A New Semester --- Two Classes

This semester I am teaching a mathematical models in biology (Math/Bio 342) as well as the first semester of Calculus with Functions (Math 231). So I expect to have entries for both of these courses showing up.

Values, Equations, and Theorems

There is some confusion about theorems. For example, consider the Mean Value Theorem: If f is continuous on [a,b] and differentiable on (a,b), then there is some value c∈(a,b) so that f'(c)=(f(b)-f(a))/(b-a).

Some students think that the ratio (f(b)-f(a))/(b-a) is the Mean Value Theorem. But it is not; it is just a value that is called the average rate of change of f between a and b. You can compute this value as long as both f(a) and f(b) exist. It has nothing to do with derivatives or continuity.

Other students think that the equation f'(c)=(f(b)-f(a))/(b-a). This is closer to the truth, but still is incorrect. First of all, what is c? Second, this statement may not be true. For example, suppose that f(x)=-1 if x<0 and f(x)=1 if x>0. Suppose that a=-2 and b=+2. Then the ratio (f(b)-f(a))/(b-a) is equal to 1/2. But f'(x)=0 everywhere except at x=0, where f'(0) does not exist.

Even closer is to say that f'(c)=(f(b)-f(a))/(b-a) for some c between a and b. This is actually the conclusion of the Mean Value Theorem. It requires the entire statement, particularly the statement that the equation is true for some c and that the value c must be between a and b.

But my example above provides an example where the conclusion of the Mean Value Theorem is false. That does not mean that the Mean Value Theorem itself is false. After all, it is a theorem, and that means that it has been proved to be true always. The part that is missing is the hypothesis for the theorem. The conclusion can only be guaranteed to be true using the theorem if the hypotheses are all satisfied. In this case, you must also check (or give a reason why) the function f is continuous and differentiable on the interval from a to b, including the endpoints for continuity.

Similarly, ∫_a^b f(x) dx/(b-a) computes the average value of a function f on an interval [a,b]. The value can be computed anytime the function is integrable over the interval [a,b]. The Mean Value Theorem for Integrals has nothing to do (in principle) with this calculation.

However, if f is continuous on [a,b], then
f(c)=∫_a^b f(x) dx/(b-a)
for some c∈(a,b). This entire statement comprises the Mean Value Theorem for Integrals. The hypothesis that must be verified to use the theorem is that f is continuous on [a,b]. The conclusion is that you are guaranteed that
f(c)=∫_a^b f(x) dx / (b-a)
for at least one value c between a and b.

Limit or Function Evaluation?

I've noticed that some students are perplexed about when they use a limit or function evaluation. I presume that the cause of this confusion is that students have learned that you evaluate a limit by plugging in a value. But this is only because nearly all functions that they work with are continuous.

You use a limit evaluation when you need to know what the value of the function should be by using information from the side of the point of interest. When using a limit, you must use limit notation: lim_x→c f(x). Then you use the appropriate rules of limits to evaluate (and hopefully, the function is continuous).

You use a function evaluation when you need to know the value of the function at an actual point. There is no limit involved, just function evaluation. You just use function notation, say f(c), and compute the value defined by the function.

For example, suppose you are calculating an instantaneous rate of change as the limit of an average rate of change. The average rate of change only makes sense when the interval of interest includes two points (endpoints of an interval). The instantaneous rate of change is found by seeing what the value of the average rate of change does when the two points move closer to each other, or more particularly, as the second point approaches the first point.

On the other hand, suppose that you know the derivative, which is itself a function. Then the instantaneous rate of change is calculated by function evaluation using the derivative function. (The limit was already used to create this new function.)

In particular, I have noticed this problem when dealing with finding extreme values of a function. When the interval of interest is an open interval, we are acting as though the domain does not include the endpoints. So, with this restricted domain, evaluation of the function is not possible (since the points are not in the domain). So we must use the information about the function immediately adjacent to the endpoints, and this is with a limit. In this context, the value found in the limit is not achieved at the endpoint, although it might be achieved somewhere else in the domain.

On the other hand, if the interval is a closed interval, then the endpoints are included in the restricted domain. If the function is continuous at these points, then evaluating the function directly is appropriate, since the point is in the domain.

Final remark, if the function is discontinuous at some point in the interval, you must also check the limits at that point for consideration when looking for extreme values.

Differential Equations Project Tips

As some questions are asked more regularly, I thought I'd provide some general discussion here.

(1) Start with the proposed form of X(t).  Compute X'(t) and X''(t) based on that form.  Then use those calculations to discover when X'' + k/m X = 0.

