SSL Notes for December 9, 2003
Notetaker today: Martin
Notetaker for Thursday: Emilie
Snackbringer for Thursday: Carl (hopefully his sister can bring
brownies)
Hal's Roto Router. Rule: Follow the arrows, rotating clockwise every time
we touch one, until we visit a new square
Contract: A summary of how the semester went. Especially useful if there
wasn't anything concrete you created. Jim can search through the
summaries in the future to see what you've done. By Dec. 31, a summary.
By Jan. 15, candid comments on how SSL went. Not sure what we'll do. In
the past, telling Jim anecdotally what worked went well, so unless that
seems like a bad idea... Melania thinks it's a good idea.
As soon as you know your Tues-Thurs afternoon schedule pretty well, send
it to Jim. (Perhaps Yahia from Jordan will be joining us?)
Recent developments and our experience with research. What are helpful
approaches? What will we pursue over the winter break and how will we
pursue?
-- What are we working on --
Jim: Trying to find a webpage he lost. There are people who have worked
on things like Farey factions and the Stern-Brocot tree in three
dimensions. The literature on 3D Farey/continued fractions might be a
good place to continue.
Jim will contact Peter Sarnak for how to analyze graphs for their
expansion properties.
Martin will finish up notes on the efficiency of calculating Fibonacci
numbers.
Paul: Newton's method for x^k. Carl figured out a, b, c for the
denominator. Paul figured out for the numerator, but then Carl had
figured it out simultaneously and sent an email. Current status is
conjecture for all the terms of the numerator and denominator.
Carl: The "why" part is all that's left. Not entirely straightforward.
"Proofish" email sent by Carl to the list. Each coefficient is a product
of two binomial coefficients, but you will not always get all the primes
from those.
Jim: The real goal is to get the formula for the coefficients. The
curious property of which primes you get is a curiosity but probably
nothing more. The full formula is most important.
Paul: Do we want to assume that our conjecture is true and prove from it,
or assume true?
Jim: Perhaps find an algebraic proof and then extend it to a combinatoric
proof. But don't just get bored with it and never find a proof of it.
Jim: Any generalizations of this result? Cubic instead of quadratic?
Doesn't work for cubic. Higher? How about a rational function which is
not a polynomial? ax+b / cx+d ? The maximum of the numerator degree and
the denominator degree; sounds kludgy but mathematically sound. Might
conceivably be nice because conceivably in the same rough setting as the
quadratic. Something to work out.
Sam: Same stuff with Newton's method. Haven't gotten a hold of the
noncommuting variable case, would like to study that and pick up on the
structure more. Plugging in 1, -1, -1 for a, b, c because things get
messy. Carl: Setting up a, b, c? Jim: Rather than taking the generic
quadratic, take a somewhat specific but still interesting quadratic: x^2 -
x - 1. Use human processing capabilities to see what's going on. Study
irrational roots: x^2 - 2. Or even rational root: x^2 - 1. Turn the
complexity knob down further. Even x^2. Fully understand the simple
cases, work your way back up.
Carl: Comment: Keeping variables for a, b, c. Doesn't think he would have
noticed the exact formula if he didn't look at all the variables. If they
had been added together, we wouldn't have seen the pattern.
Paul: Exact conjecture depends on powers of a, b, and c a lot.
Hal: Haven't done done much in the past week (due to working on applets).
Things he mentioned on Tuesday: Figure out how to test automorphism
groups, reimplement code in C. Needs to finish applets by January.
Jim: We should write up data on the girth.
Martin will forward email about girth data to Jim.
Emilie: 30-60-90 lattice. On website: From 1,1,1 triangle, moving the
degree 4 point to its nearest 12 neighbors, which ones are valid triples,
and so on. Move the 6 one way, move the 4 in the other direction. 1
resulted in a triple, 3 didn't.
Jim: Restricting to label-preserving points?
Emilie: Moving to vertices of the same degree.
Jim: May not correspond to automorphisms? Some triples work, some don't.
At the same time, some of these trading operations correspond to linear
transformations of the plane which transfer a point to a point of the same
label, and some don't.
Example: 4, 4, 8 lattice
----C----
| /|\ |
| / | \ |
| / | \ |
|/___|___\|
A B
[ x ] -> [ x + 2y]
[ y ] -> [ y ]
[ x ] -> [ 1 2 ] [ x ]
[ y ] -> [ 0 1 ] [ y ]
C'________
|\ | /|
| \ | / |
| \ | / |
| \|/ |
|----X----|
| /|\ |
| / | \ |
| / | \ |
|/___|___\|
A B
Linear maps preserve colinearity.
We want to know that the linear transformation is invertible and that it
preserves all the points.
Abby: Mainly comprehended Carl's email. Went to Emilie's website.
Couldn't find a link. Trying to apply what he (Neil Harriot) had done
with the 8-4 triangular lattice.
-- What works well for research --
Jim: We should write up what Neil did and understand it better. But now
let's talk about research experience, what works well and what doesn't.
