When Least Is Best: How Mathematicians Discovered Many Clever Ways to Make Things as Small (or as Large) as Possible - Softcover

Nahin, Paul J.

Softcover

ISBN 10: 0691130523 ISBN 13: 9780691130521

Publisher: Princeton University Press, 2007

Synopsis

What is the best way to photograph a speeding bullet? Why does light move through glass in the least amount of time possible? How can lost hikers find their way out of a forest? What will rainbows look like in the future? Why do soap bubbles have a shape that gives them the least area?


By combining the mathematical history of extrema with contemporary examples, Paul J. Nahin answers these intriguing questions and more in this engaging and witty volume. He shows how life often works at the extremes--with values becoming as small (or as large) as possible--and how mathematicians over the centuries have struggled to calculate these problems of minima and maxima. From medieval writings to the development of modern calculus to the current field of optimization, Nahin tells the story of Dido's problem, Fermat and Descartes, Torricelli, Bishop Berkeley, Goldschmidt, and more. Along the way, he explores how to build the shortest bridge possible between two towns, how to shop for garbage bags, how to vary speed during a race, and how to make the perfect basketball shot.


Written in a conversational tone and requiring only an early undergraduate level of mathematical knowledge, When Least Is Best is full of fascinating examples and ready-to-try-at-home experiments. This is the first book on optimization written for a wide audience, and math enthusiasts of all backgrounds will delight in its lively topics.

About this title

About the Author

Paul J. Nahin is Professor Emeritus of Electrical Engineering at the University of New Hampshire. He is the author of many books, including the bestselling "An Imaginary Tale: The Story of the Square Root of Minus One", "Duelling Idiots and Other Probability Puzzlers", and "Dr. Euler's Fabulous Formula: Cures Many Mathematical Ills" (all Princeton).

From the Back Cover

"This is a delightful account of how the concepts of maxima, minima, and differentiation evolved with time. The level of mathematical sophistication is neither abstract nor superficial and it should appeal to a wide audience."--Ali H. Sayed, University of California, Los Angeles

"When Least Is Best is an illustrative historical walk through optimization problems as solved by mathematicians and scientists. Although many of us associate solving optimization with calculus, Paul J. Nahin shows here that many key problems were posed and solved long before calculus was developed."--Mary Ann B. Freeman, Math Team Development Manager, Mathworks

Excerpt. © Reprinted by permission. All rights reserved.

When Least Is Best

How Mathematics Discovered Many Clever Ways to Make Things as Small (or as Large) as PossibleBy Paul J. Nahin

Princeton University Press

Copyright © 2003 Princeton University Press
All right reserved.
ISBN: 978-0-691-13052-1

Chapter One

Minimums, Maximums, Derivatives, and Computers

1.1 Introduction

This book has been written from the practical point of view of the engineer, and so you'll see few rigorous proofs on any of the pages that follow. As important as such proofs are in modern mathematics, I make no claims for rigor in this book (plausibility and/or direct computation are the themes here), and if absolute rigor is what you are after, well, you have the wrong book. Sorry!

Why, you may ask, are engineers interested in minimums? That question could be given a very long answer, but instead I'll limit myself to just two illustrations (one serious and one not, perhaps, quite as serious). Consider first the problem of how to construct a gadget that has a fairly short operational lifetime and which, during that lifetime, must perform flawlessly. Short lifetime and low failure probability are, as is often the case in engineering problems, potentially conflicting specifications: the first suggests using low-cost material(s) since the gadget doesn't last very long, but using cheap construction may result in an unacceptable failure rate. (An example from everyday life is the ordinary plastic trash bag-how thick should it be? The bag is soon thrown away, but we definitely will be unhappy if it fails too soon!) The trash bag engineer needs to calculate the minimum thickness that still gives acceptable performance.

