[Fig. 1.7]
The next page is going to reveal one of the key ideas behind calculus. The discussion is just about numbers—functions and slopes can wait. The numbers are not even special, they can be any numbers. The crucial point is to look at their differences:
| Suppose the numbers are | ƒ = | 0 2 6 7 4 9 |
| Their differences are | v = | 2 4 1 −3 5 |
The differences are printed in between, to show 2−0 = 2 and 6−2 = 4 and 7−6 = 1.
Notice how 4 − 7 gives a negative answer -3. The numbers in f can go up or down, the differences in v can be positive or negative. The idea behind calculus comes when you add up those differences:
2 + 4 + 1 − 3 + 5 = 9
The sum of differences is 9. This is the last number on the top line (in ƒ). Is this an accident, or is this always true? If we stop earlier, after 2 + 4 + 1, we get the 7 in ƒ. Test any prediction on a second example:
| Suppose the numbers are | ƒ = | 1 3 7 8 5 10 |
| Their differences are | v = | 2 4 1 −3 5 |
The ƒ's are increased by 1. The differences are exactly the same—no change. The sum of differences is still 9. But the last ƒ is now 10. That prediction is not right, we don't always get the last ƒ.
The first ƒ is now 1. The answer 9 (the sum of differences) is 10 − 1, the last ƒ minus the first ƒ. What happens when we change the ƒ's in the middle?
| Suppose the numbers are | ƒ = | 1 5 12 7 10 |
| Their differences are | v = | 4 7 −5 3 |
The differences add to 4 + 7 − 5 + 3 = 9. This is still 10 − 1. No matter what ƒ's we choose or how many, the sum of differences is controlled by the first ƒ and last ƒ. If this is always true, there must be a clear reason why the middle ƒ's cancel out.
The sum of differences is (5 − 1) + (12 − 5) + (7 − 12) + (10 − 7) = 10 − 1.
The 5's cancel, the 12's cancel, and the 7's cancel. It is only 10 − 1 that doesn't cancel. This is the key to calculus!
1B The differences of the ƒ's add up to (ƒlast − ƒfirst).
| The numbers grow linearly: | ƒ = | 2 3 4 5 6 7 |
| Their differences are constant: | v = | 1 1 1 1 1 |
The sum of differences is certainly 5. This agrees with 7 − 2 = ƒfast − ƒfirst. The numbers in v remind us of constant velocity. The numbers in f remind us of a straight line ƒ = vt + C. This example has v = 1 and the ƒ's start at 2. The straight line would come from ƒ = t + 2.
| The numbers are squares: | ƒ = | 0 1 4 9 16 |
| Their differences grow linearly: | v = | 1 3 5 7 |
1 + 3 + 5 + 7 agrees with 4² = 16. It is a beautiful fact that the first j odd numbers always add up to j². The v's are the odd numbers, the ƒ's are perfect squares.
Note The letter j is sometimes useful to tell which number in f we are looking at. For this example the zeroth number is f0 = 0 and the jth number is ƒj =j². This is a part of algebra, to give a formula for the ƒ's instead of a list of numbers. We can also use j to tell which difference we are looking at. The first v is the first odd number v1= 1. The jth difference is the jth odd number vj = 2j − 1.(T hus v4 is 8 − 1 = 7.) It is better to start the differences with j = 1, since there is no zeroth odd number v0.
With this notation the jth difference is vj = ƒj − ƒ j-1. Sooner or later you will get comfortable with subscripts like j and j − 1, but it can be later. The important point is that the sum of the v's equals ƒlast − ƒfirst. We now connect the v's to slopes and the ƒ's to areas.
[Fig. 1.7]
Figure 1.7 shows a natural way to graph Example 2, with the odd numbers in v and the squares in ƒ. Notice an important difference between the v-graph and the ƒ-graph. The graph of ƒ is "piecewise linear." We plotted the numbers in ƒ and connected them by straight lines. The graph of v is "piecewise constant." We plotted the differences as constant over each piece. This reminds us of the distance-velocity graphs, when the distance ƒ(t) is a straight line and the velocity v(t) is a horizontal line.
Now make the connection to slopes:
The slope of the ƒ-graph is distance up ⁄ distance across = change in ƒ ⁄ change in t = v
Over each piece, the change in t (across) is 1. The change in ƒ (upward) is the difference that we are calling v. The ratio is the slope v⁄1 or just v. The slope makes a sudden change at the breakpoints t = 1, 2, 3, .... At those special points the slope of the ƒ-graph is not defined—we connected the v's by vertical lines but this is very debatable. The main idea is that between the breakpoints, the slope of ƒ(t) is v(t).
