# My Favorite One-Liners: Part 46

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s one-liner is something I’ll use after completing some monumental calculation. For example, if $z, w \in \mathbb{C}$, the proof of the triangle inequality is no joke, as it requires the following as lemmas:

• $\overline{z + w} = \overline{z} + \overline{w}$
• $\overline{zw} = \overline{z} \cdot \overline{w}$
• $z + \overline{z} = 2 \hbox{Re}(z)$
• $|\hbox{Re}(z)| \le |z|$
• $|z|^2 = z \cdot \overline{z}$
• $\overline{~\overline{z}~} = z$
• $|\overline{z}| = |z|$
• $|z \cdot w| = |z| \cdot |w|$

With all that as prelude, we have

$|z+w|^2 = (z + w) \cdot \overline{z+w}$

$= (z+w) (\overline{z} + \overline{w})$

$= z \cdot \overline{z} + z \cdot \overline{w} + \overline{z} \cdot w + w \cdot \overline{w}$

$= |z|^2 + z \cdot \overline{w} + \overline{z} \cdot w + |w|^2$

$= |z|^2 + z \cdot \overline{w} + \overline{z} \cdot \overline{~\overline{w}~} + |w|^2$

$= |z|^2 + z \cdot \overline{w} + \overline{z \cdot \overline{w}} + |w|^2$

$= |z|^2 + 2 \hbox{Re}(z \cdot \overline{w}) + |w|^2$

$\le |z|^2 + 2 |z \cdot \overline{w}| + |w|^2$

$= |z|^2 + 2 |z| \cdot |\overline{w}| + |w|^2$

$= |z|^2 + 2 |z| \cdot |w| + |w|^2$

$= (|z| + |w|)^2$

In other words,

$|z+w|^2 \le (|z| + |w|)^2$.

Since $|z+w|$ and $|z| + |w|$ are both positive, we can conclude that

$|z+w| \le |z| + |w|$.

QED

In my experience, that’s a lot for students to absorb all at once when seeing it for the first time. So I try to celebrate this accomplishment:

Anybody ever watch “Home Improvement”? This is a Binford 6100 “more power” mathematical proof. Grunt with me: RUH-RUH-RUH-RUH!!!

# My Favorite One-Liners: Part 34

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Suppose that my students need to prove a theorem like “Let $n$ be an integer. Then $n$ is odd if and only if $n^2$ is odd.” I’ll ask my students, “What is the structure of this proof?”

The key is the phrase “if and only if”. So this theorem requires two proofs:

• Assume that $n$ is odd, and show that $n^2$ is odd.
• Assume that $n^2$ is odd, and show that $n$ is odd.

I call this a blue-light special: Two for the price of one. Then we get down to the business of proving both directions of the theorem.

I’ll also use the phrase “blue-light special” to refer to the conclusion of the conjugate root theorem: if a polynomial $f$ with real coefficients has a complex root $z$, then $\overline{z}$ is also a root. It’s a blue-light special: two for the price of one.

# My Favorite One-Liners: Part 17

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Sometimes it’s pretty easy for students to push through a proof from beginning to end. For example, in my experience, math majors have little trouble with each step of the proof of the following theorem.

Theorem. If $z, w \in \mathbb{C}$, then $\overline{z+w} = \overline{z} + \overline{w}$.

Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z + w} = \overline{(a + bi) + (c + di)}$

$= \overline{(a+c) + (b+d) i}$

$= (a+c) - (b+d) i$

$= (a - bi) + (c - di)$

$= \overline{z} + \overline{w}$

$\square$

For other theorems, it’s not so easy for students to start with the left-hand side and end with the right-hand side. For example:

Theorem. If $z, w \in \mathbb{C}$, then $\overline{z \cdot w} = \overline{z} \cdot \overline{w}$.

Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$

$= \overline{ac + adi + bci + bdi^2}$

$= \overline{ac - bd + (ad + bc)i}$

$= ac - bd - (ad + bc)i$

$= ac - bd - adi - bci$.

A sharp math major can then provide the next few steps of the proof from here; however, it’s not uncommon for a student new to proofs to get stuck at this point. Inevitably, somebody asks if we can do the same thing to the right-hand side to get the same thing. I’ll say, “Sure, let’s try it”:

$\overline{z} \cdot \overline{w} = \overline{(a + bi)} \cdot \overline{(c + di)}$

$= (a-bi)(c-di)$

$= ac -adi - bci + bdi^2$

$= ac - bd - adi - bci$.

