[i^2=-1\]

- Tags Calculus, Complex numbers

## i Cycle, the Sum of Powers

- Post author By Amir
- Post date
- No Comments on i Cycle, the Sum of Powers

\[1+i+i^2+i^3+…+i^{57}\] Do I need to say what the question is?

“Find the sum!” is the immediate question that comes to mind when you start mathematics. When you gain more experience in mathematics, you learn to ask a deeper question:

**What is the pattern?**

You know, 57 is irrelevant. The only important thing here is the **general** question of the finding the sum without actually doing all the additions. However, to get to the point of finding the sum without actually doing all the additions, we should experiment with actual additions first! For example, let us find the sum of the first seven terms.

**Please find it before continuing reading. **

Did you notice that to have 7 terms, the last power in the sum should be 6 and not 7, and that means we should calculate \[1+i+i^2+i^3+i^4+i^5+i^6\] and not \[1+i+i^2+i^3+i^4+i^5+i^6+i^7\].

This was not a trick. In fact, thinking in terms of the number of terms is the key to the pattern that we are looking for.

What is the sum of the first two terms?

What is the sum of the first three terms?

What is the sum of the first four terms?

What is the sum of the first five terms? The first six? The first seven?

Let us also include “the sum of the first term”. It is silly, but you see why it is useful. Here are the results starting from “the sum of the first term”, following by “the sum of the first two terms”, then, the sum of the first three term”, and so on.

\[1,1+i, i, 0, 1, 1+i, i, 0, …\]

See, the sum of the first four terms is 0, the sum of the first eight terms is 0, and so on. And then, after each 0, we have 1, 1+i, and i. Now if we want to know what is \(1+i+i^2+i^3+…+i^{57}\) we only need to know how many fours goes into 58 (that is the number of terms). This number of fours only matters to us, as long as it helps us to find how many terms remain. In this case, no terms remains. So, the sum is 0.

\[1+\frac{1}{i}+\frac{1}{i^2}+\frac{1}{i^3}+…+\frac{1}{i^{2019}}\]

Do I need to say what the question is?

- Tags Calculus, Complex numbers

This post has nothing to do with Michael Artin Algebra! Artin of this post is my son, and this is the story of him learning algebra. He is 11 years old now, and I guess one of few students on the planet that still has a positive mentality towards mathematics!

We are reading together Gelfand’s’ Algebra, page by page, solving its problems one by one. The section “Letters in algebra” includes the following “Magic trick”.

Choose the number you wish. Add 3 to it. Multiply the result by 2. Subtract the chosen number. Subtract 4. Subtract the chosen number once more.

I asked Artin to choose a number without revealing his number to me. It was his first encounter with this kind of algebraic trick. I read the rule, and at the end, I knew the result is 2. Interestingly, when I asked what was his number, I realized that even right from start he had tried to “beat me” by choosing a big number, one million. Of course, It didn’t work.

Artin: Does it work for *every* number?

Then he said, “okay, I beat you” and chose another number, that was 1.5, and again the result was 2.

Artin: It also works with *these*. God damn it!

I knew that it takes time until his *every* number embraces different categories of numbers that he knows. Fortunately, he automatically was choosing different kind of numbers. Then, he chose a fraction, and this time, the result wasn’t 2. As I mentioned in the paper Specularity in Algebra, in this case, getting the result wrong is a better algebraic opportunity than getting it right. First of all, I said that I am sure he had some miscalculations, and that is why he hasn’t got 2 as the final result. To prove my point, I used the idea of specularity. I asked him to write his number on a piece of paper. I folded the paper to avoid seeing the number and touching it (using it in calculations). Then, we followed the rule, keeping the number intact. Here something amazing happened that is the reason that I am writing this post. You know, it is well-known that one of the difficulties of moving from arithmetic thinking to algebraic thinking is accept something like \( x+3 \) as it is. To see the difficulty, try to think of say \( 4+3 \) without thinking of 7. In arithmetic we are pushed into the result, but in algebra, the operation is the result. Simply speaking, you cannot do anything with\( x+3 \).

Artin’s number was written on a folded piece of paper (let us suppose that \( [1/4] \) denotes his number in the folded paper) and he could not do anything with it. The first rule change \( [1/4] \) to \( [1/4]+3 \) . Suddenly, Artin got to the heart of algebra, but at the same time, *found it strange and arithmetically useless. *

Artin: You have the answer right here (pointing to \( [1/4]+3 \) ). What is the point of this if this \( [1/4]+3 \) is the answer?

## Read Euler, Read Euler!

- Post author By Amir
- Post date
- 2 Comments on Read Euler, Read Euler!

“Read Euler, read Euler, he is the master of us all” written by Robin Wilson or “Euler: the master of us all” written by William Dunham are to show us how great and multifaceted Euler was as a mathematician. Indeed, he was. In this post, I want to write how great he was as an educator.

