# Deciphering recommendation engines

From the video’s description: “Data scientist Cathy O’Neil provides a glimpse of the methods that Netflix, Google, and others apply to recommend or offer to users selections based on their apparent interests.” This is a non-intuitive but real application of linear algebra.

# How to Avoid Thinking in Math Class: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. Recently, Math With Bad Drawings had a terrific series on how students try to avoid thinking in math class.

Part 1: Introduction: “In teaching math, I’ve come across a whole taxonomy of insidious strategies for avoiding thinking. Albeit for understandable reasons, kids employ an arsenal of time-tested ways to short-circuit the learning process, to jump to right answers and good test scores without putting in the cognitive heavy lifting. I hope to classify and illustrate these academic maladies: their symptoms, their root causes, and (with any luck) their cures.”

Part 2: Students’ natural desire to mindlessly plug numbers into a formula without conceptual understanding.

Part 3: The importance of both computational proficiency and conceptual understanding.

Part 4: Fears of word problems.

Part 5: What happens when students get stuck getting started on a problem.

Part 6: Is only getting the right answer important?

# Engaging students: Deriving the Pythagorean theorem

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 comes from my former student Belle Duran. Her topic, from Geometry: deriving the Pythagorean theorem.

How can technology be used to effectively engage students with this topic?

Using the video in which the scarecrow from The Wizard of Oz “explains” the Pythagorean theorem, I can get the students to review what the definition of it is. Since the scarecrow’s definition was wrong, I can ask the students what was wrong with his phrasing (he said isosceles, when the Pythagorean theorem pertains to right triangles). Thus, I can ask why it only relates to right triangles, starting the proof for the Pythagorean theorem.

What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

While Pythagoras is an important figure in the development of mathematics, little is truly known about him since he was the leader of a half religious, half scientific cult-like society who followed a code of secrecy and often presented Pythagoras as a god-like figure. These Pythagoreans believed that “number rules the universe” and thus gave numerical values to many objects and ideas; these numerical values were endowed with mystical and spiritual qualities. Numbers were an obsession for these people, so much so that they put to death a member of the cult who founded the idea of irrational numbers through finding that if we take the legs of measure 1 of an isosceles right triangle, then the hypotenuse would be equal to sqrt(2). The most interesting of all, is the manner in which Pythagoras died. It all roots back to Pythagoras’ vegetarian diet. He had a strong belief in the transmigration of souls after death, so he obliged to become a vegetarian to avoid the chance of eating a relative or a friend. However, not only did he abstain from eating meat, but also beans since he believed that humans and beans were spawned from the same source, hence the human fetal shape of the bean. In a nutshell, he refused access to the Pythagorean Brotherhood to a wealthy man who grew vengeful and thus, unleashed a mob to go after the Brotherhood. Most of the members were killed, save for a few including Pythagoras (his followers created a human bridge to help him out of a burning building). He was meters ahead from the mob, and was about to run into safety when he froze, for before him stretched a vast bean field. Refusing to trample over a single bean, his pursuers caught up and immediately ended his life.

How has this topic appeared in the news?

Dallas Cowboys coach, Jason Garrett recently made it mandatory for his players to know the Pythagorean theorem. He wants his players to understand that “’if you’re running straight from the line of scrimmage, six yards deep…it takes you a certain amount of time…If you’re doing it from ten yards inside and running to that same six yards, that’s the hypotenuse of the right triangle’” (NBC Sports). Also, recently the Museum of Mathematics (MoMath) and about 500 participants recently proved that New York’s iconic Flatiron building is indeed a right triangle. They measured the sides of the building by first handing out glow sticks for the participants to hold from end to end, then by counting while handing out the glow sticks, MoMath was able to estimate the length of the building in terms of glow sticks.

The lengths came out to be 75^2 + 180^2 = 38,025. After showing their Pythagorean relationship, MoMath projected geometric proofs on the side of the Flatiron building.

References

http://www.geom.uiuc.edu/~demo5337/Group3/hist.html

http://profootballtalk.nbcsports.com/2013/07/24/jason-garrett-wants-the-cowboys-to-know-the-pythagorean-theorem/

# Different ways of computing a limit (Part 5)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #5. Another geometric approach. The numbers $x$ and $\sqrt{x^2+1}$ can be viewed as two sides of a right triangle with legs $1$ and $x$ and hypotenuse $\sqrt{x^2+1}$. Therefore, the length of the hypotenuse must be larger than the length of one leg but less than the sum of the lengths of the two legs. In other words,

$x < \sqrt{x^2+1} < x+1$,

or

$1 < \displaystyle \frac{\sqrt{x^2+1}}{x} < \displaystyle 1+\frac{1}{x}$.

Clearly $\displaystyle \lim_{x \to \infty} 1 = 1$ and $\displaystyle \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right) = 1$. Therefore, by the Sandwich Theorem, we can conclude that $\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = 1$.

# Different ways of computing a limit (Part 4)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #4. The geometric approach. The numbers $x$ and $\sqrt{x^2+1}$ can be viewed as two sides of a right triangle with legs $1$ and $x$ and hypotenuse $\sqrt{x^2+1}$. So as $x$ gets larger and larger, the longer leg $x$ will get closer and closer in length to the length of the hypotenuse. Therefore, the ratio of the length of the hypotenuse to the length of the longer leg must be 1.

