Thursday, October 23, 2008

Fun with Sequences

We started the chapter on Sequences and Series in my calc II class last week. This is the chapter I have been looking forward to the most, even though, of course, all of the chapters are interesting. I decided to use one class period mostly talking about things that didn't matter for the class at all, because they are irresistibly fun.

I started with some limits of recursive sequences, which actually are (a small) part of the class. Things like
$\sqrt{2}$, $\sqrt{2\sqrt{2}}$, $\sqrt{2\sqrt{2\sqrt{2}}}$, $\ldots\rightarrow 2$

$a_1=1, a_{n+1}=1+1/a_n$

which converges to the golden ratio. I resisted talking too much about the golden ratio and the Fibonacci sequence, because I figured many students had probably already seen it, or could easily learn about it online (which I suggested they do). Of course, at that point, you might as well write out, without any simplifying, the first couple of terms in that sequence and start talking about continued fractions, which I did.

Then I jumped to the (unsolved) 3n+1 problem: Consider the piecewise rule that defines $a_{n+1}$ to be $a_n/2$ if $a_n$ is even, and $3a_n+1$ if $a_n$ is odd. Now given any starting number $a_1$, the question is: does the sequence starting at $a_1$ and following the piecewise rule always eventually end up at the cycle 1, 4, 2, 1, 4, 2, 1...? I really love stating these simple problems (simple to state, that is) for my class, and pointing out that they are unsolved. I've also already mentioned the problem of finding odd perfect numbers.

From there, it's probably time to draw some pictures. The Koch snowflake is a fun example to do. I only talked about how it has an infinite perimeter, and will return later to finding the area, once we've done a little more with series.

Once you're on fractals, it's hard to resist the Mandelbrot set. This one's a little harder to talk about in a calc II class, because it involves complex numbers. But you can still talk about taking a function $f$ and a starting value $a_1$, and considering the sequence $a_1, f(a_1), f(f(a_1)), f(f(f(a_1))),\ldots $. This sequence is what you consider to determine if a point is in the Mandelbrot set, using the function $f(z)=z^2+a_1$ (a different function for each point you consider). So I showed a couple printed out pictures of the set, and encouraged my students to go online and look for more pictures.

I regret, just a little, missing the Cantor set in my discussion. Perhaps another day.

That's all I got to in class. Of course, there's still several weeks left in the semester, and lots of fun things that can be said with sequences and series. When we get to Taylor series, I'll probably tell them a little about my research, which is analogous to the series they'll be looking at. Since I'll be talking about my research, I'll also talk a bit about topology.

While I was wandering around wikipedia, looking for more fun things to talk about, I ran across the page on the ? function. I'd not heard of it before, but it looks pretty interesting. I should probably look at it some more sometime.

Wednesday, October 8, 2008

A Common Tangent

For what value of $a$ ($a>1$) does $a^x$ have a common tangent with its inverse, $\log_a x$?

Let $c$ be the $x$-coordinate of the point of tangency, so that the $y$-coordinate is $a^c$ and $\log_a c$. Setting these $y$-coordinates equal doesn't seem to get us anywhere.

So let's move to the tangent line. The derivative of $a^x$ is $a^x\cdot \ln a$, while the derivative of $\log_a x$ is $1/(x\cdot \ln a)$. Plugging in our common point, $x=c$, and setting the two slopes equal, we get $a^c\cdot \ln a=1/(c\cdot \ln a)$. Again, not obviously useful.

What gets me through this problem is thinking about it graphically. Since $a^x$ and $\log_a x$ are inverse functions, each is the mirror image of the other across the line $y=x$. This means that their common tangent line will, in fact, be that same line, $y=x$.

Therefore, the slope of the common tangent line will be 1. Our two ways to represent the slope are $a^c\cdot \ln a$ and $1/(c\cdot \ln a)$. Setting the second equal to 1 and solving for $c$, we get $c=1/(\ln a)$, which we could also write as $(\ln e)/(\ln a)=\log_a e$. The first way we represented the point of tangency was $(c,a^c)$, but since it is on the line $y=x$, this is also $(c,c)$, so $c=a^c$.

We're almost there. In $c=a^c$, replace the left-hand side with $1/(\ln a)$, from our slope calculations. On the right-hand side, replace $c$ with $\log_a e$. Then we get $1/(\ln a)=a^{\log_a e}=e$, so $1/e=\ln a$, and we see that $a=e^{(1/e)}$.

