Parabolas from String Art (Part 8)

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

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

Finding the curve traced by the strings is a two-step process:

• For a fixed value of $x$, find the value of $s$ that maximizes $y$.
• Find this optimal value of $y$.

Previously, we showed using only algebra that the optimal value of $s$ is $s = \displaystyle \frac{x}{2}$, corresponding to an optimal value of $y$ of $y = \displaystyle \frac{x^2}{16} - x + 8$.

For a student who knows calculus, the optimal value of $s$ can be found by instead solving the equation $\displaystyle \frac{dy}{ds} = 0$ (or, more accurately, $\displaystyle \frac{\partial y}{\partial s} = 0$):

$\displaystyle \frac{dy}{ds} = -\frac{2s}{4} + \frac{x}{4}$

$0 = \displaystyle \frac{-2s+x}{4}$

$0 = -2s + x$

$2s = x$

$s = \displaystyle \frac{x}{2}$,

matching the result that we found by using only algebra.

Parabolas from String Art (Part 7)

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.

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 $s + 8$.

In the previous post, we established that the equation for string $s$ is

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

This has the appearance of a quadratic equation, but it’s actually a linear equation in $x$ for a fixed value of $s$. For example, if s = 5, we find that the equation of string 5 is

$y = -\displaystyle \frac{25}{4} + \frac{5x}{4} - x +8 = 0.25x+1.75$,

matching the equation of the blue string we found in a previous post in this series.

To prove that the strings trace a parabola, we now determine which string $s$ maximizes the value of $y = -\displaystyle \frac{1}{4}s^2 + \frac{x}{4}s - x + 8$ for a given value of $x$. Algebra students can determine this maximum by recalling that a quadratic function $y = as^2 + bs + c$ is maximized (for negative $a$) when $s = \displaystyle -\frac{b}{2a}$. Therefore, the string with largest $y$-coordinate for a given value of $x$ is

$s = \displaystyle -\frac{x/4}{2 \cdot (-1/4)} = \frac{x}{2}$.

For example, if $x = 4$, then string $s = 4 / 2 = 2$ has the largest $y$-coordinate, matching our previous observations.
To complete the proof that the strings above trace a parabola, we substitute $s = \displaystyle \frac{x}{2}$ into $y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8$ to find the value of this largest $y-$coordinate:

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

$= -\displaystyle \frac{x^2}{16} + \frac{x^2}{8} - x + 8$

$= \displaystyle \frac{x^2}{16} - x + 8$,

matching the result that we found earlier in this series.

There’s also a bonus result. We further note that, for every $x$, there is only one string $s = \frac{x}{2}$ that intersects the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$. Since each $x$ is associated with a unique string $s$ and vice versa, we conclude that each string intersects the parabola at exactly one point. In other words, string $s$ is tangent to the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ when $x=2s$.

We note that all of the above calculations were entirely elementary, in the sense that calculus was not used and that only techniques from algebra were employed. That said, the word “elementary” in mathematics can be a bit loaded — this means that it is based on simple ideas that are perhaps used in a profound and surprising way. Perhaps my favorite quote along these lines was this understated gem from the book Three Pearls of Number Theory after the conclusion of a very complicated multi-page proof in Chapter 1:

You see how complicated an entirely elementary construction can sometimes be. And yet this is not an extreme case; in the next chapter you will encounter just as elementary a construction which is considerably more complicated.

In the next post, we take a second look at this derivation using techniques from calculus.

Parabolas from String Art (Part 6)

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.

As discussed previous posts, we begin our explorations with string art connecting evenly spaced points on line segments $\overline{AB}$ and $\overline{BC}$ with endpoints $A(0,8)$, $B(8,0)$, and $C(16,8)$. We will call these colored line segments “strings.” We then found the string with the largest $y-$coordinate at $x = 2, 4, 6, \dots, 14$, resulting in the following picture:

In previous posts, we discussed three different ways of establishing that the colored points lie on the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$.

Unfortunately, checking that a statement is true for a few points (in our case, $x= 0, 2, 4, \dots, 14, 16$) does not constitute a complete proof for all points. Furthermore, it’s conceivable that “fuller” string art with additional strings, like the picture below, may identify a new string with a higher $y-$coordinate than a colored point.

To prove that the string art indeed traces a parabola, we study 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 $s + 8$.

