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Assignment 1

Exercise 1 (Computing Cube Roots) Can you modify the babylonians procedure which found approximates of \(\sqrt{2}\) to instead find rational approximates of \(\sqrt[3]{2}\)?

Here, instead of starting with a rectangle of sides \(x,y\) let’s start with a three dimensional brick with a square base (sides \(x\) and \(x\)), height \(y\), and area \(2\). Our goal is to find a “closer to cube” shaped brick than this one, and then to iterate. Propose a method of getting “closer to cube-shaped” and carry it out: what are the side lengths of the next shape in terms of \(x\) and \(y\)?

Start with a simple rectangular prism of volume \(2\) and iterate this procedure a couple times to get an approximate value of \(\sqrt[3]{2}\). How close is your approximation?

Exercise 2 (Irrationality) Following analogous logic to that of Pythagoras, prove that \(\sqrt{3}\) is irrational. Generalize this to prove that \(\sqrt{6}\) is irrational.

Exercise 3 (Irrationality, II) Show where the argument you give above for \(\sqrt{6}\) fails if you instead consider \(\sqrt{9}\) (which is rational!)

Definition 1 (The Babylonian Algorithm and Number Theory) Because \(\sqrt{2}\) is irrational, there is no pair of integers \(p,q\) with \(p^2=2q^2\). Good rational approximations to \(\sqrt{2}\) will almost satisfy this equation, so the closer \(p^2\) is to \(2q^2\) the closer \(p/q\) is to \(\sqrt{2}\). If \(p\) and \(q\) are integers, the best we can hope for is these values to be one apart. Thus, we will call \(p/q\) an excellent approximation to \(\sqrt{2}\) if \[p^2=2q^2+1\]

Exercise 4 (The Babylonian Algorithm and Number Theory) Prove that all approximations produced by the babylonian sequence starting from the rectangle with sides \(1\) and \(2\) are excellent, by induction.

In class we looked at Archimedes’ argument for determining the perimeter of a circle by circumscribing it with \(n\)-gons. The idea was that since the \(n\) gons got closer and closer to the circle, their perimeters should get closer and closer to the circle s perimeter, so we could use them to approximate the circumference.

The exercise below shows a way this type of argument can fail! We will look at a sequence of polygons which are converging to a triangle (the difference in area between them and the triangle they approximate is getting smaller and smaller), even though the perimeter of these polygons is not at all approaching the perimeter of the triangle.

Exercise 5 (Convergence to the Diagonal) Consider the right triangle whose legs are length 1. Define a sequence \(T_n\) of polygons approximating this triangle as follows. Let \(T_0\) denote the unit square, and \(T_n\) denote the polygon formed by \(n\) equal “steps” approximating the triangle:

A sequence of right angled polygons converging to a triangle.

• 1. Prove that as \(n\) goes to infinity the area of the polygons \(T_n\) do converge to the area of the triangle (Hint: can you write down a formula for the total error between \(T_n\) and the triangle, as a sum of the areas of small triangles that you can calculate?)
• 2. prove that the length of the zig-zag diagonal side of the \(T_n\) has length \(2\) always, independent of \(n\), and so the perimeter of each polygon \(T_n\) in the sequence is 4. Thus, following the style of argument of archimedes, we should be tempted to conclude the perimeter of the limiting shape is also 4.

This is a problem! The pythagorean theorem tells us that the hypotenuse of the triangle must be \(\sqrt{1^2+1^2}=\sqrt{2}\), so the perimeter should be \(2+\sqrt{2}\). Thus, if we believe both our argument inspired by Archimedes calculation and Pythagoras’ argument, we are forced to conclude that \(4=2+\sqrt{2}\), or \(2=\sqrt{2}\) which is false!

Now we turn to look at the practical details of Archimedes’ and Zu Chongzi’s estimates of \(\pi\). We saw in class there is a nice formula for the perimeter \(P_n\) of the \(n\)-gon circumscribing the circle in terms of trigonometry: \[P_n = 2n\tan\frac{180}{n}\]

But this is useless to us in practice unless we can actually compute the values of the tangent function! Archimedes’ big idea was that he could simplify things alot if he considered doubling a polygons number of sides. That is, the goal is to compute a formula for the number \(P_{2n}\), assuming you already know the number \(P_n\).

Exercise 6 Let \(t_n=\tan(180/n)\).

• 1. Show that \(t_n\) satisfies the recurrence relation \[t_{2n}=\sqrt{1+\frac{1}{t_n^2}}-\frac{1}{t_n}\]

Hint: you’ll need some trig identities reviewed in the text to write everything in terms of tangent!

• 2. Use this to find a recurrence relation for \(P_n\).
     
• 3. Find the perimeter of the square circumscribing the unit circle. Use this and your recurrence formula to find the perimeter of the octagon and 16 gon that circumscribe the unit circle.

Exercise 7 In his argument solving the Basel problem, Euler crucially uses that if we know

• all the zeroes of a function
• the value of that function is 1 at \(x=0\)

then we can factor the function as an infinite polynomial in terms of its zeroes. This implies that a function is completely determined by its value at \(x=0\) and its zeroes (because after all, once you know that information you can just write down a formula like Euler did!) This is absolutely true for all finite polynomials, but it fails spectacularly in general.

Show that this is a serious flaw in Euler’s reasoning by finding a different function that has all the same zeroes as \(\sin(x)/x\) and is equal to \(1\) at zero (in the limit)!