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Tuesday, March 28, 2006



One perquisite of this job is that I have access to Adobe Illustrator. I discovered recently that it can make regular polygons on demand, and this led me to renew my interest in tesselations. One thing the program won't do is make said polygons with sides of a specified length; instead it makes them with a specified radius, which is the distance from the center point to a vertex. So in order to make different polygons with sides the same length, I had to do a little bit of trig first to approximate the radii.

I am specifically interested in tesselations of the plane using equilateral triangles, squares, and regular hexagons. Each of these can tesselate the plane if used alone, but there are a lot of interesting tesselations using combinations of them. There are infinitely many such tesselations, but I've been trying to find a way to catalogue the repeating ones. My current approach is to start with certain "base shapes" formed by taking a polygon, adding another polygon to each of its sides, and also adding any polygons that are definitely determined by these. For instance, if you have a 90 degree angle, the only thing that can go there is a square. I believe I have drawn all of these, and there are 10 with triangles in the center, 15 with squares and 28 with hexagons. (I'm not counting reflections and rotations.) It surprised me that only 6 of these 53 shapes have no lines of symmetry. There were some possibilities that I didn't draw because they wouldn't allow for eventual tesselation; namely, a hexagon and a square cannot be placed on adjacent edges of a hexagon, because the angle between them would be 30 degrees.

The next step will be to see which of these basic shapes can tesselate by themselves.

Friday, March 24, 2006



Yesterday was interesting. I took the bus to meet Mr. Nickel to go do classroom observation at the private school, but he never met me at the bus station. Normally I go there on Tuesdays, but since there was no class on Tuesday I assumed we'd do it on Thursday, the same as last week. He failed to make this assumption. So, I took the 9:15 bus back, leaving me with about 2 1/2 hours to kill before Mr. Eby was expecting me back. Since he wasn't in the office, I left my bookbag there with the secretary and used the time to explore the town a bit, get a haircut, and eat brunch. It took me a long time to find a place to get a haircut, and I apparantly found the only barbershop in town. It being a tourist town, they also sell antiques, and the inside does not look like a barber shop. The chair you sit in to get a haircut is itself an antique; it is over 100 years old, but still in good enough condition that they use it every day. The lady said some people are horrified to learn they're using it as opposed to putting it behind glass, but she said after all that's what it's made for, adding that they'll sell it if they run into an interested buyer willing to pay a lot of money. I noticed that they had gold stars for sale that were shaped like stellated dodecahedrons. I have a Christmas ornament that looks similar in design but is a stellated icosahedron.

Sunday, March 19, 2006



Okay, more about that Fibonacci variant. You've got an infinite sequence, right? It looks like this:

1, 1, 1, 2, 3, 4, 6, 9, 13, 19....

For any given number other than the first three, f(n) = f(n-1) + f(n-2). So I was wondering: Is it possible to tack numbers onto the left, so that the pattern continues to hold? Consider:

X, 1, 1, 1....

For the pattern to hold with the third 1, we need 1 + X = 1. Which gives X = 0. So now we have:

0, 1, 1, 1, 2, 3, 4, 6....

Tack on another one, and we get:

Y, 0, 1, 1...

So Y + 1 = 1, yielding Y = 0, and:

0, 0, 1, 1, 1, 2, 3....

Carry this process out a few more terms and you have:

....-1, 5, -3, -2, 3, 0, -2, 1, 1, -1, 0, 1, 0, 0, 1, 1, 1, 2, 3, 4, 6, 9, 13, 19....

Doesn't it seem weird that on the righthand side the sequence is monotonically increasing, probably with all positive derivatives, but on the lefthand side it's so chaotic?

The same sequence that's being generated off to the left can be formed in its own right with the following definition:

f(n) = f(n-3) - f(n-2), n > 2

Thus for the first few terms we get:

1, 1, 1, 0, 0, 1, 0, -1, 1, 1, -2, 0, 3, -2, -3, 5, 1....

