Category: Math Posts

Parabola-a-Go-Go, Part 3

Olly / / Math Posts

This is my 50th daily post! Thanks for reading, everyone.

Today I watched the second and third episodes of the Lockdown Math series by 3blue1brown. They are the first two parts of a three-part subseries dedicated to the connections between trigonometry and complex numbers.

This evening I am finally ready to continue my exposition of the parabola problem. Below are two inverse trigonometric identities needed for the proof of the two remaining cases. I found these in an online table of identities, but I proved them myself.

Lemma: When $xy<1$, $$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}\text{,}$$ and when $xy>1$, $$\arctan x + \arctan y = \pi + \arctan \frac{x+y}{1-xy}\text{.}$$

Proof: Let us take as given the sum formula for tangent that is found in every trigonometry textbook: $$\tan (a+b)=\frac{\tan a + \tan b}{1-\tan a\tan b}\text{ when }\tan a \tan b\neq 1\text{.}$$

Using $\arctan x$ and $\arctan y$ as the two angles being added, it follows that $$\tan(\arctan x + \arctan y)=\frac{\tan(\arctan x)+\tan(\arctan y)}{1-\tan(\arctan x)\tan(\arctan y)}=\frac{x+y}{1-xy}\text{ when }xy\neq 1\text{.}$$

According to the definitions of the tangent, sine, and cosine functions, this implies that, when $xy\neq 1$, $\sin(\arctan x + \arctan y)=\frac{x+y}{h}$ and $\cos(\arctan x + \arctan y)=\frac{1-xy}{h}$ for some $h>0$.

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Range Of Artangent

Now consider the range of the arctangent function, shown in red above. It dictates that $\frac{-\pi}{2}<\arctan x<\frac{\pi}{2}$ and $\frac{-\pi}{2}<\arctan y<\frac{\pi}{2}$ and therefore that $-\pi<\arctan x+\arctan y<\pi$.

Given any non-zero tangent $t$, there are two angles on the interval $(-\pi,\pi)$ with tangent $t$, $\theta$ and $\theta +\pi$, as shown below. Only one of these, however, is within the range of the arctangent function, and it is that one that will be found when the arctangent of $t$ is taken.

Angles With Same Tangent

Since $\cos(\arctan x + \arctan y)=\frac{1-xy}{h}$ for some $h>0$, $\cos(\arctan x+\arctan y)$ is positive when $xy<1$ and $\cos(\arctan x+\arctan y)$ is negative when $xy>1$. Notice that a positive cosine corresponds to being within the range of the arctangent function, while a negative cosine corresponds to being outside it.

It follows that, when $xy<1$, $\arctan x+\arctan y$ will be the angle found by taking the arctangent of $\frac{x+y}{1-xy}$, while when $xy>1$, the angle found will be $(\arctan x+\arctan y)-\pi$.

Thus, when $xy<1$, $$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}\text{,}$$ and when $xy>1$, $$\arctan x + \arctan y = \pi + \arctan \frac{x+y}{1-xy}\text{.}$$

Addendum: Apropos of the identity at the beginning of this proof, note that when $\tan a \tan b=1$ is when $a+b=\pm\frac{\pi}{2}$, the points at which tangent is undefined.

When $\tan a \tan b=1$, $\tan a$ and $\tan b$ are reciprocals. Since tangent can be interpreted as the slope of the terminal side of an angle, this means that the terminal sides of $a$ and $b$ are either the reflections of one another across the line $x=1$ or else a terminal side and the “other end” of its reflection, which is the reflection rotated by $\pi$. The diagram below shows the two cases.

Reciprocal Pairs 2

That they sum to $\pm\frac{\pi}{2}$ can be shown algebraically:
$a-\frac{\pi}{4}=\frac{\pi}{4}-b$ or $a-\frac{\pi}{4}=(\frac{\pi}{4}-b)+\pi$
$a+b=2(\frac{\pi}{4})=\frac{\pi}{2}$ or $a+b=2(\frac{\pi}{4})+\pi=\frac{3\pi}{2}=-\frac{\pi}{2}$

Computer-Assisted Post

Olly / / Math Posts

Today I spent an hour on trigonometry exercises and another half hour reading The Mathematical Tourist by Ivars Peterson, which I’m considering as my next light reading book. It’s an explanation for the lay reader of some of the developments in mathematics current in the mid 1980s. Its first section is about the proof of the four-color theorem in 1976, which was controversial because it was one of the first proofs to be partly generated by computer. Proofs of this kind are less controversial now; the proof of the ternary Goldbach conjecture that I mentioned a few weeks ago is another example. A little outside reading suggests there is still some dissent, though, regarding whether proofs that cannot be fully verified by a human being should be considered valid.

