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TODAY’S TOPICS:
Generating functions, traces, and eventually recurrent sequences
Popoviciu’s lemma and reciprocity
Orthogonality
Discuss:
Theorem: Define the generating function
T(t) = sum_{n geq 1} Tr(A^n) t^n.
Then
T(t) = – t Q’(t) / Q(t),
where Q(t) = det(I – tA).
Why didn’t we include an n=0 term in the sum?
Note that we can know Tr(A^n) for all integers n not equal to 0
and still not know Tr(A^0).
Also note that the sequence f(0),f(1),f(2),...
whose generating function is – t Q’(t) / Q(t) for all non-negative n
does NOT satisfy a linear recurrence relation
valid throughout the full range of definition of f(n).
This is important
for solving one of the homework problems correctly!
Theorem A: A sequence f(0),f(1),... that satisfies
a linear recurrence relation of the form
(*) f(n+d) + a_1 f(n+d-1) + a_2 f(n+d-2) + ... + a_d f(n) = 0
for all n \geq 0
has a generating function of the form
sum_{n \geq 0} f(n) x^n = P(x)/Q(x)
where Q(x) = 1 + a_1 x + a_2 x^2 + ... + a_d x^d
and P(x) is a polynomial in x of degree less than d.
Note that –t Q’(t) / Q(t) does NOT satisfy this property.
Theorem B: A sequence f(0),f(1),... that satisfies f(n)=0
for all but finitely many values of n
has a generating function of the form
sum_{n \geq 0} f(n) x^n = P(x)
where P(x) is a polynomial in x
whose degree is exactly the largest value of n
with f(n) non-zero.
Every sequence is a sum of a sequence of Type A
and a sequence of Type B.
We can see this easily on the level of polynomials,
via the division algorithm.
Theorem: Let f:N->C and suppose that
sum_{n geq 0} f(n) x^n = P(x)/Q(x),
where P,Q in C[x].
Then there is a function g:N->C that
agrees with f outside of a finite set of values
(the “exceptional set”)
such that sum_{n geq 0} g(n) x^n = R(x)/Q(x)
where deg R < deg Q;
if the exceptional set is non-empty,
its largest element is deg P – deg Q.
In fact, the generating function for f(n) – g(n)
is a polynomial of degree deg P – deg Q.
We say that such a function f is “eventually recurrent”.
For f(n)=Tr(A^n),
there is a unique way to define f(0)
so that it satisfies (*) for n=0 as well as n>0, namely: ...
f(0) = the number of non-zero eigenvalues of A.
Does the number of non-zero eigenvalues of A
have some sort of combinatorial significance?
I don’t know!
Something else that you might find useful for the homework
(which brings us to our next topic, reciprocity):
Popoviciu’s Lemma: Let d be a positive integer
and a_1,...,a_d be complex constants with a_d non-zero.
Suppose f:Z->C is a function satisfying
f(n+d) + a_1 f(n+d-1) + ... + a_d f(n) = 0
for ALL n in Z.
Then F(x) = sum_{n \geq 0)} f(n) x^n
and G(x) = sum_{n > 0} f(-n) x^n
are both rational functions
and satisfy G(x) = – F(1/x).
Example 1: f(n) = c^n for fixed non-zero c.
F(x) = 1 + cx + c^2 x^2 + ... = 1/(1-cx)
G(x) = c^{-1} x + c^{-2} x^2 + ...
= (x/c)/(1-x/c)) = 1/(c/x-1) = -1/(1-c/x) = -F(1/x).
Example 2: f(n) = (n choose k) for fixed k.
(n choose k) - (n-1 choose k) = (n-1 choose k-1)
Multiply by x^n and sum over all non-negative n:
F_k – x F_k = x F_{k-1}
F(x) = x^k / (1-x)^{k+1} (check!).
g(n) = (-n choose k) = (-1)^k (n+k-1 choose k)
Multiply by x^n and sum over all positive n:
G(x) = (-1)^k x / (1-x)^{k+1}
Check: -F(1/x) = -(1/x^k) / (1-1/x)^{k+1}
= -x / (x-1)^{k+1} = (-1)^k / (1-x)^{k+1}.
