proof – The Renewal Equation

Less Fun with Logarithms

Olly / May 13, 2024August 19, 2024 / Math Posts

Today I did more exercises and reading in my calculus book. I also proved the converse of the proposition from last week’s post Fun with Logarithms. You can see the proof below. As I expected, all that was required was a little algebra. Finally, I listened to some of the BBC radio series A Brief History of Mathematics.

It was struggling to listen to that series at the beginning of this year that inspired me to try doing math with the aid of CBD gummies (as described in Take Drugs, Do Math). The series is intended for a broad audience and is not intellectually demanding. Yet even a short time listening to it left me completely exhausted. That served as strong evidence that the overwhelming difficulties I was still having when I tried to do math were at least partly emotional, as opposed to cognitive. Before, it had never been so clear. The new certainty, combined with an overall improvement in my mental health around that time, led me to seek new ways to overcome the emotional block.

Today I was able to enjoy the series with no ill effects, even without CBD gummies, which I am now using infrequently.

Proposition: Given two natural numbers $a$ and $b$ that are both greater than $1$, if both $a$ and $b$ are powers of a natural number $c$, then $\log_a b$ is rational.

Suppose that $a$ and $b$ are natural numbers greater than $1$ that are both powers of a natural number $c$.

Then, by definition, $a=c^m$ and $b=c^n$ for some natural numbers $m$ and $n$.

Since $a\neq 1$, it follows that $m\neq 0$.

And since $m\neq0$, $a^{\frac{n}{m}}=(c^m)^{\frac{n}{m}}=c^n=b$.

Thus $\log_a b = \frac{n}{m}$, which is rational, and the proposition is proven.

Fun with Logarithms

Olly / May 9, 2024August 19, 2024 / Math Posts

My concentration is better today, and I spent three and a half hours ironing out the wrinkles in my logarithm proof. I’ve had the general outline written down for some time, but there turned out to be a lot of details that needed attention, so that I think I’m justified in counting the effort as my study for the day, even though I don’t usually include time spent blogging.

The proof uses several supporting facts that I’ve listed below as Notions 1-3. I haven’t decided yet whether I want to prove them, but I may at some point.¹ I’ve also stated and used the Fundamental Theorem of Arithmetic without proving it. The course I took my first semester at college in which we developed the tools to prove that theorem, was one of the turning points in my life. I can no longer remember how it’s done, however.

I’m not entirely happy with this proof. I think it’s sound, but unclear in places. I had fun making it, though, and I hope someone enjoys it.

Definition: Two integers are relatively prime if they have no common factors besides $1$.

Notion 1: Every rational number can be written as the ratio of two integers that are relatively prime and at least one of which is positive.

Notion 2: If $a>1$, then $a^n>1$ if and only if $n$ is positive.

Notion 3: If $a$ divides $bc$ and $a$ and $b$ are relatively prime, then $a$ divides $c$.

The Fun(damental) Theorem of Arithmetic: Every integer greater then $1$ can be expressed as a product of primes in exactly one way, ignoring order.

Proposition: Given two natural numbers $a$ and $b$ that are both greater than $1$, if $\log_a b$ is rational, then both $a$ and $b$ are powers of a natural number $c$.

Suppose that $a$ and $b$ are natural numbers greater than $1$ and that $\log_a b$ is rational.

Then, using Notion 1, there exist integers $n$ and $m$ such that $a^{\frac{n}{m}}=b$, $n$ and $m$ are relatively prime, and $n$ or $m$ is positive.

A bit of algebra yields $a^n=b^m$.

Since both $a$ and $b$ are greater than $1$ and $n$ or $m$ is positive, the equal expressions $a^n$ and $b^m$ are both greater than $1$ by Notion 2. It follows, again by Notion 2, that both $n$ and $m$ are positive.

Consequently, because the integers are closed under multiplication, the value $a^n=b^m$ is an integer.

Let it be called $d$.

By the Fundamental Theorem of Arithmetic, $a$, $b$, and $d$ can all be expressed as a product of primes in exactly one way.

If the prime factorization of $a$ is $(a_1)(a_2)…(a_i)$, with some values possibly equal, then since $d=a^n$, the prime factorization of $d$ is $(a_1)^n(a_2)^n…(a_i)^n$.

Likewise, if the prime factorization of $b$ is $(b_1)(b_2)…(b_j)$, then since $d=b^m$, the prime factorization of $d$ is $(b_1)^m(b_2)^m…(b_j)^m$.