Now make the connection to areas:
The total area under the v-graph is ƒlast − ƒfirst
This area, underneath the staircase in Figure 1.7, is composed of rectangles. The base of every rectangle is 1. The heights of the rectangles are the v's. So the areas also equal the v's, and the total area is the sum of the v's. This area is ƒlast − ƒfirst.
Even more is true. We could start at any time and end at any later time—not necessarily at the special times t = 0, 1, 2, 3, 4. Suppose we stop at t = 3.5. Only half of the last rectangular area (under v = 7) will be counted. The total area is 1 + 3 + 5 + ½ • 7 = 12.5. This still agrees with ƒlast − ƒfirst = 12.5 - 0. At this new ending time t = 3.5, we are only halfway up the last step in the ƒ-graph. Halfway between 9 and 16 is 12.5
1C The v's are slopes of ƒ(t). The area under the v-graph is ƒ(tend) − ƒ(tstart).
This is nothing less than the Fundamental Theorem of Calculus. But we have only used algebra (no curved graphs and no calculations involving limits). For now the Theorem is restricted to piecewise linear ƒ(t) and piecewise constant v(t). In Chapter 5 that restriction will be overcome.
Notice that a proof of 1 + 3 + 5 + 7 = 4² is suggested by Figure 1.7a. The triangle under the dotted line has the same area as the four rectangles under the staircase. The area of the triangle is ½ ⋅ base ⋅ height = ½⋅4⋅8, which is the perfect square 4² When there are j rectangles instead of 4, we get ½⋅j⋅2j =j² for the area.
The next examples show other patterns, where ƒ and v increase exponentially or oscillate around zero. I hope you like them but I don't think you have to learn them. They are like the special functions 2t and sin t and cos t—except they go in steps. You get a first look at the important functions of calculus, but you only need algebra. Calculus is needed for a steadily changing velocity, when the graph of ƒ is curved.
The last example will be income tax—which really does go in steps. Then Section 1.3 will introduce the slope of a curve. The crucial step for curves is working with limits. That will take us from algebra to calculus.
Start with the numbers ƒ = 1,2,4,8,16. These are "powers of 2." They start with the zeroth power, which is 2º = 1. The exponential starts at 1 and not 0. After j steps there are j factors of 2, and ƒj equals 2j. Please recognize the difference between 2j and j² and 2j. The numbers 2j grow linearly, the numbers j² grow quadratically, the numbers 2j grow exponentially. At j = 10 these are 20 and 100 and 1024. The exponential 2j quickly becomes much larger than the others.
The differences of ƒ = 1,2,4,8,16 are exactly v = 1,2,4,8.. We get the same beautiful numbers. When the ƒ's are powers of 2, so are the v's. The formula vj = 2j−1 is slightly different from ƒj = 2j, because the first v is numbered v1. (Then v1 = 2º = 1. The zeroth power of every number is 1, except that 0º is meaningless.) The two graphs in Figure 1.8 use the same numbers but they look different, because ƒ is piecewise linear and v is piecewise constant.
[Fig. 1.8]
Where will calculus come in? It works with the smooth curve ƒ(t)= 2t. This exponential growth is critically important for population and money in a bank and the national debt. You can spot it by the following test: v(t) is proportional to ƒ(t).
Remark The function 2t is trickier than t². For f = t² the slope is v = 2t. It is proportional to t and not t². For ƒ = 2t the slope is v = c2t, and we won't find the constant c = .693 ... until Chapter 6. (The number c is the natural logarithm of 2.) Problem 37 estimates c with a calculator—the important thing is that it's constant.
We have seen a forward-back motion, velocity V followed by −V. That is oscillation of the simplest kind. The graph of ƒ goes linearly up and linearly down. Figure 1.9 shows another oscillation that returns to zero, but the path is more interesting.
The numbers in ƒ are now 0,1,1,0,−1,−1,0. Since ƒ6 = 0 the motion brings us back to the start. The whole oscillation can be repeated.
The differences in v are 1,0,−1,−1,0,1. They add up to zero, which agrees with ƒlast − ƒfirst. It is the same oscillation as in ƒ (and also repeatable), but shifted in time.
The ƒ-graph resembles (roughly) a sine curve. The v-graph resembles (even more roughly) a cosine curve. The waveforms in nature are smooth curves, while these are "digitized"—the way a digital watch goes forward in jumps. You recognize that the change from analog to digital brought the computer revolution. The same revolution is coming in CD players. Digital signals (off or on, 0 or 1) seem to win every time.
The piecewise v and ƒ start again at t = 6. The ordinary sine and cosine repeat at t = 2π. A repeating motion is periodic—here the "period" is 6 or 2π. (With t in degrees the period is 360—a full circle. The period becomes 2π when angles are measured in radians. We virtually always use radians—which are degrees times 2π/360.) A watch has a period of 12 hours. If the dial shows AM and PM, the period is .
