Tag: algebra (basic)

Work Smart

Olly / / Math Posts

Today I finished reading a section of my calculus book and worked on some of the associated exercises. I’ve decided not to skip section exercises entirely, but to be selective with them. It still feels hard to know which exercises will be most instructive, without doing them. Yet trying to evaluate that might be a useful exercise in itself. One thing that is important in tutoring is the ability to look at a problem and quickly outline the steps to its solution in your mind so that you can guide the student.

One of the problems I worked on today was finding the equation for a parabola given three points on that parabola. I made a start on it, but I think a different approach may be needed. In particular, I think the simultaneous equations involved will probably be easier to solve if I couch them in the $y=ax^2+bx+c$ form rather than the $(y-k)=a(x-h)^2$ one (or the $(x-h)^2=4p(y-k)$ one). Tomorrow I’m going to think a bit about how these relate to one another. For instance, where does the $p$, so useful in finding the focus and directrix, end up in $y=ax^2+bx+c$?

Functioning and Functions

Olly / / Math Posts

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

  1. To do. ↩︎

Parabola-a-Go-Go, Part 4

Olly / / Math Posts

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.

A Greater Than 2p Annotated 1

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

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 $\alpha$.

Therefore, $$\alpha = \arctan\frac{2p}{a}\text{.}$$

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 $B$.

Thus, $$\beta+\alpha=\arctan\frac{4ap}{a^2-4p^2}$$$$\beta=\arctan\frac{4ap}{a^2-4p^2}-\alpha=\arctan\frac{4ap}{a^2-4p^2}-\arctan\frac{2p}{a}\text{.}$$

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.

A Less Than 2p Annotated 2

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.

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.

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. ↩︎

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. ↩︎

Rational-a-Rama

Olly / / Math Posts

Today I will be sharing my proof that the decimal representations of rational numbers must repeat. This proof is based on a simple idea, but turned out to be pretty involved, with several parts.

The fundamental idea can be illustrated using long division. Consider the rational number $\frac{8}{7}$. We can find its decimal representation using the long division calculation below.

Long Division Remainders

Each step of this calculation begins with the remainder from the previous step (circled in red). Since the remainder on division by 7 must be an integer between 0 and 6, there are only 7 possible remainders. Thus, the remainder must repeat within no more than 8 steps. Once it repeats, the calculation will begin to cycle, causing the decimal representation it produces to repeat, as this one does.

Now let’s prepare for the more formal proof by proving two lemmas. I think the first of these is quite interesting and not immediately obvious. The second one is obvious, but I wanted to be assured of it, since it is vital to the main proof.

Lemma 1: Given a rational number $x$, if there exist two distinct digits $d_1$ and $d_2$ in the decimal representation of $x$ such that the sequence of digits following $d_1$ is the same as the sequence of digits following $d_2$, then the decimal representation of $x$ repeats.

Assume that two such digits exist and let the one which appears first be called $d_1$. Then the decimal representation of $x$ begins as follows: $s_1.s_2d_1s_3d_2$, where $s_1$ is a sequence of at least one digit and $s_2$ and $s_3$ are sequences of zero or more digits.

Now, since the sequence of digits following $d_2$ is the same as the sequence of digits following $d_1$, the decimal representation must continue: $s_1.s_2d_1s_3d_2s_3d_2$. Yet this further specifies the sequence of digits following $d_1$, so the same digits must appear again in the sequence following the initial appearance of $d_2$, yielding $s_1.s_2d_1s_3d_2s_3d_2s_3d_2$. This process (illustrated below) will continue infinitely, so $x=s_1.s_2d_1\overline{s_3d_2}$.

Thus, the lemma holds.

Two Digits With Sequences

(I realize the notation was a little sloppy here, with $d_2$ representing both a digit and a digit with its position, but I couldn’t find a good way to fix it.)

Lemma 2: Let $x$ be a non-negative rational number such that $x=\frac{n}{m}$, where $n$ and $m$ are non-negative integers and $m\neq 0$. There is no more than one way to represent $x$ as a mixed number $q\frac{r}{m}$ such that $q$ and $r$ are non-negative integers and $r<m$.

Suppose $\frac{n}{m}=q_1\frac{r_1}{m}$ and $\frac{n}{m}=q_2\frac{r_2}{m}$ where $q_1$, $q_2$, $r_1$, and $r_2$ are non-negative integers and $r_1,r_2<m$.

It follows that $q_1+\frac{r_1}{m}=q_2+\frac{r_2}{m}$ and thence that $q_1-q_2=\frac{r_1-r_2}{m}$.

Since $0<r_1<m$ and $0<r_2<m$, $-1<\frac{r_1-r_2}{m}<1$.

Thus $-1<q_1-q_2<1$, from which it follows that $q_1=q_2$, since $q_1$ and $q_2$ are integers.

Hence $\frac{r_1-r_2}{m}=0$, so $r_1=r_2$.

