Notes 20040330, by Hal
Present were: Jim, Stephen, Hal, Emilie, Brendan, Carl, John.
Thursday's notetaker: Emilie.
Snacks on Thursday: Carl
John ran 10 miles this morning.
John wore a shirt that says "for a good prime, call 5552357!"
Tshirts with slogans are dumb.
Everyone is being paid.
Writeups:
Brendan is updating the Newton's Recurrence (NR) paper
[http://www.sit.wisc.edu/~bcyounger/]
No need to CVS to do versioning.
Brendan has written up to the "fractional linear transformation" part.
What if the roots are the same in NR? Can we just appeal to
continuity?
(Aside: if f(x)=a(xr)^2, then N^n(x)=r(rx)/2^n. Neat!)
Jim says:
You can show that successive Fibonacci numbers are relatively prime by
induction.
x \mapsto p(x)/p'(x)
p(x) = x^2x1
x \mapsto (x^2+1)/(2x1)
1/1 \mapsto 2/1 \mapsto 5/3 \mapsto 34/21 \mapsto ...
(1,1,2,3,5,21,34,... are fibb. numbers)
m/n \mapsto (m^2+n^2)/(n(2mn)) (*)
and the m/n formulation comes from the x formulation, but it will also
give back the x formulation.
Jim says something about affine vs. projective. x vs. m/n
P_n and Q_n are relatively prime. Can we prove that?
Is it true that m/n reduced gives (m^2+n^2)/(n(2mn)) in reduced form?
x \mapsto 1+1/x
gives
1/1 \mapsto 2/1 \mapsto 3/2 \mapsto 5/3 \mapsto ...
every fib. number! Is equiv to:
m/n \mapsto (m+n)/m
THEOREM: gcd(m,n)=1 ==> gcd(m^2+n^2,n(2mn))=1
Jim gave an partial proof by contradiction. Stephen claims that the
easier way to prove it would be to use arithmetic mod p. Jim's proof
begins:
Assume gcd(m,n)=1. Assume exists a prime $p$ that divides m^2+n^2 and
n(2mn). We want to show that such a prime must also divide m and n,
violating our assumption.
Since pm^2+n^2 and p2mnn^2, then pm^2+2mn=m(m+2n) (the
difference). Then, pm OR pm+2n. Proceed on a casebybase basis.
Break.
I note that (*) gives:
for F_0=0 and F_1=1,
F_{2n+1} = F_{n+1}^2 + F_{n}^2
and
F_{2n} = F_n * ( 2 * F_{n+1}  F_{n} )
Switch gears to Somostype sequences:
Lifting is hard. We have a 2d lift for the sequence family Emilie
and Paul discovered (EP). Need a combinatorial model for the EP
sequences.
Analogy to
s_n * s_{n4} = s_{n1} * s_{n3} + s_{n2}^2
which Hal says result from a hexagonshaped matching 
but Jim meant to create an analogy with the slightly
but crucially different recurrence
s_n * s_{n4} = s_{n1} * s_{n3} + s_{n2}
(not that there's a nonquadratic term). What does the
graph look like? It's in Reid Barton's paper; someone
should check.
Refer to Paul's recursion:
s_n * s_{n5} = s_{n1} + s_{n2} * s_{n3} + s_{n4}
[Note: During the meeting, I (Jim) called this the "Bernard LeCloitre
recursion", but his first name is Benoit, his last name is Cloitre,
and his recursion is
s_n * s_{n3} = s_{n1} * s_{n2} + s_{n1} + s_{n2}
so I was triply wrong!]
BC can be lifted to a 2d lattice:
. . . . . . . .
. . . . . . . .
. . D E F . . .
. . A B C . . .
. . . . . . . .
F = (B+C*D+E)/A
or
a_{i,j} * a_{i2,j1} = a_{i2,j} * a_{i,j1} + a_{i1,j} + a_{i1,j1}
This recurrence lifts both BC and EP.
I expect that there are ways to extend this recurrence to one of the form
a_{i,j} * a_{i2,j1} = a_{i2,j} * a_{i,j1} + U a_{i1,j} + V a_{i1,j1}
where U and V carry subscripts (linear functions of i and j), in such
a way that we get not just faithful polynomials but homogeneous faithful
polynomials. These subscripted U and V variables should correspond to
edges of a graph. At least, that's the hope!
U and V are edge variables. Insert values here to find out something
about the edges in the resulting graph.
\ /
u\ /v if degrees of u,v,and w all add
\ / up to 1 in each term, then the
Y picture probably looks like
 that (they share a vertex)!
w

Edge variables are better than face variables.
We recess to the other room.