Since the equations of $\overline{AB}$ and $\overline{BC}$ are $y=-x+8$ and $y=x-8$, respectively, the $y-$coordinates of $P$ and $Q$ are $-s+8$ and $(s+8)-8 = s$, respectively. For example, if $s = 5$, the coordinates of $P$ are $(s,8-s)=(5,3)$ and the coordinates of $Q$ are $(s + 8, s) = (13, 5)$, matching the endpoints of the blue string in the first figure.
We now use standard algebraic techniques to find the equation of string $s$. Its slope is

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

The coordinates of either $P$ or $Q$ can now be used to find the equation of string $s$ via the point-slope formula. As it turns out, the coordinates of $P$ are simpler to use:

$y-y_1 = m(x-x_1)$

$y-(8-s) = \displaystyle \frac{s-4}{4}(x-s)$

$y = \displaystyle \frac{(s-4)(x-s)}{4} + (8-s)$

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

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

to finally arrive at the equation of string $s$:

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

This has the appearance of a quadratic equation, but it’s actually a linear equation in $x$ for a fixed value of $s$. For example, if s = 5, we find that the equation of string 5 is

$y = -\displaystyle \frac{25}{4} + \frac{5x}{4} - x +8 = 0.25x+1.75$,

matching the equation of the blue string we found in a previous post in this series.

We are now almost in position to prove that the string art traces a parabola. We demonstrate this in the next post.

Parabolas from String Art (Part 5)

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.

As discussed previous posts, we begin our explorations with string art connecting evenly spaced points on line segments $\overline{AB}$ and $\overline{BC}$ with endpoints $A(0,8)$, $B(8,0)$, and $C(16,8)$. We will call these colored line segments “strings.” We then found the string with the largest $y-$coordinate at $x = 2, 4, 6, \dots, 14$, resulting in the following picture:

However, perhaps it’s clearer to plot these points on a separate graph, without the clutter of the strings:

These points are definitely following some kind of curve. In the previous posts, we established algebraically that the curve is the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$.

A more modern way of convincing students that the points lie on a parabola is by using technology: specifically, quadratic regression. First, we can input the nine points into a scientific calculator.

Then we ask the calculator to perform quadratic regression on the data.

The calculator then returns the result:

The best quadratic fit to the data is $y = 0.0625x^2 - x + 8$, or $y = \displaystyle \frac{x^2}{16} - x + 8$ as before. The line $R^2 = 1$ indicates a correlation coefficient of 1, meaning that the points lie perfectly on this parabola. A parenthetical note: If the $R^2$ line does not appear using a TI-83 or TI-84, this can be toggled by using DiagnosticOn:

I’ve presented three different ways that algebra students can convince themselves that the nine points generated by the above string art indeed lie on a parabola. I don’t suggest that all three methods should be used for any given student; as always, if one technique doesn’t appear to work pedagogically, then perhaps a different explanation might work.

However, our explorations aren’t done yet. Any of these three techniques may convince algebra students that the strings above trace a parabola. Unfortunately, checking that a statement is true for a few points (in our case, $x= 0, 2, 4, \dots, 14, 16$) does not constitute a complete proof for all points. Furthermore, it’s conceivable that “fuller” string art with additional strings, like the picture below, may identify a new string with a higher $y-$coordinate than a colored point.

So we have more work to do to prove our assertion that string art traces a parabola. We begin this next phase of our investigations in the next post.

Parabolas from String Art (Part 4)

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.

As discussed previous posts, we begin our explorations with string art connecting evenly spaced points on line segments $\overline{AB}$ and $\overline{BC}$ with endpoints $A(0,8)$, $B(8,0)$, and $C(16,8)$. We will call these colored line segments “strings.” We then found the string with the largest $y-$coordinate at $x = 2, 4, 6, \dots, 14$, resulting in the following picture:

However, perhaps it’s clearer to plot these points on a separate graph, without the clutter of the strings:

These points are definitely following some kind of curve. In the previous post, we established that the curve is a parabola by using the vertex form of a parabola $y = a(x-h)^2+k$.

In this post, we use the other general form. If the curve is a parabola, then the equation of the curve must be $y = ax^2 + bx + c$ for some values of $a$, $b$, and $c$. Since there are three unknowns, we need to have three equations to solve for them. This can be done by plugging in three $(x,y)$ pairs into this equation. While we can pick any three pairs that we wish, it seems convenient to use the points $(0,8)$, $(8,4)$ and $(16,8)$:

$a(0)^2+b(0)+c = 8$

$a(8)^2 + b(8) + c = 4$

$a(16)^2 + b(16) + c =8$

This simplifies to the $3\times 3$ system of linear equations

$c = 8$

$64a+8b+c=4$

$256a+16b+c=8$

In general $3\times 3$ systems of linear equations can be challenging for students to solve. However, while this is technically a $3\times 3$ system, it’s clear that $c =8$, and so this reduces to a $2\times 2$ system

$64a+8b+8=4$

$256a+16b+8=8$

or

$64a+8b=-4$

$256a+16b=0$

or

$16a+2b=-1$

$16a+b=0$.

