$$ \newcommand{\RR}{\mathbb{R}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\EE}{\mathbb{E}} \newcommand{\HH}{\mathbb{H}} \renewcommand{\SS}{\mathbb{S}} \newcommand{\DD}{\mathbb{D}} \newcommand{\pp}{^{\prime\prime}} \newcommand{\p}{^\prime} \newcommand{\proj}{\operatorname{proj}} \newcommand{\area}{\operatorname{area}} \newcommand{\len}{\operatorname{length}} \newcommand{\acc}{\operatorname{acc}} \newcommand{\ang}{\sphericalangle} \newcommand{\map}{\mathrm{map}} \newcommand{\SO}{\operatorname{SO}} \newcommand{\dist}{\operatorname{dist}} \newcommand{\length}{\operatorname{length}} \newcommand{\uppersum}[1]{{\textstyle\sum^+_{#1}}} \newcommand{\lowersum}[1]{{\textstyle\sum^-_{#1}}} \newcommand{\upperint}[1]{{\textstyle\smallint^+_{#1}}} \newcommand{\lowerint}[1]{{\textstyle\smallint^-_{#1}}} \newcommand{\rsum}[1]{{\textstyle\sum_{#1}}} \newcommand{\partitions}[1]{\mathcal{P}_{#1}} \newcommand{\erf}{\operatorname{erf}} \newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}} \newcommand{\smat}[1]{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)} $$

Assignment 2

Problems

Square Roots

Exercise 1 (The Square Root of 3) Read carefully the geometric proof of Theorem thm-sqrt2-irrational, which proves \(\sqrt{2}\) is irrational by showing its impossible to make two integer side-length squares where one has twice the area of the other.

Construct a similar argument showing that it is impossible to find two integer side-length equilateral triangles where one has three times the area of the other.

Hint: try to mimic the argument in the book, but now use the diagram below for inspiration

Archimedes

In Measurement of the Circle, Archimedes measure the circumference of a circle by approximating it with a polygon, and taking th number of sides to infinity. He was right to be careful in his argument: this does not always work!

Exercise 2 (Convergence to the Diagonal) Consider a simpler analog of Archimedes’ situation, where instead of trying to measure a curve using straight lines, we are trying to measure a straight diagonal line using only horizontal and vertical segments. The following sequence of paths converges pointwise to the diagonal of the square, but what happens to the lengths?

If you believed that because this sequence of curves limits to the diagonal, its sequence of lengths must limit to the length of the diagonal, what would you have conjectured the pythagorean theorem to be?

Exercise 3 Use the result of last week’s problem Exercise Inscribing an Equilateral Triangle (that you can inscribe an equilateral triangle with half the side lengths) to produce an alternative proof of Archimedes sum \[\sum_{n=0}^\infty\left(\frac{1}{4}\right)^n=\frac{4}{3}\] By dividing up a triangle instead of a square. Draw some nice pictures (its pretty!)

Exercise 4 Construct an argument in the same spirit as Archimedes’ geometric series to show the following equality: \[\sum_{n=1}^\infty \left(\frac{1}{3}\right)^n=\frac{1}{2}\] Can you cut something iteratively into thirds? It may not be as pretty as Archimedes’, but thats ok!

Fractals

The final two problems involve the Koch Snowflake fractal. In these problems you should still explain why things you are doing are valid geometrically, but you do not need to prove every thing you do from the axioms. We are getting ourselves ready for a calculus mindset!

This shape is the limit of an infinite process, starting at level \(0\) with a single equilateral triangle. To go from one level to the next, every line segment of the previous level is divided into thirds, and the middle third replaced with the other two sides of an equilateral triangle built on that side.

The Koch subdivision rule: replace the middle third of every line segment with the other two sides of an equilateral triangle.

Doing this to every line segment quickly turns the triangle into a spiky snowflake like shape, hence the name. Denote by \(K_n\) the result of the \(n^{th}\) level of this procedure.

The first six stages \(K_0,K_1,K_2,K_3,K_4\) and \(K_5\) of the Koch snowflake procedure. \(K_\infty\) is the fractal itself.

Say the initial triangle at level \(0\) has perimeter \(P\), and area \(A\). Then we can define the numbers \(P_n\) to be the perimeter of the \(n^{th}\) level, and \(A_n\) to be the area of the \(n^{th}\) level..

