The rudiments of the calculus ought to have been compulsory for me. One ought, after all, to be initiated into the life of the world in which one is going to have to live.
The calculus is one of the characteristic expressions of the modern Western genius.
— philosopher Arnold Toynbee
Calculus has its limits.
— anonymous, probably a calculus student
Two things are clear: First, calculus has a notorious reputation as a difficult subject. And second, it erects a barrier to further study of mathematics for far too many students, leaving our adult community with a serious gap in its understanding of this subject. I have problems with both of those propositions. I believe first that the poor reputation of calculus is undeserved. The rudiments of calculus, those you will meet in this chapter, are, I suggest, not nearly as difficult as are many of the basic concepts of school mathematics. And I believe, sadly, that this barrier to progress is too often erected by insensitive teaching of this important subject. It is my hope then that this chapter will clarify what calculus is about.
Having made those claims, I must add that this chapter should not be considered a substitute for college courses in calculus; rather, it is a brief introduction to the subject’s fundamental concepts. Hopefully it will provide a basis for further study of this important subject for those who will be meeting it for the first time, and an organizing tool for those who have already studied it. As one student who had completed a year and a half of calculus coursework remarked to me after reading a draft, “So that’s what they’ve been talking about.”
I also add, however, that I believe this chapter can stand on its own, serving you even if you study no further calculus. Hopefully you will finish it with a general understanding of this important subject and, while you will not have the techniques for solving derivatives and integrals mastered (or many even addressed), you will know more about where those results come from, what those results mean, and how you can interpret them. This is, I claim, important because, as Toynbee says quite rightly in that quoted passage that began this chapter, “calculus is one of the characteristic expressions of the modern Western genius.”
6.1 Calculus: a perfect title
So nat’ralists observe, a flea Hath smaller fleas that on him prey,
And these have smaller fleas that bite ’em, And so proceed ad infinitum.
— Jonathan Swift
Etymology informs us that the word calculus is derived from the Latin for pebble91 or, to better serve my purposes here, a grain of sand. Think of those tiny particles that get between your toes when you walk along a beach. There are, of course, quite a few of them there. Let’s take a brief detour to consider just how many there really are. Fine sand is usually defined as having an average diameter of less than 0.2 millimeters. That means that you could get, even loosely packed, 5 x 5 x 5 = 125 grains in a single cubic millimeter. Since 1 meter = 1000 millimeters, you have about 125 x 1000 x 1000 x 1000 = 125,000,000,000 grains of sand in a cubic meter. That’s 125 billion and you have only taken a bite of sand slightly larger than a cubic yard. Eight of those cubic meters would give us a trillion grains and you’re already talking about the kind of numbers discussed when considering the United States’ national debt.92
But think about this for a minute. Size is relative. Proportionally, our planet Earth is even less in size relative to the size of the Universe or even a small fraction of our Solar System. Indeed, a quick calculation suggests that you could squeeze the volume of more than ten trillion Earth-sized planets into a sphere whose radius is the distance from the Earth to the Sun.
The point of those two examples is that removal of that grain of sand has virtually no effect on that beach and removal of our Earth would have virtually no effect on the volume of the Universe. You have a situation where you can at least consider the equation:
B + s = B, when s is very small relative to B, but still not zero.
That is, however, not an equation with which you should be comfortable, so mathematicians have a formal way of expressing this idea. They write:
(6.1.1)
which is read: “The limit, as s approaches zero, of B + s is equal to B.”
Why go to all that trouble? There are reasons, as you will soon see. In fact the entire calculus enterprise rests on this idea. It is worth, however, considering a more interesting example before we move on. Suppose you wish to graph this equation:
If you do so with a computer or calculator, you will obtain a result that looks like this:
Figure 6.1.1.
Appearance of the graph of the equation y = 
That graph should be familiar to you. It appears the same as the graph of y = 2 and you might have simplified the equation above to lead you to the same conclusion:
The graph and that series of simplifications may convince you that these two equations are equivalent:
There is, however, a seemingly insignificant but still important difference between them. And that difference occurs when x = 1. If you let x = 1 in the second equation where x doesn’t even appear, you find that y = 2, just as it does for all x values. On the graph this value would be plotted as the point (1, 2). But in the first equation substituting x = 1 and simplifying you have:
and you are in trouble, because one of the basic commandments in mathematics is: “Thou shalt not divide by zero.” The reason you cannot divide by zero is that it
would make our entire mathematical enterprise come crashing down. We know, for example, that 5 x 0 = 0 and that 6 x 0 = 0. Then, since both equal zero, you would have 5 x 0 = 6 x 0. If you allowed division by zero, you would then have 5 = 6. By similar means you could prove all numbers equal, not the kind of thing you want in our system. You might still want to make an exception for 0/0: since they are alike, why not have 0/0 = 1? But then that equation 5 x 0 = 0 would be “equivalent to” 5 = 0/0 and you would have 5 = 1 as well.
Thus, that graph that appeared in Figure 6.1.1 is just like the graph of y = 2, except that the single point (1,2) is missing.
Of course, an individual point is so small that you would never notice its absence from that line. To stress that the one point is missing, mathematicians often use a device like that of Figure 6.1.2 to show its absence:
Figure 6.1.2.
Modified graph of the equation y = 
Once again, mathematicians get around this problem through the use of limits:
(6.1.2)
You will be given an opportunity to familiarize yourself a bit more with limits by working the exercises, but I must enter a warning here. A few purists among mathematicians would write
but most would simply write
just as you have since you first studied algebra. To make this acceptable, you should join those less formal mathematicians to consider the equal sign as meaning “equal for all values for which the expressions are defined.” Once you accept that approach, you don’t have to add those restrictions as you would for the simple statement tan x = sin x/cos x. If you did not accept that convention, you would have to write that statement as:
The clear warning of all this is: Watch out for situations where the denominator of a fraction can be zero. We will make use of this problem and its solution through the use of limits soon.
Earlier mathematicians treated ideas like this rather cavalierly until an 18th century English cleric, Bishop George Berkeley, tired of having mathematicians criticize religion, struck back with his 1734 book, The Analyst: a Discourse Addressed to an Infidel Mathematician. At the time Berkeley wrote his book, mathematicians were referring to tiny quantities, like that tiny s in B + s = B with which we began this discussion, as infinitesimals, that is, infinitely small quantities. In the bishop’s presentation, he derided this designation, calling each of those infinitesimals “a ghost of a vanishing cipher,” thus relegating them to the world of fairies and witches.
Indeed, mathematicians found themselves reaching unacceptable conclusions when they were not careful in handling ideas like these. For example, long division will show that
and now, since that sum on the right is unending, you have 1/0 = (read as “one divided by zero equals infinity”), which seems even worse than having 0/0 lead to 5 = 6.
Even if you let x = −1 in that equation, you have the unacceptable result:
with the subtotals of right side of the equation jumping back and forth between 1 and 0, neither of which has the same value as the left side.
Mathematicians had to clean up their act and they have done so with great care. Today advanced math students study in an upper division course called Analysis the details of that careful foundation laid under the idea of limits. Too often students are asked to do so in beginning calculus courses as well. So-called €-δ (epsilon-delta) proofs underlie many of these ideas. We will bypass them here in order to retain our focus on the ideas of calculus.
There is just one idea that you need to take away from this discussion. In informal terms, it is that there are mathematical expressions that include particular points that are undefined. We can, however, often deal with those points by considering other nearby points.
Exercises 6.1
(6.1.1) Simplify each of the following and then tell what value of x would not be allowed in the calculation:
(6.1.2) A limit that turns out to be very important in developing calculus relationships is:
(a) If you substitute 0 for x in that fraction, what is the result?
(b) Set your calculator MODE to Radian measure and calculate values for 
for x = 1, .1, .01, .001, .0005 and .0001.
(c) What does this series of values suggest is the value of lim
?
(d) The graph y = sin x appears to produce a continuous line. What point (x,y) does not appear on this graph?
(6.1.3) Here is a “proof” that 2 = 1. I set a = b. Then this series of steps follows:
(a) Give an algebraic reason that justifies each step in this argument.
(b) Surely 2 1= Where then is the error?
(6.1.4) Here is another slightly more complicated proof that 2 = 1. You learned in school algebra (and may have forgotten) that (x − y)2 = x2 − 2xy + y2. If you substitute values in that form, you have two instances: (2−
)2 = 4 − 6 +
and (1−
)2 = 1 − 3 + 
. Those equations are used in the fourth step of the following:
(a) Give an algebraic reason that justifies each step in this argument.
(b) Again, you know that 2 ≠ 1 Where then is the error
(6.1.5) I claimed that you could fit more than a trillion Earths into a sphere centered on the Sun and with the Earth-Sun distance as radius. Check this with a calculation of your own. Some useful information: the volume of a sphere is V =
, the radius of the earth is about 4000 miles and the distance from the Earth to the Sun is about 93 million miles. You will save calculations if you set up the problem as
6.2 Slope and velocity
Much is made in school mathematics of the idea of slope. Rightly so, for it is a very important mathematical concept and calculus builds on it. When you consider an equation like y = 
x + 1, you designate that 
, the coefficient of x, as the slope of the equation. The + 1 that completes the right side of the equation is a kind of starting point at the y-axis. When x = 0, you begin with y = +1, and, since that point is on the y-axis, it is also called the y-intercept.93
Figure 6.2.1.
Graph of the equation
Recall how that slope, a well-chosen designation, tells us about the steepness of the graph of y = 
x + 1. Quite reasonably, a line with zero slope is horizontal, increasing positive slope values cause the graph to tip up steeper and steeper from left to right, and negative values cause it to tip down from left to right.
