Friday, October 24, 2008

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.

Friday, October 10, 2008

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, ....

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 limx → 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.

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:
limx → 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:
limx → c g(x) = limx → 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 "..."

Monday, October 6, 2008

Intermediate Value Theorem

A theorem is a statement that is always true because it has been proved. Theorems are usually stated as implications. That is, they usually are stated as "If [something is true], then [something else is true]." However, this does not mean that the hypothesis (what appears as [something is true]) is actually true. Nor does it mean that the conclusion (the statement instead of [something else is true]) is true. It means that you are guaranteed to know that the conclusion is true whenever the hypothesis is true.

When applying a theorem, it is your task to establish that the hypothesis is true. Then, by stating the theorem, you are allowed to state that the conclusion is also true.

Here is an example using the Intermediate Value Theorem. Recall that the theorem states that if you have a function f that is continuous on a closed interval [a,b] (where a and b can be any numbers with a < b), then for any y-value C between the values f(a) and f(b), you are guaranteed to be able to find a value x such that a < x < c and f(x) = C.

Here is a hypothetical situation. My car holds 12 gallons of gasoline. (That is not the hypothetical part -- I have actually filled the tank :-) I have installed an automated gas-tank tracking system that records the amount of gas as a function of the car's mileage. (Yep, that's the hypothetical part) If you ask me how much gas I had when the car was at 97,034 miles, then I can tell you it had exactly 5.93 gallons of gasoline.

Last week, I filled up my tank when the car was at 98,012 miles. This morning, I checked my car and it now records the tank as having 1.45 gallons and 98,143 miles. (All figures are also hypothetical, including mileage) So here is a question: will I actually be able to identify a mileage on the car when between that last fill up and today when the car contained exactly 4.7 gallons?

Hmm. Let's see. Imagine that we use the variable x to represent the mileage on the car. Also, let f be a function that measures the gallons in the car f(x) when the mileage is x. We know that f(98,012) = 12 and f(98,143) = 1.45. So C=4.7 is between f(98,012) and f(98,143). What does the Intermediate Value Theorem say?

Now, before you go on, I need to tell you a story. On Friday, I needed to mow the lawn. My backyard is pretty large, so it takes a while. Funny thing! I ran out of gas. I knew I had recently filled the car, so I found my gas siphon and pumped a gallon out of the car's tank and into my gas can. Phew! Glad that was available! Finished the lawn with nary a problem.

So what did you answer?

(Extra credit toward quiz grade if you answer correctly this week by e-mail: waltondb at jmu dot edu)

Thursday, October 2, 2008

Common Limit Issues

Part of the challenge of mathematics is learning the language of mathematics. Mathematics is meant to be spoken, kind of like poetry. But in my class, I'm not wanting a cinquain. Scattered thoughts that are related, but not directly connected in sentences, do not a coherent message provide.

So, here are the top ten problems in limits:

1. Dropping "lim" suddenly. If you have two expressions f(x) = g(x) where g(x) is a simplified version of f(x) (cancelled something), you should write lim f(x) = lim g(x) and NOT lim f(x)=g(x). That "lim" doesn't apply to both sides of the equal sign.

2. Writing "lim" too many times. Just because you don't want to forget to write "lim" doesn't mean you write it in front of everything. You keep writing "lim" while you massage the formula into a form where you can decide the limit. As soon as you are allowed to "plug-in" the value, you have just "taken the limit" and you should stop writing "lim".

Example: f(x) = (x^2-4)/(x-2) and g(x) = x+2. We know that f(x)=g(x) for x ≠ 2. So
limx → 2(x^2-4)/(x-2) = limx → 2(x+2) = 2+2 = 4.

3. Lonely "lim". "lim" is not simply an abbreviation for the word "the limit". It is an operator wanting to do something to a formula. It needs a formula next to it at all times. It is without a formula. So when students write "lim = 3", clearly intending to say, "the limit is 3", they are really saying, "the limit of is 3". Of what? And that is the problem. The limit is lonely and has nothing to act on.

4. Stopping at a limit form. Just because you see a zero (0) in the denominator in the limit form does not mean you are done. If the limit has form 0/0, you must try to factor and cancel. If the limit has form L/0, you must identify the sign of the function to decide whether it is going to +&infty; or -&infty;.

5. Writing "=" for undefined values. (Don't do that!) Use a limit form notation to indicate that the denominator is 0 or terms go to infinity.

6. Piecewise using x=a. A piecewise function that has a formula when x=a is a distractor for limits. Remember, a limit always determines what the function would predict if you came from the sides. So a limit never checks at x=a.

7. Writing f(a) instead of lim f(x). This is another piecewise function issue. To check the value predicted by the two sides, you need to say you are checking the sides (meaning limit). So you must write that it is a limit.

8. Using rules for x going to infinity at a. When x goes to infinity, we can ignore any terms that look like 1/x since those terms go to zero as x goes to infinity. However, when x goes to a, those terms are still numbers other than zero. Don't just forget about them (and don't even factor out the dominant terms).

9. Step-by-step when not required. Unless I explicitly ask you to show that the limit has a value using the elementary limit rules, you should just compute the limit. You don't need to spend time showing the step-by-step justification unless asked.

10. No work at all. Often you can see the limit from a graph (say on a calculator). But you need to show a reason on the paper based on mathematics that gives your answer. At the very least, say you looked at a calculator to motivate your answer.

