Download MATH 4707 PROBLEM SET 3 1. Required problems

MATH 4707 PROBLEM SET 3
FINAL VERSION; DUE WEDNESDAY 3/11
1. Required problems
(1) Complete the example from class: the sequence Gn is defined for n ≥ 0 by G0 = 2,
G1 = 5 and Gn+2 = 5Gn+1 − 6Gn . Find an explicit formula for Gn .
• Exercises from the text: 4.3.2 (note that there is a solution in the back of the book,
but it contains several errors; so if you copy it unthinkingly, the results will not be
good!), 4.3.11, 4.3.13
(2) Let Hn = n · Fn , where Fn is the nth Fibonacci number.
P (So H3 = 3 · F3 = 6, for
example.) Compute the generating function H(x) := n≥0 Hn xn . [Hint: if F (x) is
d
F (x)?
the generating function for Fibonacci numbers, what are the coefficients of dx
You might try comparing to the third solution of Problem 2.5.2 given in the solutions
to P-set 1 posted on the website.]
• Exercises from the text: 7.1.6, 7.2.5, 7.3.9, 7.3.12, 8.2.2, 8.5.2, 8.5.3, 8.5.4, 7.3.10.
(3) In exercise 7.1.6, you are asked to count labeled graphs, i.e., those in which vertices
are labeled and so distinct; let’s say that the correct answer to this question is Ln .
Let Un be the number of graphs with n unlabeled vertices. Obviously Un ≤ Ln , since
every labeled graph gives rise to an unlabeled graph if you erase the vertex labels.
Prove that Un ≥ Ln!n by showing that in this erasing process, at most n! labeled graphs
can give rise to the same unlabeled graph. [Not part of the exercise are the following
two interesting facts: first (easy), Ln is much larger than n! when n is large, so the
result of this theorem is potentially interesting. Second (hard), when n is large in
fact the bound in this theorem is very good, and Un will be very close to Ln /n! (in
the sense that the ratio of the two will be close to 1).]
2. Optional problems
• Text: 4.2.5, 4.3.3, 4.3.4, 4.3.9, 4.3.14, 7.1.2, 7.1.7, 7.2.3, 7.2.7, 7.2.9, 7.3.13, 8.3.1
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