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!!!!!!!