(2) The question "mean physically about the mass on a spring" is not asking you to think about the mass (as in measurement) but is asking you to think about what the statement X(0)=1 means about the state of the mass at time t=0 and what the statement X'(0)=0 means about the state of the mass at time t=0.

(3) X(0) is a constant and has derivative of d/dt[X(0)] = 0.  Recall that dX/dt > 0 implies that X is increasing, dX/dt < 0 implies that X is stationary (instantaneous rate = 0)

(4) An arbitrary quantity A is proportional to some other quantity B if it is always that case that A = k B for some constant value k (the constant of proportionality).  Now interpret the statements to identify what pieces of the equation are proportional to what.

(5) Although you have studied e^x in precalculus, you will not use the logarithm at all in this work.  Instead, I just want you to consider some function that has the special property that exp' = exp.  (This sentence is analogous to sin'=cos and cos'=-sin.)  However, you do need to think about the chain rule: X(t) = A exp(rt)  (Since the argument is not simply t, you must use the chain rule.)  This problem is exactly analogous to Step 1.

(6) You will get something like

dX/dt = "formula involving X and a, b, and m"

X is increasing when "formula" > 0, decreasing when "formula" < 0, and stationary when "formula" = 0.  So use your skills with algebra (think sign analysis) to find conditions when these are the case.

(7) You need to understand the relationship between a rate of change and an actual change.  To understand this as well as possible, see the section we skipped in Chapter 3 (last section).  But what you essentially need is that we will follow the tangent line for the time increment Δt.  How much change is there when the rate of change and the duration of time are both known?

(8) The new version of Excel has some unanticipated differences from what I had when I wrote the project.  The labels are not assigned from a menu anymore.  Instead of the 3-step process that is described, you just click in the label field in the header section of Excel and type in the new label and then hit enter.

The calculations you see in the first few lines should exactly match your hand calculations in part (7).  

Do not print the spreadsheet (it takes WAY too many pages).  That is why I ask you to submit your spreadsheet on Blackboard as part of the project.

(9) I must receive a print out of the graph -- hand drawn figures are not acceptable.   Ideally, this entire project report would be typed (perhaps using Equation Editor for the equations), with the figures naturally fitting in.

(10) Make hypotheses and test your hypotheses.

Exponential Project Tips

As some questions are asked more regularly, I thought I'd provide some general discussion here.

(1) exp is the name of the function, just as sin and cos are names of functions. From calculus, you learn that sin'=cos and cos'=-sin.  This step shows that exp_b' = ln b * exp_b. (That is, it leaves the function alone except for a constant multiple. (But be careful where the chain rule is needed!)

(2) You do not need to use the limit definition (epsilons and deltas).  Instead, for perhaps the easiest solution, you should think about how to finish the statement:

lim b^x = lim [(b^x-1)/x ... ]

That is, if you start with (b^x-1)/x, what do you do to that expression to leave only b^x.  Then use elementary limit rules to compute your resulting limit.

(3) One method is to use the method of substitution for limits (change of variables) and then use an identity for the function so that the result of Step 2 is applied --- this method mimics what is done to show that sin x is continuous everywhere.  A second method is to use a general theorem that makes continuity an obvious conclusion of the results from Step 1.

(4) You must start with a statement like:

ln (1/b) = lim_{x → 0} [(1/b)^x - 1]/x

There are two easy approaches: (1) Find a common denominator to rewrite this as a simple fraction before continuing or (2) Think of (1/b)^x as b to some appropriate power and then use a limit substitution.

(5) Since you do not know the derivative of ln x, it is incorrect to use the Mean Value Theorem applied to the logarithm.  Instead, you should apply the MVT to the function exp_b(x) on an interval so that b^a and b^b are incredibly easy and where it is clear which value is larger (so that you know if the average rate of change is positive or negative).  You may use the fact that b^x is positive for all values of x.

(6) The function fb(x) is a linear function. You should write it in slope-intercept form (e.g., mx+b).

(7) Do not attempt to solve the equation fb(x) = exp_b(x).  There is one obvious solution from the definition: x=0.  But the formulas themselves do not explain where there would not be more solutions.  Instead, you should define a function (perhaps g) so that

g(x) = exp_b(x) - fb(x).

You know that g(0) = 0.  You need to show that g(x)>0 for all x ≠ 0.  My hint suggested Rolle's theorem, but I have since found that the Mean Value Theorem helps even more.  Use the Mean Value Theorem to show that for x>0, the average rate of change between 0 and x must be positive.  What about x<0? x="0?">

(8) You may not use a limit form of the type b^∞. You may take a limit of the function fb(x) because that is of a form we know how to work with. Then you should use the result of (7) to conclude what the limit of exp_b(x) must be.

(9) and (10) put all of the previous steps together to perform analysis similar to Sections 4.2 and 4.4 to understand the graph.

Spotlight: Math Games

The other day before class, I introduced a little game called Sprouts. I found the rules summarized at the MAA website. There is also a nice discussion on the Science News website. I find it interesting that such a simple game can be analyzed using mathematical properties.

One of my favorite "math" games is a game called Eleusis. This game was invented Robert Abbott as an analogy of the scientific method. So perhaps we should call this a "science" game. The game is played with a deck (or multiple decks) of playing cards. I turn over the first card and then think of a pattern that would start with that card. Now, the remaining players take turns attempting to choose a card from their hands that they believe would be a valid next card in the pattern. If they are correct, I leave it there. If they are incorrect, I move the card out of the sequence and below the card they tried to follow (for future reference).

The goal for the players is to eventually arrive at a hypothesis that they believe explains the pattern. By playing cards, they attempt to critically assess whether their hypothesis is a complete explanation of the pattern. This mimics the scientific method because we see patterns in how nature functions, and through experiment we attempt to see if controlled efforts are consistent with or contradict our acting hypotheses.

Try it out? Let me know how the game goes.

Mathematical Induction

The principle of mathematical induction is a topic that our textbook unfortunately skips over. It is used when we want to prove a rule that applies to positive integers. Often, it is the argument that is needed when you want to say, "See! It works the same way for this case and that case, so the pattern will just keep repeating." But to say that a pattern keeps repeating is exactly what we should attempt to make more precise.

The natural numbers are the set of all positive integers: 1, 2, 3, 4, .... It is the dot-dot-dot that creates the problem. Using "..." attempts to tell us that the pattern continues. But what exactly is the pattern? For the natural numbers, the pattern is that you just add one to the previous number. So here is one way of describing the natural numbers, and it is what motivates the principle of mathematical induction.

• 1 is in the set
• For every number that is in the set, call it n, we also have n+1 in the set.

We could restate this using an implication:

• 1 is in the set
• If n is in the set, then (n+1) is in the set.

And that is what we do for all applications of mathematical induction. We provide a starting point (such as 1 is in the set). Then we establish an implication that if a statement is true for one value (n is in the set), then it must also be that the statement is true for the next value (n+1 is in the set).

Here is an example from our past that should have used induction.

Theorem: xⁿ is continuous for n=1, 2, 3, ....

Scratchwork:
Before proving this statement, we should think how we might attempt this without induction. Well, f(x) is really a product xⁿ= x x x ... x, where there are n factors of x. (See how the "..." allows us to hand-wave our notation?) Well, we know that the limit of each factor x will just go to the value c, so the limit must be lim_{x → c} f(x) = cⁿ. That use of "..." keeps us from clearly stating how we used the limit of a product, other than again referring to a pattern: "Use the limit of a product n-1 times." The use of induction makes this precise.

Proof:
We prove by induction.
1) First, we show that f(x) = x¹ = x is continuous. (This is the starting point)
But this is already known: lim_{x → c} x = c.
So f is continuous at any point c.
The statement is true when n=1.
2) Second, we assume that f(x) = xⁿ is continuous and now show that this implies that g(x) = xⁿ⁺¹ is also continuous. (This is the inductive step)
So assume that f is continuous.
g(x) = x f(x) (Relate the new function in terms of what is assumed)
So using the limit of a product:
lim_{x → c} g(x) = lim_{x → c} x f(x) = c f(c) = c cⁿ = cⁿ⁺¹ = g(c)
So xⁿ⁺¹ is continuous whenever xⁿ is continuous.

So by induction, since the statement is true for n=1, and whenever the statement is true for one value n it is also true for the next value n+1, the statement is true for all integers starting with 1.
(End of Proof)

Sometimes induction is compared to reaching different rungs on a ladder. The first statement is what allows you to climb onto the first step. The inductive implication is what says that if you have already reached one rung, then you can move to the next rung. Putting the two together, you first climb on the ladder's first rung. Then you know that you can climb from the first to the second rung, from the second rung to the third rung, from the third to the fourth, and so on forever. The implication, in one fell swoop, justifies climbing each step from the previous. The principle of mathematical induction replaces the uncertainty in "..."