Tag: to-do

Puzzling Parameters

Olly / / Math Posts

I was tired today and struggled to concentrate. I only spent an hour on textbook exercises, even though I had hoped to finish the section on parametric curves. That will have to happen tomorrow. I also spent time on some experiments with parametric curves that I started yesterday. My thoughts about them are too numerous and disorganized for me to write them up here, but I have decided to share the Desmos notebook I’ve been using, to which I’ve added a bit of commentary. I hope some of you may find it interesting. You can view it here.1

  1. To do. ↩︎

Wild Functions

Olly / / Math Posts

Today went well. I’m nearly finished with the exercises for the section I’m working on.

One of the exercises I did today reminded me what a walled garden I am playing in when working from a textbook. The functions encountered there are always tame ones susceptible to the techniques being taught. Today one of the exercises gave me a glimpse of a wild function, though.

$f(x)=3+x^2+\tan(\frac{\pi x}{2})$ for $-1<x<1$

I was asked to find $f^{-1}(3)$. It’s not too difficult to find that particular value, but I could find no way to do it by the first method I tried: by deriving a formula for the inverse, $f^{-1}(x)$. I don’t know how to solve $x=3+y^2+\tan(\frac{\pi y}{2})$ for $y$. That’s what makes this a wild function. It cannot be bidden the way most of the functions used in textbook exercises can.

(I believe the function does have an inverse, though I would like to come up with a better test for that than I have.1 Right now, I have the word of the textbook and an examination of the graph, which you can see here and which does not show any two $x$ values with the same $y$ value.)

  1. To do. ↩︎

Olly Returns

Olly / / Math Posts

Hello again, faithful readers. I spent today’s study time reading the next section of my calculus book and investigating some of the ideas it presented. In particular, I was interested in what pairs of functions, $f(x)$ and $g(x)$, have the property that $f(g(x))=g(f(x))$.1 I made some minor discoveries, but I am too tired this evening to write them up. Perhaps tomorrow.

  1. To do. ↩︎

Parabola Magic

Olly / / Math Posts

I worked on more calculus exercises today, then did some experimenting with graphs using Desmos. What I discovered is that adding a line to a parabola will always produce a congruent parabola. So far, I haven’t been able to find any other type of curve that has this property. For all other curves, adding a line also adds “tiltiness” to the graph, so that the result is not congruent to the original.

Here is an example using a parabola. The original parabola appears in green, the line in red, and their sum in blue. Click here for an interactive version of this figure.

Parabolic Function

We know that these parabolas are congruent because the coefficient of the $x^2$ term does not change. (It is called $a$ in the interactive graph.) This coefficient is equal to $\frac{1}{4p}$, where $p$ is the distance of the vertex of the parabola from both the focus and the directrix. It is this distance that determines the curvature of the parabola.

(To see that $a$ does equal $\frac{1}{4p}$, try converting the general equation for a parabola $(x-h)^2=4p(y-k)$ into the general equation $y=ax^2+bx+c$. That’s another thing I did today.)

Below is a gallery of some of the other curves I tried. Again, the original graphs are in green, the line in red, and their sum in blue. You can see how each result has greater tiltiness than the original in one way or another.

I haven’t yet been able to explain why parabolas have this congruence property while other types of curve do not nor tried to understand what other implications it may have, but I find it magical.1

  • Cubic Function
    Cubic function
  • Quartic Function
    Quartic function
  • Root Function
    Root function
  • Elliptical Function 1
    Ellipse
  • Hyperbolic Function 1
    Hyperbola
  • Rational Function
    Rational function
  • Exponential Function
    Exponential function
  • Trig Function
    Trig function
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Fun with Logarithms

Olly / / 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.]

  1. To do. ↩︎

Sphere versus Cube

Olly / / Math Posts

Today I was able to spend more than an hour on calculus exercises. (Yay!) Now I am pondering the following:

The volume of a sphere of radius $r$ is given by $V=\frac{4}{3}\pi r^3$. The surface area of the same sphere is given by $A=4\pi r^2$, the derivative of the formula for the volume. This makes sense, as a small change in volume involves adding a thin layer to the surface of the sphere. In the limit, that layer will be identical to the sphere’s surface.

On the other hand, the surface area of a cube of side length $x$ is given by $A=6x^2$. This is not the derivative of $V=x^3$, the formula for the cube’s volume. Why?1

  1. To do. ↩︎

Trying

Olly / / Math Posts

I didn’t feel very well today, either. I read some of Infinite Powers and worked on my logarithm problem. I am not ready to present a proof, but I have satisfied myself that if $a$ and $b$ are natural numbers greater than $1$, then they must be powers of the same base in order for $\log_a b$ to be rational. I’m not sure how to begin thinking about other types of numbers, and it’s possible I will lay that question aside for now.1

I am still not making much progress on my calculus review. It’s an important part of this project, because I hope it will prepare me to earn some money tutoring. The truth is, though, that a lot of the time I am just not hale of mind. I’m doing my best.

  1. To do. ↩︎

Return to the Triangle of Doom

Olly / / Math Posts

I felt better today. I worked for about an hour on exercises in Analysis with an Introduction to Proof and for another half hour on a question about logarithms that occurred to me while doing one of the exercises. I also spent quite a bit of time on a puzzle I call the Triangle of Doom. (It is more widely known as the Hardest Easy Geometry Problem.) I first encountered and worked on it in 2006, but before today I hadn’t looked at it in a long time.

The puzzle is to find the the value of $x$ in the figure below without using trigonometry.1 (Image source)

Doom Plain

I didn’t solve the Triangle of Doom today, but I had some ideas.

The measures of many of the angles are easy to find using familiar geometry: vertical angles, supplementary angles, and the fact that the angles of a triangle sum to 180 degrees.

Doom With Angles Measures

Adding the fact that if two angles of a triangle are equal, then the sides that subtend them are also equal, reveals that the whole triangle is isosceles and that it also contains two smaller isosceles triangles, as highlighted below in blue and green. I suspect these isosceles triangles will have some role in the solution.

Doom Isosceles Triangles 1

Something I’m not sure the significance of is that there also appear to be two similar triangles in the figure, as highlighted below in yellow. I cannot yet prove that these are similar—that would be to solve the puzzle—but tests using the Desmos geometry tool left me nearly sure. (You can check out my interactive drawing here for however long the link lasts.) The relationship might be a coincidence, though, since it is not preserved when three isosceles triangles overlap in the same way but with different angle measures.

Doom Similar Triangles
  1. To do. ↩︎

The Proof at Last

Olly / / 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.

Bare Figure

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 $\overline {CE}$.

Screenshot
Annotated Figure Alt

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 $\frac{2\pi}{3}$.

Figure With Highlighting

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.

Triangle Property

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.

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