I've carried it out some distance using a calculator, and the numbers continue to fluctuate between positive and negative. The also seem to be generally increasing in absolute value, but it periodically dips back close to 0 as well, which means that it may have convergent subsequences as well as divergent ones in both directions.



I went to a church this morning that looks like it will be where I continue to attend for the rest of my time here. I've never been in the position of having any difficulty choosing where to go to church before. I went with the couple I'm staying with the past two weeks and wasn't sure whether I should continue doing so, but had been hesitant about looking around because I hated the idea of going a different place every week and trying to compare and decide. However, the church I went to today is a good place for me for a few different reasons. One of the three elders (and I think the one who does most of the Sunday morning teaching) is a graduate of The Master's Seminary, which is connected (organizationally, not geographically) to The Master's College, where I went to school as an undergraduate. Another guy who attends there was actually in my graduating class at TMC, and lived in the same dorm as me. It's also a fairly small church, about 100 people, which is what I'm the most comfortable with. It makes it easier to get to know people, and easier to get involved in things other than just Sunday morning. Another factor making it easier for me to get involved is that it's in the town I'm close to, rather than the one 20 miles away. And, well, it felt like coming home. It is a church family I can really be a part of.

Saturday, March 18, 2006



Okay, so I'm going to try and start this up again. I have officially relocated to Washington state, and am employed in writing a solutions manual for Mathematics: A Human Endeavor by Harold R. Jacobs, which is a pre-algebra text. The job provides room and board, and I live in an apartment which is attached to a house; one of the guys I'm working for and his wife live downstairs. They also own a vacation home rental business. On Tuesdays (or sometimes Thursdays) and Fridays I do classroom observations, where the curriculum I'm writing solutions for is being field-tested in classrooms (some at a private school and some with homeschoolers), in preparation for filming video lectures. The rest of the time I work from my apartment.

Today I got to give the last half of the lecture to the third homeschool class, which consists of two girls taking prealgebra. I was trying to help them review for a test, and we were going over conic sections. Not having seen the lectures on this material, I was mostly guessing from what was in the review section (no pun intended). In trying to explain the basic concept of conic sections, I used a styrofoam coffee cup as a prop. It's not quite a cone, but its convex hull is a truncated cone and imagination can supply the rest.

Chapter 2 of this Jacobs text is about number sequences. Earlier I was revisiting a number sequence that I've played with in the past. It's a Fibonacci variant which seems to me to be pretty obvious, but I've never seen it mentioned anywhere:

f(0) = f(1) = f(2) = 1
f(n) = f(n-1) + f(n-3), n > 2

1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 278, 407, 596, 874, 1281...

Compare this to the original Fibonacci sequence, also carried far enough to hit four digits:

f(0) = f(1) = 1
f(n) = f(n-1) + f(n-2), n > 1

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597....

One interesting thing about the new sequence is that any two consecutive numbers after the first three are generated by adding two completely different pairs of earlier terms. The sequence can actually be generated by a story problem similar to the original one with the rabbits. Fibonacci's rabbits took one month to mature, and reproduced a new pair every month thereafter. The new rabbits take two months to mature, but still reproduce at the same rate after doing so. Using * for a pair of baby rabbits, r for a pair of juvenile rabbits, and R for a pair of adult rabbits, the first few iterations look like this:

R *
R r *
R R r *
R R R r * *
R R R R r r * * *
R R R R R R r r r * * * *
R R R R R R R R R r r r r * * * * * *

Again, compare this with the original problem, where the intermediate stage does not exist:

R *
R R *
R R R * *
R R R R R * * *
R R R R R R R R * * * * *
R R R R R R R R R R R R R * * * * * * * *
R R R R R R R R R R R R R R R R R R R R R * * * * * * * * * * * * *

It is of course possible to make other, similar modified problems of the form:

f(0) = f(1) = ... = f(k) = 1
f(n) = f(n-1) + f(n - k - 1), n > k

For the original Fibonacci sequence, k = 1, and for the modification shown above, k = 2. Any such sequence would model a rabbit population where the time required grow to maturity is k times the length of a reproduction cycle. It's late now so the rest of this will have to wait.

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