For my second example, let me take you back to May 1961, to the morning the astronaut Alan Shepard lay on his back atop the rocket that would make him America's first man in space. He was very brave to be there, as previous unmanned launches of the same type of rocket had shown a disturbing tendency to explode into stupendous fireballs. When asked what he had been thinking just before blastoff, he replied "I was thinking that the whole damn thing had been built by the lowest bidder."

This book is a math history book, and the history of minimums starts centuries before the time of Christ. So, soon, I will be starting at the beginning of our story, thousands of years in the past. But before we climb into our time machine and travel back to those ancient days, there are a few modern technical issues I want to address first.

First, to write a book on minimums might seem to be a bit narrow; why not include maximums, too? Why not write a history of extremas, instead? Well, of course minimums and maximums are indeed certainly intimately connected, since a maximum of y(x) is a minimum of -y(x). To be honest, the reason for the book's title is simply that I couldn't think of one I could use with extrema as catchy as is "When Least Is Best." I did briefly toy with "When Extrema Are xxx" with the xxx replaced with exotic, exciting, and even (for a while, in a temporary fit of marketing madness that I hoped would attract Oprah's attention), erotic. Or even "Minimums Are from Venus, Maximums Are from Mars." But all of those (certainly the last one) are dumb, and so it stayed "When Least Is Best." There will be times, however, when I will discuss maximums, too. And now and then we'll use a computer as well.

For example, consider the problem of finding the maximum value of the rather benign-looking function

y(x) = 3 cos(4[pi]x - 1.3) + 5 cos(2[pi]x + 0.5).

Some students answer too quickly and declare the maximum value is 8, believing that for some value of x the individual maximums of the two cosine terms will add. That is not the case, however, since it is equivalent to saying that there is some x = [bar]x such that

4[pi] [bar]x - 1.3 = 2[pi]n 2[pi] [bar]x + 0.5 = 2[pi]k,

where n and k are integers. That is, those students are assuming there is an [bar]x such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], n and k integers.

Thus,

2n]pi] + 1.3 = 4[pi]k - 1,

or

2.3 = 4[pi]k - 2[pi]n = 2[pi](2k - n),

or

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

But if this is actually so, then as n and k are integers we would have [pi] as the ratio of integers, i.e., [pi] would be a rational number. Since 1761, however, [pi] has been known to be irrational and so there are no integers n and k. And that means there is no [bar]x such that y([bar]x) = 8, and so [y.sub.max](x) < 8.

Well, then, what is [y.sub.max](x)? Is it perhaps close to 8? You might try setting the derivative of y(x) to zero to find [bar]x, but that quickly leads to a mess. (Try it.) The best approach, I think, is to just numerically study y(x) and watch what it does. The result is that [y.sub.max](x) = 5.7811, significantly less than 8. My point in showing you this is twofold. First, a computer is often quite useful in minimum studies (and we will use computers a lot in this book). Second, taking the derivative of something and setting it equal to zero is not always what you have to do when finding the extrema of a function.

An amusing (and perhaps, for people who like to camp, even useful) example of this is provided by the following little puzzle. Imagine that you have been driving for a long time along a straight road that borders an immense, densely wooded area. It looks enticing, and so you park your car on the side of the road and hike into the woods for a mile along a straight line perpendicular to the road. The woods are very dense (you instantly lose sight of the road when you are just one step into the woods), and after a mile you are exhausted. You call it a day and camp overnight. When you get up the next morning, however, you've completely lost your bearings and don't know which direction to go to get back to your car. You could, if you panic, wander around in the woods indefinitely! But there is a way to travel that absolutely guarantees that you will arrive back at your car's precise location after walking a certain maximum distance (it might take even less). How do you walk out of the woods, and what is the maximum distance you would have to walk? The answer requires only simple geometry-if you are stumped the answer is at the end of this chapter.