Epsilons, Deltas and Limits... Oh My!

Yes, Toto, writing proofs of limits can be as scary as the wicked witch from the east! But do not fear, with the right direction, we can squash those problems with ease.

The first step is to realize that we are proving a limit based on its definition. Suppose we need to prove the statement written in its general form:
limxa f(x) = L
Notice that this really is saying that when x is a value close to a, the value of f(x) is close to L. The mathematical statement of this says:
"For any ε > 0, there exists a value δ > 0 so that if 0 < |x-a| < δ then |f(x)-a| < ε."

Now just because the Scarecrow is flapping in the breeze, we don't need to be afraid of this complicated looking formula. Our task is to be able to find a formula for δ in terms of ε so that once you know that the value of x is within δ of the value a (δ says how close), then the value f(x) is within ε of the value L.

To reach that fabled wizard of mathematics called a proof, we just need to follow the yellow brick road outlined below. The proof will always take a form involving four steps corresponding to the four parts of the definition:

1. For any ε > 0: We need to create a proof that works for any ε > 0. So that we have a value to work with, we start with any ε with the requirement ε > 0. So the first statement of the proof is something like, "Given ε > 0" or "Suppose ε > 0" or "Let ε > 0."

2. ... there exists δ > 0 such that: The second step is that we need to provide a recipe for how to provide δ > 0 that will make the rest of the statement true. Unfortunately, by the time we reach this step of the proof, we don't yet know what the right recipe is. Personally, I just write, "Let δ=____" and leave enough space to fill in later.

3. ... if 0 < |x-a| < δ ...: We are starting to prove an implication (if...then... statement). We are successful if we can show the conclusion is true whenever the hypothesis is true. So, to accomplish this, we assume that the hypothesis is true and see what happens. I write, "Assume 0 < |x-a| < δ."

4. ... then |f(x)-L| < ε: This is the conclusion of the implication. And this is also the hardest part of the proof. In the middle of completing the work in this proof, we will discover the recipe needed for δ. At that time, we can go back and fill in the missing pieces.

So the 4th step is the hard one. Don't be cowardly like the lion and give up; there is a method to this as well. For the polynomials that we work with, the value of |f(x)-L| will always factor into something of the form:
|f(x)-L| = |x-a| |"stuff"|
We know from step 3 that |x-a|< δ. We want |"stuff"| to be less or equal to a number, which for convenience in discussion we'll call k. Once we find that number k, then we know:
|f(x)-L| < δ k
So part of our recipe will be to make sure that δ≤ε/k. If the recipe requires no other parts, we can even just use δ=ε/k. With this knowledge, we will have found:
|f(x)-L| < δ k ≤ (ε/k)k = ε

All would be well, except that the wicked professor of the west hasn't yet told you how to find k. Let's step back a moment to the |"stuff"| factor. If you don't, I'll send my winged monkeys to bring you back :-). When we're lucky (and f(x)=mx+b), the factor is already a number. But for any other problem, there will be a formula that still involves x. In these cases, without knowing more about x, we won't know how big the extra "stuff" can become. In order to keep a handle on this "stuff" we are going to require for our recipe that δ itself never gets too large. For the simplest cases, we can require δ ≤ 1. And this means that we can take advantage of knowing that x will be between a-1 and a+1.

In a general problem, we could find the largest value for |"stuff"| using values of x between a-1 and a+1. This might take a bit of work. But for f(x) that is quadratic, "stuff" is going to be another linear looking term. We want that "stuff" to involve x-a, so use x = (x-a) + a.

For example, when a=2, the term x+1 can be rewritten x+1 = (x-2)+2+1 = (x-2)+3. The awesome Triangle Inequality then tells us:
|x+1| = |(x-2)+3| ≤ |x-2| + 3
But we know that |x-2|<δ and we required δ ≤ 1 for our recipe. So |x+1|< 4.

For another example, suppose that f(x)=x2-x, a=3, and L=6. We assume 0<|x-3|<δ and rewrite
|f(x)-L| = |x2-x-6| = |x-3||x+2|
We know |x-3|<δ and we need to find a number k so that |x+2|≤k. Since x is in "stuff", we require δ ≤ 1 and use the triangle inequality:
|x+2| = |(x-3)+3+2| ≤ |(x-3)|+5 < δ+5 ≤6
So |x+2| < 6 (This is our value k=6). Thus we also want to use δ = ε/6 in our recipe. Both requirements are taken care of by the formula δ = min(1, ε/6). So now we know
|x2-x-6| = |x+2||x-3| < 6 δ ≤ 6(ε/6) = ε

We have arrived at the emerald city of our desire and proved the limit statement
limx → 3 x2-x = 6.

But the proof needs to be in the right order:
Given ε>0.
Let δ = min(1, ε/6).
Assume 0 < |x-3| < δ. |x2-x - 6| = |x-3||x+2|
|x+2| = |x-3 + 5|
|x+2| ≤ |x-3| + 5
|x+2| < δ + 5 and δ ≤ 1 so |x+2| < 6
|x2-x-6| = |x-3||x+2| < 6δ and δ ≤ ε/6
So |x2-x-6| < ε
Thus, for all ε>0, there exists δ>0 so that if 0 < |x-3| < δ, then |(x2-x)-6| < ε.
Therefore, limx → 3 x2-x = 6