$\square$

I call working with both the left and right sides to end up at the same spot the Diamond Rio approach to proofs: “I’ll start walking your way; you start walking mine; we meet in the middle ‘neath that old Georgia pine.” Not surprisingly, labeling this with a catchy country song helps the idea stick in my students’ heads.

Though not the most elegant presentation, this is logically correct because the steps for the right-hand side can be reversed and appended to the steps for the left-hand side:

Proof (more elegant). Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$

$= \overline{ac + adi + bci + bdi^2}$

$= \overline{ac - bd + (ad + bc)i}$

$= ac - bd - (ad + bc)i$

$= ac - bd - adi - bci$

$= ac -adi - bci + bdi^2$

$= (a-bi)(c-di)$

$= \overline{(a + bi)} \cdot \overline{(c + di)}$

$\overline{z} \cdot \overline{w}$.

$\square$

For further reading, here’s my series on complex numbers.

# Engaging students: Adding, subtracting, multiplying, and dividing complex numbers

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Daniel Adkins. His topic, from Algebra: adding, subtracting, multiplying, and dividing complex numbers.

How has this topic appeared in pop culture?

Robot chicken aired a television episode in which students were being taught about the imaginary number. Upon the instructor’s completion of his definition of the imaginary number, one student, who understands the definition, immediately has his head explode. One student turns to him and says, “I don’t get it. No wait now I-“, and then his head also explodes.

This video can be used as a humorous introduction that only takes a few seconds. It conveys that these concepts can be difficult in a more light-hearted sense. At the same time it satirically exaggerates the difficulty, and therefore might challenge the students. All the while the video provides the definition as well.

How did people’s conception of this topic change over time?

The first point of contact with imaginary numbers is attributed to Heron of Alexandria around the year 50 A.D. He was attempting to solve the section of a pyramid. The equation he eventually deemed impossible was the sqrt(81-114). Attempts to find a solution for a negative square root wouldn’t reignite till the discovery of negative numbers, and even this would lead to the belief that it was impossible. In the early fifteenth century speculations would rise again as higher degree polynomial equations were being worked out, but for the most part negative square roots would just be avoided. In 1545 Girolamo Cardono writes a book titled Ars Magna. He solves an equation with an imaginary number, but he says, “[imaginary numbers] are as subtle as they would be useless.” About them, and most others agreed with him until 1637. Rene Descartes set a standard form for complex numbers, but he still wasn’t too fond of them. He assumed, “that if they were involved, you couldn’t solve the problem.” And individuals like Isaac Newton agreed with him.

Rafael Bombelli strongly supported the concept of complex numbers, but since he wasn’t able to supply them with a purpose, he went mostly unheard. That is until he came up with the concept of using complex numbers to find real solutions. Over the years, individuals eventually began to hear him out.

One of the major ways that helped aid with people eventually come to terms with imaginary numbers was the concept of placing them on a Cartesian graph as the Y-axis. This concept was first introduced in 1685 by John Wallis, but he was largely ignored. A century later, Caspar Wessel published a paper over the concept, but was also ignored. Euler himself labeled the square root of negative 1 as i, which did help in modernizing the concept. Throughout the 19th century, countless mathematicians aided to the growing concept of complex numbers, until Augustin Louis Cauchy and Niels Henrik Able make a general theory of complex numbers.

This is relevant to students because it shows that mathematicians once found these things impossible, then they found them unbelievable, then they found them trivial, until finally, they found a purpose. It encourages students to work hard even if there doesn’t seem to be a reason behind it just yet, and even if it seems like your head is about to blow.

How has this topic appeared in high culture?

The Mandelbrot set is a beautiful fractal set with highly complex math hidden behind it. However it is extremely complicated, and as Otto von Bismarck put it, “laws are like sausages. Better not to see them being made.”

Like most fractals, the Mandelbrot set begins with a seed to start an iteration. In this case we begin with x2 + c, where c is some real number. This creates an eccentric pattern that grows and grows.

For students, this can show how mathematics can create beautiful patterns that would interest their more artistic senses. Not only would this generate interest in complex numbers, but it might convince students to investigate recurring patterns.

