Predicate Logic and Popular Culture (Part 255): The Nightmare Before Christmas

Let p be the statement “I can see it,” and let $latex q” be the statement “I can believe it.” Translate the logical statement

\sim ( \sim p \Longrightarrow \sim q)

This matches a line from “Jack’s Obsession” in the movie “The Nightmare Before Christmas.” (It’s also a good exercise in using DeMorgan’s Laws.)

green line

Context: This semester, I taught discrete mathematics for the first time. Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.

In an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.

When I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.

Predicate Logic and Popular Culture (Part 254): The Rolling Stones

Let T be the set of all times, let H be the set of all things, let W(x,t) be the statement “I need x at time t,” and let G(x,t) be the statement “I get x at time t.” Translate the logical statement

\sim \forall t \in T \forall x \in H (W(x,t) \Longrightarrow G(x,t))

This matches the title and chorus of “You Can’t Always Get What You Want” by the Rolling Stones.

green line

Context: This semester, I taught discrete mathematics for the first time. Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.

In an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.

When I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.

Parabolas from String Art (Part 10)

Recently, I announced that my paper Parabolic Properties from Pieces of String had been published in the magazine Math Horizons. The article had multiple aims; in chronological order of when I first started thinking about them:

  • Prove that string art from two line segments traces a parabola.
  • Prove that a quadratic polynomial satisfies the focus-directrix property of a parabola, which is the reverse of the usual logic when students learn conic sections.
  • Prove the reflective property of parabolas.
  • Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. In this series of posts, I’d wanted to expand on the article with some pedagogical thoughts about connecting string art to parabolas for algebra students. After all, most mathematical studies of string art curves — formally known as “envelopes” — rely on differential equations or at least limits and calculus.

However, string art is simple enough for a young child to construct, and so this study was inspired by the quest of explaining this phenomenon using only simple mathematical tools.

The article linked above has further thoughts on this problem, including a calculus-free way of deriving the reflective property of parabolas. However, I think the article pretty much has all of my thoughts on this matter, and so I don’t think I need to elaborate upon them here.

This series of posts is dedicated to an inspired and inspiring Algebra I student who wanted to understand string art curves using tools that she could understand… even though she progressed much further into the mathematics curriculum by the time my article was published and this series of posts appeared on my blog.

Parabolas from String Art (Part 9)

Recently, I announced that my paper Parabolic Properties from Pieces of String had been published in the magazine Math Horizons. The article had multiple aims; in chronological order of when I first started thinking about them:

  • Prove that string art from two line segments traces a parabola.
  • Prove that a quadratic polynomial satisfies the focus-directrix property of a parabola, which is the reverse of the usual logic when students learn conic sections.
  • Prove the reflective property of parabolas.
  • Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. While all the necessary mathematics is in the article, I think the article is somewhat lacking on how to sell the idea to students. So, in this series of posts, I’d like to expand on the article with some pedagogical thoughts about connecting string art to parabolas.

We have shown in the last couple of posts that if the three points that generate the Our explorations of string art led us to consider an arbitrary string \overline{PQ} depicted below. For brevity, this string will be called “string s,” matching the (possibly non-integer) x-coordinate of its left endpoint P. Since P is s units to the right of A, the right endpoint Q must correspondingly be s units to the right of B. Therefore, the x-coordinate of Q is s + 8.

Previously, we established that the equation for string s is

y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8.

We also obtained a bonus result that we obtained using only algebra: string s is tangent to the parabola y = \displaystyle \frac{x^2}{16} - x + 8, which is traced by the strings, when x=2s. Of course, tangent lines are usually obtained using calculus, and so calculus should be able to confirm this result. The derivative of this function is

y' = \displaystyle \frac{x}{8} - 1,

so that the slope of the tangent line when x=2s is m = \displaystyle \frac{s}{4} - 1 = \frac{s-4}{4}. We observe that this matches the slope of line segment \overline{PQ} in the above picture:

slope = \displaystyle \frac{s - (s-8)}{(s+8) - 8} = \frac{2s-8}{8} = \frac{s-4}{4}.

Therefore, to show that \overline{PQ} is the tangent line, it suffices to show that either P or Q is on the tangent line.

At x = 2s, the y-coordinate of where the tangent line intersects the curve is

y = \displaystyle \frac{(2s)^2}{16} - 2s + 8 = \frac{s^2}{4} - 2s + 8.

Using the point-slope formula for a line, the equation of the tangent line is thus

y-y_1 = m(x-x_1)

y-\displaystyle \left( \frac{s^2}{4} - 2s + 8 \right) = \frac{s-4}{4} (x-2s)

y = \displaystyle \frac{s-4}{4} (x-2s) +  \frac{s^2}{4} - 2s + 8.

We now check to see if P(s,8-s) is on the tangent line. Substituting x =s, we find

y = \displaystyle \frac{s-4}{4} (s-2s) +  \frac{s^2}{4} - 2s + 8

= \displaystyle \frac{s-4}{4} (-s) +  \frac{s^2}{4} - 2s + 8

= \displaystyle \frac{(s-4)(-s) + s^2}{4} - 2s + 8

= \displaystyle \frac{-s^2+4s + s^2}{4} - 2s + 8

= \displaystyle \frac{4s}{4} - 2s + 8

= s - 2s + 8

= -s + 8

Therefore, the point (s,8-s) is on the tangent line, thus confirming that P is on the tangent line and that \overline{PQ} is the tangent line.