It is common in mathematics to use the Greek capital letter ∆ (Delta) to represent change, and the slope of a graph is equal to 
, the change in y over the change in In Figure 2.1, the graph of our example, y = 
x + 1, two points are highlighted on the line: (0, 1) and (2, 2). Between those two points ∆y is 1 and ∆x is 2, so 
between them is 
, and indeed that is the slope, m, that you read from the equation itself when it is in the form y = mx + b. It is important to realize that this ratio is true for any pair of points on this line.
This idea is so central to our concerns here that I restate it formally:
For any linear equation, y = mx + b, the slope is m = 
.
Recall from Chapter 2 how equations in which y is a function of x can also be expressed with function notation with f (x) replacing y. Thus in the case of any linear function, you have f (x) = mx + b and for our particular example you have: f (x) = 
x + 1. Notice that you also have such function values as f (0) = 1 and f (2) = 2. It is of particular importance to what follows in the next section to translate our representation of slope as 
into functional notation. Figure 6.2.2 displays that relationship for the graph of y = f (x).
Figure 6.2.2.
Function Notation for points (x, f (x)) and (x + ∆x, f (x + ∆x))
Since y = f (x), you replace y with f (x) where appropriate. Thus the point (x, y) becomes (x, f (x)). This point is located above an arbitrary x-value chosen on the x-axis. A second x-value is designated by moving along the x-axis a distance ∆x (recalling that ∆x means “change in x.”) This new x-value is then x + ∆x. The coordinates of the point on the graph corresponding to this x + ∆x are then (x + ∆x, f (x + ∆x)). Thus the change in y or ∆y is f (x + ∆x) f (x). This gives us the translation of 
that you need for what lies ahead:95
Before continuing, it will be useful to apply what has already been said to a widely used example of a linear equation: the familiar distance formula. You should recall from school mathematics and physics that distance = rate × time or, more appropriately for our purposes, distance = velocity time. You probably also recall the kinds of simple applications: If you drive a car at a velocity 96 of 30 miles per hour for 2 hours, how far would you travel? The answer is, of course, 60 miles. (That’s the kind of question you wish they would ask on exams.) If you use s for distance,97 you have the simple linear equation s = vt.
Compare the two equations, s = vt and y = mx to be sure you see that the roles of s and y are the same, as are the roles of t and x and, most important, v and m. Thus, if you graph the distance equation, with t along the horizontal (more often x) axis, and s along the vertical (usually y) axis, you would have a graph with v as its slope. In this case you have v = 
replacing m = 
and we have
which is exactly the same form as m = 
. This comparison may seem trivial but too many otherwise reasonable people miss it and think that they are dealing with entirely different ideas. Of course, linear equations arise in a great many other settings as well.
Warwick Sawyer, an English mathematician who had an important influence on my own studies of this subject, had this to say about the role of velocity in the history of calculus:
Calculus [began] with an apparently simple and harmless question, “What is speed and how can you calculate it?” This question arose very naturally round about the year 1600 A.D.,98 when all kinds of moving objects – from planets to pendulums – were being studied. Men were then just starting to study the material world intensively. From that study the modern world has developed, with the knowledge of stars and atoms, of machines and genes, that we have today, for good and for ill. One might have expected the study of speed to have very limited applications – to machinery, to falling objects, to the movements of the heavenly bodies. But it has not been so. Practically every development in science and mathematics, from 1600 to 1900, was connected with calculus. From this single root, in a most unexpected way, knowledge grew out in all directions. You find calculus applied to the theory of gravitation, heat, light, sound, electricity, magnetism; to the flow of water and the design of airplanes. Calculus enabled Maxwell to predict radio twenty years before any physicist could demonstrate radio experimentally; calculus still plays a vital role in Einstein’s theory of 1916 and in the atomic theories of the nineteen twenties. Apart from these, and many other applications in science, calculus stimulates the appearance of interesting new branches of pure mathematics. No person intending to study mathematics seriously could possibly leave calculus out.
Exercises 6.2
(6.2.1) Find the slope for each of the following equations:
(a) y = 5x + 6 (b) 2x + y = 5 (c) y − 23 = 0 (d) x = 5
(6.2.2) What is the problem with 2.1 (d)? What does the graph look like?
(6.2.3) In the real world we associate slope with steepness.
(a) Does that apply in the math world as well? What happens when you increase the value of m in the equation y = mx + b, when m > 0?
(b) What about when m < 0? How does slope compare with uphill and downhill?
(6.2.4) While we’re at it, let’s get something straight that often confuses students:
(a) What is the equation for the x-axis?
(b) What is the equation for the y-axis?
6.3 The derivative
The ultimate ratio of vanishing quantities is to be understood not as the ratio of quantities before they vanish or after they have vanished, but the ratio with which they vanish.
– Isaac Newton
When I went to school more than a half century ago, calculus was organized differently from the way it is today. The first semester was called Differential Calculus and the second semester, Integral Calculus. The derivative was the central feature of that first course and you will meet it now. You will meet the integral in the next section.
But wait a minute. You have already met the derivative. It is that m or v of those equations y = mx + b and s = vt. The contribution of differential calculus is to extend that straightforward idea of slope to equations that are not straight lines.
And surely this is a reasonable undertaking. Take that distance example. The next time you drive a car at a steady 30 miles per hour for two straight hours (without speed control), please let me know. I once had to drive a survey route99 supposedly maintaining a speed of 20 mph and I had great difficulty keeping even close to that requirement. Every time my attention wandered, I found that I was going 18 mph or 23 mph and had to make an adjustment. And think about other distance situations: throw a baseball, fire a gun, set a rocket on its path to Mars. In all of them the velocity (a.k.a. the slope) is changing. Calculus provides the mathematical concepts to deal with those and many more situations, and mechanical devices based on calculus, like your car speedometer, are built to implement them.
For that distance example you had s = 30t for 2 hours. A graph of that function would appear as Figure 6.3.1.
Figure 6.3.1.
Graph of the equation s = 30t
But your real trip would probably have produced a graph more like Figure 6.3.2 with its steeper and less steep sections and even with horizontal sections where no distance is gained over time and thus representing full stops. You might even have taken a wrong turn and had some downturned sections representing backing up.
Figure 6.3.2.
A more reasonable graph of a trip with average velocity 30 mph
Figure 6.3.3.
Velocity at three points on Figure 6.3.2
I hope that you can convince yourself that the slope of each of those tangents (marked with v’s) represents the velocity at the points of tangency. Those tangents are straight lines just like the lines that represent linear equations and they have the same kind of slope. Steeper tangents: faster speeds; more nearly horizontal tangents: slower speeds; horizontal tangents: stops; downward sloping tangents: backing up. Velocity at particular times is exactly what those tangent slopes indicate.
The dashed line on Figure 6.3.3 represents the average velocity of 30 mph. (Its slope is 30.) At about one hour the velocity (that first v) is steeper than the dotted line so you are going faster than 30 mph. A few minutes later the second v is not so steep; your velocity is less than 30 mph. And finally, as you reach the end of the two hours, you have “speeded up” and again your velocity is greater than 30 mph.
What you want to do is find a way to calcuate those v’s along a curved line. The speedometer in your car does exactly that. I will now show you the concepts behind its operation.
To get at this idea I will consider the slope at points on a more familiar type of curve. For the parabola of Figure 6.3.4, I could draw tangents like those of Figure 6.3.3 and calculate their slope by taking individual points on them and computing
, but it shouldn’t take too much thought to convince you that this is an impractical approach. Instead I will show you a way to find that slope (m or v) by a different, more systematic approach that has many advantages.101
Figure 6.3.3.
Slope between (x, f(x)) and (x+∆x, f(x+∆x))
In Figure 6.3.4 I want to show you how to find the slope of that parabola at the point (x, f (x)) at the left side of the figure. The parabola is the graph of f (x) = 4x x2, but you can also consider it more generally as a representative of the graph of any function, f (x). I have included two points on this graph and the chord between them, designating the second of those points (x + ∆x, f (x + ∆x)), just as I did on Figure 6.2.2. That heavy line represents the average slope between those two points and you know how to find that slope. You know that in general:
Of course, the slope line of Figure 6.3.4 certainly doesn’t represent the slope at (x, f (x)) that you would like to find. It isn’t even a close approximation of the steep tangent at that point. But you can get better approximations by decreasing ∆x as in Figure 6.3.5.
Figure 6.3.3.
Slope between (x,f(x)) and (x+∆x,f(x+∆x) as ∆x decreases.
It is important to see what is happening as I reduce ∆x. First, those slope lines102 are getting closer and closer to the tangent line whose slope we seek.
I will now apply this general idea to the point (.6, 2.04) on the curve of this figure, y = 4x – x2. These are shown in Figure 6.3.6, together with other points as in Figure 6.3.5.
Figure 6.3.3.
Like Figure 6.3.5, Figure 6.3.6 is another very “busy” diagram, that is, it contains a great deal of information. I urge you to look at it carefully in order to understand what the labels represent. Each of the points on the graph is calculated using the x-value from the x-axis with the corresponding y calculated from the equation y = 4x − x2.
With those point values I can calculate the slopes, m1, m2 and m3, of those darkest lines. Those slopes are:
The problem with those computations is that all that arithmetic gets in the way of our seeing what is happening. And that’s a perfect example of a case where Panel 6.3.1 will help us by carrying out these messy computations.
Panel 6.3.1.
This panel calculates the slope m between any two distinct points (a,b) and (c,d) for the equation y = 4x – x2. It takes as input an x-coordinate and a ∆x value. (Thus to calculate the value of m1, ∆x is 3.1 .6 = 2.5.) There is a reason for using ∆x instead of x = c: we want to see what happens when ∆x is decreased.
I urge you to recalculate m1, m2 and m3 using this panel.. To do so you will retain x = .6 and use ∆x successively equal to 2.5, 1.8 and .6.
But you can go further. Continuing to use x = .6, with ∆x = -.1 you get m = 2.7, with ∆x = .01 you get m = 2.79, with ∆x = -.001 you get m = 2.799, with ∆x = -.00001 you get m = 2.79999. You seem to be getting closer and closer to m = 2.8.