Exercise 5 (The Koch Snowflake Length) What are the perimeters \(P_1,P_2\) and \(P_3\)? Conjecture (and prove by induction, if you’ve had an intro-to-proofs class) a formula for the perimeter \(P_n\).

Explain why as \(n\to\infty\) this diverges (using the type of reasoning you would give in a calculus course): thus, the Koch snowflake fractal cannot be assigned a length!

Before doing the next problem: ask yourself what happens to the area of an equilateral triangle when you shrink its sides by a factor of 3? Can you draw a diagram (similar to that from last week’s Exercise Inscribing an Equilateral Triangle but larger) to see what the ratio of areas must be?

Exercise 6 (The Koch Snowflake Area) What are the areas \(A_1,A_2\) and \(A_3\) in terms of the original area \(A\)?

Find an infinite series that represents the area of the \(n^{th}\) stage \(A_n\) (if you’ve taken an intro to proofs class or beyond - prove it by induction!). Use calculus reasoning to sum this series and show that while the Koch snowflake does not have a perimeter, it drtoes have a finite area!

Solutions

\(\sqrt{3}\) (Katie Forrest)

Suppose for the sake of contradiction there exists an equilateral triangle with integer side lengths that has an area equal to the sum of the area of 3 smaller equilateral triangles with integer side lengths.

Note that since the total area of the 3 smaller triangles is equal to the area of the big triangle, that means geometrically if we placed these 3 triangles on top of the big one they should fill up all of the big triangle’s area. Unfortunately, only 4 triangles with one fourth the area of the big one will fill up the big triangle with no overlap. (We proved this fact in Homework 1.) Thus, the 3 triangles with one third the area of the big triangle will have some overlapping areas and an empty section.

But since the areas of the triangles equal the big one, that means that the overlapping areas equal the empty area. The area of the purple triangle equals the sum of the orange triangles.

Therefore, we can repeat this geometric process with our new big triangle and smaller triangles. Note, that since an integer minus an integer must be another integer, these new triangles also have integer side lengths.

If we continuously repeat this geometric process, we will gather an infinite sequence of smaller triangles. Due to this, the lengths of these triangles will limit to 0 as they grow smaller and smaller. This means that at some point one of the triangles side lengths will be smaller than 1. This is a contradiction as we have stated before that our triangle side lengths are integers. Therefore, we can’t make an integer sized equilateral triangle with 3 times the area of another equilateral triangle with integer side lengths.

2: Convergence to the Diagonal (Jade Cao)

3: The Series \(\sum_{n=0}^\infty \frac{1}{4^n}\) (Quinn Shapiro)

If we let the area of the large triangle to be 1, then, since all of the inscribed triangles have equal side length and therefore equal area, each inscribed triangle has an area of \(\frac{1}{4}\). If we continue inscribing triangles in the center triangle, we divide the center triangle into four equal parts each time. Then, if we say our Level 0 triangle has area one and our Level 1 triangles have area \(\frac{1}{4}\), then our level n triangles have area \(\frac{1}{4^n}\).

Repeatedly dividing an equilateral triangle into fourths, infinitely many times.

Then, if we sum a triangle at each level starting at level one, we get \(\sum_{n=1}^{\infty} \frac{1}{4^n}\).\

The sum \(\sum\frac{1}{4^n}\) realized by choosing one triangle from each level.

If we do this two more times, we see that \(\sum_{n=1}^{\infty} \frac{1}{4^n}+\sum_{n=1}^{\infty} \frac{1}{4^n}+\sum_{n=1}^{\infty} \frac{1}{4^n}=1\) as this fills the entire area of the Level 0 triangle, which is 1.

Doing this three times fills the triangle, so the sum must be exactly 1/3 the triangle

So \(3\sum_{n=1}^{\infty} \frac{1}{4^n}=1\) which implies \(\sum_{n=1}^{\infty} \frac{1}{4^n}=\frac{1}{3}\). Adding the Level 0 Triangle (one whole triangle), we have \(1+\sum_{n=1}^{\infty} \frac{1}{4^n}=\sum_{n=0}^{\infty} \frac{1}{4^n}=\frac{4}{3}\)

Exercise 4: Freya

5: Koch Snowflake Length (Daniel)