SSL Notes for meeting #5 (9/23)
Today's note-taker: Carl (Abby next time)
Today's snack-provider: Paul (Emilie next time)
MEETING PREVIEWS: Should they be sent to everyone?
Vote was taken, approximately one hand was raised in favor.
Jim says preview could be sent to everyone, although the
Ice Breakers are supposed to be surprise...
(Ice Breaker topic changed: was previously "name a real
mathematician with a strikingly silly name")
MATH JOKE TIME
(everyone's task was to produce a (preferably bad) math joke)
JIM: Doctor, lawyer, & mathematician interviewed -- asked whether
spouse or paramour is preferred. Doc & lawyer give reasons for each.
Mathematician says "both" (to the others' surprise), "that way I can
tell the spouse I'm with the paramour, and tell the paramour I'm with
the spouse, while I just go to get work done in the office."
(Jim apologizes for giving a good math joke.)
Jim's joke sparks discussion over appropriateness of genderlessness
in math jokes. Opinions seemed to vary.
MARTIN: (Q) What's yellow and equivalent to the Axiom of Choice?
(A) Zorn's Lemma
HAL: Engineer's television is on fire - must figure out how to put
it out - uses bucket of water. Mathematician later encounters same
situation - ponders thusly, "I'll just reduce it to a problem
already solved."
STEVEN: (Q) How can you tell an outgoing mathematician?
(A) He looks at _your_ shoes when talking to you.
EMILIE: (Q) Why do mathematicians go to gentlemen's clubs?
(A) To see Mobius Strip
CARL: Engineer, physicist, & mathematician locked up by mad scientist,
each given ample supply of non-perishable canned goods, but nothing to
open them. Mad scientist returns one month later. Engineer: using
proper materials creates explosive and escapes. Physicist: calculates
angle of trajectory to break open cans, and survives in cell.
Mathematician: found already long dead, having arranged the cans to say,
"Hypothesis: if I do not find a way to open these cans, I will die.
Proof: assume the contrary..."
PAUL: (Q) What is [the derivative of] natural log cabin?
(A) one over cabin d cabin
SAM: 3 mathematicians think women are bad at calc. Bet made with
waitress. One mathematician goes to bathroom, other two plot to
fool him, telling waitress, "Just answer x squared." First one
returns. "What's the antiderivative of 2x?" She answers, "x^2 + C"
ABBY: Did you know 2 and 1 is 4? 2_|_
(on blackboard, she draws the gangsta' two-fo) |
JOHN: (Q) What do you feed a baby mathematician?
(A) Formula.
--------------------
BUSINESS
To avoid excess lugging, snacks can be stored in Jim's office.
(Remember, he has office hours 2:15-2:45 on Tues)
About the Undergraduate Projects Lab: their linux machines can't
provide us Maple or Mathmatica access, but it turns out that this
doesn't really matter, as everyone has a copy of at least one already.
SSLers show initiative by filing weekly reports / hours. (Keep up the
good work!) As for those undergrads getting paid: Give hours to Vicky
Whelan (214VV?), the clerk, whom University Regulations prohibit from
telling you what forms of ID to bring.
--------------------
MATH
Unfinished business:
Is the Fibonacci progression rule true for a straight snake-head stuck
onto anything?
The picture (1):
___A'___
| |
------o--o--o
blob $ | | (where blob is anything, and two vertices belonging
------o--o--o to blob are marked by $)
|__A__|
|____A"_____|
We decided "yes", agreeing that:
.
M(A") <--> M(A') U M(A)
giving us: |M(A")| = |M(A') U M(A)| = |M(A')| + |M(A)|
Is the arithmetic progression rule true for a bent snake-head stuck
onto anything?
Also, does it matter if the snake turns left instead of right?
Or can it keep turning one way and the rule still apply?
o--o--o o--o--o
| | | | | |
o--o--o o--o--o
| | vs | |
o--o--o o--o--o
| | | | | |
o--o--o o--o--o
What about
o
/ \
o o--o
\ / /
o----o--o
| / \ \
o--o o--o
\ /
o
Can we still use the arithmetic progression rule,
numbering the cells [2][3][4][5][6]?
We counted and it seemed to work, then we
attempted to prove it more generally...
The picture (2):
___A"___
| |
o--o
| |
------o--o --
blob $ | A'
------o--o --
|_A-_| (where A- is A, with the vertices marked by $ deleted)
|__A__|
BREAKDOWN OF M(A")
Case I: Matchings of A' with extra edge attached:
o--o
------o--o --
blob $ | A'
------o--o --
Case II: Matching of A- with 3 edges stuck on:
o o
| |
------o o
blob $
------o--o
|_A-_|
So, we can see that:
.