You can also use negative values of ∆x. If you choose ∆x = -.1 you get m = 2.9, with ∆x = -.01 you get m = 2.81 and with ∆x = -.00001 you get m = 2.80001. Once again, you approach m = 2.8. In fact, you are now squeezing the slope m = 2.8 between values to the right and left on your diagram.
Indeed, the slope of the curve y = 4x x2 at the point (.6, 2.04) does turn out to be 2.8.
Now I hope you see why I took you through all that trouble about missing points. The problem with calculating the slope at the point (.6, 2.04) is that it is undefined for that point. That is because there is no change in x from a point to itself as there was between two distinct points and we have ∆x = 0. This represents a serious problem, because in the formula for slope we have a zero denominator and remember: you cannot divide by 0.. Thus you can only approach in this case x = .6, you can never reach it exactly.. The limit statement for this particular example is:
Although approaching the tangent at a particular point in this way, even using Panel 6.3.1, still requires much calculation, it gives you the idea of the calculus approach to that tangent representing the slope. You take the slope of those chords for smaller and smaller ∆x to give better and better approximations to the tangent.
This, however, is where I use the basic idea of differential calculus. Instead of creeping up on the value, as I did with the program, I simply take the limit as ∆x approaches 0. Doing that defines the slope of that tangent and, in the process, the definition of the derivative. Formally, this is what we do:
mtangent at(x,f(x)) = 
(6.3.2)
And that’s it! You have reached the top of the differential calculus mountain. If you have followed this development, everything is appropriately downhill from here. All the rest of this subject depends on this relationship.
Unfortunately, this is also the first place where students initially meeting this idea get confused. For a variety of mostly historical reasons beginning with the fact that Isaac Newton in England and Gottfried Leibniz in France separately “invented” calculus,103 there are many notations for what we have called here mtangent, the slope of the curve. Here they are: Dxy, y,, y˙, f ,(x) and 
. Each of those notations represents exactly the same concept and I choose the last one for what follows. It relates well with the notation we have used for average slope, 
. (Read 
as “d y over d x” or less formally as “d y d x.”) The d’s in that fraction retain the idea of differences that the ∆’s designated, but now apply that idea to the limiting value when ∆x = 0.
What you now have then is the defining formula for the derivative and the key to differential calculus,104 the one that appears in every calculus textbook in the world:105
Older texts use the abbreviated form of this statement:
which clarifies the passage from secant differences (∆’s) to tangent differences (d’s) in the limiting case. The formula of (6.3.3), however, is easier to use in specific cases, as the following example will demonstrate.
I will show how that formula works for the curve of Figure 6.3.6. Recall that its equation is f (x) = 4x – x2. Knowing f (x), we can substitute x + ∆x for x everywhere x occurs in that function. Doing this, we have:
f (x + ∆x) = 4(x + ∆x) − (x + ∆x)2
Simplifying this with care we have successively:106
f (x + ∆x) = 4x + 4∆x − (x2 + 2x∆x + (∆x)2)
= 4x + 4∆x − x2 − 2x∆x − (∆x)2
and we can enter these values for f (x) and f (x+∆x) into equation (3.3), our expression for the derivative:
That’s certainly still a messy expression, but notice that good fortune serves us here as several terms drop out. The 4x terms cancel each other as do the x2 terms. That leaves us with:
And here you should see the power of the limit. Although ∆x is small, it is still non-zero and we can cancel ∆x’s when we are dealing with the limit process. (This is exactly why we spent all that time developing the idea of limits.) Doing this we have:
But what about that last term with its ∆x? When we finally take that limit, it is, like those sand particles on the beach, so small we can discard it and we have determined the derivative of the function y = 4x – x2, or in another way of expressing this, we have differentiated that function to arrive at:
All that algebraic processing may have taken your attention away from the import of what we have done here, and we will soon see (happily) that that algebra will not always be necessary every time you want to find a derivative. What is important to recognize is that you now have a result that tells you the slope of the tangent, m = 4 − 2x, for any point on that graph y = 4x − x2. Here are some examples: At the point (1.5, 3.75) the slope is 4 − 2 · 1.5 = 1. At the point (2, 4) the slope is 4 − 2 · 2 = 0. At the point (3, 3) the slope is 4 − 2 · 3 = −2. These are displayed in Figure 6.3.7.
Figure 6.3.7.
Now we can return to the example of Figure 6.3.6 and the slope at that point (.6, 2.34). For that point our calculation, m = 4 – 2x, gives us 4 2(.6) = 4 – 1.2 = 2.8, exactly the value that those lengthy calculations approached.
Is that all there is to the differential calculus?
(1) If y = c, with c a constant, then 
= 0
(2) If y = cxn, with c a constant, then 
= cnxn−1
(3) If y = u ± v, with u and v functions, then = 
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- If y = c, with c a constant, then dy= 0
This confirms that a horizontal line has slope 0.
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- If y = cxn, with c a constant, then dy= cnxn−1
This translates to “multiply the coefficient by the power and reduce the power by one.”
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- If y = u ± v, with u and v functions, then = du± dv
This last tells us that we can find the derivative of functions term by term.108
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The value of those short-cuts is that they allow you to differentiate any polynomial function. Take the one I have used as my example, y = 4x x2. Recalling that 4x = 4x1
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and x0 = 1, for that first term we have 4 1x1−1 = 4x0 = 4, and for the second term, x2,
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we have 2x2−1 = 2x1 = 2x. When we combine those results, we have dy = 4 2x, just
as I did working out all that algebraic simplification. You will have a chance to get used to applying these short-cuts in the exercises.
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I close this section with one application of this concept.109 I’ll extend Galileo’s investigation at the Leaning Tower of Pisa. We climb the 294 steps to the top of the 190-foot tower and from there shoot a stone up with a slingshot, allowing it to fall past us down to the ground below. By taking measurements of the height of the stone at various times, we come up with an equation for its height in feet as related to time in seconds after we release it. That equation is: s = 190 + 64t 16t2. What does the differential calculus tell us about the path of this stone?
It tells us a great deal as Figure 6.3.8 will show.
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We can find the derivative of that function using our short-cuts. It is ds = 64 32t. What do we know from that? Recall that in distance equations, the slope is velocity. Thus we have v = 64 32t, which tells us the velocity of the stone after t seconds. After one second it is still going up at 32 feet per second. Of more interest, after two seconds its velocity is zero. That means it has risen as high as it will go. We can find out how high that is by plugging t = 2 into our equation for s. That gives us s = 190 + 64 2 16 22 = 254 so the highest the stone reaches is 254 feet above the ground.
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We can also tell how fast the stone is traveling when it finally reaches the ground. To find this, we solve the distance equation for s = 0. We can see from the graph that the stone will hit the ground at approximately t = 6. Entering that value into our equation for velocity, v = 64 32t, gives us v 128, so the stone’s downward speed is 128 feet per second when it finally hits the ground.
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In fact, we can confirm Galileo’s important contribution to our understanding of moving objects. We know that velocity is the rate of change of distance over time, which we have found is ds . What about the rate of change of velocity? It is
called acceleration and by the same reasoning we applied to ds , it is dv . We have
dt dt
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the equation for v (v = 64 32t) and need only differentiate it in order to find this value. We do so and find that the acceleration a = dv = 32, a constant. Indeed, the acceleration due to gravity is a constant 32 feet per second per second downward — usually recorded as -32ft/sec2.111
There in a nutshell then, I have given you the basis for the differential calculus.
We will turn in the next section to the equally straightforward integral calculus.
Exercises 6.3
Panel 4.2.3.
Determine Present Value of an Ordinary Annuity
(6.3.1) Use Panel 6.3.1 to determine the slope of y = 4x − x2 at x = 2.
(6.3.2) How does your answer to 6.3.1 compare with the graph of Figure 6.3.7?
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(6.3.3) Panel 6.3.1 worked only for the function y = 4x x2. Panel 6.3.2 generalizes this so that you can approximate the slope for any polynomial at a given point, x.
Panel 6.3.2. Polynomial Slope near a given x
Use Panel 6.3.2 to find the slope of the following functions at the given value of x.
(a) f (x) = 2x2 − 5x + 7 at x = 2
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(b) y = x3 − 2x2 − x + 4 at x = 2 (c) s = −16t2 + 10t + 253 at t = 5
(6.3.4) Use the rules on pages 143-144 to find the derivative dy or ds for each function and
dx dt
then evaluate that slope or velocity for the given value:
(a) f (x) = 2x2 − 5x + 7 at x = 2
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(b) y = x3 − 2x2 − x + 4 at x = 2 (c) s = −16t2 + 10t + 253 at t = 5
(6.3.5) Your answers to exercises 6.3.3 and 6.3.4 should be the same. Are they?
6.4 The integral calculus
I now invite you to consider an entirely different problem, one that certainly appears to be completely unrelated to what was introduced in the last section. It is, however, still called calculus so there ought to be some connection. Historically, mathematicians before Newton and Leibniz did not identify that connection. And no wonder: the two will seem entirely different.
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Much of the integral calculus is about area and volume, although its applications extend well beyond that limitation. Thus, just as the derivative extended the concept of slope, the integral too relates to math you have studied earlier. You already know how to calculate the area and volume of many figures: among them in two dimensions, the area of rectangles, triangles and circles; in three dimensions, the volume of rectangular solids, pyramids and cones. Hopefully, some of you even recall the formulas for the surface area S of a sphere, S = 4πr2, and its volume V as well, V = 4 πr3.