# Different ways of computing a limit (Part 3)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #3. A trigonometric identity. When we see $\sqrt{x^2+1}$ inside of an integral, one kneejerk reaction is to try the trigonometric substitution $x = \tan \theta$. So let’s use this here. Also, since $x \to \infty$, we can change the limit to be $\theta \to \pi/2$:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\tan^2 \theta+1}}{\tan \theta}$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\sec^2 \theta}}{\tan \theta}$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{ \sec \theta}{\tan \theta}$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{ ~~\displaystyle \frac{1}{\cos \theta} ~~}{ ~~ \displaystyle \frac{\sin \theta}{\cos \theta} ~~ }$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{ 1}{\sin \theta}$

$= 1$.

# Different ways of computing a limit (Part 2)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #2. Using L’Hopital’s Rule. The limit has the indeterminant form $\infty/\infty$, and so I can differentiate the top and the bottom with respect to $x$:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{d}{dx} \left( \sqrt{x^2+1} \right) }{\displaystyle \frac{d}{dx} \left( x \right)}$

$= \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{1}{2} \left( x^2+1 \right)^{-1/2} \cdot 2x }{1}$

$= \displaystyle \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}}$.

Oops… it looks like I just got the reciprocal of the original limit! Indeed, if I use L’Hopital’s Rule again, I’ll just return back to the original limit.

So that doesn’t look very helpful… except it is. If I define the value of this limit to be equal to $L$, then I’ve just shown that $L = 1/L$ (assuming that the limit exists in the first place, of course). That means that $L = 1$ or $L = -1$. Well, clearly the limit of this nonnegative function can’t be negative, and so we conclude that the limit is equal to $1$.

# Different ways of computing a limit (Part 1)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #1. The straightforward approach, using only algebra:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{\sqrt{x^2}}$

$= \displaystyle \lim_{x \to \infty} \sqrt{1 + \frac{1}{x^2}}$

$= \sqrt{1 + 0}$

$= 1$.

# Different ways of solving a contest problem (Part 3)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If $3 \sin \theta = \cos \theta$, what is $\sin \theta \cos \theta$?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Yesterday, I presented a solution using a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. Again, I begin by squaring both sides.

$9 \sin^2 \theta = \cos^2 \theta$

$9 (1 - \cos^2 \theta) = \cos^2 \theta$

$9 - 9 \cos^2 \theta = \cos^2 \theta$

$9 = 10 \cos^2 \theta$

$\displaystyle \frac{9}{10} = \cos^2 \theta$

$\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta$

Yesterday, I used the Pythagorean identity again to find $\sin \theta$. Today, I’ll instead plug back into the original equation $3 \sin \theta = \cos \theta$:

$3 \sin \theta = \cos \theta$

$3 \sin \theta = \displaystyle \frac{3}{\sqrt{10}}$

$\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}$

Unlike the example yesterday, the signs of $\sin \theta$ and $\cos \theta$ must agree. That is, if $\cos \theta = \displaystyle \frac{3}{\sqrt{10}}$, then $\sin \theta = \displaystyle \frac{1}{\sqrt{10}}$ must also be positive. On the other hand, if $\cos \theta = \displaystyle -\frac{3}{\sqrt{10}}$, then $\sin \theta = \displaystyle -\frac{1}{\sqrt{10}}$ must also be negative.

If they’re both positive, then

$\sin \theta \cos \theta = \displaystyle \left( \frac{1}{\sqrt{10}} \right) \left( \frac{3}{\sqrt{10}} \right) =\displaystyle \frac{3}{10}$,

and if they’re both negative, then

$\sin \theta \cos \theta = \displaystyle \left( -\frac{1}{\sqrt{10}} \right) \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}$.

Either way, the answer must be $\displaystyle \frac{3}{10}$.

This is definitely superior to the solution provided in yesterday’s post, as there’s absolutely no doubt that the product $\sin \theta \cos \theta$ must be positive.

# Different ways of solving a contest problem (Part 2)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If $3 \sin \theta = \cos \theta$, what is $\sin \theta \cos \theta$?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Yesterday, I presented a solution using triangles. Here’s a second solution that I received: begin by squaring both sides and using a Pythagorean trig identity.

$9 \sin^2 \theta = \cos^2 \theta$

$9 (1 - \cos^2 \theta) = \cos^2 \theta$

$9 - 9 \cos^2 \theta = \cos^2 \theta$

$9 = 10 \cos^2 \theta$

$\displaystyle \frac{9}{10} = \cos^2 \theta$

$\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta$

We use the Pythagorean identity again to find $\sin \theta$:

$\displaystyle \frac{9}{10} = \cos^2 \theta$

$\displaystyle \frac{9}{10} = 1 - \sin^2 \theta$

$\sin^2 \theta = \displaystyle \frac{1}{10}$

$\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}$

Therefore, we know that

$\sin \theta \cos \theta = \displaystyle \left( \pm \frac{1}{\sqrt{10}} \right) \left( \pm \frac{3}{\sqrt{10}} \right) = \displaystyle \pm \displaystyle \frac{3}{10}$,

so the answer is either $\displaystyle \frac{3}{10}$ or $\displaystyle -\frac{3}{10}$. However, this was a multiple-choice contest problem and $\displaystyle -\frac{3}{10}$ was not listed as a possible answer, and so the answer must be $\displaystyle \frac{3}{10}$.

For a contest problem, the above logic makes perfect sense. However, the last step definitely plays to the fact that this was a multiple-choice problem, and the concluding step would not have been possible had $\displaystyle -\frac{3}{10}$ been given as an option.