I'm not sure why I first asked this question. I was teaching calc 1 at the time, and probably looking for interesting homework problems. I seem to recall chickening out in assigning this as a problem. Perhaps some other semester. Either way, I thought it was cute.

Update 20081229: Apparently (according to the penguin dictionary of curious and interesting numbers) this value, $e^{(1/e)}$ is also the value of $x$ for which $x^{(1/x)}$ is a maximum. This isn't entirely correct, as the maximum occurs at $x=e$, and therefore $e^{(1/e)}$ is the maximum value. Wikipedia has a page.

Friday, October 3, 2008

Origami Numbers

Today I gave a (brief) talk for the Graduate Seminar in the math department here at UVA. This is an informal setting, which is only for grad students (no professors heckling the speaker). They're generally an hour, but I only went slightly over half of that time. All the same, it was a pretty comprehensible talk. At the very least, it was comprehensible material - who knows how the talk went.

The title for my talk was Origami Numbers. I walked through the first 5 of the Huzita-Hatori(-Justin-...) axioms [wikipedia], which are enough to do standard straight edge and compass constructions. The first 4 are pretty straight forward, and it's with the 5th that you start making interesting things - parabolas. Then with the 6th axiom, we are able to obtain cubic roots, showing that origami is more powerful than a straight edge and compass. A great reference, and my starting point, for all of this, is the book Project Origami by Thomas Hull (which has lots of other goodies).

One thing I found while preparing my talk was identified as Lill's method in this paper by Alperin and Lang. It's a graphical method for finding roots of polynomials, and I'd never seen it before. The paper 'Geometric Solutions of Algebraic Equations' by Riaz talks about it, as does this site, which I just found (so I should take another look at). Lill's method draws a piecewise linear path (starting, say, at O, and ending at T), where the lengths of pieces correspond to coefficients of your chosen polynomial. Then you are supposed to draw another path from O that bounces around off the lines you made and ends at T. This line shows you one of the roots for the original polynomial (see the above references for more details). It doesn't seem particularly practical if you want to find roots (I'd go for Newton's method, if all else failed), but it's fun to have a new graphical way to think about things.

Wednesday, October 1, 2008

Homework Helper

I've been a little bit frustrated with the way the discussion session for my calculus class has been going recently. That time is set aside as a time for students to ask whatever questions they have, without me lecturing on any new content. Of course, generally the questions they have are 'can you do this homework problem?' or 'I got stuck on this problem, can we go through it?'. Generally, those are fine questions that I'm happy to answer. However, I'm getting the impression that many of the students have not yet looked at the assignment yet, and are just waiting for me to do half of it for them. Of course, this'll come back to bite them on the exam, but it's fairly frustrating all around. So I've been trying to decide what to do about it.

It occurred to me today that even just answering those questions asked by the students who have looked at the assignment isn't very efficient. If they've already started, but gotten stuck, it'd be fairly quick for me to sit down with them individually, find their error, and send them on their merry way. Even there, though, that's not what I should be doing. It's easy for me to spot errors, generally. Especially when I've already looked at the problem with several other students. But it would be hugely valuable for students to be able to find their own mistakes. It can be maddening trying to find your own mistakes, of course, but it's an important skill to have.

A good way to practice finding mistakes, even if they get all of their own problems correct, would be to help identify mistakes in other people's work. Of course, this process can be ironed out a little online. I am envisioning a system where students can go and enter the work they have on a problem, up to the point where they got stuck. Then other students could go and try to find errors in people's work. This way people that get stuck can get help whenever it's convenient for them (as opposed to waiting for office hours or something), and students can practice finding errors in work.

It seems there should be some sort of credits system involved. At the beginning of the semester, students have, say... 3 credits, or 5 or something. A credit gives you permission to ask a question. To earn credits, you submit a bug report on another person's question. Perhaps a bug report just identifies what line the error occurs on, without identifying the error. And I guess answers would need to be verified before credit is added to the person who submitted the answer. Perhaps the person asking the question verifies it?

That's about as far as I've taken the idea today. Clearly there'd have to be an easy way to enter work, perhaps with some sort of graphical formula editor. Also probably some anonymity, so you can see questions, but not who submitted them (nor who answered them?). It also seems like what might happen is that the people who have the most questions might have a hard time spotting other people's mistakes in order to earn credits to ask more questions. So perhaps there's a way to account for that. Something like... if you earn lower than an N on the exam, each point less than that gets you a free credit?

Anyway, that's a day's thought on the idea. What do you all think? Do you know of a system that does something like this already? Could something like the above idea be worthwhile and helpful? How could it fail? Where does it need improvement? What additional policies might you use?