M(A") <--> M(A') U M(A-)
giving: |M(A")| = |M(A')| + |M(A-)|
BREAKDOWN OF M(A')
Case I: Case II:
------o o ------o--o
blob $ | blob $
------o o ------o--o
|__A__| |_A-_|
So, |M(A')| = |M(A)| + |M(A-)|
We now have:
(1) |M(A)| = |M(A)| (+ 0|M(A-)| )
(2) |M(A')| = |M(A)| + |M(A-)|
(3) |M(A")| = |M(A')| + |M(A-)|
= |M(A)| + 2|M(A-)|
Note: this is _not_ actually a bijection, but it seemed to convince us.
--------------------
BREAK
Paul's Cookies are fantastic.
Emily suggests that Paul email cookie recipe to list.
Jim and Steven are talking: "What's the archetypical contribution of a
mathematician? Noticing a new symmetry."
--------------------
BACK TO MATH
Note that these are nice LINEAR relations.
Now we're going to want to study non-linear relations.
But first, a loose end:
Someone conjectured that we could replace every left-turn by a right-turn
and vice versa. Is this true? (We seemed to say yes)
What about:
_ _ _ _ _ _
|_|_|_| |_|_|_|
_ _ _ --> _ _|_|
|_|_|_| |_|_|_|
Observe:
n 1 2 3 4 5 6 ...
F_n 1 2 3 5 8 13 ...
(F_2n)-(F_n)^2 1 1 4 9 25 64
Conjecture: F_2n = F_{n}^2 + F_{n-1}^2
Kuo's condensation formulas
Proof? (discuss in small groups, then re-convene)
Removable light/dark vertex pairs:
+--+--+--+ +--+--+--x--o--+--+--+
| | | | | | | | | | | |
x--o--+--+ +--+--+--o--x--+--+--+
| |
+--+--o--x (G1) (G2)
| | | |
+--+--+--+
graph #matches on #of matches with #with both
original one pair removed pairs removed
------------------------------------------------------
G1 29 10, 4, 10, 4 4
G2 34 15, 9, 15, 9 9
No one seems to be seeing a pattern yet: gather more data for Thursday.
Another loose end: weights. Re-do the Fibonacci recurrence with weights,
and use this to study F_{2n+2} = 3 F_{2n} - F_{2n-2}.
Convention: y for vertical connections, x for horizontal ones
o
y| Weight sum(1): y
o
x
o--o
y| y| Weight sum(2): x^2 + y^2
o--o
x
x x
o--o--o
y| y| y| Weight sum(3): yx^2 + yx^2 + y^3 = 2yx^2
o--o--o
x x
x x x x
o--o--o--o--o
y| y| y| y| y| F_n(x,y) = F_{n-1}(x,y) - F_{n-2}(x,y)
o--o--o--o--o
x x x x
Multiply to both sides:
A(x,y) F_n(x,y) = F_{n-1}(x,y) - F_{n-2}(x,y)
B(x,y) F_{n-1}(x,y) = F_{n-2}(x,y) - F_{n-2}(x,y)
C(x,y) F_{n-2}(x,y) = F_{n-3}(x,y) - F_{n-3}(x,y)
Look at weight sum(2): yx^2 + y^3 + yx^2 = y(x^2 + y^2) + x^2(y)
*** F_n(x,y) = y F_{n-1}(x,y) + x^2 F_{n-2}(x,y)
This retains some structure of the actual matchings (!)
B(x,y) = y A(x,y)
x^2 B(x,y) + y C(x,y) = 0
||
\/
A(x,y) = 1
B(x,y) = y
C(x,y) = -x^2
||
\/
F_n(x,y) - x^2 F_{n-2}(x,y) =
x^2 F_{n-2}(x,y) + y^2 F_{n-2}(x,y) - x^4 F_{n-4}(x,y)
*** F_n(x,y) + x^4 F_{n-4} = (2x^2 + y^2) F_{n-2}(x,y)
Moral: Sometimes putting in weights can give insights.
Now we can play around with weighted versions of all our old matchings:
_ _ _ _ _ _ _ _ _ _ _ _
|M_xy(|_|_|_|_|_|_|_|)| + |M_xy( _ |_|_|_| _ )| =
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
|M_xy(|_|_|_|_|_| _ )| + |M_xy( _ |_|_|_|_|_|)| + |M_xy(| |_|_|_|_|_| |)|
Also, observe F_3(3,2):
/\ /\
o==o==o
|) |) |)
o==o==o
\/ \/
_ _
is just like |_|_|, but with 3 possibilities at each horizontal connection
and 2 at each vertical connection...
Homework: Figure out weighted version of Kuo.
Also, take a look again at the relation F_{n}^2 = F_{n-1}F_{n+1} + (-1)^n
This time, see if weighting offers any new insights.