Notes for 12/11/03
Notetaker today: Emilie
Notetaker next time: Old Man Winter
Snack bringer today: Carl
Snack bringer next time: Bucky Badger (wouldn't that be cool?!)
We start the meeting without Jim because he is at another meeting. We are
told he will come around 4:30.
Steven: Have we gotten anywhere in Newton's method since Tuesday?
Hal: Please explain where you are at because it serves as good review for you,
and we learn too!
Carl: Since tuesday we have only made "fluff progress". (i'm not sure what
that means)
Basic idea-
f(x,y) permutes a function and keeps iterating. you get a polynomial
over a polynomial with cool coefficients.
Carl: We found a formula for each coefficient in the top and bottom polynomials.
Steven: Let's write it on the board.
P(n+1) = a*P(n)^2 + c*Q(n)^2 P(0) = x
Q(n+1) = Q(n)*(2a*P(n) + b*Q(n)) Q(0) = y
P(n) is top polynomial after nth iteration. Q(n) is bottom polynomial after nth
iteration.
Their conjecture:
In Q(n) each term looks like this:
z(a^r)(b^s)(c^t)(x^k)(y^m)
where z = (-1)^t*(2^n)(s+t)
( k )( s ) (the parenthesis over parenthesis
is a binomial coefficient)
and k=r-t
and r+s+t+1 = k+m = 2^n
k,m,r,s,t >= 0
a,b,c are coefficients in f(x) = ax^2+bx+c
R(0) = x/y
R(n+1) = R(n)-(f(R(n))/f'(R(n)))
Steven: So do these polynomials factor?
Sam: What would that do?
Steven: I'm not sure, it was just a suggestion. Maybe something cool?
Sam: Have you plugged in 1 and -1 for a, b, c?
Carl: no
Paul: I did 1, 0, -1 and 1, 0, 1 and got binomial coefficients, like every
other binomial coeff in a row of Pascals triangle.
Sam: Plugging in numbers may give a better/more clear combinatorial
interpretation.
Steven: It's probably good to get combinatorial proof as well as inductive
proofs of these coefficients.
Hal gets up and begins to write:
Q(n)(x,y) =
Sum(Sum((-1)^t*(2^n)(m-1-t)*a^(2^n-m-t)*b^(m-1-2t)*c^(t)*x^(2n-m)*y^(m),
( m )( t )
t = 0..??), m = 1..2^n)
note: you can pull the (2^n choose m) and x and y terms out of the inner sum.
Now was the time to break off into groups. Emilie, Abby, Steven, Hal, Martin
went into the computer lab. Paul, Carl, Sam stayed in the room and worked on
a combinatorial proof on the board.
Emilie and Abby worked on finding the triangles that correspond to the third
Markoff brother in the 30-60-90 triangular lattice.
That's all. We're done for the semester.
SEE Y'ALL NEXT SEMESTER!!!!!!!