Month: February 2024

Dancing Graphs

Olly / / Math Posts

Today I was tired again and not feeling mathematically creative, so I worked on exercises in the review of trigonometry rather than doing more work on the parabola problem. A few of the exercises involved graphing trigonometric functions. I’ve always found something enchanting in the graphs of sine and cosine, especially when they dance together as shown below.

Simultaneous Sine And Cosine

I’ve also always wondered if there is a name for the leaf-like shape formed between them

Shape Between

(And while we’re on the subject, check out this lovely animation I found of sine and cosine graphed side by side.)

Sine And Cosine

Source

Parabolic Reflections

Olly / / Math Posts

Today I spent all of my study time investigating the parabola problem from last week. There are still details to be worked out, but I’m fairly sure the approach I’m using will allow me to prove that the angles I conjectured might be equal actually are. I thought today I’d give a better explanation of what my conjecture was.

A well known principle in physics, called the Law of Reflection, is that when light reflects from a flat surface, the angle of incidence (that is, the angle at which the light hits the surface) is equal to the angle of reflection (that is, the angle at which the reflected light leaves the surface). This is shown in the diagram below.

Angles Of Incidence And Reflection 1

A well known property of parabolas is that, if light is emitted from the focus of the parabola ($F$ in the diagram below), then it will reflect from the surface of the parabola in a direction parallel to the parabola’s orientation (vertical, in the diagram). This is why parabolic mirrors are used in car headlights to reflect all light outward from a bulb at the focus.1

Reflecting Parabola

My question is basically how this property of parabolas relates to the Law of Reflection. It is not obvious how one would measure the angle at which light hits or reflects from a curved surface. What would that mean? One possibility is to measure the angle between the path of the light and the line tangent to the curve at the point of reflection. (In the diagram, the point of reflection is called $P$ and the blue line is tangent to the parabola at $P$.2)

My conjecture is that the a generalization of the Law of Reflection holds true for light emitted from the focus of the parabola and reflected at $P$, if one interprets the angles of incidence and reflection as the angles formed with the line tangent to the parabola at $P$. That is, I conjectured that the angle between the blue tangent line in the diagram and the vertical line through point $P$ is the same as the angle between the blue tangent line and the segment $\overline{PF}$.

  1. Moving in the other direction, if light hits the surface of the parabola from a direction parallel to the parabola’s orientation, it will reflect along a line that passes through the focus of the parabola. This is why parabolic mirrors are used in power generation to concentrate all incoming light on a collector a the focus. ↩︎
  2. The line tangent to a parabola at a point intersects the parabola at that point and no other, and its slope is sometimes described as the slope of the parabola at that point. ↩︎

Rubber Ball

Olly / / Math Posts

I was still very tired for most of today, but I rallied in the late afternoon and managed to do 90 minutes of trig review and 30 of investigation into the parabola problem. I think I’m making progress, though I haven’t started trying to work out all the algebra and trig yet.

I will probably need some inverse trigonometric identities. I was not certain such things existed, but apparently they do. Most of the exercises I did today concerned the regular kind: either proving them or using them to turn one expression into another. I remember doing a lot of the latter kind of problem in my tenth-grade precalculus class. I always thought they were kind of fun.

A Slow Day

Olly / / Updates

Today I felt almost normal again. I also felt terribly weary. I nearly didn’t do any math at all, but pushed myself to read a few chapters of The Joy of X after dinner. It would be a shame to have kept up with my daily study through all the physical and mental turbulence brought on by the change in my medication, only to falter when things started to smooth out again.

The Joy of X is all right. It deals shallowly with a wide range of topics, and I often find myself wishing for more substance. It is useful on days when I am not up to other types of review, however, and it has also pointed out some interesting avenues of inquiry.

Unfocus Meets Focus

Olly / / Math Posts

Today was rocky, though I am feeling a little better now. My math activity for the day was more reading in The Joy of X. The section I read included discussion of an imagined elliptical pool table on which a ball shot from one focus will always fall into a hole at the other focus. Since I know that pool balls (at least ideal ones) bounce off the cushion at the same angle at which they hit it, that had me wondering whether line segments like those shown below in red must always form equal angles with the line tangent to the ellipse at the point where they intersect. (The tangent line is shown in blue.) This seems likely, but I’m not sure how I would go about investigating it.1

Ellipse With Tangent

Naturally, I then wondered a similar thing about parabolas, so useful for reflecting and focusing light, which behaves in the same way as ideal pool balls.

Parabola With Tangent Two

(Sorry the explanation of these questions is a bit sketchy. I know it may be difficult for those without a lot of math background to understand, when this is a topic that should be accessible to them. This is the best I can do at the moment, though.)

  1. To do. ↩︎

More Digit Sums

Olly / / Math Posts

Today I read more about trigonometry, both in the appendix of my calculus textbook and in OpenStax’s Algebra and Trigonometry. (The calc book does not discuss the Unit Circle, the Law of Sines, the Law of Cosines, or for some mysterious reason, the half-angle trig identities.) I also started on the calc book’s review exercises. Depending on how I feel after finishing those, I may do some of the exercises from Algebra and Trigonometry. It has a lot of applied problems that look as if they might be interesting.