Sources:

History of imaginary numbers:

http://rossroessler.tripod.com/

Mendelbrot sets:

https://plus.maths.org/content/unveiling-mandelbrot-set

# Exponents and the decathlon

During the Olympics, I stumbled across an application of exponents that I had not known before: scoring points in the decathlon or the heptathlon. From FiveThirtyEight.com:

Decathlon, which at the Olympics is a men’s event, is composed of 10 events: the 100 meters, long jump, shot put, high jump, 400 meters, 110-meter hurdles, discus throw, pole vault, javelin throw and 1,500 meters. Heptathlon, a women’s event at the Olympics, has seven events: the 100-meter hurdles, high jump, shot put, 200 meters, long jump, javelin throw and 800 meters…

As it stands, each event’s equation has three unique constants — $latex A$, $latex B$ and $latex C$— to go along with individual performance, $latex P$. For running events, in which competitors are aiming for lower times, this equation is: $latex A(BP)^C$, where $latex P$ is measured in seconds…

$B$ is effectively a baseline threshold at which an athlete begins scoring positive points. For performances worse than that threshold, an athlete receives zero points.

Specifically from the official rules and regulations (see pages 24 and 25), for the decathlon (where $P$ is measured in seconds):

• 100-meter run: $25.4347(18-P)^{1.81}$.
• 400-meter run: $1.53775(82-P)^{1.81}$.
• 1,500-meter run: $0.03768(480-P)^{1.85}$.
• 110-meter hurdles: $5.74352(28.5-P)^{1.92}$.

For the heptathlon:

• 200-meter run: $4.99087(42.5-P)^{1.81}$.
• 400-meter run: $1.53775(82-P)^{1.88}$.
• 1,500-meter run: $0.03768(480-P)^{1.835}$.

Continuing from FiveThirtyEight:

For field events, in which competitors are aiming for greater distances or heights, the formula is flipped in the middle: $latex A(PB)^C$, where $latex P$ is measured in meters for throwing events and centimeters for jumping and pole vault.

Specifically, for the decathlon jumping events ($P$ is measured in centimeters):

• High jump: $0.8465(P-75)^{1.42}$
• Pole vault: $0.2797(P-100)^{1.35}$
• Long jump: $0.14354(P-220)^{1.4}$

For the decathlon throwing events ($P$ is measured in meters):

• Shot put: $51.39(P-1.5)^{1.05}$.
• Discus: $12.91(P-4)^{1.1}$.
• Javelin: $10.14(P-7)^{1.08}$.

Specifically, for the heptathlon jumping events ($P$ is measured in centimeters):

• High jump: $1.84523(P-75)^{1.348}$
• Long jump: $0.188807(P-210)^{1.41}$

For the heptathlon throwing events ($P$ is measured in meters):

• Shot put: $56.0211(P-1.5)^{1.05}$.
• Javelin: $15.9803(P-3.8)^{1.04}$.

I’m sure there are good historical reasons for why these particular constants were chosen, but suffice it to say that there’s nothing “magical” about any of these numbers except for human convention. From FiveThirtyEight:

The [decathlon/heptathlon] tables [devised in 1984] used the principle that the world record performances of each event at the time should have roughly equal scores but haven’t been updated since. Because world records for different events progress at different rates, today these targets for WR performances significantly differ between events. For example, Jürgen Schult’s 1986 discus throw of 74.08 meters would today score the most decathlon points, at 1,384, while Usain Bolt’s 100-meter world record of 9.58 seconds would notch “just” 1,203 points. For women, Natalya Lisovskaya’s 22.63 shot put world record in 1987 would tally the most heptathlon points, at 1,379, while Jarmila Kratochvílová’s 1983 WR in the 800 meters still anchors the lowest WR points, at 1,224.

FiveThirtyEight concludes that, since the exponents in the running events are higher than those for the throwing/jumping events, it behooves the elite decathlete/heptathlete to focus more on the running events because the rewards for exceeding the baseline are greater in these events.

Finally, since all of the exponents are not integers, a negative base (when the athlete’s performance wasn’t very good) would actually yield a complex-valued number with a nontrivial imaginary component. Sadly, the rules of track and field don’t permit an athlete’s score to be a non-real number. However, if they did, scores might look something like this…

# Difference of Two Powers (Index)

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on getting students to discover the formula for factoring $x^n - y^n$.