Notice that the prime factorization of $d$ contains the same factors as the prime factorization of $a$ and that the prime factorization of $d$ also contains the same factors as the prime factorization of $b$. It follows that the prime factorizations of $a$ and $b$ contain the same factors as one another, as well, though not necessarily the same number of times.

[This relationship between $a$ and $b$ is one I found very early on. I thought there was probably a name for it, but inquiries on Math Stack Exchange only turned up the notion of a radical, which is the product of one copy of every one of an integer’s prime factors. In those terms, what I’m saying is that $a$ and $b$ have the same radical.]

Now consider $p$, an arbitrary prime factor of $a$, $b$, and $d$. Let the number of occurrences of $p$ in the prime factorizations of each number be called $O_{a,p}$, $O_{b,p}$, and $O_{d,p}$, respectively.

Because an exponent represents repeated multiplication by the same value, it multiplies the number of occurrences of each factor. Thus, the relationships among $a$, $b$, and $d$ mean that $(O_{a,p})(n)=O_{d,p}=(O_{b,p})(m)$.

By Notion 3, since $(O_{a,p})(n)=(O_{b,p})(m)$ and $n$ and $m$ are relatively prime, $m$ divides $O_{a,p}$ and $n$ divides $O_{b,p}$. Thus, for the arbitrary prime factor $p$, $\frac{O_{a,p}}{m}=\frac{O_{b,p}}{n}$ is an integer. Furthermore, since all values involved are positive, it is a positive integer.

Therefore, it is possible to construct a natural number $c$ such that, for each prime factor $p$ in the prime factorizations of $a$ and $b$, the prime factorization of $c$ contains $\frac{O_{a,p}}{m}=\frac{O_{b,p}}{n}$ occurrences of $p$.

Now, by construction, the prime factorization of $c$ contains the same factors as the prime factorization of $a$.

Furthermore, by the reasoning used above, the prime factorization of $c^m$ contains the same factors as the prime factorization of $c$. Thus the prime factorization of $c^m$ also contains the same factors as the prime factorization of $a$.

Since an exponent multiplies the number of occurrences of each factor, given any prime factor $p$ in the prime factorization of $a$, the number of occurrences of $p$ in the prime factorization of $c^m$ is $\frac{O_{a,p}}{m}(m)=O_{a,p}$.

Recall that this is the number of occurrences of $p$ in the prime factorization of $a$. Hence $c^m$ has the same prime factors as $a$ with the same number of occurrences for each factor. It follows that $c^m=a$.

The same reasoning can be used to show that $c^n=b$, so the proposition is proven.

[Phew.]

To do. ↩︎

The Proof at Last

Olly / April 25, 2024August 19, 2024 / Math Posts

More low-level depression and concentration problems today, but I did do a bit of watching and reading. I also finished writing up the proof below, which I know you’ve all be waiting for. As I explained yesterday, I’ve tried to make it a bit more readable than other proofs I’ve posted recently.

Proposition: If $\triangle ABC$ is an equilateral triangle inscribed in a circle and $D$ is a point on the circle between $A$ and $C$, then $AD+CD=BD$.

Place a point $E$ on $\overline{BD}$ so that the distance from $D$ to $E$ is the same as the distance from $D$ to $C$, as shown below.

Draw the segment $\overline {CE}$.

Since the original triangle, $\triangle ABC$, is equilateral, its three angles must be equal by the proposition from the post Easy Lemmas.

The sum of the measures of the three angles of a triangle is $\pi$ radians. Thus, when the three angles are equal, each one must measure $\frac{\pi}{3}$. It follows that the angles of $\triangle ABC$ have this measure: $\angle BAC=\angle ABC=\angle ACB=\frac{\pi}{3}$.

Because they are both subtended by arc $BC$, by the second proposition from the post Angles and Arcs, $\angle BAC$, the leftmost of the angles of the original equilateral triangle, is equal to $\angle BDC$, the rightmost of the angles with their verticies at $D$.

Similarly, because they are both subtended by arc $AB$, $\angle ACB$, the rightmost of the angles of the original equilateral triangle, is equal to $\angle ADB$, the leftmost of the angles with their verticies at $D$.

It follows that the two angles with their verticies at $D$, $\angle BDC$ and $\angle ADB$, both measure $\frac{\pi}{3}$.