In algebra, students are taught multiple ways of solving $2\times 2$ systems of linear equations, and any of these techniques can be used at this point to solve for $a$ and $b$. Perhaps the easiest next step is subtracting the two equations:

$(16a + 2b) - (16a + b) = -1 - 0$

$b = -1$

Substituting into $16a+b=0$, we see that

$16a - 1 = 0$

$16a = 1$

$a =\displaystyle \frac{1}{16}$.

We conclude that $a = \displaystyle \frac{1}{16}$, $b = -1$, and $c = 8$, so that, if the points lie on a parabola, the equation of the parabola must be

$y = \displaystyle \frac{x^2}{16} - x + 8$.

By construction, this parabola passes through $(0,8)$, $(8,4)$, and $(16,8)$. To show that this actually works, we can substitute the other six values of $x$:

At $x =2$: $y = \displaystyle \frac{(2)^2}{16} - 2 + 8 = 6.25$

At $x =4$: $y = \displaystyle \frac{(4)^2}{16} - 4 + 8 = 5$

At $x =6$: $y = \displaystyle \frac{(6)^2}{16} - 6 + 8 = 4.25$

At $x =10$: $y = \displaystyle \frac{(10)^2}{16} - 10 + 8 = 4.25$

At $x =12$: $y = \displaystyle \frac{(12)^2}{16} - 12 + 8 = 5$

At $x =14$: $y = \displaystyle \frac{(14)^2}{16} - 14 + 8 = 6.25$

Therefore, the nine points in the above picture all lie on the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$.

In the next post, we’ll discuss a third way of convincing students that the points lie on this parabola.

Parabolas from String Art (Part 3)

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.

As discussed previous posts, we begin our explorations with string art connecting evenly spaced points on line segments $\overline{AB}$ and $\overline{BC}$ with endpoints $A(0,8)$, $B(8,0)$, and $C(16,8)$. We will call these colored line segments “strings.” We then found the string with the largest $y-$coordinate at $x = 2, 4, 6, \dots, 14$, resulting in the following picture:

However, perhaps it’s clearer to plot these points on a separate graph, without the clutter of the strings:

These points are definitely following some kind of curve. Let’s suppose that the curve is a parabola. The vertex form of a parabola is

$y = a(x-h)^2+k$.

If the curve is a parabola, then clearly the vertex will be the lowest point on the axis of symmetry. By inspection, this point is $(8,4)$, which is labeled $V$ in the above picture. So, if it’s a parabola, the equation has the form

$y = a(x-8)^2+4$.

To find the value of $a$, we note that the point $(x, y) = (16, 8)$ must be on the parabola, so that

$8 = a(x-8)^2 + 4$

$8 = 64a + 4$

$4 = 64a$

$a = \displaystyle \frac{1}{16}$.

Therefore, the equation of the conjectured parabola is

$y = \displaystyle \frac{1}{16}(x-8)^2 + 4$

$= \displaystyle \frac{1}{16} (x^2 - 16x + 64) + 4$

$= \displaystyle \frac{x^2}{16} - x + 4 + 4$

$= \displaystyle \frac{x^2}{16} - x + 8$.

So, if the curve is a parabola, it must follow the function this curve. By construction, this parabola passes through $(8,4)$ and $(16,8)$. To show that this actually works, we can substitute the other seven values of $x$:

At $x =0$: $y = \displaystyle \frac{(0)^2}{16} - 0 + 8 = 8$

At $x =2$: $y = \displaystyle \frac{(2)^2}{16} - 2 + 8 = 6.25$

At $x =4$: $y = \displaystyle \frac{(4)^2}{16} - 4 + 8 = 5$

At $x =6$: $y = \displaystyle \frac{(6)^2}{16} - 6 + 8 = 4.25$

At $x =10$: $y = \displaystyle \frac{(10)^2}{16} - 10 + 8 = 4.25$

At $x =12$: $y = \displaystyle \frac{(12)^2}{16} - 12 + 8 = 5$

At $x =14$: $y = \displaystyle \frac{(14)^2}{16} - 14 + 8 = 6.25$

Therefore, the nine points in the above picture all lie on the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$.

