Notes for SSL Meeting of 4/22/2004
Notetaker:
Today: Martin
Tuesday: Sam
Thursday: Emilie
Snackbringer:
Today: Paul
Tuesday: Martin
Start: Hal was looking at the non-commuting case. He wrote:
P_{n+1} = aP_n^2 - cQ_n^2
Q_{n+1} = 2aP_nQ_n + bQ_n^2
P_0 = x, Q_0 = y.
The belief is that the non-commuting case is a "one liner" using the
q-binomial theorem.
Jim believes that instead of doing research we should focus on writing.
Sam had recently finalized his writeup.
We looked at Emilie and Paul's latest writeup, at:
http://www.sit.wisc.edu/~eahogan/folder/writup.pdf
We looked at it section by section.
In the early sections, Jim considered using notation x_{_v_} where _v_ is
a vector. If _v_ = (2, -3) then x_{_v_} = x_1^2 x_2^{-3}.
In section 2, the b-recurrence was removed.
In section 3, we examined the presence or absence of the bar.
In section 4, some equations are too long. Possibilities:
* Creative formatting using arrays
* Break up equations using English (Hal: "There's always English")
* Shrink margins and font: Start the document this way:
\documentclass[11pt]{article}
\usepackage{fullpage}
\begin{document}
Section 5 seems too short. Suggest merging with 6 or adding stuff.
In section 6, be more explicit about the integrality property.
Jim digressed about the proper use of the circled multiplication sign:
Let's say you have two rings/modules M and N. Take expressions of he form
a (X) b where a is in M and b is in N. Examine expressions of the form
Sum(a_i (X) b_i)/n.
If you have multiplication on one side, you can move it to the other side.
In other words, ka (X) b = a (X) kb.
If you take the integers tensor with themselves, you get 1 (X) 1, 1 (X) 2,
2 (X) 3, etc. But, 2 (X) 3 = 6 (X) 1 because we can take the factor 3 and
move it over to the left hand side.
Tensor multiplication is distributed over addition.
1 (X) 2 + 1 (X) 3 = 1 (X) (2+3) = 1 (X) 5
(The symbol (X) is written as \otimes in LaTeX.)
So Z (X) Z is just Z again. This is clearly not what we have in mind!
Rather, we are looking at the Cartesian product of Z with itself,
treating Z as just an indexing set, not a ring or module. So the right
notation to use is Z \times Z, not Z \otimes Z. (Getting the "Z" to
look right is another story...)
Anyway, back to the write-up:
The passage about ZxZ gives the thought process we went through when we were
trying to lift the 1D recurrence. But the lifted 2D recurrence can be pulled
out of a hat. The positive part can be stated separately. Can we separate
the conjecture into multiple pieces? The semi-heuristic procedure... some of
this stuff isn't necessary for the conjecture, which is that if you have a
family of quantities defined by certain initial conditions.. we derive the
recurrence and on the next page we say what the initial conditions are.
Actually, section 7 doesn't contain the initial conjecture. I think it
requires initial conditions that are tilted.
Paul: Not for this way of saying it... this is before the tilt. It would
just be if n < 0.
Jim: If we just put those initial conditions in, that seems ok.
A priori, you might not think that it lifts to a 2D recurrence with the
Laurent property, but it does.
( (a_{n-1}^2 + 1)/a_{n-2} for n odd
a_n = <
( (a_{n-1}^3 + 1)/a_{n-2} for n even
So, state the 2D recurrence Laurentness conjecture, and show that it
implies the first half of conjecture 2.
Nice to know where you can't extend the conjecture, but doesn't mean that
the negative part is equally important as the positive part.
Hal: Would prefer instead of ``faithful'' polynomials, to say {\em
faithful} polynomials when defining faithful.
Jim:
Reid Barton: a_n = (a_{n+1}^2 + a_{n-1})/a_{n-2}
Dana Scott: a_n = (a_{n-1}a_{n-3}+a_{n-2})/a_{n+1}
No combinatorial interpretation of DS. But mention DS as special case.