Consider the newly formed triangle $\triangle CDE$. Two of its sides, $\overline{CD}$ and $\overline{DE}$, are equal by construction (i.e. because of the way point $E$ was chosen). Thus, the two angles subtended by those sides, $\angle CED$ and $\angle DCE$, are equal to one another by the lemma from the post Revenge of the Squares, Part 1.

Furthermore, since $\angle CDE$, the other angle of $\triangle CDE$, is one of the angles that measures $\frac{\pi}{3}$, it follows that $\angle CED$ and $\angle DCE$ together must measure $\pi-\frac{\pi}{3}=\frac{2\pi}{3}$ and that each one individually must measure $\frac{\pi}{3}$.

Hence the three angles of $\triangle CDE$ are all equal, from which it follows, by the proposition in the post Easy Lemmas, that $\triangle CDE$ is equilateral.

Now consider $\angle BEC$, the angle supplementary to $\angle CED$, the uppermost of the angles of $\triangle CDE$. Since the angles of $\triangle CDE$ all measure $\frac{\pi}{3}$, the supplementary angle $\angle BEC$, must measure $\frac{2\pi}{3}$.

Note that, since each one measures $\frac{\pi}{3}$, the sum of $\angle ADB$ and $\angle BDC$, the two equal angles with their verticies at $D$, is also $\frac{2\pi}{3}$. Thus, the measure of $\angle ADC$, the large angle they form, is also $\frac{2\pi}{3}$.

Yet that means that the two triangles highlighted above in yellow, $\triangle BCE$ and $\triangle ACD$, are congruent by the side-side-angle property, which applies when the angle in question is greater than or equal to $\frac{\pi}{2}$. The triangles’ large angles, $\angle BEC$ and $\angle ACD$, are equal, as just shown, and measure $\frac{2\pi}{3}$; their long sides, $\overline {BC}$ and $\overline {AC}$, are equal, since both are sides of the original equilateral triangle $\triangle ABC$; and another pair of their sides, $\overline {CE}$ and $\overline {CD}$, are equal because both are sides of the new, smaller equilateral triangle $\triangle CDE$.

It follows that the remaining sides, $\overline {BE}$ and $\overline{AD}$, are equal, as well. Hence $BE=AD$.

As above, $DE=CD$ by construction.

Thus $AD+CD=BE+DE$. Yet $\overline {BE}$ and $\overline {DE}$ together make up the larger segment $\overline {BD}$. Therefore $AD+CD=BD$ and the theorem of is proven.

With the addition of SSA for right and obtuse angles, I think I might have a complete set of triangle congruence properties to prove.¹

I could almost have done this proof without assigning measures to the angles. Equality between $\angle BEC$ and $\angle ADC$ could be established using the property of triangles illustrated below, and I think I could have shown that they are obtuse as well. Without angle measures, though, I couldn’t think of a way to show that, with two sides of $\triangle CDE$ equal by construction, not only must the angles they subtended be equal to one another, but both must be equal to the third angle.

Let me know what you thought of this verbose-mode proof. I found it a little difficult to write without an audience clearly in mind.

To do. ↩︎

Angles and Arcs

Olly / April 23, 2024August 19, 2024 / Math Posts

My back is somewhat better today. I spent an hour on exercises in Analysis with an Introduction to Proof and I plan to read some more of Infinite Powers this evening, as well.

I’m also ready to prove the proposition about inscribed angles from my post Groundwork. It turns out that proposition falls right out of another that reader Tim McL suggested starting with when I said I knew little about the geometry of circles.

Proposition: If a central angle and an inscribed angle are subtended by the same arc of a circle, then the measure of the central angle is twice that of the inscribed angle.

Let $A$ and $B$ be points on a circle with center $D$.

Draw the central angle subtended by arc $AB$.

Draw an inscribed angle also subtended by arc $AB$ and let its vertex be called $C$.

Join $\overline {CD}$.

Now suppose that the sides of the two angles intersect only at $A$ and $B$ (as in the figure above).

Since $\overline{BD}$ and $\overline{CD}$ are both radii of the circle, $BD=CD$, from which it follows that $\angle CBD = \angle BCD$ (as I’ve proven previously).

Similarly, $\overline{AD}$ and $\overline{CD}$ are both radii of the circle, so $AD=CD$ and $\angle CAD = \angle ACD$.