1.2 When Derivatives Don't Work

Here's another example of a minimization problem for which calculus is not only not required, but in fact seems not to be able to solve. Suppose we have the real line before us (labeled as the x-axis), stretching from -[infinity] to +[infinity]. On this line there are marked n points, labeled in increasing value as [x.sub.1] < [x.sub.2] < ... < [x.sub.n]. Let's assume all the [x.sub.i] are finite (in particular [x.sub.1] and [x.sub.n]), and so the interval of the x-axis that contains all n points is finite in length. Now, somewhere (anywhere) on the finite x-axis we mark one more point (let's call it x). We wish to pick x so that the sum of the distances between x and all of the original points is minimized. That is, we wish to pick x so that

S = |x - [x.sub.1]| + |x - [x.sub.2]| + ... + |x - [x.sub.n]|

is minimized. I've used absolute-value signs on each term to insure each distance is non-negative, independent of where x is, either to the left or to the right of a given [x.sub.i]. Those absolute-value signs may seem to badly complicate matters, but that's not so. Here's why.

First, focus your attention on the two points that mark the ends of the interval, [x.sub.1] and [x.sub.n]. The sum of the distances between x and [x.sub.1], and between x and [x.sub.n], is

|x - [x.sub.1]| + |x - [x.sub.n]|

and this is at least |[x.sub.1] - [x.sub.n]|. If x > [x.sub.n], or if x < [x.sub.1] (i.e., if x is outside the interval), then strict inequality holds, but if x is anywhere inside the interval (i.e., [x.sub.1] [greater than or equal to] x [greater than or equal to] [x.sub.n]) then equality holds. Thus, the minimum value of |x - [x.sub.1]|+|x-[x.sub.n]| is achieved by placing x anywhere between [x.sub.1] and [x.sub.n].

Next, shift your attention to the two points [x.sub.2] and [x.sub.n-1]. We can repeat the above argument, without modification, to conclude that the minimum value of |x - [x.sub.2]| + |x - [x.sub.n-1]| is achieved when x is anywhere between [x.sub.2] and [x.sub.n-1]. Note that this automatically satisfies the condition for minimizing the value of |x - [x.sub.1]| + |x - [x.sub.n]|, i.e., placing x anywhere between [x.sub.2] and [x.sub.n-1] minimizes |x - [x.sub.1]| + |x - [x.sub.2]| + |x - [x.sub.n-1]| + |x - [x.sub.n]|. You can now see that we can repeat this line of reasoning, over and over, to conclude

|x - [x.sub.3]| + |x - [x.sub.n-2]| is minimized by placing x anywhere between [x.sub.3] and [x.sub.n-2],

|x - [x.sub.4]| + |x - [x.sub.n-3]| is minimized by placing x anywhere between [x.sub.4] and [xn.sub.-3],

and finally, if we suppose that n is an even number of points, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

So, we simultaneously satisfy all of these individual minimizations by placing x anywhere between [x.sub.n/2] and [x.sub.(n/2)+1] (if n is even), and this of course minimizes S.

But what if n is odd? Then the same reasoning as for even n still works, until the final step; then there is no second point to pair with [x.sub.(n+1)/2]. Thus, simply let x = [x.sub.(n+1)/2] and so |x - [x.sub.(n+1)/2]| = 0, which is certainly the minimum value for a distance. Thus, we have the somewhat unexpected, noncalculus solution that, for n even, S is minimized by placing x anywhere in an interval, but for n odd there is just one, unique value for x (the middle [x.sub.i]) that minimizes S.

1.3 Using Algebra to Find Minimums

As another elementary but certainly not a trivial example of the claim that derivatives are not always what you want to calculate, consider the fact that ancient mathematicians knew that of all rectangles with a given perimeter it is the square that has the largest area. (This is a special result from a general class of maximum/minimum questions of great historical interest and practical value called isoperimetric problems, and I'll have more to say about them in the next chapter.) Ask most modern students to show this and you will almost surely get back something like the following. Define P to be the given perimeter of a rectangle, with x denoting one of the two side lengths. The other side length is then (P - 2x)/2, and so the area of the rectangle is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

A(x) is maximized by setting d A/dx = 1/2 P -2x equal to zero, and so x = 1/4P, which completes the proof. Using only algebra, however, an ancient mathematician could have argued that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

since (x - [(P/4)).sup.2] [greater than of equal to] 0 for all x. That is, A is never larger than the constant [P.sup.2]/16 and is equal to [P.sup.2]/16 if and only if (a useful phrase I will henceforth write as simply iff) x = P/4, which completes the ancient, noncalculus proof.