Part 1: A numerical way of discovering the formula for $x^2 - y^2$.

Part 2: A geometric way of discovering the formula for $x^2 - y^2$.

Part 3: Pedagogical thoughts on the importance of students discovering the formula for $x^3 - y^3$.

Part 4: A geometric way of discovering the formula for $x^3 - y^3$.

Part 5: Guiding students to the formula for $x^n - y^n$.

# The antiderivative of 1/(x^4+1): Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the computation of

$\displaystyle \int \frac{dx}{x^4+1}$

Part 1: Introduction.

Part 2: Factoring the denominator using De Moivre’s Theorem.

Part 3: Factoring the denominator using the difference of two squares.

Part 4: The partial fractions decomposition of the integrand.

Part 5: Partial evaluation of the resulting integrals.

Part 6: Evaluation of the remaining integrals.

Part 7: An apparent simplification using a trigonometric identity.

Part 8: Discussion of the angles for which the identity holds.

Part 9: Proof of the angles for which the identity holds.

Part 10: Implications for using this identity when computing definite integrals.

# Diamond Rio and Proofs

Sometimes it’s pretty easy for students to push through a proof from beginning to end. For example, in my experience, math majors have little trouble with each step of the proof of the following theorem.

Theorem. If $z, w \in \mathbb{C}$, then $\overline{z+w} = \overline{z} + \overline{w}$.

Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z + w} = \overline{(a + bi) + (c + di)}$

$= \overline{(a+c) + (b+d) i}$

$= (a+c) - (b+d) i$

$= (a - bi) + (c - di)$

$= \overline{z} + \overline{w}$

$\square$

For other theorems, it’s not so easy for students to start with the left-hand side and end with the right-hand side. For example:

Theorem. If $z, w \in \mathbb{C}$, then $\overline{z \cdot w} = \overline{z} \cdot \overline{w}$.

Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$

$= \overline{ac + adi + bci + bdi^2}$

$= \overline{ac - bd + (ad + bc)i}$

$= ac - bd - (ad + bc)i$

$= ac - bd - adi - bci$.

A sharp math major can then provide the next few steps of the proof from here; however, it’s not uncommon for a student new to proofs to get stuck at this point. Inevitably, somebody asks if we can do the same thing to the right-hand side to get the same thing. I’ll say, “Sure, let’s try it”:

$\overline{z} \cdot \overline{w} = \overline{(a + bi)} \cdot \overline{(c + di)}$

$= (a-bi)(c-di)$

$= ac -adi - bci + bdi^2$

$= ac - bd - adi - bci$.

$\square$

I call working with both the left and right sides to end up at the same spot the Diamond Rio approach to proofs: “I’ll start walking your way; you start walking mine; we meet in the middle ‘neath that old Georgia pine.” Not surprisingly, labeling this with a catchy country song helps the idea stick in my students’ heads.

Though not the most elegant presentation, this is logically correct because the steps for the right-hand side can be reversed and appended to the steps for the left-hand side:

Proof (more elegant). Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$

$= \overline{ac + adi + bci + bdi^2}$

$= \overline{ac - bd + (ad + bc)i}$

$= ac - bd - (ad + bc)i$

$= ac - bd - adi - bci$

$= ac -adi - bci + bdi^2$

$= (a-bi)(c-di)$

$= \overline{(a + bi)} \cdot \overline{(c + di)}$

$\overline{z} \cdot \overline{w}$.

$\square$

# How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction
Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by $\sec^2 x$, and the substitution $u = \tan x$.
Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution $u = \tan \theta/2$.
Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle
Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter $a$, the magic substitution $u = \tan \theta/2$, and partial fractions.
Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter $a$, the magic substitution $u = \tan \theta/2$, and contour integration using the real line and an expanding semicircle.
Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

# Is 2i less than 3i? (Index)

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on why it doesn’t make sense to say one complex number is less than another complex number (using the usual definition of “less than”).

Part 1: A sketch of a direct proof based on the order axioms proving that $i$ is not an element of a number system that has the usual definition of inequality.

Part 2: An indirect proof.

Part 3: Discussion about the lexicographic ordering, which almost works.

Part 4: Two other partial orderings which almost work.