As with so much of math, the conceptual basis for integration was around for cen- turies before it was ever named or studied systematically. Thus we find Archimedes solving an area problem by methods that fall well within the field of integral calcu- lus. The problem he addressed was finding the area included between a parabola
and a chord intersecting it. We too will solve that kind of problem in this chapter. The story of Archimedes’ solution has come down to us in a book he wrote called The Method. But that is a story in itself. (See exercise 6.4.6)
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To get us started I will introduce some notation that will support your study of this interesting subject. One of the problems that non-mathematicians have is understanding some of the symbols that are used regularly by mathematicians and two of those symbols are going to come up in this section. They are and . The basic thing for you to remember is that both stand for “sum”, that is, both are shorthand signals for you to add what follows them.112 The symbol is the Greek
Sigma and S both stand for sum. (Even mathematicians sometimes use reasonable
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abbreviations.)
The two symbols are usually accompanied by indices. I’ll take first. Here is an example of a summation represented by that symbol:
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�( 1 n − 1)
You read that statement as “the sum from n = 2 to5 of 1 n − 1,” and what it means is
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1 n − 1 evaluated for n = 2, 3, 4, and 5 and the results added. Thus we have:
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�( 1 n − 1) = ( 1 · 2 − 1) + ( 1 · 3 − 1) + ( 1 · 4 − 1) + ( 1 · 5 − 1)
and, if you carry out the arithmetic, you have:
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�( 1 n − 1) = 0 + 1 + 1 + 1 1 = 3
That’s all you need to know about the useful symbol . The exercises will give you practice using it.
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The symbol is a bit more complicated. deals with discrete addends, as you saw in the example separate values for n = 2, n = 3 and so on. on the other hand deals with continuous “addends.” As you will see in section 6.9, the difference
between discrete and continuous is conveyed by the difference between stepping
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stones and a sidewalk. You’re “adding” all those values between the limits set by the indices.
Also, the symbol usually is accompanied by a dx, which is the same kind of dx
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you met in the denominator of the derivative dy. I will show you where this comes
from in a moment, but for now I offer an example of what I am talking about. It should help you see what is going on. Suppose we have the symbol:
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5
x=2
( 1 x 1)dx (6.4.1)
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2
Now you might think that, since I said that it represented the “continuous sum,” this symbol would be the sum of the y-values for all the points between x = 2 and x = 5 as on Figure 6.4.1 (those marked and many many more). If we did that, however, even though each point would contribute only a small amount, the sum would be infinite. That’s why I added those quotes around “addends” and “adding.”
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2
Figure 6.4.1. A wrong interpretation for { 5
5
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( 1 x − 1)dx
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Instead, we define this sum as the shaded area between the graph of our function y = 1 x 1, and the x-axis as shown in Figure 6.4.2. Of course, the way of thinking about this sum in Figure 6.4.1 and Figure 6.4.2 is very different. We will see in time, however, that there are connections between the two approaches to the meaning of
.
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2 1
y=
2
Figure 6.4.2. The area interpretation for { 5
5
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( 1 x − 1)dx
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In this case you can easily calculate that area and assign a value to that integral. The indicated area is that of a triangle with base 3 and height 1.5. You can use the triangle area formula A = 1 bh to obtain A = 2.25. Thus you have:
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5
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x=2
( 1 x − 1)dx = 2.25
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and you have evaluated your first integral.
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Until now I have not said how we read this symbol, . It is called an integral and that equation in the last paragraph is read: “The integral from x = 2 to 5 of 1 x − 1
is 2.25.” (Notice that, although you should include dx in the expression, you do not need to read it when you name the integral.)
Exercises 6.4
(6.4.1) Evaluate the following sums by writing out the terms:
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(a) 3 (3x − 15) (b) 4 (x2) (c) 9
(20 − 2x)
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(6.4.2) Write out some of the terms of ∞k=1 1 and suggest what sum you seem to be
approaching.
(6.4.3) I’m sure you would not wish to add all the terms of
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100
(x2 3x 15)
X=1
and that kind of calculation is a perfect task for a computer. You can find the sum of any polynomial by Panel 6.4.1.
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H
x=L
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Use this panel to find the following sums: (a) 100 (x2 − 3x − 15)
f (x) for Polynomial Functions
23
x=5
- 3
(x3 − 23x2 − 5x + 6)
- (Hint: The degree for the function f (x) = 3 is 0.)
x=−3
(6.4.4) Check your solution to 6.4.3(c) by writing out the terms and adding.
(6.4.5) Use what you already know about area to evaluate the following integrals. Drawing
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the graphs should help. Remember that { b f (x)dx is the area under the curve y = f (x)
between x = a, x = b and the x-axis.
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- { 4
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- { 6
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- { 6
x − 1 dx 6 − x dx 5 dx
(6.4.6) The story of Archimedes’ book, Method, continues today. Search for it on the web and summarize the current activity related to this important historical document.
6.5 The definite integral and area
Now you have met what is called the calculus definite integral. What I introduced in Section 6.4 is important enough that I will restate what you have seen in formal terms.
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The definite integral { b f (x) dx is the area
bounded by x = a, x = b, the graph of f (x) and the x-axis.
Figure 6.5.1 shows this for a general f (x).
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You should immediately see that we have a problem here. Just as when we were finding derivatives, there was no difficulty when we had straight lines. We knew then that the derivative dy was the same as the slope m. Equally, we have no problem when we integrate a linear function. We can use area formulas for rectangles and triangles as needed to determine such an area, just as we did for our example.
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When we have curved lines as in Figure 6.5.1, however, we cannot do this. Before I offer a more formal approach to what we do instead, I offer an example of how the approach works. We will consider the area under the function y = 1 x2 − x + 3
between x = 0 and x = 4. Figure 6.5.2 shows this region.
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Figure 6.5.2. The function y = 1 x2 − x + 3 between x = 0 and x = 4.
I will now show you how to calculate the area using only rectangles. We will employ only that simplest of all area formulas: A = lw.
First, I will approximate the area we wish to calculate with one rectangle as in Figure 6.5.3.
Figure 6.5.3. A first area estimate with one rectangle.
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Since that is a 3 4 rectangle, I think that you will agree that our first area estimate is 12. It isn’t a very good estimate, of course. As you can see, it is far too much for it includes all that area above the curve. But we can improve on that estimate by taking two rectangles as in Figure 5.6.4.
Figure 6.5.4. A second area estimate with two rectangles.
Notice how I chose those rectangles. The first had its vertical side the segment from 0 to 3; the second had its vertical segment from 0 to 2, and, of course, the width of each rectangle is 2. Thus the area of the rectangles is 6 and 4 and we have improved our area estimate from 12 to 10.
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What is important for you to see is the fact that we can calculate the lengths of those vertical sides by substituting their x-value into the function. Thus we have, for x = 0, 1 (0)2 − 0 + 3 = 3 and, for x = 2, 1 (2)2 − 2 + 3 = 2.
Our area estimate is still not good, but we can continue by using more rectangles as in Figure 6.5.5.
Figure 6.5.5. Third and fourth area estimates with four and eight rectangles.
Of course, the problem that faces us is calculation. While the area formula is easy once we have length and width, the problem is finding those dimensions. In fact, the width isn’t bad. We just take the width of the entire figure and divide it by the number of rectangles, in this case simply 4/n.
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It is the height that requires more calculation. Consider, for example, the fourth rectangle on the fourth estimate. Its height rises from x = 1.5 = 3 , so you must substitute that value into the function equation to calculate the height. We have 1 ( 3 )2 − 3 + 3 = 33 . And you have to calculate eight of those heights.
Thank goodness, the computer loves to carry out such mundane computations and we can assign them to Panel 6.5.1.
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Panel 6.5.1. Area under y = 1 x2 − x + 3 between x = 0 and x = 4 for N rectangles.
Using that panel you will find that the area estimate is 9.5 for four rectangles and
9.38 for eight. What is great about such a program is the fact that you can use it with 50 or 100 rectangles without having to stay up all night calculating.113 When you employ this panel in doing the exercises, you should appreciate the computations you are avoiding.
Formalizing the Integration Process
In the remainder of this section I will formalize what we have done in that example. I hope you will follow the argument, but, if you understand the example, you understand what is going on.
Returning to the general case of Figure 6.5.1, Figure 6.5.6 shows the area under the curve f (x) between x = a and x = b divided into five rectangles to illustrate the case when you could use any number of rectangles.
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I hope that it is clear what has been done on this figure. The x-axis between x = a and x = b has been divided into parts, each of whose width is ∆x, with ∆x =(b a)/n, when n is the number of rectangles. The lower left-hand corner of the first of these rectangles (we’re working from left to right) has the value x = a. The point on the curve y = f(x) above that point is f(a). That f(a) is also the height of that rectangle, since that is how f(a) is located. Thus the area of that first rectangle is f(a) ∆x.
Now consider the second rectangle. The x-coordinate of its lower left corner is
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a + ∆x, its height then f(a+∆x), and its area f(a+∆x) ∆x.
You can continue calculating these rectangles until you have an area estimate, A:
A = f(a)∆x + f(a+∆x)∆x + f(a+2∆x)∆x + f(a+3∆x)∆x + f(a+4∆x)∆x
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You can represent the sum of equation (5.2) by using L notation in this way:
A = f (a + k∆x)∆x
k=0
and you can improve on this estimate by increasing the number of rectangles, in the process, of course, making ∆x shorter. Figure 6.5.7 shows this for twice as many rectangles.
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Figure 6.5.7. Improved estimate for { b
f (x)dx
And now we have:
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9
A = f (a + k∆x)∆x
k=0
We need not stop increasing the number of rectangles here and, in general, we have, letting the number of rectangles be n, the relationship:
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n−1
A = f (a + k∆x)∆x (6.5.1)
k=0
Notice that we know the value of ∆x in terms of a, b and n. Each ∆x is the length of the segment b − a divided by n, that is: b−na , so we have in the most general form:
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n−1 (
b − a \ b − a
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Don’t let that expression put you off simply because you are still not comfortable with the notation. All it is saying is that we are adding up n rectangles, each of whose height is the f (x) at a given point on the x-axis and each of whose width is
∆x = b−na .
Now you have a means of finding the approximate area under various functions.
