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Logical warmup
This is a donkey made of toothpicks
Move exactly one toothpick to make an identical donkey oriented in another direction
Second-Order Linear Homogeneous Relations with Constant Coefficients Second-order linear homogeneous relations
with constant coefficients are recurrence relations of the following form: ak = Aak-1 + Bak-2 where A, B R and B 0
These relations are: Second order because they depend on ak-1 and ak-2
Linear because ak-1 and ak-2 are to the first power and not multiplied by each other
Homogeneous because there is no constant term Constant coefficients because A and B are fixed
values
Why do we care?
I'm sure you're thinking that this is an awfully narrow class of recurrence relations to have special rules for
It's true: There are many (infinitely many) ways to formulate a recurrence relation Some have explicit formulas Some do not have closed explicit formulas
We care about this one partly for two reasons1. We can solve it2. It lets us get an explicit formula for Fibonacci!
Pick 'em out
Which of the following are second-order linear homogeneous recurrence relations with constant coefficients?
a) ak = 3ak-1 + 2ak-2
b) bk = bk-1 + bk-2+ bk-3
c) ck = (1/2)ck-1 – (3/7)ck-2
d) dk = d2k-1 + dk-1dk-2
e) ek = 2ek-2
f) fk = 2fk-1 + 1g) gk = gk-1 + gk-2
h) hk = (-1)hk-1 + (k-1)hk-2
Characteristic equation
We will use a tool to solve a SOLHRRwCC called its characteristic equation
The characteristic equation of ak = Aak-1 + Bak-2 is: t2 – At – B = 0 where t 0
Note that the sequence 1, t, t2, t3, …, tn satisfies ak = Aak-1 + Bak-2 if and only if t satisfies t2 – At – B = 0
Demonstrating the characteristic equation
We can see that 1, t, t2, t3, …, tn satisfies ak = