Review for Final Exam (Days 78-79)

Today starts the review for the final exams -- a sure sign that the semester is almost over.

Of course, I'm still trying to work out what exactly the finals will look like. Recall that the Calculus teacher at the main high school will be giving a 30-question final next week -- and that's at a school where the each finals block is only around 80 minutes. My own school has the traditional two-hour blocks, so working out the proportion, I ought to give 45 questions -- and that makes sense because the actual AP exam gives that many multiple choice questions.

But after the students took longer than I was hoping to finish the quiz on Tuesday, I'm not so eager to give them a 45-question test. So I'll end up giving a final of the same length as the flagship teacher -- and they'll have the full two hours to complete it. Let's hope that will be enough time for them.

I also gave the students the survey comparing my teaching methods to those of the flagship teacher. As I expected, the most popular method by the veteran teacher was giving more time to go over questions from the homework. I definitely want to implement this -- ideally, I will ask each student to name one question from the HW, and I'll try to go over as many as I can. If I can't do all of the requested problems, I will at least try to set them up so they can finish, just like the veteran teacher.

It was also split between my individual marker practice and the veteran teacher's group work. I might try stick to marker practice, but whenever I can find a group activity, use it. Perhaps I'll start with the group project that I tried to give on Tuesday but ran out of time. Not enough students supported radical changes to the homework or quiz structures for me to change them -- and recall that I wasn't too keen to make such radical changes to this class anyway. That's why I'm saving the new HW and quiz structures for Trig only.

Speaking of which, in Stats class, when I was discussing the veteran teacher and his methods, the special ed student and his aide endorsed my old quiz method -- at least the idea of doing quizzes at the end of the period, after reviewing the lesson earlier in the period. This is tricky -- on one hand, whenever I need to go over the quiz before giving it, it means that I didn't teach the lessons properly. On the other hand, he's a special ed student, and reviewing on quiz day appears to be beneficial to special ed students.

But that's enough about second semester -- we still have a first semester final to get through. I'm leaning towards making the Stats final 30 questions, just like the Calculus test.

For both Calculus and Stats, the final review consists of a battery of study guide worksheets. Then Assignments #64 and #65 in Calc, and #41 in Stats, are all review worksheets. I get the Calc questions from a Barron's study book that I purchased earlier, while the Stats questions are odds from our text.

Meanwhile, I'm afraid that my Ethnostats students aren't ready for a final yet. I'm almost starting to regret not giving them that Chapter 10 Quiz yesterday. Perhaps the students would have been ready for a quiz had I just made tomorrow the quiz day in the first place. (Hmm -- technically, I still can.)

As I promised, let me now discuss some Putnam problems -- at least the ones I discussed in class earlier this week. We'll start with Problem A1 -- typically the easiest problem on the Putnam. I gave this problem to my Stats class.

(The video is from the Art of Problem Solving website, where Putnam problems are generally posted every year.)

https://artofproblemsolving.com/community/c7h2731212_putnam_2021_a1

A1. A grasshopper starts at the origin in the coordinate plane and makes a sequence of hops. Each hop has length , and after each hop the grasshopper is at a point whose coordinates are both integers; thus, there are  possible locations for the grasshopper after the first hop. What is the smallest number of hops needed for the grasshopper to reach the point ?

And here is the answer given by the poster "blacksheep2003":

The answer is . We can achieve this by making  hops in the direction ,  hops in the direction ,  hop in the direction , and  hop in the direction . We also see this is the minimum since the sum of the coordinates of  is , and the grasshopper can only increase the sum of the coordinates of its position by at most  after each hop, so it must make at least  hops 

In discussing this question with my Stats class, I actually tied it to the upcoming Trig class -- in particular, the Pythagorean Theorem (or Distance Formula), which explains why the directions (3, 4) and (4, 3) have length 5. (One of my students immediately recognized the 3-4-5 right triangle.)

I also mentioned a simpler yet similar problem -- what's the smallest number of football possessions needed to score 80 points, given that each possession ends in either a touchdown plus an extra point (7 points) or a field goal (3 points)? The answer is 12 possessions -- 11 TD's and a field goal -- and this is minimal since the least integer not less than 80/7 is 12.

Indeed, to make this analogy complete, the original question becomes, what's the smallest number of football possessions -- offensive and defensive (assuming the defense is on the field first) -- to reach 2021 points, given that each possession ends in either a touchdown plus an extra point or a field goal (which is now worth 5 points for some reason)?

Note that on the Putnam, the year number usually appears in at least one question. This year, two problems mention the year 2021. In the other 2021 problem, knowing that 2021 = 43 * 47 goes a long way in solving it. This is why I recommend that before taking the Putnam, you should always know what the prime factorization of the current year is. (For next year, 2022 = 2 * 3 * 337.)

In my Calculus class, I gave them Problem A2 instead of A1. That's because Question A2 involves limits, which these students have learned back in Chapter 2:

A2. For every positive real number , let
Find .

I solved this problem on Saturday night when it was first posted, but the solution I came up with -- and the one I showed my Calculus class -- is a bit different from the posted solutions. Notice that the original poster uses the binomial theorem (that is, with infinitely many terms), and I also used it in a slightly different way.

g(x) = lim(x^(r + 1) + (r + 1)x^r + ... - x^(r + 1))^(1/r)

       = lim((r + 1)x^r + ...)^(1/r)

Then I proceed a bit similarly to mathisawesome, by performing the outer limit (that is, dividing by x and then taking the limit as x approaching infinity). Like mathisawesome, I write x as (x^r)^(1/r) so that I can divide by x^r inside the parentheses. (I use a to denote the desired limit, the final answer.)

a = lim(g(x)/x) = lim(lim(((r + 1)x^r + ...)/x^r)^(1/r))

And like mathisawesome, I interchange the limits. Originally, the outer limit is x -> infinity and the inner limit is r -> 0, but after the interchange, the outer limit is r -> 0 and the inner is x -> infinity. In the inner limit, since x is approaching infinity, we can ignore all terms in the numerator except x^r, which matches the denominator (hence the ellipsis):

a = lim(1 + r)^(1/r) (as r ->0)

All of the posters in the thread reach this same limit. In the thread, all posters then conclude a = e, either because this is itself a well-known limit, or by making the substitution h = 1/r and then taking the limit (1 + 1/h)^h as h approaches infinity, which is also known to be e.

But my Calculus students don't know these limits. Instead, our text defines e by the following limit:

lim (e^h - 1)/h (as h -> 0)

which is convenient in beginning Calculus to prove that e^x is its own derivative.

Hmm. What if I were to ignore limits and just proceed algebraically:

a = (1 + r)^(1/r)

a^r = 1 + r

a^r - 1 = r

(a^r - 1)/r = 1

And then taking the limit as r->0, we see that the left side matches our definition of e, so a = e. But technically, not all of these steps are valid using limits -- in particular, we can't divide by r, since r is approaching 0. (And some of the earlier steps are equally sketchy anyway.)

The only completely solutions in that thread that are rigorous enough to gain points on the Putnam involve L'Hopital's Rule in some way (since after all, the limit on the last line is 0/0). But my students won't see L'Hopital until Chapter 4 (after winter break). So I did need to do some hand-waving there.

I remind my students that while they won't see this Putnam problem on their final exams, they will see limits as x approaches infinity of rational functions. (Technically, the functions that appeared in the previous problem aren't rational functions, because in x^r, r is a real number, so x^r isn't a monomial.)

When I started reviewing the students for the final, we were all confused with what happens when x is approaching minus infinity, and the numerator degree is greater than the denominator degree. The explanation in the text is confusing.

I believe what happens here is that we ignore all terms except the leading terms in the numerator and denominator, and then the sign depends on the degrees and the signs of the leading coefficients. After all, this is what would happen if we plug in numbers like 10, 100, 1000 for x approaching infinity, and -10, -100, -1000 for x approaching minus infinity. At any rate, I'll continue to look into this and tell my students about this on Monday.

My Ethnostats classes have the weakest students, so even A1 might be too tricky. Instead, I go back to last year's Putnam (delayed to February 2021 due to the coronavirus) and do Problem B2. Here I cut and paste a description of it from the old blog:

B2. Let k and n be integers with 1 < k < n. Alice and Bob play a game with k pegs in a line of n holes. At the beginning of the game, the pegs occupy the k leftmost holes. A legal move consists of moving a single peg to any vacant hole that is further to the right. The players alternate moves, with Alice moving first. The game ends when the pegs are in the k rightmost holes, so whoever is next to play cannot move and therefore loses. For what values of n and k does Alice have a winning strategy?