Not good enough. Mathematicians and other scientists want two additional things:
- the exact value of such areas, and (2) a way of finding them without all that calculation and programming. (Recall that calculators and computers that do this work for us have only been available for about 60 years.) I address the first of those problems The second I will address in the next section.
First, just as I did with the slope to find the derivative, I define the value of the definite integral simply by taking the limit as the width of those rectangles, ∆x, approaches zero. We have then:
b
f (x)dx = lim
n−1
f (a + k∆x)∆x (6.5.3)
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x=a
∆x→0 k=0
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which, just as the limit provided us the exact slope in the differential calculus, here we have the exact area under a curve. Notice how the process at arriving at this result was similar as well. in both cases we let ∆x 0, knowing that, as it does so, the area approximation improves.
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There is another aspect to this as well: when the width of the rectangles decreases, the number of rectangles increases, so we have: lim∆x→0 = limn→∞. Thus we have an alternate version of the preceding equation:
b
f (x)dx = lim
n−1
f (a + k ·
b − a \ ·
b − a
(6.5.4)
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x=a
n→∞ k=0 n n
We will be using equation (6.5.3) in what follows, but you should realize that (6.5.3) and (6.5.4) are two ways of showing the same thing.
Okay, that is the formal development and I hope you have followed it. But remember: the basic idea of the integral is a simple one. You find the area between a curve and the x-axis by adding up rectangles. The process involves much computation. Happily, it will turn out in the next section that there are short-cuts to avoid all that computation.
Exercises 6.5
(6.5.1) Using Panel 6.5.1, approximate the area of Figure 6.5.2 for:
(a) N = 10.
(b) N = 100
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(c) The exact area under this curve is 91 . Do your answers in (a) and (b) support this?
As I did with derivatives, I can generalize the technique used in Exercise 6.5.1 to apply it to any polynomial function and any range of x=values. This is done in Panel 6.5.2.
Panel 6.5.2. Approximate area between polynomial f (x), the x-axis, x = A and x = B
(6.5.3) For each of the following functions estimate the area using Panel 6.5.2:
(a) y = x2 − x + 2, y = 0, x = −2 and x = 2.
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(b) f (x) = 1 x3 + 1, the x-axis, x = −1 and x = 3. (c) y = 1 x4 − 1 x2 + 1, y = 0, x = −2 and x = 3.
(6.5.4) Use Panel 6.5.2 to estimate the value of:
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- 3
x=−1
(x2 − 2x + 3) dx
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- { 5 ( 1x3 + 2) dx (It is common to omit the x = for the )
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- 6 5 dx. Note: y = 5 is a 0-degree polynomial function.
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- Use the graph of y = 5 to explain your answer to (c).
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(6.5.5) What is the value of a(27x5 − 35x + 7) dx? (Trick question: Notice the limits.) (6.5.6) Those equations (6.5.3) and (6.5.4) can be confusing. There are many symbols, some
new to you, and that makes things difficult, but these equations are important so you need to make sense of them. Here are some questions designed to help you understand them. Refer to Figures 6.5.6 and 6.5.7 to see the geometry:
- What do a and b represent?
- What is the role of f(x)? Could it stand for 3x − 5 or sin x or √x − 7?
- What is the role of ∆x?
- How is ∆x related to a, b and n?
- Why is lim∆x→0 equivalent to limn→∞?
6.6 The remarkable connection
Much of what I have described to this point was already known before the time of Newton and Leibniz. The extremely important idea that Newton and Leibniz contributed was the important connection between the derivative and the integral. It turns out that, just as subtraction “undoes” addition and division “undoes” mul- tiplication, so too integration “undoes” differentiation. This is summed up in the following statement about the definite integral:114
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, b
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Here, as is so common in mathematics, notation hides a straightforward idea. You probably don’t recall that I pointed out in Section 6.3 that there are different notations for the derivative. And in equation (6.6.1) one of them is used. Here F,(x) is another way of saying dy when y = F(x). For example, suppose F(x) = 3x2 − 5x, then, using those short-cuts for finding the derivative, F,(x) = dy = 6x 5.
But in equation (6.6.1) you are going in the reverse direction. You start with
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F,(x) = dy = 6x − 5 and end up with F(x) = 3x2 − 5x. That is what is meant by
“integration undoing differentiation.” Once you have carried that out, you evaluate
F(x) for x = b and x = a and find their difference.
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I’ll use this same example with F,(x) = dy = 6x 5 and its integral F(x) = 3x2 5x to show how this works. Suppose you have the following integral to evaluate:
5
Integrating that 6x − 5 gives you 3x2 − 5x, so you have:
F(x) = 3x2 − 5x
Using the designated limits x = 5 and x = 2, F(5) = 50 and F(2) = 2, you now have
F(5) − F(2) = 48
and your solution:
5
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(6x 5) dx = 48
2
That is exactly the procedure you should follow to evaluate any definite integral. In simple terms: Find the function that produces that derivative, evaluate it at the integral limits and subtract.
You are, of course, left with one problem. How do you find that so-called anti- derivative? As I have throughout this chapter, I will restrict us to polynomials115 and it is reasonably easy to undo their derivatives.
Here in tabular form is a comparison between differentiation and integration for simple polynomial functions.116 Note that u‘, v‘ and y‘ are the derivatives of u, v and y.
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Differentiation Integration
{ u + { v
These rules allow us to differentiate and integrate polynomials, all we need to be concerned with in this introduction to the ideas of calculus. To show the power of this process, consider an example. Suppose we wish to know the area under the curve y = .5x3 − 5x + 8 between x = 1 and x = 5, as shown in Figure 6.6.1.
40 | ||||||
30 | ||||||
y | = .5x3 | – 5x + | 8 | |||
20 | ||||||
10 | x=5 | |||||
x=1 | ||||||
2 | 4 |
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Figure 6.6.1. Area under y = 1 x3 − 5x + 8 between x = 1 and x = 5
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In calculus terms we are finding the value of { 5( 1 x3 − 5x + 8)dx. To find the value of
that integral we need the function F(x) that corresponds to the f (x) = 1 x3 − 5x + 8. Using the techniques of the table, we have F(x) = 1 x4 − 5 x2 + 8x. (Be sure you see how
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these terms were calculated.) Given that information, we now calculate F(5) − F(1)
for our answer. F(5) = 1 (5)4 − 5 (5)2 + 8 · (5) = 55.625 and F(1) = 1 (1)4 − 5 (1)2 + 8 · (1)
8 2 8 2
= 5.625. Thus F(5) − F(1) = 50.
That connection between differentiation and integration is so important to calculus that the equation (6.6.1) on page 156 has been designated the Fundamental
Theorem of Calculus. 117 All you need to take away from this brief introduction, however, is the idea that differentiation and integration are effectively inverse processes, one undoes the other. If anyone asks you what the Fundamental Theorem of Calcu- lus is, just tell them that the derivative and the integral undo each other. The person asking may carp at your language, but she will know that you have the right idea.
I cannot overestimate the importance of this theorem. We rightly celebrate the names of Isaac Newton and Gottfried Leibniz118 for their having separately found this connection within a few months in the late 17th century. You now share their famous discovery.
There is, however, one more important issue that I want to address. In that table comparing derivatives and integrals, I omitted the fact that integrating 0 you get a constant (since differentiating a constant you get 0) but in my discussion of the definite integral no constant appeared. Two questions arise: (1) Why did we not need to consider that constant? and (2) Even if we don’t need to consider it when dealing with the definite integral, are there situations when we do need to do so?
A simple example should respond to the first question: We seek to determine
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5 x dx. Our rules tell us that the function whose derivative is x is 1 x2; thus our
answer would be 1 · 52 − 1 · 22, which is easily simplified. However 5 x dx is certainly
equal to 5(x + 0) dx so it would seem that we need to include a constant in our calculations. And in fact the general function whose derivative is x is 1 x2 + c, for some constant, c. Don’t we then need to say 5 x dx = 1 x2 + c which is then evaluated
by taking the difference between that function evaluated at x = 5 and x = 2?
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But notice what happens when you carry out that evaluation. You get: 1 52 + c
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( 1 22 + c) and the c’s cancel out. Thus we do not need to bother with those constants in dealing with the definite integral.
That answers the first question but the second question leads us to situations in which that constant does play an important role. And that is the subject of the next section.
Exercises 6.6
(6.6.1) State in words how you integrate the term bxn. (See the table on page 153.)
(6.6.2) Use that table to determine the function resulting when each of the following are integrated:
(a) x3 + x2 + x + 1 (b) 5x2 − 3x + 7 (c) 7 − 2x4 (d) 23
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(6.6.3) In the text we found that { 5(6x − 5) dx = 48. Use Panel 6.5.2 to approximate and thus
support this answer.
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(6.6.4) Use Panel 6.5.2 to confirm the other text example: { 5( 1 x3 − 5x + 8)dx = 50.
(6.6.5) Some notation can help keep track of the process of integration, as the following example shows:
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r 4 2
4 3 3 3
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That left square bracket keeps track of the numbers to be substituted into the integrated terms. Using this model, evaluate the following:
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(a) { 2(x + 1) dx (b) { 3(3x2 + 5x + 6) dx (c) { 2 x3 dx
(6.6.6) Check your answers to exercise (6.6.5) using Panel 6.5.2.
6.7 The indefinite integral
We paid no attention to the fact that integrating 0 produces a constant when performing definite integration. In other settings that constant turns out to be important. It does so in what is called indefinite integration.119 Consider indefinite integration through an example I used earlier.
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Recall our example of the Tower of Pisa back in Figure 6.3.8 when you first met the derivative. That example is much better handled as an indefinite integration problem, beginning (rather than ending) with the fact that the acceleration of an object due to gravity120 in feet per second per second (ft/sec2) is 32.
Acceleration is the rate of change of velocity which, expressed as a derivative, is
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dv , thus we have dv = −32. We want to undo that derivative so we want to integrate it.