[Fig. 1.9]
The next example is a car that is driven fast for a short time. The speed is V until the distance reaches ƒ = 1, when the car suddenly stops. The graph of ƒ goes up linearly with slope V, and then across with slope zero:
| v(t)= | { | V | up to | t = T | ƒ(t)= | { | Vt | up to | t = T |
| 0 | after | t = T | 1 | after | t = T |
This is another example of "function notation." Notice the general time t and the particular stopping time T. The distance is ƒ(t). The domain of ƒ (the inputs) includes all times t≥0. The range of ƒ (the outputs) includes all distances 0≤ƒ≤1.
Figure 1.10 allows us to compare three cars—a Jeep and a Corvette and a Maserati. They have different speeds but they all reach ƒ = 1. So the areas under the v-graphs are all 1. The rectangles have height V and base T = 1 ⁄ V.
[Fig. 1.10]
Optional remark It is natural to think about faster and faster speeds, which means steeper slopes. The ƒ-graph reaches 1 in shorter times. The extreme case is a step function, when the graph of ƒ goes straight up. This is the unit step U(t), which is zero up to t = 0 and jumps immediately to U = 1 for t>0.
What is the slope of the step function? It is zero except at the jump. At that moment,which is t = 0, the slope is infinite. We don't have an ordinary velocity v(t)—instead we have an impulse that makes the car jump. The graph is a spike over the single point t = 0, and it is often denoted by δ—so the slope of the step function is called a "delta function." The area under the infinite spike is 1.
You are absolutely not responsible for the theory of delta functions! Calculus is about curves, not jumps.
Our last example is a real-world application of slopes and rates—to explain "how taxes work." Note especially the difference between tax rates and tax brackets and total tax. The rates are v, the brackets are on x, the total tax is ƒ.
Suppose you are single with taxable income of x dollars (Form 1040, line 37—after all deductions). These are the 1991 instructions from the Internal Revenue Service:
The first bracket is 0 ≤ x ≤ $20,350. (The IRS never uses this symbol ≤, but I think it is OK here. We know what it means.) The second bracket is $20,350 ≤ x ≤ $49,300. The top bracket x ≥ $49,300 pays tax at the top rate of 31%. But only the income in that bracket is taxed at that rate.
Figure 1.11 shows the rates and the brackets and the tax due. Those are not average rates, they are marginal rates. Total tax divided by total income would be the average rate. The marginal rate of.28 or .31 gives the tax on each additional dollar of income—it is the slope at the point x. Tax is like area or distance—it adds up. Tax rate is like slope or velocity—it depends where you are. This is often unclear in the news media.
[Fig. 1.11]
Question What is the equation for the straight line in the top bracket?
Answer The bracket begins at x = $49,300 when the tax is ƒ(x) = $11,158.50. The slope of the line is the tax rate .31. When we know a point on the line and the slope, we know the equation. This is important enough to be highlighted.
1D For x in the top bracket the tax is ƒ(x) = $11,158.50 + .31(x &minus $49,300).
Section 2.3 presents this "point-slope equation" for any straight line. Here you see it for one specific example. Where does the number $11,158.50 come from? It is the tax at the end of the middle bracket, so it is the tax at the start of the top bracket.
Figure 1.11 also shows a distance-velocity example. The distance at t = 2 is ƒ(2)= 40 miles. After that time the velocity is 60 miles per hour. So the line with slope 60 on the ƒ-graph has the equation
ƒ(t) = starting distance + extra distance = 40 + 60(t − 2).
The starting point is (2,40). The new speed 60 multiplies the extra time t − 2. The point-slope equation makes sense. We now review this section, with comments.
Central idea Start with any numbers in ƒ. Their differences go in v. Then the sum of those differences is ƒlast − ƒfirst.
Subscript notation The numbers are ƒ0, ƒ1, ... and the first difference is v1 = ƒ1 − ƒ0. A typical number is ƒj and the jth difference is vj = ƒj − ƒj-1. When those differences are added, all ƒ's in the middle (like ƒ1) cancel out:
v1 + v2 + … + vj = (ƒ1 − ƒ0) + (ƒ2 − ƒ1) + … + (ƒj − ƒj-1) = ƒj − ƒ0
Examples ƒi = j or j² or 2j. Then vj = 1 (constant) or 2j -1 (odd numbers) or 2j-1.
Functions Connect the ƒ's to be piecewise linear. Then the slope v is piecewise constant. The area under the v-graph from any tstart to any tend, equals ƒ(tend) − ƒ(tstart).
Units Distance in miles and velocity in miles per hour. Tax in dollars and tax rate in (dollars paid) ⁄ (dollars earned). Tax rate is a percentage like .28, with no units.
