Exercises and Entertainments

The exercises are an important part of the course. Do not expect to be able to do all of them the first time you try them, but you should understand them after they have been discussed in class. Some important theorems will be proved in the exercises. There are hints for some of the questions in appendix A, but you should not look for a hint unless you have made some effort to answer a question.

Sections whose titles are marked by an asterisk (e.g. section 2.6) are not used later in the notes, and may be omitted. Hovever they contain really neat material, so you will not want to omit them.

In addition to the exercises, there are some questions and statements with the label ``entertainment''. These are for people who find them entertaining. They require more time and thought than the exercises. Some of them are more metaphysical than mathematical, and some of them require the use of a computer or a programmable calculator. If you do not find the entertainments entertaining, you may ignore them. Here is one to start you off.

0.1   Entertainment (Calculation of $\pi$.) . The area of a circle of radius 1 is denoted by $\pi$. Calculate $\pi$ as accurately as you can.

Archimedes showed that $\pi$ is half of the circumference of a circle of radius 1. More precisely, he showed that the area of a circle is equal to the area of a triangle whose base is equal to the circumference of the circle, and whose altitude is equal to the radius of the circle. If we take a circle of radius 1, we get the result stated.

You should assume Archimedes' theorem, and then entertainment 0.1 is equivalent to the problem of calculating the circumference of a circle as accurately as you can. An answer to this problem will be a pair of rational numbers $b$ and $c$, together with an argument that $b < \pi$ and $\pi < c$. It is desired to make the difference $c - b$ as small as possible.

This problem is very old. The Rhind Papyrus[16, page 92] (c. 1800 B.C.?) contains the following rule for finding the area of a circle:

RULE I: Divide the diameter of the circle into nine equal parts, and form a square whose side is equal to eight of the parts. Then the area of the square is equal to the area of the circle.

The early Babylonians (1800-1600BC) [38, pages 47 and 51] gave the following rule:

RULE II: The area of a circle is 5/60th of the square of the circumference of the circle.

Archimedes (287-212 B.C.) proved that the circumference of a circle is three times the diameter plus a part smaller than one seventh of the diameter, but greater than 10/71 of the diameter[3, page 134]. In fact, by using only elementary geometry, he gave a method by which $\pi$ can be calculated to any degree of accuracy by someone who can calculate square roots to any degree of accuracy. We do not know how Archimedes calculated square roots, but people have tried to figure out what method he used by the form of his approximations. For example he says with no justification that

$${265 \over 153} < \sqrt{3} < {1351 \over 780}$$

and

$$\sqrt{3380929} < 1838{9\over 11}.$$

By using your calculator you can easily verify that these results are correct. Presumably when you calculate $\pi$ you will use a calculator or computer to estimate any square roots you need. This immediately suggests a new problem.

0.2   Entertainment (Square root problem.) Write, or at least describe, a computer program that will calculate square roots to a good deal of accuracy. This program should use only the standard arithmetic operations and the constructions available in all computer languages, and should not use any special functions like square roots or logarithms. An answer to this question must include some sort of explanation of why the method works.

Zu Chongzhi (429-500 A.D.) stated that $\pi$ is between 3.1415926 and 3.1415927, and gave 355/113 as a good approximation to $\pi$.[47, page 82]

Here is a first approximation to $\pi$. Consider a circle of radius 1 with center at $(0,0)$, and inscribe inside of it a square $ABCD$ of side $s$ with vertices at $(1,0),(0,1),(-1,0)$ and $(0,-1).$ Then by the Pythagorean theorem, $s^2 = 1^2 + 1^2 = 2$. But $s^2$ is the area of the square $ABCD$, and since $ABCD$ is contained inside of the circle we have

$$2 = \mbox{Area of inscribed square} < \mbox{Area of circle} = \pi.$$

Consider also the circumscribed square $WXYZ$ with horizontal and vertical sides. This square has side 2, and hence has area 4. Thus, since the circle is contained in $WXYZ$,

$$\pi = \mbox{area of circle} < \mbox{area}(WXYZ) = 4.$$

It now follows that $2 < \pi < 4.$

A number of extraordinary formulas for $\pi$ are given in a recent paper on How to Compute One Billion Digits of Pi[12]. One amazing formula given in this paper is the following result

$${1 \over \pi} = {\sqrt{8} \over 9801} \sum_{n=0}^\infty\frac{(4n)!}{(n!)^4} \frac{ [1103 + 26390n]}{396^{4n}},$$

which is due to S. Ramanujan(1887-1920)[12, p 201,p 215]. The reciprocal of the zeroth term of this sum i.e.

$$\frac{9801}{1103\sqrt{8}}$$

gives a good approximation to $\pi$ (see exercise 0.4).

0.3   Exercise. A The formulas described in RULES I and II above each determine an approximate value for $\pi$. Determine the two approximate values. Explain your reasoning.

0.4   Exercise. Use a calculator to find the value of

$${9801\over 1103\sqrt{8}},$$

and compare this with the correct value of $\pi$, which is $3.14159265358979\ldots$.