In the next couple of posts, we’ll discuss a couple of different ways of establishing that the points lie on this parabola.

Parabolas from String Art (Part 2)

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.

As discussed in the previous post, we begin our explorations with string art connecting evenly spaced points on line segments $\overline{AB}$ and $\overline{BC}$ with endpoints $A(0,8)$, $B(8,0)$, and $C(16,8)$. We will call these colored line segments “strings.”

We now ask the following two questions:

• For each of $x = 2, 4, 6, 8, 10, 12,$ and $14$, which string has the largest $y-$coordinate?
• For each of these values of $x$, what is the value of this largest $y-$coordinate?

Evidently, for $x=4$, the brown string that connects $(2,6)$ to $(10,2)$ has the largest $y-$coordinate. This point is marked with the small brown circle. From the lines on the graph paper, it appears that this brown point is $(4,5)$.

For $x=8$, the horizontal green string appears to have the largest $y-$coordinate, and clearly that point is $(8,4)$.

For $x=12$, the pink string that connects $(6,2)$ to $(14,6)$ has the largest $y-$coordinate. From the lines on the graph paper, it appears that this point is $(12,5)$.

Unfortunately, for $x=2$, $x=6$, $x=10$, and $x=14$, it’s evident which string has the largest $y-$coordinate, but it’s not so easy to confidently read off its value. For this example, this could be solved by using finer graph paper with marks at each quarter (instead of at the integers). However, it’s far better to actually use the point-slope formula to find the equation of the colored line segments.

For example, for $x=2$, the red string has the largest $y-$coordinate. This string connects the points $(1,7)$ and $(9,1)$, and so the slope of this string is $\displaystyle \frac{1-7}{9-1} = -\frac{6}{8} = -0.75$. Using the point-slope form of a line, the equation of the red string is thus

$y - 7 = -0.75(x - 1)$

$y-7= -0.75x + 0.75$

$y = -0.75x + 7.75$

Substituting $x =2$, the $y-$coordinate of the highest string at $x=2$ is $y = -0.75(2) + 7.75 = 6.25$.

Similarly, at $x=6$, the equation of the orange string turns out to be $y=-0.25x+5.75$, and the $y-$coordinate of the highest string at $x=6$ is $y=-0.25(6)+5.75=4.25$.

At $x=10$, the equation of the blue string is $y=0.25x+1.75$, and the $y-$coordinate of the highest string at $x=10$ is $y=0.25(10)+1.75=4.25$.

Finally, at $x=14$, the equation of the purple string is $y=0.75x-4.25$, and the $y-$coordinate of the highest string at $x=14$ is $y=0.75(14)-4.25=6.25$.

The interested student could confirm the values for $x=4$, $x=8$, and $x=12$ that were found earlier by just looking at the picture.

We now add the coordinates of these points to the picture.

However, perhaps it’s clearer to plot these points on a separate graph, without the clutter of the strings:

These points are definitely following some kind of curve. In the next post in this series, I’ll discuss a way of convincing students that the curve is actually a parabola.

Parabolas from String Art (Part 1)

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.

To begin, we use graph paper to sketch to draw coordinate axes, the point $A(0,8)$, the point $B(8,0)$, the point $C(16,8)$, line segment $\overline{AB}$, and line segment $\overline{BC}$.

Along $\overline{AB}$, we mark the evenly spaced points $(1,7)$, $(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$, $(6,2)$, and $(7,1)$.

Along $\overline{BC}$, we mark the evenly spaced points $(9,1)$, $(10,2)$, $(11,3)$, $(12,4)$, $(13,5)$, $(14,6)$, and $(15,7)$.

Next, we draw line segments of different colors to connect:

• $(1,7)$ and $(9,1)$
• $(2,6)$ and $(10,2)$
• $(3,5)$ and $(11,3)$
• $(4,4)$ and $(12,4)$
• $(5,3)$ and $(13,5)$
• $(6,2)$ and $(14,6)$
• $(7,1)$ and $(15,7)$

The result should look something like the picture below:

It looks like the string art is tracing a parabola. In this series of posts, I’ll discuss one way that talented algebra students can convince themselves that the curve is indeed a parabola.