Not quite yet. Before we do that integration we multiply both sides of that equation
by dt (recalling that, although dt is small, it is not zero121) to have dv = −32 dt. Now
the equation is in a form ready for integration. We have { dv = −32 dt.
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Now recall that in this kind of integration, we must generate constants. Thus, integrating we get v + a = 32t + b, with not necessarily equal constants (called a and b here) generated on each side of that equation. If we then subtract a from each side,
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we have v = 32t + b a, and we simply set b a = c. You may be uncomfortable with doing this at first, but recall that constants are simply numbers that are easily combined. Suppose, for example, that in this case a = 2 and b = 5. Then b a = 3 and c = 3.
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In any case, we now have v = 32t + c and we need to evaluate c. We can do so with information from the original problem. We shot the stone up with a slingshot with an initial upward velocity of 64 feet per second. That means that at t = 0, v = 64. We can substitute those values in the equation v = 32t + c to find that c = 64. Therefore, our equation is v = 32t + 64.
We also know that v = ds , the rate of change of distance over time, so we can
dt ( (
substitute this, multiply by dt and integrate once again. We have ds = −32t+64 dt,
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which, both sides integrated, gives us s = 16t2 + 64t + C, with C another constant. Recalling that we started 190 feet up in the Tower of Piza, we can evaluate C by using the fact that at time t = 0, s = 190. Substituting these values we have C = 190 and the equation for s is s = 16t2 + 64t + 190, which matches the equation of Figure
6.3.8 on page 145.
There are many applications of indefinite integration. You will find a number of significant applications of just one of them in another chapter. Others are introduced in the exercises.
This integration constant is the subject of one of the rarest of all humorous stories, a math joke. You must know what we have just described in order to appreciate it.
On their way to a conference two college mathematicians — we’ll call them Professor A and Professor B — stop to have lunch at a neighborhood tavern. Professor A spends the entire lunchtime complaining about the poor quality of math teaching. “It has declined so much,” he claims, “that the general public knows virtually nothing about the subject.”
Professor B disagrees and while his companion is visiting the men’s room, he calls over the waitress who has served them. He offers her $5 if she will help play a joke on Professor A. When you bring our dessert, he tells her, I’ll ask you a question and all you have to do is answer simply, “One-third x cubed.” The waitress agrees, takes the $5 and goes off mumbling “One-third x cubed” to herself.
When Professor A returns, Professor B proposes a $20 bet. To prove that many common folk do know some math, he will ask their waitress to solve an integral. If she responds correctly, he will win the bet; if not, Professor A will win. Of course, Professor A agrees. What chance is there that a waitress in this neighborhood tavern will know any calculus?
When the waitress next returns to the table, Professor B asks her, “What is the integral of x squared?”
“One-third x cubed,” she replies and walks off.
But as Professor B is collecting his $20 from Professor A, the waitress turns to add, “plus a constant.”
Exercises 6.7
(6.7.1) A sky-diver leaps out of an airplane at 12,000 feet and free-falls toward the earth before opening his parachute.
- Using the acceleration due to gravity as -32 ft/sec2, derive the equation for his (Remember that at t = 0 his velocity was v = 0.)
- From your answer to (a) derive an equation for s, representing his height above earth as he
- From your answer to (b), determine how long he has before he must pull the ripcord to open his parachute, if he must do so when he is at least 1,000 feet above the
(6.7.2) The slope of the tangent line to a particular curve, y = f (x), at any point on the curve is x + 1.
- Integrate to find the equation that satisfies this It will include a constant.
Identify the specific curve if it passes through the point (2,5).
6.8 Tying up loose ends
First, let us review what we have done — all in a single paragraph:
We generalized the idea of slope of a line to slope of a curve, that is, the derivative. Then we found that the area under a curve, the definite integral, involved the process of undoing the derivative, that is, integrat- ing. Thus we tied together the basic concepts of the differential and the integral calculus.
And that’s it? Certainly not. There is far more to calculus than what we have done here. There are many applications of these ideas to a wide range of problems. Most of the applications you meet in a calculus course apply to mathematics itself and to physics, but the ideas extend far beyond those subjects as well: to astronomy, to business, to biology, to weather, to the environment — the list is virtually endless.
What you should take from this development are the following ideas:
- Whenever you see a discussion of change, think derivative. Your cup of cocoa cools: think derivative. Debts (with their associated interest rates) accumulate: think derivative.
- Whenever you hear of something maximized or minimized, best or worst, don’t just think derivative, think derivative = 0, recalling how that slope was zero at the top or bottom of a What time was it coldest last night? Think temperature derivative = 0.
- When things accumulate or add up, think integral.
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- Recognize the mathematical symbols when they appear, as they occasionally
do, in standard text. means sum and t means sum over a period of time. When you see a prime (the symbol ‘ as in f ‘) associated with a function, it often
— but unfortunately not always — means the derivative of that function.
You don’t have to be able to calculate the centroids and volumes of solids or astronomical orbits to have an idea what calculus is about. If you have followed this chapter, you now have a general understanding of what is arguably the most powerful tool in all of science.
Hopefully, this chapter will also whet your appetite and make you consider exploring this subject further.
In the foregoing you have dealt with the ideas about calculus developed through simple examples. As I pointed out, students in a calculus course derive many formulas beyond those shown here. Nowadays, computers and some advanced calculators will carry out symbolic differentiation and integration. Symbolic here means giving you a formula. If, for example, you want the derivative of ax, your computer would tell you ax ln a. And, if you asked it to integrate secant2 x, it would tell you tan x, expecting you to add a constant in both cases if you are working with indefinite integrals.
Exercises 6.8
(6.8.1) You are asked to tell someone what the calculus is about. Without looking back at the text, write a paragraph that tells what you have learned about it.
(6.8.2) Using the techniques you learned earlier in this chapter:
- Find the derivative of y = 3x2 − 5x +
- Integrate y = 3x2 − 5x +
As the table back on page 153 shows, finding the derivative and integrating polynomial terms is quite formulaic. Panel 6.8.1 will find the derivative of a polynomial term for you.
Panel 6.8.1. Derivative of a Polynomial Term and Panel 6.8.2 will integrate a polynomial term.
Panel 6.8.2. Integrate a Polynomial Term
(6.8.3) Using these panels:
- Find the derivative of y = 3x2 − 5x +
- Integrate y = 3x2 − 5x +
- Do your answers conform with those of exercise (6.8.2)?
6.9 Discrete Mathematics
You have now met calculus, one of the most powerful mathematical tools of modern science. You may even think, as I did when I first studied it, that calculus is going to solve all advanced problems. But just as calculus has its limits, it also has its limitations. Calculus addresses problems involving what are called continuous processes: processes like time, motion, weather and growth. Discrete mathematics on the other hand considers processes that have separate states: especially processes that involve counting. You “count” your money, so money is discrete.
If that sounds overly technical, think of continuous in the form of a hiking path or a sidewalk, discrete math as a series of separate stepping-stones. Think of a number line as continuous, but the ruler marks on it as discrete. Thus the real numbers are continuous, the integers are discrete. Any time you talk about a population – the students in your class, the number of cats in your county, even germs in a culture – you are dealing with discrete individuals and therefore discrete mathematics applies with its different armory of techniques.
Sometimes, however, the distinction between continuous and discrete is blurred. For example, when dealing with large populations like those cats or germs, there are enough to consider them almost continuous so that you can apply calculus techniques. But, unlike King Solomon, you still cannot talk about 3.62 cats.122
What are some of the techniques of discrete mathematics? You have already met some of them in the Money Matters chapter where you worked with arithmetic and geometric sequences and series. As an example of the activities of discrete mathematics, we’ll return to those matters now, but from a different perspective.
Sequences
A sequence is an ordered list of objects, usually but not always numbers. You have certainly seen hundreds of sequences because they are used often as intelligence tests in the form of “What should appear next?” (Psychologists consider pattern recognition an aspect of intelligence.) Here are three examples of sequences with a
? inviting your guess for what term should appear. Please answer each now. (a) {1, 5, 25, 125, ?, …} (b) {4, 14, 5, 15, 6, ?, …} (c) {A,C,E,G,?,…}
Before considering values of those ?s, consider something about those sequences. Each of them has three dots (called an ellipsis123) following the last term given. That
means that the sequence continues without bound and thus is infinite. Sequences that have a specific number of terms are called finite.
Now consider your values for those missing terms. Did you answer 625, 16 and I? If you did, you gave the answers most of us would and that would please those psychologists. In the first case you probably saw that each number is five times its predecessor. In the second case you probably saw pairs of numbers, 4 and 14, 5 and 15, and thus 6 and 16. (You would expect 7 and 17 to follow.) In the third case you probably saw every other letter of the alphabet and you would continue I,K,M,O,…. Surely, however, there are some musicians among you who saw (c) differently.
You might well continue the sequence B,D,F,A,…., thinking of the notes in the scale. Aha! There is nothing whatsoever wrong with your thinking even though you did not come up with the standard “expected” answer. You know the expression: you do not fit the mold. In fact, that is a general problem with sequence questions like that. The really creative answers, like the one you musicians came up with, count against you.
People love to give puzzle challenges in the form of strangely defined sequences. I offer two here, both finite. Don’t waste much of your time on them as they are trick questions:
(d) {6, 6, 7, 9, 8, 6, ?} (e) {42, 33, 28, 23, 14, ?}
Some of you might have thought of the answer to (d) because there are exactly seven entries. Those are the number of letters in the words for the days of the week beginning with Sunday. The answer to (e) is Astor Place. I doubt if anyone, even among those of you living in New York City, would guess that answer. Those are the Lexington Avenue subway stops from Grand Central Station to New York University.124 Be assured, however, that each of those is a well-defined sequence.