This is a a very nice game to set up in the classroom. We divide the students into pairs, and they can take turns playing Alice and Bob in order to discover the winning strategy. It's also good for the teacher to play Bob against a student to play the role of Alice. (The student should be Alice since, after all, the questions asks us to find her winning strategy, not Bob's.)

We might start with, say k = 3 and n = 8. A simple game might look like this:

1. 2-7 3-4

2. 7-8 1-2

3. 2-5 4-6

4. 5-7A

This means, on the first move, Alice moves a peg from hole 2 to hole 7 and Bob moves a peg from hole 3 to hole 4. On the fourth move, Alice moves a peg from hole 5 to hole 7 and wins (A for Alice).

An easy way to find a strategy is to consider simple cases first. For example, suppose k = 1 -- that is, that there's only one peg. It's now trivial to find a winning strategy for Alice:

1. 1-8A

And this obviously works for any value of n -- just move the peg to the last hole in one move. So we see that Alice has a winning strategy for k = 1 and any value of n.

Let's see what happens when k = 2. When n = 8, we can see that Alice has no way to win -- if Bob plays correctly, he can force a win. Let's try it:

1. 1-3 2-4

2. 3-5 4-6

3. 6-7 5-8B

Bob wins by mimicking Alice -- whenever she moves one peg, he moves the other. He has mentally divided the board into pairs of holes -- 1/2, 3/4, 5/6, 7/8. Whichever hole Alice moves her peg to, Bob moves the other peg to the paired hole. Sooner or later, Alice moves a peg into either hole 7 to 8, thus allowing Bob to move the other peg into the other hole and win.

Of course, Bob can only make these pairs if n is even. And k doesn't have to be 2 -- it can also be any even number. Then both the pegs and holes are paired -- whenever Alice makes a move, Bob moves the paired peg (so if Alice moves peg 3, Bob moves peg 4 and vice versa) into the paired hole.

This suggests that Alice has a winning strategy if either k or n is odd -- that would prevent Bob from making pairs from the start. For example, if k = 2 and n = 7, Alice should start with the following move:

1. 1-3

Now with pegs in holes 2 and 3, Alice can divide the rest of the board into pairs -- 4/5 and 6/7. Now she can do to Bob what he does to her in the k, n even case -- whenever Bob moves a peg, Alice makes the paired move and wins:

1. .. .. 3-5

2. 2-4 4-7

3. 5-6A

If k is a larger even number like 4 and n is odd, Alice should begin with the following move:

1. 1-5

which allows her to pair all the remaining holes starting with 6/7, 8/9, and so on.

If k is odd, it's actually helpful for Alice to imagine that there is an invisible (k+1)st peg in the (n+1)st hole, so that there are now an even number of pegs to be paired. Alice's first move is then to pair this invisible peg with one of the real pegs. If both k and n are odd -- for example, k = 3, n = 7 -- then Alice should begin with:

1. 3-7

Now when Bob moves, he breaks the 1/2 pair, allowing Alice to make a pair on her next move. (Notice that even though both k + 1 and n + 1 are even, this invisible peg doesn't produce a win for Bob. That's because this invisible peg begins unpaired, so Alice can pair it. On the other hand, in the k, n even case, all the pegs begin paired, so Alice can only break a pair that Bob can repair on his move.)

If k is odd and n is even -- for example, k = 3, n = 8 like the original problem I gave above -- then Alice should pair the first peg with the invisible peg:

1. 1-8

Then the hole pairs are 2/3, 4/5, up to n/invisible peg.

So Bob wins with k, n both even, and Alice wins with at least one of k, n odd. This is a fascinating problem that I hope our students can enjoy.

And so indeed, I did this problem in Ethnostats. There's one thing I did this week that perhaps I should have done differently -- in the case where both k, n are odd, I often used k = 3, n = 9. After seeing the case where k, n are both even, many guessed that k = 3, n = 9 had something to do with multiples of three (which is a good guess, albeit a wrong one), Perhaps k = 3, n = 7 might be better if we want the students to discover the case when both k, n are odd.