Futhermore, since the sum of the angles of a triangle is $\pi$ radians, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.

It is also true that $\angle BCD + \angle ADC + \angle ADB = 2\pi$.¹

Combining these equations will yield $\angle ADB = 2(\angle BCD + \angle ACD)$.

But $\angle ACB = \angle BCD + \angle ACD$, so $\angle ADB = 2\angle ACB$, and the theorem is proven for this case.

Suppose instead that sides of the angles intersect other than at $A$ and $B$, as shown above. (Note that this argument does apply in the edge case where $\overline{BC}$ intersects $\overline{AD}$ at $D$.)

Then, $\angle ACB =\angle ACD-\angle BCD$ and $\angle ADB = \angle BDC – \angle ADC$.

As above, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.

Combining these equations will yield $\angle ADB = 2(\angle ACD – \angle BCD)$.

Hence $\angle ADB = 2\angle ACB$, and the theorem is proven for this case, as well.

After proving this proposition, I looked up how Euclid did it. His solution was largely the same. It was slightly more elegant, using the property of triangles illustrated below to cut out some of the algebra, but not so elegant as to dispense with the cases.

Proposition: If two inscribed angles are subtended by the same arc of a circle, then they are equal.

I’m just going to sketch a proof of this. By the proposition above, the blue and green inscribed angles subtended by arc $AB$ both have half the measure of the red central angle subtended by the same arc. Therefore, they must be equal to each other.

Footnotes

Do I need to prove this? I have an idea of how I would do it, but it seem so fundamental. ↩︎

Easy Lemmas

Olly / April 20, 2024August 19, 2024 / Math Posts

Today I worked on exercises in Analysis with an Introduction to Proof and did a little mathematical puttering.

I’m also ready today to start sharing my work on the proposition about the inscribed equilateral triangle. I’ll start by proving that a triangle is equilateral if and only if its three angles are equal.

First, recall from my post Revenge of the Squares, Part 1 that the angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal. The proof of the following related lemma is very similar.

Lemma: If two angles of a triangle are equal, then the sides that subtend them are also equal.

Consider the triangle $\triangle ABC$ with $\angle ACB\cong\angle ABC$.

Draw a line bisecting $\angle BAC$ and extend it to intersect $\overline{BC}$. Let the point of intersection be called $D$.

Now, by the angle-angle-side property, $\triangle ACD\cong\triangle ABD$, since $\angle ACB\cong\angle ABC$, $\angle CAD\cong\angle BAD$, and $\overline{AD}$ is shared.

Therefore, $AC=AB$. $\overline {AC}$ and $\overline {AB}$ are the sides that subtend the equal angles $\angle ACB$ and $\angle ABC$, so the lemma is proven.

(Note that I have added AAS to my list of triangle congruence properties to prove.¹)

Proposition: A triangle is equilateral if and only if its three angles are equal.

Consider a triangle $\triangle ABC$.

Assume that $\triangle ABC$ is equilateral.

Then, since $AB=AC$, $\angle ACB\cong\angle ABC$, by the theorem cited above. Similarly, since $AC=BC$, $\angle ABC\cong\angle BAC$.

It follows that $\angle ACB\cong\angle ABC\cong\angle BAC$.

Thus, if $\triangle ABC$ is equilateral, then its three angles are equal.

Assume instead that the three angles of $\triangle ABC$ are equal.

Then, by the lemma, since $\angle ACB\cong\angle ABC$, $AB=AC$. Similarly, since $\angle ABC\cong\angle BAC$, $AC=BC$.

It follows that $AB=AC=BC$.

Thus, if the three angles of $\triangle ABC$ are equal, then it is equilateral.

This proves the equivalence.

Stay tuned tomorrow for some circle geometry.

To do. ↩︎

Parabola-a-Go-Go, Part 4

Olly / March 10, 2024August 19, 2024 / 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.

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.

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 3

Olly / March 9, 2024August 19, 2024 / 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 $h>0$.

—

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.

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.

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 $a+b=2(\frac{\pi}{4})+\pi=\frac{3\pi}{2}=-\frac{\pi}{2}$

Parabola-a-Go-Go, Part 2

Olly / March 4, 2024August 19, 2024 / 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.

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.

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 / March 3, 2024August 19, 2024 / 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.

The relationships can bee seen in the figure below.

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.

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.

More Digit Sums

Olly / February 22, 2024August 19, 2024 / 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 $m$.