I thought that today I would share the other fruit of my investigation of digit sums. I actually proved this proposition before the one from yesterday, but I didn’t have the energy to write it up. I originally found it in my notebook stated for base 10 only, but it occurred to me that it applied to other bases, as well. (Other than that, it’s not all that interesting.)

Proposition: In base $n>1$, if $m$ is a natural number then the digit sum of $m$ is less than or equal to $m$.

If $m$ is a natural number, then $m=n^md_m+n^{m-1}d_{m-1}+…+nd_1+d_0$ where $d_m, d_{m-1},…,d_1, d_0$ are the digits, in order, of $m$ expressed in base $n$.

The digit sum of $m$, expressed in base $n$, is $d_m+d_{m-1}+…+d_1+d_0$.

Since $n>1$, $d_p<n^pd_p$ for all $p\in\mathbb{N}$.

It follows that $d_m+d_{m-1}+…+d_1+d_0\leq n^md_m+n^{m-1}d_{m-1}+…+nd_1+d_0\text{.}$

Therefore, the digits sum of $m$ is less then or equal to $m$.

Elegance and Regret

Olly / / Math Posts

I didn’t do any trigonometry today, but instead read more of Steven Strogatz’s The Joy of X and worked on the propositions about digit sums that I found in my notebook.

The most interesting tidbit from The Joy of X was Strogatz’s comment regarding a particular proof of the Pythagorean Theorem that, “The proof does far more than convince; it illuminates. That’s what makes it elegant.” That caught my attention and had me wondering whether all elegant proofs must be illuminating. I haven’t come to a conclusion yet.

To be illuminating is certainly not the only requirement for an elegant proof. The main proposition I worked on today was “the sum of the digits of any multiple of 9 is also a multiple of 9.” I came up with the outline of a proof by induction, the meat of which was a description of what happens to the digit sum when you add 9 to any natural number. It was full of cases and loops and was decidedly not elegant. I do think it gave a good sense of what was happening, though, and it could have been reworked to prove the general theorem that, in base $n$, the digit sum of any multiple of $n-1$ is also a multiple of $n-1$.

I was sure there must be a better proof, however, and I ended up searching for guidance on the Internet. I regret that, as I would rather have discovered the answer I found for myself. I did get to generalize it, at least, since it applied only to two-digit numbers. I’m not sure I would say that the result is illuminating. I don’t think it reveals why the theorem is true, except in as much as “why” is that 9 is one less than the base of the number system being used. It is simple, though. Does that make it elegant?

Proposition: For all $n\in\mathbb{N}$, the digit sum of $9n$ is a multiple of 9.

When $n$ is a natural number, $9n = 10^md_m+10^{m-1}d_{m-1}+…+10d_1+d_0$, where $d_m, d_{m-1},…,d_1, d_0$ are the digits of $9n$, in order.

$$10^md_m+10^{m-1}d_{m-1}+…+10d_1+d_0=$$ $$\left[\left(10^m-1\right)d_m+\left(10^{m-1}-1\right)d_{m-1}+…+(10-1)d_1\right]+\left[d_m+d_{m-1}+…+d_1+d_0\right]$$

Thus, $$d_m+d_{m-1}+…+d_1+d_0=9n-\left[\left(10^m-1\right)d_m+\left(10^{m-1}-1\right)d_{m-1}+…+(10-1)d_1\right]\text{.}$$

Note that the left-hand side of the equation is the digit sum of $9n$.

Note also that, for any natural number $x$, $$10^x-1=\sum_{i=0}^{x-1}10^i(9)=9\sum_{i=0}^{x-1}10^i\text{.}$$

[For instance, $10^3-1=1000-1=999=9(111)$.]

Hence, because each term of the expression is a multiple of 9, $$9n-\left[\left(10^m-1\right)d_m+\left(10^{m-1}-1\right)d_{m-1}+…+(10-1)d_1\right]=9p$$ for some integer $p$.

It follows that the digit sum of $9n$ is a multiple of 9 for all $n\in\mathbb{N}$.

This proof could also be rewritten to cover all bases. In addition, the same argument could be used to prove that, in base 10, the digit sum of any multiple of three is also a multiple of three.

(And it occurs to me as I write this that you could also use it to prove the analogous fact, in base $n$, for any factor of $n-1$. That’s very cool and makes me feel like I have come up with something worthwhile on my own today.)

Extra Extra

Olly / / Math Posts

Sometimes, during my review, my mind will supply facts before they are mentioned or even in contexts where they never are mentioned. Other times, I will read something an think, “Wait, did I ever know that?”

Today I read about trigonometry, both in the appendix to my calculus text and in the OpenStax free textbook Algebra and Trigonometry. It turns out that, when an angle’s vertex is located at the center of a circle, the measure of that angle in radians is the ratio between the length of the arc subtended by the angle and the radius of the circle. That’s not just a property of radians; that’s what a radian is. Surely I must have known that at some point, but it came as total news to me.

Radians 1

(I’m feeling better today, though still not very good. The fun math revelations were helpful. All that I did remember about radians now makes a lot more sense.)