Carl: It's on and off. Can be consuming, in a fun kind of way.
Sometimes we don't see where it's going.
Sam: My main feeling has been frustration with not having enough time
to pursue what I want. In the future, wants to make more time for
things. By the wayside, ideas which are never pursued. Which is why
everything gets document. Also, unlike the classroom experience, no
guidelines if what you are trying to prove is true or not. Freeing but
frustrating.
Jim: At times, the opposite. Don't know what to work on, as opposed to
too much. My goal is to make it so that there's always too much.
Melania: Intuition. A sense of where you are at.
Sam: I personally don't have any control over flashes of understanding.
They will come at times. It takes work, mulling it over for a long time,
leaving it, then coming back with a fresh look.
Melania: Intuition is the idea of "where to go from there", e.g., if you
have a proof how would you extend it and so forth. I've seen
mathematicians who need a theorem to prove. Other mathematicians come up
with conjectures. Some people are more famous for conjectures than
proofs: they have a sense of where the field is going.
Jim: There are people with good intuitions but no strategy for how to
develop a line of inquiry. They don't know what the next question is.
Other people know where to go but don't know what is going on. "In that
theory, this is what happen next, so maybe there is an analog in this
theory." Instincts are different than intuitions.
Melania: Your advisor can guide you but at some point, you are the one.
Jim: When you learn that, you get a Ph. D.
Carl: Control over intuition?
Jim: Intuition is like climate: It changes slowly, but it does change.
Things intuitive 10 years ago can stop being intuitive. Your own work can
be confusing.
Any other thoughts of a general nature? Anything surprising?
Frustrating (pure work)? I had a student whose major benefit from Reach
was learning that he didn't want to be a mathematician.
Emilie: Maybe sometime it will have something to do with the world?
Carl: One thing for me, when working with a lot of stuff... The answer to
things comes a lot more gradually than I would have expected. More and
more experimentation.
Paul: Anything you see right away is not worth looking at.
Jim: Working alone versus working with others?
Emilie: Sometime alone is better since you can think more clearly when
staring at something. Sometimes bouncing ideas is helpful.
Paul: Working alone is better to find a direction, but once you have it,
working together is better.
Hal: I hate working alone. I tend to get off track.
Jim: Sometimes I get the wrong idea and I spend two hours working on it,
when five minutes worth of thought would have convinced me that it was a
bad avenue to pursue.
How about computers?
Paul: Computers are helpful. The third iteration of this one problem took
half an hour by hand.
Jim: Sometimes you'll discover something by mistyping. If you enter the
wrong formula, it could still have some neat properties. One of the
Somos-type sequences was discovered when somebody forgot to square a term.
The element of happenstance from a coded representation.
What's worked and what hasn't worked?
Emilie: If a problem is too complex. For example, when we looked at the
circles inscribed in triangles. There were 9 variables or something.
Something with that many variables is gonna be pretty hard.
Jim: Another feature of that problem was only weakly motivated. Even
with an answer, it was unclear if you would get something good like the
Soddy relation. But so far nobody has seen anything good out of it.
Circle curvatures in Hal's pictures?
Hal: The curvatures are rationals but not integers? I dunno. My
pretty pictures were just for the sake of prettiness, and also I
learned some math.
Jim: With the circle thing, we didn't have a sharply defined conjecture.
Did we prove that you could get Laurent polynomials? Should we go back
and prove that? But we never said what we wanted.
Hal: Given a multivariable quadratic equation, what are the necessary and
sufficient conditions to get Laurent behavior? (Markoff, Soddy... at
most quadratic in each variable and we have n variables.) Notes on my
webpage. The Herwitz equation. Soddy doesn't look like Markoff but it
has some of the same properties, and it looks like a cluster algebra. If
it is a true cluster algebra, we know we get Laurent phenomena. We know
this from Fomin's paper. Never got far enough to algebraically justify
things. A very algebraic question but it may lead to combinatorics.
Stephen: Soddy doesn't obey the exchange relation. I checked. But the
Laurent phenomena is easy?
Jim: Can prove integrality but not Laurentness. Did I ever give a proof
that the Soddy equation gives Laurent polynomials? That's something
crisply defined. Which coefficients are positive and which are negative?
If we can find a rule for finding which coefficients are positive and
which are negative...
A problem is good if it gives nontrivial resistance when you try to solve
it, but ultimately yields.
-- BREAK --
Stephen explained the group properties of the Markoff relation: if we take
two Markoff triples, there is exactly one triple colinear to them.
Paul et al examined the nth iteration of x^2 - 1:
P(n) := n -> sum(binomial(2^n, 2*k)*x^(2*k), k=0..2^n);
Q(n) := n -> sum(binomial(2^n, 2*k+1)*x^(2*k+1), k=0..2^n-1);
Martin realized that using the linear method to compute the nth Fibonacci
number would take O(n^2) but is fairly certain that the matrix
multiplication method would take O(n*log n*log log n) using Strassen
multiplication. He will write this up.