Today I continued to play with the problem from my post Puzzling Primes. Among other things, I installed the computer algebra system SageMath, which I can use to test quickly whether numbers are prime. With that tool, I might be able to solve the problem by brute force. I can’t imagine that this would be chosen as a problem of the week if there were not a better way, though.
So far I have been able to prove (a) that a superprime may contain at most two occurrences of the digits 1 or 7, (b) that all the other digits must be 3 or 9, except the first digit, which may also be 2 or 5, and (c) that if the first digit is 2 or 5, there will be no occurrences of 1 or 7.
All this means that, if there were some factor limiting the number of trailing 3s and 9s a prime could have, I would at least be able to prove a limit on the length of superprimes, and therefore that there must be a largest one. So far I’ve had no luck with that avenue of inquiry, though.
I will share the actual proofs of these claims in the future, but I’m still feeling unwell today. In the meantime, here is a chart of the ways that potential superprimes can be constructed. Red represents potential starting digits while blue represents digits that may be added.
Today was another bad mental health day, but I did read some of The Mathematical Tourist in the evening. Something written there reminded me of the problem I shared in my post Puzzling Primes, and I ended up working on that for a while, as well. I didn’t make any progress, though.
Today I read more of The Mathematical Tourist. I’ve acquired an ebook version, which I can read using text-to-speech software, my preferred reading method for text-heavy books such as this. (As I think most current readers know, I’m dyslexic.) It’s a big improvement on my attempts to read my hard copy in the ordinary way.
I’ve decided to take a three-day break from my regular study schedule and from updating the blog. There is a lot going on for me during the first half of the coming week. Expect me back on Thursday.
Today I read the first section of the first chapter of my calculus book and worked on the associated exercises. (Yippee!) My concentration, which was such a problem on Thursday, was fine today, and I’m feeling encouraged.
The section I read was about functions and discussed the vertical line test. According to the test, a graph represents a function of $x$ if and only if no vertical line intersects the graph more than once. The way I remember first being taught the test, though, the “of $x$” condition was not included. It always bothered me that $y=x^2$, an upward facing parabola, should be a function, while $x=y^2$, a rightward facing parabola, should not. It is, of course. It’s just a function of $y$ rather than a function of $x$. This has me wondering whether, say, the graph of a diagonally facing parabola could be interpreted as representing a function, and what it would be a function of.1
A busy week caught up with me today, and I ended up resting and sleeping for most of the day. I didn’t make another attempt on the first chapter of my calc book, but I did read some of The Mathematical Tourist, my light reading book.
Happy Pi Day, readers! Don’t forget also to celebrate International Pi Approximation Day, which falls on the 22nd of July. (That’s 22/7, according to the more common method of abbreviating dates.)
Today I started reading the first chapter of my calculus book, which provides a review of functions. I struggled badly to concentrate on today’s reading, for reasons I’m not sure of. Possibly it is just that I slept poorly last night. Better luck in the morning, I hope.
I spent most of my study time today conducting experiments with shadows, inspired by the trigonometry problems I discussed yesterday. In the two OpenStax exercises, the only measurement that could conceivably be related to the angle of the spotlight was the height of the shadow. Yet it seemed to me that the height of the shadow was determined by the other measurements. Since those were clearly independent of the angle of the spotlight, I suspected that the height of the shadow was as well.
I decided to test this experimentally. As shown in the photos below, I glued a small upright representing a human to the floor of a box and cut an aperture at floor level in front of it. I then shone a light through the aperture to cast the upright’s shadow on the opposite wall of the box.
The tools I had for my experiments were imperfect, but after multiple trials with different light sources, I am nearly certain that the height of the shadow cast by the upright is independent of the angle of the light. Below are two photos. The first was taken with the light at a low angle and the second with it at a high angle. As you can see, the height of the shadow is unchanged.
The only thing that changed the shadow was moving the light source. This led to some difficulty, since with the tools I had, it was hard to change the angle of the light without also changing its location. (For instance, in this video, in which I tilt a phone light from zero to about 45 degrees and back, I attribute the slight changes in the shadow to changes in the distance of the light from the aperture.) Still, this gives insight into how the problem I brought up at the end of yesterday’s post differs from the ones I was confused by. The “angle” of the sun depends on its position, while the angle of the spotlight does not.