Today was my first session of a course in number theory. I had to start with “sums of two squares“. Like most other things in introductory number theory, this one also starts with a very simple observation. Some numbers like 5 ( \( 2^2+1^2 \) ) can be written as a sum of two squares, and some like 7 cannot be written as such. The natural question is which number is which. The students I had to work with had studied some number theory in the previous semester, but not primitive roots and quadratic residues, and of course, not the Gaussian integers. Basically, they hadn’t got the tools that are usually used to study (i.e., to prove useful facts about) the problem of our interest. Fortunately, Euler was there with two amazingly exploratory and yet rigorous enough pieces of work (as ever) on the subject: first work published in 1758, “On numbers which are the sum of two squares“, and the second work published in 1760, “Proof of Fermat’s theorem that every prime number of the form \(4n+1\) is the sum of two squares“. The two works are lengthy by today’s standard where using the “right” tools, there is even “a one-sentence proof” (due to Zagier) for what Euler proves in the first 12 pages of the first paper (its English version is 24 pages) plus the second paper (that is 11 pages). You can re-write all those 23 pages in just one page, calling it Euler’s proof by infinite descent. By doing so, you would have a nice and presentable proof of Fermat’s theorem (for an elementary number theory course as mine). However, you would miss most of the mathematical insights that your students (that mathematically might be as naïve as mine) could gain from the original text. A simple example of such insights is the mere fact that Euler calls his first argument (given in 1758) “Attempt at a Proof” not “A Proof”. This distinction would be just one of the lessons for your students. There would be many of such lessons as I try to convince you now. First, a few words about my plan is in order since it is quite related to what we might get from Euler.

The first step was to ask the students to list “those numbers which arise from the sums of two squares” (Paper I), say up to 50 (Euler himself listed them up to 200 without any use of calculators, mobiles, computers and so on!). The next step was to let them see and come up with any general statement that might be true for all such numbers, or for the numbers which cannot be written as the sum of two squares. I believed whatever they come up with could be found somewhere in Euler’s texts, perhaps not the exact things, but for sure, something quite related. Thus, after examining the statements found in the class, I could direct them to the Euler’s text where he addresses the same statement or something similar. It was the plan and it went better than I expected.

The first observation* was that powers of 2 can be written as a sum of two squares:

\( 2=1^2+1^2, 4=2^2+0^2, 8=2^2+2^2, 16=4^2+0^2, … \)

We called our first observation Theorem 1.

**Theorem 1**: Any power of two can be written as a sum of two squares.

The students had some experience in “university mathematics”, but most of them felt no need to prove our first theorem since it was true for all the examples on the blackboard! Here again, Euler comes in rescue, saying “in this class (dissertation) of many such statements (propositions), which until now have been accepted without proofs, we (I) will furnish proofs of their truth” (Paper I). Thus, I asked students to prove Theorem 1. To my surprise, they chose to prove it by mathematical induction (perhaps, because they had a lot of such proofs in the first semester). This is the way they did it:

Base: \( 2=1^2+1^2 \)

Assume that \( 2^k \) can be written as a sum of two squares.

\( 2^(k+1)=2. 2^k \) . Thus,\( 2^(k+1) \)can be written as a sum of two squares.

It was not easy for students to see why the product (here, \( 2^(k+1) \) does not automatically inherit the property of its factors (here, the factors are 2 and \( 2^k \) and the property is being a sum of two squares). Again Euler has something to say about this inheriting phenomenon. For example, a number that is a sum of two squares but neither of its factors is a sum of two squares. I used a silly example, giving that 3 and 5 are prime numbers (the property here is being a prime) but their product is not a prime number! This discussion put forward three options:

(i) Ignore proving Theorem 1 (that indeed wasn’t an option)

(ii) Choose a different direction to prove it.

(iii) Amend our failed proof.

The students chose the last option and this brought us to our next theorem.

**Theorem 2**: If \( m \) and \( n \) are two numbers, each of which is the sum of two squares, then their product \( mn\) will also be the sum of two squares.

Interesting, Euler suggests a number of simpler propositions that are special cases of Theorem 2 before giving the general form, because by it (special cases) this (the general case) will be more easily observed (Paper I). In fact, one of his lemmas (special cases) was enough to complete our proof by induction: If a number \( m \) is a sum of two squares, then so will be \( 2m \). However, we decided to proceed by proving the general case. Then, in addition to correcting our mathematical induction, we wrote 65 and 1105 (both suggested by Euler) as sums of two squares, using our proof of Theorem 2.

Euler has it all. He plays with examples, draw conclusions, warn you not to rely on them, seek proofs, guess when you might think wrong or overgeneralize, discuss them, give counterexamples and so on.

I haven’t yet distributed his papers in the class. This post was just about the first session. I’ll complete this story.

*Euler himself considered the square numbers first.

## Structure, Structure, Structure!

- Post author By Amir
- Post date
- 4 Comments on Structure, Structure, Structure!

Look at this equality:

\( (a + b) + c = a + (b + c) \) , or this one:

\( a . (b + c) = a.b + a.c \) .

They are true **structurally**. In principle, You can just replace one side of the equality with the other side without any extra comment. However, we usually like to describe these in terms of processes. For the first equality, we might say that *adding* the first two and then the sum to the third is the same as adding the first to the sum of the last two. For the second equality, we might say that from left to right it is *multiplying* through a set of parentheses, and from right to left, it is *factoring* something out. The italics are in the language of processes. Thus, when does the structure matter? Here is a strange example that I observed when teaching differential equations.