As a final comment on this result, which again illustrates the intimate connection between minimum and maximum problems, we can restate matters as follows: of all rectangles with a given area, the square has the smallest perimeter. This is the so-called dual of our original problem and, indeed, all isoperimetric problems come in such pairs. I'll prove this particular dual in section 1.5. Another useful isoperimetric result that seems much like the one just established-one also known to the precalculus, ancient mathematicians-is not so easy to prove: of all the triangles with the same area, the equilateral has the smallest perimeter. See if you can show this (or its dual) before I do it later in this chapter.

We can use the previous result-of all rectangles with a fixed perimeter, the square has the maximum area-to solve without calculus a somewhat more complicated appearing problem found in all calculus textbooks. Suppose we wish to enclose a rectangular plot of land with a fixed length of fencing, with the side of a barn forming one side of the enclosure. How should the fencing now be used? We could, of course, use calculus as follows: let x be the length of each of the two sides perpendicular to the barn wall, and l - 2x be the length of the side parallel to the barn wall (l is the fixed, total length of the fencing). Then the enclosed area is

A = x (l - 2x) = xl - [2x.sup.2]

and so

dA/dx = l - 4x,

which, when set equal to zero, gives x = 1/4 l. Thus, l - 2x = 1/2 l, which says the enclosed area is maximized when it is twice as long as it is wide. But this solution is far more sophisticated than required. Simply imagine that we enclose another rectangular area on the other side of the barn wall. We already know that, together, the two rectangular plots should form a square, and so each of the two rectangular plots are half of the square, i.e., twice as long in one dimension as in the other.

Our ancient mathematician's trick of completing the square is a very old one, and some historians claim that it can be found implicit in Euclid's Elements (Book 6, Proposition 27), circa 300 B.C. There, the problem discussed is equivalent to that of dividing a constant into two parts so that their product is maximum. So, if the constant is ITLITL, then the two parts are x and C - x, with the product

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Thus, as [(x - (ITLITL/2)).sup.2] [greater than or equal to] 0 for all x, then M is never larger than [ITLITL.sup.2]/4 and is equal to [ITLITL.sup.2]/4 iff x = ITLITL/2.

Stated this way, Euclid's problem surely seems rather abstract, but in 1573 the Dutch mathematical physicist Christiaan Huygens gave a nice physical setting to the calculation. Suppose we have a line and two points (A and B) not on the line. Where should the point ITLITL be located on the line so that the sum of the squares of the distances from ITLITL to A and ITLITL to B, [(AC).sup.2] + [(BC).sup.2], is minimum? With no loss in generality we can draw the geometry of this problem as shown in figure 1.1, with A on the y-axis. The figure shows A and B on the same side of the line, and places ITLITL between A and B, but as the analysis continues you'll see that these assumptions in no way affect the result.

In the notation of the figure we are to find the value of x that, with a, b, and c constants, minimizes [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Now,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Thus, we need to minimize the product x(x - b); but we already know from Euclid how to do that-set x = 1/2b. That is, ITLITL is midway between A and B. If you redraw figure 1.1 so that either x > b or x < 0, and then write the expression for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], you'll see that the result is unchanged.

(Continues...)

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Edition details

• Publisher: Princeton University Press
• Publication date: 2007
• ISBN 10: 0691130523
• ISBN 13: 9780691130521
• Binding: Paperback
• Language: English
• Number of pages: 392