Emilie: Mentioned in introduction. Different from the others.
Jim: Include numerical evidence for faithfulness. Say "We have tested
this for blah, blah..." or "The evidence is striking for blah blah..."
Math club Monday.
High schoolers visiting.
Hyping up CURL (Collaborative Undergraduate Research)
- Break -
Snacks included Diet Coke and Krispy Kreme FUN-D Raising glazed donuts.
Discussion for choice of restaurants leaned toward Japanese (Takara) or
Italian (Tutto Pasta). A final decision was postponed to include the
input of more members.
Paul says that he will have a first draft writeup on the Cube Snake
recurrence by Tuesday. This is for straight cube snakes. The primary
thing not fully shown is the weighted version of the generating function.
Examining Sam's writeup:
http://www.sit.wisc.edu/~sslachterman/newton_proof.pdf
Examining the end of the document (page 3).
Various snippets of conversation:
* Treat a, b, c as formal variables?
* Things become irrational.
* Relatively prime.
You should be able to check what the highest degree contribution is.
P_n has larger degree than Q_n.
P_1 = ax^2 +... 1
P_2 = a(a^2 x^4) + ...3
P_3 = a(a^6 x^8) + ...7
...
If you write P_n as a polynomial with the highest degree first, the
leading term is a^{2^n-1}x^{2^n}.
k is 2^n.. k goes from 0 to 2^n... Something is wrong, appears to be no
contribution.
Carl: With Brendan, I saw that relative primality ends up being a lot
simpler.
Jim: If i is 0, ... Something is wrong, the - should be a + or something
like that...
Sam: It's the formula we've been using all along. I have proven its
correctness. Brendan has been replacing the i - 1. I don't fully
understand why Maple doesn't like that expression. We have to replace the
bottom choose index with n-k.
Jim: That surprises me.
Hal: I have it in Maple in a version that kind of works in my most recent
email on the subject.
Jim re-enters the formula into Maple.
Sam: Maple won't like i-1 on the bottom.
Hal: Don't use binomial.
Jim: Seems fine to me.
...
Does this look right?
Sam: You want to choose the top thing minus i-1.
We've never had i-1 as a choosing thing. It doesn't work for some
strange reason we've never been able to figure out.
Jim: These two versions appear to be the same.
Sam: Some terms are missing. It's missing the leading term.
Jim: I'm using the standard definition of binomial coefficients, where -1
choose 3 is -1. You need it redefined.
Hal: I think i should start at 0 and go to where it was before, 2^n - k -
1. I think the proof is wrong.
Sam: Look at the proof and tell me what's wrong.
Jim: There is a standard way of interpreting binomial coefficients. You
should not deviate from the standard or people may misinterpret your
theorem out of context.
x choose 1 = x
x choose 2 = x(x-1)/2!
x choose n = prod_{i=0}^{n-1}(x-i)/n!
Even if x is negative.
Sam: Pointing to equation (6)... the i=0 term...
There's nothing wrong in the logic until you shift the index.
Jim: There's nothing choose -1 yet. Where's the first time? (7)?
Hal: There's nothing until I told you to change the index in the formula.
We were just missing a term.
Jim: Let's go back to the beginning. It doesn't say what range k has in
(2). Oh, the binomial theorem, everyone knows k from 0.
Hal: We didn't get these problems until I told you to change the limits.
Sam: But they're the formulas we've had since the beginning.
Examining (10), we don't want to kill the correct term.
Jim has to go at this point.
Jim: Go by section, verify the formulas using Maple. Ideally import the
formulas programmatically (quick check does not reveal a converter). I've
heard rumors that Maple does not handle binomial coefficients correctly.
Sam: We need to see whether it's a small issue or something more
fundamental. I can't work on it this weekend, as I'm going out of town.
Carl: I guarantee I will be able to resolve it.
Sam: When I was writing it up, I wanted to..
Carl: It's wrong, Sam. The formula's wrong. It's OK.
- End of meeting -