There is a technique in which you multiply both sides of a first order equation by something called “integrating factor”. The point is to write one side of the equation as the derivative of a function. For that, you need the product rule for derivatives: The derivative of a product of two functions is the derivative of the first times the second, plus the derivative of the second times the first. In symbols, we can write the product rule as \( (fg)’=f’g+g’f \) or many other ways. When moving from left to right we are *doing* something, we *find* the derivative, we *multiply*, we *add*. However, it seems that when moving from right to left, we are not doing anything. We just replace \( f’g+g’f \) with \( (fg)’ \) . Indeed, the mere fact that we can just replace the right side of the product rule with the left side was quite unacceptable for one of the best students of my class (hence he couldn’t see the logic behind the integrating factor). I would like to see more examples in which understanding structure really matter.

- Tags Advanced Mathematics, Algebra

## Plan for the Unexpected!

- Post author By Amir
- Post date
- No Comments on Plan for the Unexpected!

- Tags Primary Mathematics

“**Plan for the mess**” is one of my favourite teaching ideas. The idea is to bring students to a point where after a heavy messy work (most of the time calculations and symbol pushing) they say “why I didn’t see that earlier!” Here is how recently I used the idea in my linear algebra class.

The students had just solved a couple of two equations with two unknowns by whatever tools they had from highschool. Many of them had no idea about any geometric interpretation of such equations (i.e., solution, if there is one, is at the meeting point of two lines), and of course, none knew the vector interpretation in which, say \( \begin {cases} 2x + 3y = 4 \\x + 2y = 3\end{cases} \), can be thought as \( x\begin {bmatrix}2\\1\end{bmatrix}+y\begin {bmatrix}3\\2\end{bmatrix}=\begin {bmatrix}4\\3\end{bmatrix} \). Of course, one should be too naive to assume that students immediately appreciate the importance of having several interpretations for the same thing. Here is where you **plan for the mess. **Thus,** **I asked them to solve a system of three equations with three unknowns such as \( \begin {cases} 2x + 3y+4z = 4 \\x + 2y-z = -1\\3x-y+2z=2\end{cases} \). It was “fun” watching students messing around with the equations, substituting this into that, sometimes eliminating this unknown, the other time eliminating that unknown until they got the answer: \( x=0, y=0, z=1 \).

I am sure you can imagine how their faces look like when they realize that they could get the answer just by “looking” at the equation as

\[ x\begin {bmatrix}2\\1\\3\end{bmatrix}+y\begin {bmatrix}3\\2\\-1\end{bmatrix}+z\begin {bmatrix}4\\-1\\2\end{bmatrix}=\begin {bmatrix}4\\-1\\2\end{bmatrix} \].

Learning from a hard way is unforgettable!

- Tags Advanced Mathematics

## Proof-Generated Definitions!

- Post author By Amir
- Post date
- No Comments on Proof-Generated Definitions!

Okay, for a long time, I didn’t know what to blog about. Now, I have decided to write about my teaching ideas that take ages to turn into a piece of research. That is why I have started with this strange title for my first blog. The title comes from the exact same phrase coined by Lakatos in his book Proofs and Refutations. Here is how I used the idea recently.

Last week, I was going to teach the definition of a convergent sequence. Previously, we had played a lot with sequences geometrically: how they are represented on a number line and how they are graphed as a function on the coordinate plane. We hadn’t done any calculations; just we drew a lot of figures to see how different sequences (i.e., convergent, divergent, bounded, increasing, decreasing) behave graphically. The question I (as the lecturer) had to answer was which “theorem” could justify and motivate the definition of a convergent sequence. Indeed, at that stage of students’ knowledge, there was perhaps only one choice; but, surprisingly that choice seemed to be a perfect fit: if a sequence converges, then its limit is unique. There was only one small problem, students have no reason to think otherwise: how on earth a convergent sequence could have two limits? Thus, I needed to raise the question and also play devil’s advocate. Here is how the plan goes in the class.

I: How do we know that this sequence has only one limit (I drew the following figure)?

Students: What? Obviously, it has only one; see, it is getting closer and closer to the red line.

I: But, what if there is another line very close to the red line that the sequence is getting close to it too (I didn’t draw that other line on the graph).

Students: It (the sequence) * is going to pass that line*; see (one of the students drew the following (blue) line):

I: What if the line I thought of is closer to the red line.

Students: * After a few other terms, it is going to pass that line as well*.

Here, suddenly one of the students said: “**What if the sequence also passes the red line**?” As a result, I changed my role from someone who was “against” uniqueness to someone who was defending it.

I: But, it can’t go that far from the red line, can it?

Then, I cheated and I told the rest of the story! After all, it was the first time that I tried to generate the definition of convergnce of a sequence via a theorem in the class. I more motivated the definition and less generated it, though, we were very close to hit the target and generated a number of key ideas that were used in the definition. Next time that I teach sequences, I’ll try to go all the way and report it in another post.

- Tags Advanced Mathematics, Calculus