An elementary proof of the insolvability of the quintic

When I was in middle school, I remember my teacher telling me, after I learned the quadratic formula, that there was a general formula for solving cubic and quartic equations, but no such formula existed for solving the quintic. This was also when I first heard the infamous story of young Galois’s death from a duel.

Using my profound middle-school logic, I took this story as a challenge to devise my own formula for solving the quintic. Naturally, my efforts came up short.

When I was in high school, with this obsession still fully intact, I attempted to read through the wonderful monograph Field Theory and Its Classical Problems. Here’s the MAA review of this book:

Hadlock’s book sports one of the best prefaces I’ve ever read in a mathematics book. The rest of the book is even better: in 1984 it won the first MAA Edwin Beckenbach Book Prize for excellence in mathematical exposition.

Hadlock says in the preface that he wrote the book for himself, as a personal path through Galois theory as motivated by the three classical Greek geometric construction problems (doubling the cube, trisecting angles, and squaring the circle — all with just ruler and compass) and the classical problem of solving equations by radicals. Unlike what happens in most books on the subject, all three Greek problems are solved in the first chapter, with just the definition of field as a subfield of the real numbers, but without even defining degree of field extensions, much less proving its multiplicativity (this is done in chapter 2). Doubling the cube is proved to be impossible by proving that the cube root of 2 cannot be an element of a tower of quadratic extensions: if the cube root of 2 is in a quadratic extension, then it is actually in the base field. Repeating the argument, we conclude that it is not constructible because it is not rational. A similar argument works for proving that trisecting a 60 degree angle is impossible. Of course, proving that duplicating the cube is impossible needs a different argument: chapter 1 ends with Niven’s proof of the transcendence of π.

After this successful bare-hands attack at three important problems, Chapter 2 discusses in detail the construction of regular polygons and explains Gauss’s characterization of constructible regular polygons, including the construction of the regular 17-gon. Chapter 3 describes Galois theory and the solution of equations by radicals, including Abel’s theorem on the impossibility of solutions by radicals for equations of degree 5 or higher. Chapter 4, the last one, considers a special case of the inverse Galois problem and proves that there are polynomials with rational coefficients whose Galois group is the symmetric group, a result that is established via Hilbert’s irreducibility theorem.

Many examples, references, exercises, and complete solutions (taking up a third of the book!) are included and make this enjoyable book both an inspiration for teachers and a useful source for independent study or supplementary reading by students.

As I recall, I made it successfully through the first couple of chapters but started to get lost with the Galois theory somewhere in the middle of Chapter 3. Despite not completing the book, this was one of the most rewarding challenges of my young mathematical life. Perhaps one of these days I’ll undertake this challenge again.

Anyway, this year I came across the wonderful article The Abel–Ruffini Theorem: Complex but Not Complicated in the March issue of the American Mathematical Monthly. The article presents a completely different way of approaching the insolvability of the quintic that avoids Galois theory altogether.

The proof is elementary; I’m confident that I could have understood this proof had I seen it when I was in high school. That said, the word “elementary” in mathematics can be a bit loaded — this means that it is based on simple ideas that are perhaps used in a profound and surprising way. Perhaps my favorite quote along these lines was this understated gem from the book Three Pearls of Number Theory after the conclusion of a very complicated proof in Chapter 1:

You see how complicated an entirely elementary construction can sometimes be. And yet this is not an extreme case; in the next chapter you will encounter just as elementary a construction which is considerably more complicated.

I believe that a paid subscription to the Monthly is required to view the above link, but the main ideas of the proof can be found in the video below as well as this short PDF file by Leo Goldmakher.

Thoughts on Numerical Integration: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show my series on numerical integration.

Part 1 and Part 2: Introduction

Part 3: Derivation of left, right, and midpoint rules

Part 4: Derivation of Trapezoid Rule

Part 5: Derivation of Simpson’s Rule

Part 6: Connection between the Midpoint Rule, the Trapezoid Rule, and Simpson’s Rule

Part 7: Implementation of numerical integration using Microsoft Excel

Part 8, Part 9, Part 10, Part 11: Numerical exploration of error analysis

Part 12 and Part 13: Left endpoint rule and rate of convergence

Part 14 and Part 15: Right endpoint rule and rate of convergence

Part 16 and Part 17: Midpoint Rule and rate of convergence

Part 18 and Part 19: Trapezoid Rule and rate of convergence

Part 20 and Part 21: Simpson’s Rule and rate of convergence

Part 22: Comparison of these results to theorems found in textbooks

Part 23: Return to Part 2 and accuracy of normalcdf function on TI calculators