Proof sketch: f( ), F( ), and G( ) are all mutually determining,
and each depends linearly on the others.
So it suffices to prove this for the case where
F(x) = 1/(1-cx)^k for some c and some k,
since (by the method of partial fractions)
these functions form a basis for the space
of rational functions of x of the form P(x)/Q(x)
with deg P(x) < deg Q(x).
This is left as an exercise to you
(not a homework assignment!).
Combination of the two examples.
(Apply binomial theorem).
For another proof, see page 206 of Stanley.
(n+k-1 choose k) equals number of ways to choose
k elements of an n element set, with repetition allowed
(denoted ((n multichoose k))).
Proof by example: n=4, k=2.
11 !!+++
12 !+!++
13 !++!+
14 !+++!
22 +!!++
23 +!+!+
24 +!++!
33 ++!!+
34 ++!+!
44 +++!!
Interpret
Number of !’s: k
Number of +’s: n-1
Number of slots to fill: n+k-1
Orthogonality formulas for Stirling numbers:
Recall that the Stirling numbers of the first and second kind, being
elements of two mutually inverse matrices, satisfy the relation that
sum_m S(n,m) s(m,k) = delta_{n,k} = 1_{n=k} =
sum_m s(n,m) S(m,k)
for all natural numbers n,k, where
S(n,m) = number of partitions of [n] into m blocks
and
s(n,m) = (-1)^{n-m} times number of permutations of [n]
with m cycles
Since we now have combinatorial interpretations of S(n,m) and
c(m,k), we have a combinatorial assertion that we’d like to prove
combinatorially. Specifically:
Define a gadget of type (n,m,k) as a pair consisting of
a partition of [n] into m blocks
and a partition of [m] into k cycles,
and assign it weight (-1)^{m-k}.
If you hold n,k fixed and let m vary,
the sum of the weights of all the gadgets of type (n,m,k)
is 1 if n=k (trivial)
and 0 if n>k (non-trivial).
Did anyone find a proof of this?
Idea #1: Focus on the case n>k.
Idea #2: We want to pair up each gadget with one of opposite sign;
we’ll do this by either increasing or decreasing m by 1, while
leaving n and k fixed. We’ll use the following general principle:
General principle: Suppose S is a set with some weight function
W( ) on S and some involution Phi:S->S (i.e., Phi(Phi(s))=s
for all s in S). If W(Phi(s))= – W(s) for all s in S, then
sum_{s in S} W(s) = 0.
Idea #3: Think of a gadget as a bunch of cycles of blocks of
elements of n. E.g., here’s a (9,6,3) gadget:
({1,3} -> {2,6} -> {9} -> {1,3})
({4} -> {5} -> {4})
({7,8} -> {7,8})
For short, we’ll write
({1,3}{2,6}{9}{1,3})({4}{5})({7,8})
Thus, S(n,m)s(m,k) is a signed enumeration of permutations with k
cycles, where the things in the cycles are the m blocks B_1,...B_m
of some partition of [n] into m blocks.
Example: n=4, k=2.
m=1: none
m=2: ({1,2})({3,4})
({1,3})({2,4})
({1,4})({2,3})
({1,2,3})({4})
({1,2,4})({3})
({1,3,4})({2})
({2,3,4})({1})
Total for m=2: 7
m=3: ... ({1,2})({3}{4})
(... six like this)
... ({1,2}{3})({4})
(... twelve like this)
Total for m=3: 18
m=4: ... ({1}{2})({3}{4})
(... three like this)
... ({1}{2}{3})({4})
(... eight like this)
Total for m=4: 11
Orthogonality: 0 – 7 + 18 – 11 = 0.
An involution is an operation which when performed twice
yields the identity element.
We need a sign-reversing involution \Phi: \gamma -> \gamma’
on the set of gadgets.
STOP HERE
Omit: In a random domino tiling of a 2-by-n rectangle,
how many vertical tiles on average? What’s the variance?
Omit: Substitution systems (Fibonacci and Catalan)
Omit: Coin tossing: average number of runs,
average time till occurrence of a pattern;
variance and covariance (give examples where
the sign of the covariance is not obvious)