Repairs are Ongoing

Olly / / Math Posts

The depressive episode I started yesterday persists, but has been much less severe today. This afternoon I finished the last few chapters of The Joy of X without too much struggle. Now I will have to find a new book for my light math reading. I have several options in my library already, but I would welcome recommendations, as well.

The final chapter of The Joy of X was about infinity and included a treatment of Cantor’s diagonal argument that the set of real numbers is uncountable. (I’m sorry, non-specialist readers, but I’m not up to explaining what that is right now, though it is fairly accessible.) The Joy of X did not account in any way for the fact that some numbers have more than one decimal representation (e.g. that $0.1\overline{0}=0.0\overline{9}$). I’m sure a more formal presentation of the argument would have. I intend to think for a while about how I would do it before looking up the accepted method. Not today, though.1

  1. To do. ↩︎

Parabola-a-Go-Go, Part 2

Olly / / Math Posts

Today has been a very busy day—I think I did every errand known to man—and I haven’t fit in any study time yet. I give my readers my solemn word to at least read some of The Joy of X before bed, however.

Today I am proving the proposition below for the case $a=2p$. The case $a=0$ was dealt with yesterday, and the cases $0<a<2p$ and $a>2p$ will be addressed tomorrow or later in the week.

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 $C$.

First, note that the equation of the parabola can be written as $y=\frac{1}{4p}x^2$.

Now suppose $a=2p$.

Then, $R$ is the point $(2p, p)$, and $B$ is the horizontal line $y=p$.

Using calculus, $y’=(\frac{1}{4p})2x=\frac{1}{2p}x$. Thus, the slope of $C$, the line tangent to the parabola at $R=(2p, p)$, is $(\frac{1}{2p})(2p)=1$.

It follows that $C$ intersects $A$, $B$, and the x-axis as show in the diagram below.

A Equals 2p Annotated 1

In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$.

Note that $\alpha$ is equal to the angle that is vertical to $\alpha$.

Since $B$ is horizontal, it is parallel to the x-axis. Thus, $\beta$ is equal to the alternate interior angle to $\beta$ that is formed by $C$ and the x-axis.

Because $A$ is perpendicular to the x-axis, the triangle formed by $A$, $C$, and the x-axis is a right triangle. Hence the ratio of the legs must be equal to the slope of $C$, which is 1. It follows that the legs are equal.

Thus, the angle vertical to $\alpha$ and the alternate interior angle to $\beta$ are also equal, because they subtend the equal sides of an isosceles triangle.

Therefore, $\alpha=\beta$ and the proposition holds when $a=2p$.

(Fun!)

Parabola-a-Go-Go, Part 1

Olly / / Math Posts

Today I will begin sharing the results of my investigation into reflections in parabolas. For a refresher on the problem, see my post Parabolic Reflections.

First of all, I have chosen to investigate only upward-facing parabolas with their verticies at the origin. If the conjecture is true for those parabolas, then it is true for parabolas in general, because the lines involved in the problem will form the same angles if translated, rotated, or reflected.

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 $C$.

The relationships can bee seen in the figure below.

A Equals 2p

I found it necessary to split my proof into cases, depending on the value of $a$. The figure above represents the case $a=2p$. In that case, the slope of $B$ is zero. The other relevant cases are $a>2p$, where the slope of $B$ is positive, $0<a<2p$, where the slope of $B$ is negative, and $a=0$, where the slope of $B$ is undefined. The last of these cases is pictured below. As you can see, the proposition is trivially true when $a=0$ because $A$ and $B$ are the same line.

A Equals 0

I will not discuss the cases where $a<0$. Because a parabola is symmetrical, it is sufficient to prove that the proposition holds for one half of it. (I did, in fact, consider those cases. I initially did the proof for the left side of the parabola instead of the right side. The math for the right side is slightly simpler, though, so that is what I will be presenting here.)

Tune in tomorrow for the proof for the case $a=2p$. It’s quite simple if you know that the slope of the line tangent to the parabola at $x$ is $\frac{x}{2p}$. Readers might enjoy seeing if they can work it out for themselves.

Working, Please Wait

Olly / / Math Posts

I spent both some wakeful time last night and all of my study time today working on my parabola problem. It’s turning out to be an ideal project for this point in my review, as I’m using all the subjects I’ve been learning—algebra, analytic geometry, conics, and trigonometry—plus a bit of remembered calculus.

One thing I did today was some research on inverse trigonometric functions. In the process, I discovered a resource called LibreTexts, which offers a largish selection of free math textbooks. I found it useful to look at a couple of different treatments of the same topic.

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.