I hope I have made my point here. If you and I want to be sure about the entries in a sequence, we should not rely on guessing. We should define what they are with care. Notice in this regard that the answers to the sequences (c), (d) and (e) are straightforward once the pattern is defined.125
Explicit and Recursive Sequence Definitions
As usual, we need some notation to talk about the terms in a sequence. It is common practice to use a subscript to represent the number of the term. Thus you can write a general sequence as: a1, a2, a3, a4, …, an, … . (In some sequences, especially those involving time, it is useful to label the first term, a0.) Think how this labeling
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would apply to the sequence in (a): 1, 5, 25, 125, … . In that sequence a1 = 1, a2 = 5, a3 = 25 and a4 = 125. Now you want to be able to tell people exactly what is going on in that sequence. You can do so in two ways.
First, you can define the sequence by telling how it starts and how you get from one term to the next. In this case you could say: “The first term is one and to get each successive term you multiply by five.” I will translate that into math lingo shortly, but now just be sure you see that it tells you exactly how to recreate that sequence. What you have done in this way is give what is called a recursive definition of that sequence.
The second way of defining any sequence is to tell how every term is formed. You are probably more familiar with this way. For our example (a) you could say: “Every term is a power of five, with the exponent one less than the number of the term. For example, the 24th term would be 523.” This is the explicit definition of that sequence.
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Now that we have an idea of what recursive and explicit sequence definitions are, we need to express them in mathematical form. The recursive form always requires two statements. The first tells the starting term. For our example (a), that is easy: a1 = 1. The second statement must tell how to get from one term to the next. In this case an = 5an−1 for n 2.
The explicit definition requires only the form of the general term. In this case
an = 5n−1 for n ≥ 1.
Two Famous Sequences
Here are two famous sequences. Each can be defined both recursively and ex- plicitly, but the explicit definition of the first is very complex and is less often given.
The Fibonacci Sequence
In 1202, the Italian mathematician Leonardo Fibonacci posed the following prob- lem: A single pair of rabbits (a male and a female) is born at the beginning of a year. The rabbit pairs are not fertile during their first month of life, but after that, each rabbit pair will give birth to a new male/female pair of rabbits at the end of every month. If none of the rabbits die, how many rabbits will there be at the end of a year?126
At the beginning, there is one pair of rabbits, so we let F0 = 1. At the end of the first month, this pair has not yet reproduced, so there is still only one pair of rabbits and F1 = 1. At the end of the second month the pair does breed so we have one additional pair and F2 = 2. We could continue in this way, but things get more and more complicated. Instead, think of the situation in subsequent months as equal
to the number of rabbits during the previous month plus the new pairs. But those new pairs are equal to the number of rabbit pairs two months ago. This leads to the famous Fibonacci sequence recursive definition, in this case requiring two starting numbers:
F0 = 1 and F1 = 1 and Fn = Fn−1 + Fn−2 for n ≥ 2 (6.9.1)
This is an example of a sequence for which the recursive definition is straightforward but the explicit definition is very difficult to derive.
You will be asked to use this definition to answer Fibinacci’s problem in the ex- ercises.
The Tower of Hanoi
In 1883, French mathematician Edouard Lucas invented the Tower of Hanoi puzzle along with the legend that accompanied it. According to his legend, a certain Hindu temple contains three thin diamond poles on one of which, at the time of creation, God placed 64 golden disks that decrease in size from the base up. The priests at the temple work unceasingly to transfer all the disks, one by one, from the first pole to one of the other poles, without ever placing a larger disk on top of a smaller one. Supposedly, when they have completed their task by moving all the disks from the first pole to the third pole, all will crumble into dust and the world will vanish in a thunderclap. Our question is, how many steps does it take to move all 64 disks from one pole to another pole?127
You have probably seen toys based on this puzzle with fewer disks stacked on pegs.
Figure 4.5.1.
A Monacan Roulette Wheel
We can consider this recursively as follows: Moving one disk from pole 1 to pole 3 takes just one move. Therefore T1 = 1. To move two disks from pole 1 to pole 3:
(1) move the smaller disk to pole 2, (2) move the larger disk to pole 3, and then (3)
move the smaller disk from pole 2 to pole 3. So, T2 = 3. To move 3 disks: (1) move the smallest disk to pole 3, (2) move the middle disk to pole 2, (3) move the smallest
disk from pole 3 to pole 2, so 2 disks are now on pole 2 and pole 3 is again empty. Next (4) move the largest disk to pole 3, and then repeat the first 3 steps in reverse
- move the smallest disk from pole 2 to pole 1, (6) move the middle disk from pole 2 to pole 3, and finally (7) move the smallest disk from pole 1 to pole 3. So, T3 =
This is starting to get complicated, so let’s jump into the middle and think this
through. To move n disks from pole 1 to pole 3, first we need to move the previous n-1 disks from pole 1 to pole 2, move the bottom disk to pole 3, then move the previous n-1 disks from pole 2 to pole 3. This can be expressed by the recurrence relation, Tn = Tn−1 + 1 + Tn−1, which simplifies to:
T1 = 1 and Tn = 2Tn−1 + 1 for n ≥ 2 (6.9.2)
We will use this recurrence definition to find an explicit definition and to solve Lucas’s problem in the exercises.
Exercises 6.9
(6.9.1) What is the value of the ? in sequence (d) in the text? (6.9.2) Are the sequences in (d) and (e) finite or infinite?
(6.9.3) Label the following as discrete or continuous as time passes:
- the population of the United States
- the height of a child
- number of white cells in a blood sample
- the amount of water flowing over Niagara Falls
- number of toys in a toddler’s toy box
- amount of money spent annually on health care in the United States
(6.9.4) To which of your answers in (6.9.3) that you classified as discrete, could calculus methods be reasonably applied.
(6.9.5) Here are a few of those trick sequences that test your creativity:
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- Find the other five terms of the sequence: 7,8,5,5,3,4,4,?,?,?,?,? . Notice that there are exactly 12 entries in this discrete
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- Find the other four terms of the finite sequence: 3,3,5,4,4,3,?,?,?,? . This time there are exactly 10
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- What appears to be happening in the sequence 1,2,3,4,5,8,7,16,9,… ? It may help to know that the next three terms are 32, 11 and
(6.9.6) Write the first five terms of each recursively defined sequence:
- a1 = 1 and an = an−1 + 7 for n ≥ 2
- b1 = 2 and bn = 3bn−1 for n ≥ 2
- c1 = 1 and cn = 4cn−1 − 2 for n ≥ 2
- d1 = 1 and dn = 2dn−1 + n for n ≥ 2
- e1 = 1, e2 = 1 and en = en−1 + 2en−2 for n ≥ 3 (6.9.7) Define each sequence recursively:
(a) {1, 4, 7, 10, 13, 16, …}
(b) {25, 21, 17, 13, 9, …}
(c) {2, 10, 50, 250, 1250, …}
(d) {1, −2, 4, −8, 16, −32, …}
(6.9.8) Define the sequences of exercise (6.9.7) explicitly.
(6.9.9) Use equation 6.9.1 to give the values of F3, F4 and so on up to F12. This answers Fibonacci’s question.
(6.9.10) Notice how in exercise (6.9.9) you had to find all those intermediate values to finally find F12. You can imagine how long it would take you to find, say, F35. The program of Panel 6.9.1 does this for you.
Panel 6.9.1. The nth Fibonacci Number, F0 = 1, F1 = 1 (Calculated Recursively)
Use it to find F12, F20 and F35.
(6.9.11) When a sequence is defined by an explicit formula, it is usually easier to calculate the value of a term using that expression. But the explicit formula for Fibonacci numbers is very complicated:
f 1+ √5 \n+1 − f 1− √5 \n+1
2
Fn =
2
√5 (6.9.3)
Equation (6.9.3) is used to construct Panel 6.9 2.
Panel 6.9.2. The nth Fibonacci Number, F0 = 1, F1 = 1 (Calculated by Formula)
uses this formula to find those numbers. Use it to find F12, F20 and F35 and compare your answers with exercise (6.9.10).
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(6.9.12) That strange fraction 1+ √5 that appears in the explicit formula for Fn is important enough to mathematicians to be given the name τ (that’s the Latin letter tau.)
- Calculate the value of τ.
- Use Panel 6.9.1 or 6.9.2 to calculate the value of Fn/Fn−1 for n = 3, 5 and 12.
- How do your answers in (b) relate to your answer in (a)? Use one of your Fibonacci programs to check this for F40/F39.
(6.9.13) For the Tower of Hanoi problem I showed you that T1 = 1, T2 = 3 and T3 = 7.
- Use the recursion formula (6.9.2) to find T4 and T5.
- How do the values that you have found relate to powers of 2?
- Use what you found in (b) to argue that the explicit formula for the Tower of Hanoi is Tn = 2n −
(6.9.14) According to Lucas’s legend, the world will end when all 64 disks are moved.
- Use your result in exercise (6.9.13)(c) to find T64. (Note that the minus one makes little difference to a number this )
- If the monks move one disk per second, how long after they started will our world end? (You should find this answer one of the least troubling predictions of our apocalypse.)
Figure 4.5.2.
The Roulette Betting Table
Notice that, even though you won three spins and lost seven, you ended up ahead. In fact, with the martingale strategy, if you win on the final spin, you will have won in dollars overall, exactly the number of winning spins. In this example, you won $3 because you won three times.
That sounds wonderful. You can sit at the gaming table all evening, bet following these simple rules, quit on a win, and end up ahead as many dollars as the roulette wheel went your way as long as you played. And remember: the spinner should win for you almost half of the time. No matter how many individual spins you lose, you need only wait for a final win to quit ahead.