Therefore, both mixed numbers are the same, and the theorem holds.

Finally, we come to the main theorem.

Proposition: If $x$ is a rational number, then the decimal representation of $x$ repeats.

For any $x$, if the decimal representation of $x$ repeats, then the decimal representation of $-x$ repeats. Thus, it is sufficient to show that the theorem holds for all non-negative $x$.

Let $x$ be a non-negative rational number. Then $x=\frac{n}{m}$ for some non-negative integers $n$ and $m$, where $m\neq 0$.

The fraction $\frac{n}{m}$ can be expressed as a mixed number $q_1\frac{r_1}{m}$ such that $q_1$ and $r_1$ are non-negative integers and $r_1<m$. By Lemma 2, there is only one way to do this.

Using algebra we have: $$q_1\frac{r_1}{m}=q_1+\frac{r_1}{m}=q_1+\frac{r_1}{m}\left(\frac{10}{10}\right)=q_1+\frac{10r_1}{m}\left(\frac{1}{10}\right)$$

As above, the fraction $\frac{10r_1}{m}$ can be expressed uniquely as a mixed number $q_2\frac{r_2}{m}$ such that $q_2$ and $r_2$ are non-negative integers and $r_2<m$.

This yields $$q_1+\frac{10r_1}{m}\left(\frac{1}{10}\right)=q_1+q_2\left(\frac{1}{10}\right)+\frac{r_2}{m}\left(\frac{1}{10}\right)\text{.}$$

Notice that $q_2$ has only one digit, since $0\leq\frac{r_1}{m}<1$ and $0\leq q_2\leq\frac{10r_1}{m}$. Notice also that $0\leq\frac{r_2}{m}\left(\frac{1}{10}\right)<\frac{1}{10}$, since $0\leq\frac{r_2}{m}<1$. This means that $q_2$ is the tenths-place digit of the decimal expansion of $x$.

The algorithm can be continued as shown below, $$q_1+q_2\left(\frac{1}{10}\right)+\frac{r_2}{m}\left(\frac{1}{10}\right)=q_1+q_2\left(\frac{1}{10}\right)+\frac{r_2}{m}\left(\frac{1}{10}\right)\left(\frac{10}{10}\right)$$ $$=q_1+q_2\left(\frac{1}{10}\right)+\frac{10r_2}{m}\left(\frac{1}{100}\right)=q_1+q_2\left(\frac{1}{10}\right)+q_3\left(\frac{1}{100}\right)+\frac{r_3}{m}\left(\frac{1}{100}\right)$$ $$=q_1+q_2\left(\frac{1}{10}\right)+q_3\left(\frac{1}{100}\right)+\frac{r_3}{m}\left(\frac{1}{100}\right)\left(\frac{10}{10}\right)=q_1+q_2\left(\frac{1}{10}\right)+q_3\left(\frac{1}{100}\right)+q_4\left(\frac{1}{1000}\right)+\frac{r_4}{m}\left(\frac{1}{1000}\right)…$$ with $q_i\frac{r_i}{m}$ for $i>1$ always the unique mixed number representation of the fraction $\frac{10r_{i-1}}{m}$ such that $q_i$ and $r_i$ are non-negative integers and $r_i<m$. By logic similar to that above, when $i>1$, $q_i$ will always be a digit of the decimal representation of $x$, with its value given by the power of $\frac{1}{10}$ by which it is multiplied.

Since, for all $i$, $r_i$ is a non-negative integer less than $m$, there are only $m$ possible values of $r_i$. This means that, among the infinite values of $r_i$, there must exist distinct $r_j$ and $r_k$ such that $r_j=r_k$. (ETA: By the Pigeonhole Principle!)

Since the mixed number representations used by the algorithm are unique, if $r_j=r_k$, then $q_{j+1}=q_{k+1}$ and $r_{j+1}=r_{k+1}$. By induction, it therefore follows that $q_{j+h}=q_{k+h}$ for all $h$. Since, when $i>1$, each $q_i$ is a digit of the decimal representation of $x$, it follows by Lemma 1 that decimal representation of $x$ repeats.

Therefore the theorem holds.

I hope this makes some measure of sense. I had a lot of trouble making this proof lucid, and eventually I had spent too much time working on. Those whose eyes did not glaze over will note that the algorithm I employed here is essentially long division. In developing it, I did not set out specifically to reproduce the long division algorithm, and I was intrigued when I noticed they are basically the same. Here is our long division problem annotated to show the relationship:

Long Division

A Week Off

Olly / / Updates

Hooray! Today, with a battery of hyperbolas, I finished the last of the repetitive graphing exercises from the review of conics.

I was busy with a lot of other tasks, as well, because I’m leaving tomorrow on a trip with my family. I hope to do some math while I am gone, but I probably won’t be updating the blog, as I am not taking my laptop. Expect me back on Monday, 12 February.