Should I contact OpenStax about this issue? Let me know what you think.
Today I worked on some of the applied problems from the OpenStax textbook Algebra and Trigonometry. Among them were the two pictured below.
These questions made me uneasy as soon as I started working on them. Clearly, how tall the human is and how close the human stands to the spotlight are not enough to determine the angle of the spotlight. Of course, we are given another pair of measurements. Yet, in the context of the problem, these are really the same measurements. The ratio between the height of the human and the distance of the human from the spotlight must always be the same as that between the height of the shadow and the distance of the wall from the spotlight. Since it’s those ratios that ought to be giving us the angle of the spotlight, I don’t see how the problems can be done.
(If $H$ is the height of the human and $S$ is the height of the shadow, the ratio between $S-H$ and the distance of the human from the wall will be the same as the other two ratios, also. Note the similar triangles in the diagram.)
On the other hand, in the situation illustrated above, the height of the human and the length of the shadow are sufficient to calculate the angle of the sun. This makes me wonder if I’m missing something in the textbook’s problems.
Today was busy, and I only fit in about an hour of study, which I dedicated to the next video in 3Blue1Brown’s Lockdown Math series. It turned out not to be closely related to the previous three videos. Instead, it was about the silly idea of an imaginary interest rate and its relationship to a real application in physics. I didn’t find the discussion as enlightening as the previous ones, and I’m afraid my mind wandered. I did learn a bit, though.
For dessert, I watched the video below, an old favorite of mine that I recently rediscovered. I think it’s delightful.
Today I watched more of 3Blue1Brown’s Lockdown Math series. (See the new Resources tab for a link.) It turns out that the subseries I was following has four installments, not three, so I may be spending another day on it. I’ve enjoyed the primer on complex numbers 3Blue1Brown provides. Although they cropped up occasionally in my mathematical education, I never learned about them in any depth. I had certainly never understood their connection to trigonometry before.
Now it is time for the proofs of the two remaining cases of the proposition that is the focus of the Parabola-a-Go-Go series. (You will find links to the rest of the series under the new Highlights tab.)
Proposition: Consider the parabola $x^2=4py$, where $0<p$. Let $a$ be a real number. Let $A$ be the vertical line $x=a$. Let $B$ be the line through $F=(0, p)$, the focus of the parabola, and $R=(a,\frac{a^2}{4p})$, the point where $A$ intersects the parabola. Let $C$ be the line tangent to the parabola at $R$. Then, for all $a$, the acute or right angle formed by $A$ and $C$ is equal to the acute or right angle formed by $B$ and $C$.
As noted in Parabola-a-Go-Go, Part 2, the equation of the parabola can be written as $y=\frac{1}{4p}x^2$ and, using calculus, $y’=(\frac{1}{4p})2x=\frac{1}{2p}x$.
Thus, the slope of $C$, the line tangent to the parabola at $R=(a,\frac{a^2}{4p})$, is $\frac{a}{2p}$.
Furthermore, the slope of $B$, the line through $F=(0, p)$ and $R=(a,\frac{a^2}{4p})$ is $$\frac{\frac{a^2}{4p}-p}{a-0}=\frac{a}{4p}-\frac{p}{a}=\frac{a^2-4p^2}{4ap}\text{.}$$
Now suppose that $a>2p$.
In that case, the slopes of $B$ and $C$ are both positive. Note that the slope of $C$ is greater than the slope of $B$, since $\frac{a}{2p}=\frac{2a^2}{4ap}$ and $2a^2>a^2-4p^2$. It follows that the lines intersect one another, line $A$, and the x-axis as shown in the diagram below.
In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$.
Consider the right triangle formed by $A$, $C$, and the x-axis. Note that one of its vertecies is vertical to $\alpha$. Note also that the ratio of its vertical leg to its horizontal leg is equal to the slope of $C$. It follows that the ratio of its horizontal leg to its vertical leg is the reciprocal of the slope of $C$. Yet that ratio is also the tangent of the angle that is vertical to $\alpha$.
