CHAPTER 8
Reggie’s Problem
Let’s say that while you’re at a party with friends, the con- versation turns to what everyone has been reading lately. When you confess that you’ve been delving into math—imaginary numbers, trig, infinite sums, and the like—and to your surprise found it more understandable and interesting than you’d expected, the hosts’ very bright but socially inept son, a 14-year-old math whiz named Reggie, is suddenly all ears. Excitedly butting into the conversation, he exclaims that he’s been learning about exactly the same things and wonders if you’d like to try working an “easy” problem on them from his favorite book, Basic Math for the Complete Genius.
Before you can beg off, he whips out a piece of paper and scribbles the following scary-looking function consisting of an infinite sum:
f(θ) = (i cos θ)2/2 + (i cos θ)4/4 + (i cos θ)8/8 + (i cos θ)16/16 + ….
“Here’s the problem,” he says. “What’s this function’s value when the variable equals pi?
“It took me two minutes to figure it out,” he adds brightly, oblivious to the blush spreading across your face. “It’s really simple. Even Jake—he’s my math club friend who isn’t as good as me at solving problems—figured it out pretty fast.”
Thinking to yourself, “What an annoying little dweeb,” you say, “Oh, I’m sure I’m not as good at math as you and Jake. But I’ll take a crack at it later. Thanks for sharing it with me.”
Picking up the piece of paper before tucking it into your pocket, you make a little show of perusing what Reggie has written so as not to seem dismissive of your hosts’ son. Then suddenly you have a flash: when you substitute π for θ in the infinite sum, every cos θ will become cos π, which is equal to −1. (In fact, while reading this book’s trig chapter recently, you learned that cos π = −1.) That means all the cosines will go away, leaving behind unscary −1’s. In addition, you recall that i2 is equal to −1 (since i is defined as the square root of −1), and so whenever “i times i” appears in any of the fractions, it can also be replaced by −1.
So you think, “Maybe this isn’t as bad as it looks.” Instead of pocketing the problem, you find yourself asking Reggie to lend you a pencil so you can sit down to fiddle with it. Everyone is astonished, including you. But then, you never did like doing what people expected you to do—beneath your facade of polite good humor lurks a wild thing. (Why else would you be reading this book?)
You start by evaluating the infinite sum’s first θ-containing term, (i cos θ)2/2, when θ is set equal to π. The i cos θ part represents i multiplied times cos θ. Thus, after plugging π in for θ, you write, using ×’s for multiplication,
(i cos π)2 = (i × cos π) × (i × cos π) [by the definition of exponents]
= (i × −1) × (i × −1) [by subbing in −1 for cos π]
= i × i × −1 × −1 [rearranging by the commutative and associative laws]
= i2 × 1 [definition of exponents, and enemy of my enemy = friend]
= −1 × 1 [i is defined as the square root of −1, and so i2 = −1]
= −1 [the enemy of my friend is someone I feel negative about].
You’re on your way: Since you’ve shown that the numerator of the fraction (i cos π)2/2 is equal to −1, you’ve proved that the fraction is just an absurdly complicated way of writing −1/2.
Now you fearlessly attack the numerator of the second term, (i cos π)4/4:
(i cos π)4 = (i × cos π) × (i × cos π) × (i × cos π) × (i × cos π)
= (i × cos π)2 × (i × cos π)2 [definition of exponents]
= −1 × −1 [since, as shown above, (i × cos π)2 = −1]
= 1.
Which means that (i cos π)4/4 equals 1/4.
After a moment’s thought, you realize that the numerator of every subsequent fraction is of the form (i × cos π), with an even exponent larger than 4, meaning that it can be reduced to an even number of terms of the form (i × cos π)2 multiplied together—this follows from the same logic used above to simplify (i cos π)4. Those terms, in turn, can be reduced to pairs of −1’s multiplied together, and since each such pair is equal to 1, what you’ve got in each numerator is just 1’s multiplied together, or simply 1. For instance,
(i cos π)8 = (i × cos π)2 × (i × cos π)2 × (i × cos π)2 × (i × cos π)2
= −1 × −1 × −1 × −1
= 1 × 1
= 1.
Thus, (i cos π)8/8 = 1/8.