Before you rush to the gaming table, however, think seriously about this algorithm. Surely there must be a catch. If there weren’t, the news would get around and the casinos would soon go bankrupt. Be assured: the casinos are doing fine and croupiers will not bat an eye if they see you betting by this (or any other) system.59
Notice something about this example. After that 9th spin (following 4 losses) you are down $13 and you have to bet $16. You need a significant amount of ready cash to use your strategy. At this point you would need to have arrived at the table with $29. One more loss and you would have been down $29 and have to bet $32. Then you would need $61 to continue. A sixth loss and you would be down $61 and have to bet $64 and you would have had to brought $125 to the casino initially to make this bet.
You may argue that those situations would involve long strings of losses: in those examples five or six losses in a row. But remember when we ran those random number tests: we found that lengthy strings do happen, and quite often. Here is what a mathematical analysis of a series of roulette spins says of the chances of that happening:60
In 73 spins, there is a 50.3% chance that you will at some point have lost at least 6 spins in a row. Similarly, in 150 spins, there is a 77.2% chance that you will lose at least 6 spins in a row at some point. And in 250 spins, there is a 91.1% chance that you will lose at least 6 spins in a row at some point.
Another roulette strategy is called antimartingale and you will see why from the algorithm. Here is the antimartingale strategy:
- Bet $1.
- If you lose, repeat step 1. If you win, bet twice the amount you won. (That is: let your bet with its payoff )
- Repeat step
Let’s see what would have happened with the earlier example if it had been played by the antimartingale strategy:
That certainly does not look very interesting, but the antimartingale player is looking for strings of wins. Consider what would have happened if we reversed the values of W and L in that table and quit after that string of wins.
This looks at least a bit better. If the antimartingale player quits after a strIng of wins, he can win quite a bit. If, for example, instead of losing on that 10th spin, he had won, he would have been $29 ahead.
Clearly, both systems have their problems. The martingale system offers a good chance of winning a little and a small chance of losing a lot; the antimargingale system offers a small chance of winning a lot against a good chance of losing a little.
You can experiment with these two strategies with the following panels:
Panel 4.7.3.
Experiment with the Martingale Strategy
Panel 4.7.4.
Experiment with the Antimartingale Strategy
My recommendation: play the games with these panels instead of at the casino or on the web.
Mathematicians did find a way to beat the casinos at the blackjack tables by using the very complex card-counting procedure already mentioned in an earlier footnote. I have, however, looked at a number of the books that purport to teach you those strategies and I come away from the experience with four thoughts:
(1) Why are these people writing books and not simply cleaning up at the blackjack table?
(2) If you are bright enough (and patient enough) to handle these techniques, you should be writing this book, not reading it.
(3) Even if you master a winning technique, you will soon be blacklisted by and excluded from all casinos, and
(4) You have better things to do with your life.
The success of the mathematicians who beat the casinos was short-lived, however, as they were soon identified and excluded.61
Exercises 4.7
(4.7.1) If you have not already done so, use a scientific calculator or computer calculator app to evaluate the probability formula of equation (4.7.1).
(4.7.2) Run the program of Panel 4.7.1 twice each with bankrolls of $10 and $50 and with $1 bets. Record your results. (Note: these sessions take quite a bit of time, but you can watch your fortunes rise and decline as each evening at the tables progresses.)
(4.7.3) Gather information from the program of Panel 4.7.2 for 11 sessions (this requires just one program run) beginning with a bankroll of $20 and making $1 bets. Compare your results with the prediction of your answer to exercise (4.7.1). It is instructive to end this example with a final comment about the results of the computation of exercise (4.7.1) and the experimentation of exercise (4.7.2) by Professor Honsberger: “You have more than four times the probability of winning with the single bet. It follows that the longer you play, the better the house likes it.”
(4.7.4) Complete a martingale table like those of this section for each of the following series of roulette results:
(a) WWWLLWLLLLLW (b) LLLLLWWWWW
(4.7.5) Complete an antimartingale table for each of the following series of roulette results:
(a) WLLLWLLLWW (b) LLLLLWWWWW
In the following exercises, you will gain comparable information by running the programs with the same bankroll and bets. You will also find that, with large bankrolls, the programs will take a long time to run unless you bet correspondingly large amounts. I suggest you run the programs with a bankroll of 20 and a bet of 1. You are welcome, of course, to experiment with other values.
(4.7.6) Run the Martingale Panel (4.7.3) for 11 sessions and record your results.
(4.7.7) Run the Antimartingale Panel (4.7.4) for 11 sessions and record your results.
(4.7.8) How do your answers compare? You may wish to pool your answers with those of other students.
4.8 Edge
An important concept that plays a role not only in gambling but in decision- making more generally is what is called edge. Specifically, an edge is special information that makes an investment have a better chance — sometimes greater than a P = .5 chance — of success.
Here is a highly simplified example of an edge. You want to bet on a horse race in which eight horses, numbered one through eight, are entered. In this fictional situation, each horse has an equal chance of winning; thus, P win = 1/8 for each horse and the odds against each horse winning would be 7:1.
You would like to think that you would then win the $7 (plus your $1) if you bet on one horse and it won. Not quite. The racetrack administration must pay its overhead and make a profit so the odds are posted as something less, say 5:1 for each horse.
But now you get an edge. A trusted tout (assuming there is such a person in the real world) informs you that the four even numbered horses are worthless nags and thus a horse with an odd number is sure to win. You still don’t know which horse will win, but your odds have improved substantially. Here is the situation now:
Assuming your additional information (your edge) is accurate, you can now bet as much money as you are willing to wager, let’s say $40, by dividing it among the prospective winning horses, betting $10 on each horse with an odd number. When one of those horses wins, three of your tickets will be worthless, but for the fourth you will be paid $60 since the racetrack odds is 5:1. Your edge has allowed you to come out $20 ahead.
In the real world rarely is there an edge as great as the one in that fictional account.62 But successful investors study international markets with great care to find opportunities — often investments in promising young companies — where they can gain such an edge.
Exercises 4.8
(4.8.1) Rework the horse racing example of this section when you have still better information:
(a) Five of the eight horses have no chance of winning.
(b) Six of the eight horses have no chance of winning.
(c) Seven of the eight horses have no chance of winning.
(4.8.2) This exercise extends exercise 4.8.1.(c). In an old comic strip the title character, Li’l Abner, twice bets Bet-a-Million Bashby who has talked him into a coin flip for which Bashby has announced, ”Heads I win, tails you lose.” Despite this seemingly sure bet against him, Abner wins both times. How do you think that happened? (This was a comic strip.)
(4.8.3) A different but related meaning of edge is the share of each bet that goes to the betting establishment — often called the house. We calculated this for one casino game and provided it for another. Look back to see what the house edge is for:
(a) roulette
(b) twenty-one
(4.8.4) Suggest some legal ways you might improve your own edge in competition:
(a) for better grades
(b) to become a member of an athletic team
(c) for a job
Math works for cheats as well
I close this chapter with a scam described by John Allen Paulos in his book, A Mathematician Plays the Stock Market:
Someone claiming to be the publisher of a stock newsletter rents a mailbox in a fancy neighborhood, has expensive stationery made up, and sends out letters to potential subscribers boasting of his sophisticated stock-picking software, financial acumen, and Wall Street connections. He writes also of his amazing track record, but notes that the recipients of his letters needn’t take his word for it.
Assume you are one of these recipients and for the next six weeks you receive correct predictions about a certain common stock index. Would you subscribe to the newsletter?
Here’s the scam. The newsletter publisher sends out 64,000 letters to potential subscribers. To 32,000 of the recipients, he predicts the index in question will rise the following week and to the other 32,000, he predicts it will decline. No matter what happens to the index the next week, he will have made a correct prediction to 32,000 people. To 16,000 of them he sends another letter predicting a rise in the index for the following week, and to the other 16,000 he predicts a decline. Again, no matter what happens the next week, he will have made correct predictions to 16,000 people.
Focusing at each stage on the people to whom he’s made only correct pre- dictions and winnowing out the rest, he iterates this procedure a few more times until there are 1,000 people left to whom he’s made six straight correct “predictions.” To these he sends a different sort of follow-up letter, pointing out his successes and saying that they can continue to receive these oracular pronouncements if they pay the $1000 subscription price to the newsletter. If they all pay, that’s a million dollars for someone who need know nothing about stock, indices, trends, or dividends.
Lest you are tempted to carry out such a fraud, Paulos adds, “If this is done knowingly, it is illegal.”
Panel 4.8.1.
Answers for Chapter 4.
References
33Because it still causes confusion for some, it is worth reminding you that % is read percent and is an abbreviation for “per 100” or “divided by 100.” Thus we have 17% = 17/100 or .17. Similarly, the cent coin is 1/100th of a dollar.
34This situation does occur. In response to the financial crash of 2008, the federal government loaned banks large sums without stating required dates for return. Many banks promptly repaid those loans, but others felt no time pressure to repay.
35Mathematicians call an equation like (4.1.4) a recursion relation, because it summarizes a sequence of functions with each term defined by the one preceding it. Any recursion relation needs a starting point and in this case, ours is P0 = P. It is the Pn that is recurring as n runs through the integers.
36You might want to ask local bankers if you would get that $1061.21 or if instead their bank would give you $1061.20. Does the bank round to the nearest cent, thus making the transaction neutral, or does it always round down, accepting the contribution of those mils in the same way that crooked programmer did?
44Unfortunately, as this is written savings accounts are paying far less than 5%. There are, of course, other investment opportunities, but usually the risk of losing your investment goes up as the rate of return goes up. A Latin phrase applies here: caveat emptor, let the buyer beware.
45Annuity programs are usually contracted with insurance companies and some have life insurance associated with them. When your annuity is completed, you retrieve the amount paid in plus the interest generated. You can either take a lump sum payment or payments over a period of time.
54Available from their website at magazine.byu.edu/article/gambling-what-are-the-odds/.
59There is a significant qualifier to this statement. A method of winning at 21 by card counting is known; however, see my comments about this system on pages 110-111.
60This is excerpted from the Wikipedia entry for “Martingale (betting system)”, where the mathematical argument is included.