Consider instead the right triangle formed by $A$, $B$, and the x-axis. By similar reasoning we see that the tangent of $\beta+\alpha$ is the reciprocal of the slope of $B$.
Since $a>2p$, $(\frac{2p}{a})^2<1$. Therefore, by the lemma from Parabola-a-Go-Go, Part 3, $$2\arctan\frac{2p}{a}=\arctan\frac{2(\frac{2p}{a})}{1-(\frac{2p}{a})^2}=\arctan\frac{\frac{4p}{a}}{1-\frac{4p^2}{a^2}}=\arctan\frac{\frac{4ap}{a^2}}{\frac{a^2-4p^2}{a^2}}=\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $\arctan\frac{2p}{a}=\arctan\frac{4ap}{a^2-4p^2}-\arctan\frac{2p}{a}$ and thence that $\alpha=\beta$.
Thus the proposition holds when $a>2p$.
Suppose instead that $0<a<2p$.
In that case, the slope of $B$ is negative and the slope of $C$ is positive. It follows that the lines intersect one another, line $A$, and the x-axis as shown in the diagram below.
In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$. An additional angle is labeled $\gamma$ and has the property that $\alpha+\beta+\gamma=\pi$.
Considering the two right triangles formed by the lines $A$, $B$, and $C$ with the x-axis, we see that the tangent of $\alpha$ is the reciprocal of the slope of $C$ and the tangent of $\gamma$ is the absolute value of the reciprocal of the slope of $B$.
Thus, $\alpha = \arctan\frac{2p}{a}$ and $\gamma=\arctan\left|\frac{4ap}{a^2-4p^2}\right|$.
Since the $0<a<2p$, $\frac{4ap}{a^2-4p^2}$ is negative. Thus $\left|\frac{4ap}{a^2-4p^2}\right|=\frac{-4ap}{a^2-4p^2}$.
It is a property of arctangent that $\arctan (-x)=-\arctan x$. (I forgot to write up a proof of this, but the proof is simple. “Left to the reader as an exercise.”) Thus $$\gamma=\arctan\left|\frac{4ap}{a^2-4p^2}\right|=\arctan\frac{-4ap}{a^2-4p^2}=-\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $$\beta =\pi – \arctan\frac{2p}{a} -\left(-\arctan\frac{4ap}{a^2-4p^2}\right)=\pi – \arctan\frac{2p}{a} +\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
Since $0<a<2p$, $(\frac{2p}{a})^2>1$. Thus, by the lemma, $$2\arctan\frac{2p}{a}=\pi+\arctan\frac{2(\frac{2p}{a})}{1-(\frac{2p}{a})^2}=\pi+\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $\arctan\frac{2p}{a}=\pi-\arctan\frac{2p}{a}+\arctan\frac{4ap}{a^2-4p^2}$ and thence that $\alpha=\beta$.
Therefore the proposition holds for $0<a<2p$ and holds in general.
This gnarly argument was fun to work out, and I got to practice all the things I’ve been reviewing: algebra, analytic geometry, conics, and trig, plus some basic geometry. That said, if you want to see a really good proof of this theorem, check out this page. I looked this up sometime after I finished my own proof, and I’ll admit I was a little discouraged to find there was a way to do the problem that was so much better. (And this definitely is. It uses the definition of the object involved directly, requires less advanced concepts, and doesn’t rely on cases.) Not everyone can be optimally clever all the time, though—one is lucky if one can be optimally clever any of the time—and, as I said, doing it my way was fun and educational. I’m not sure whether either approach can be adapted to ellipses. That is a project for the future.
For those wondering, the diagrams for the Parabola-a-Go-Go series are based on images from the graphing program Desmos. I added additional details in GIMP and, in some cases, annotated the results by hand using a stylus and the photo app of my tablet. Most of my diagrams (the ones with the cream quad-ruled backgrounds) are made on my tablet using the stylus and a freehand note-taking app.
I know this series of posts was not very accessible. I want to explain more of what I do in a way non-specialists can understand; that’s a task that has always appealed to me. After doing my study each day, though, I only have so much energy and time left for the blog, and it tends to be more efficient to write at a higher level. Perhaps some of the competition problems I hope to try will have fairly short solutions that lend themselves to more complete explanations.