And so on.
Putting it all together, you write Reggie’s original equation with π plugged in for θ,
f(π) = (i cos π)2/2 + (i cos π)4/4 + (i cos π)8/8 + (i cos π)16/16 + …
and then below that write the partial solution to the problem that you’ve now figured out:
f(π) = −1/2 + 1/4 + 1/8 + 1/16 + ….
Looking on, Reggie says in his piercing way, “That’s not the answer. You have to figure out what the infinite series is equal to.”
“Thanks for pointing that out,” you reply, miraculously sounding as if you mean it. By this time, your hosts and other friends are crowded around to see what you’re doing. Obviously you can’t stop now.
Fortunately, you’re on a roll: You recall running across an infinite sum of fractions that looked a lot like the one you’re now facing. Let’s see, when was that? Then it comes to you—it was the time you read about Zeno’s paradox.*
How did it go? Oh yes—an ancient Greek philosopher named Zeno pictured a runner on a racetrack. First, the runner went half of the way to the finish line, then half of the remaining distance (equal to a fourth of the way), then half of the remainder after that (an eighth of the way), and so on. It appeared that he couldn’t possibly get to the finish given the infinite number of segments he had to complete.
But it seemed there was a flaw in Zeno’s argument. The Greek philosopher implied that it should take an infinite amount of time for the runner to complete the race because he had to cross an infinite number of racetrack segments, each of which would have required some amount of time to negotiate. But Zeno apparently wasn’t aware of the fact that an infinite sum of fractions can be finite if the successive fractions dwindle toward zero. In fact, the sum of racetrack segments that the runner had to cross (1/2 + 1/4 + 1/8 + …) equals 1, not infinity. Your source on Zeno had shown you how to demonstrate that by drawing a square, 1 unit on a side, divided into subparts—as shown in Figure 8.1.
FIGURE 8.1
The total area of this square is 1, since the area of any rectangle equals length times width, and both the length and width are both 1 in this case. Now, when half of its area is added to half of the remaining area, which is one-fourth of the square, and then half of the area remaining after that (which is one-eighth) is added to the two previously summed areas, and so on, the sum of areas will get ever closer to the total area of the square. In fact, the difference between the sum and the total area can be made ever smaller by adding enough fractional areas to the sum. In other words, the difference can be reduced to a reasonable facsimile of zero, which means that the infinite sum and the total area are actually equal.
This implies that one-half plus one-fourth plus one-eighth and so on forever add up to one. (And that means, by the way, that if the runner goes at a constant speed of one racetrack length per x minutes, it will take him exactly x minutes to cross the infinite sum of fractions to get to the finish, since they add up to one racetrack length.)
Now nothing can stop you. You recreate the square on the piece of paper for everyone to see. Then you triumphantly exclaim, “The areas of all the subparts of this square add up to the total area of the square, which we know is 1. This shows that
1/2 + 1/4 + 1/8 + … = 1.
“If we add −1 to both sides of this last equation, we get
−1 + 1/2 + 1/4 + 1/8 + … = −1 + 1
and since the −1 + 1/2 on equation’s left side is equal to −1/2, and the −1 + 1 on its right side is equal to 0, we can rewrite the equation as
−1/2 + 1/4 + 1/8 + … = 0.
“Remember that we earlier found out that f(π) is equal to an infinite sum of fractions:
f(π) = −1/2 + 1/4 + 1/8 + ….
“Notice that this infinite sum is the same infinite sum that we just proved is equal to 0. In other words, f(π) equals an infinite sum that is equal to zero, and so f(π) itself is equal to zero. Thus, the answer to Reggie’s problem is
f(π) = 0.
“There you go, Reggie. QED and all that. You were right, it isn’t very hard.”
* Zeno, the infinity-invoking puzzle poser mentioned in Chapter 3, came up with several related paradoxes that have stirred debate for about 2,500 years. Some mathematicians and philosophers have argued that the paradoxes are based on naïve or faulty premises and thus aren’t truly paradoxical. But others have argued that purported solutions to the paradoxes have merely glossed over deep issues they raise. In any case, the lack of universal agreement about how to deal with Zeno’s conundrums suggests that their power to perplex still hasn’t been definitively quashed.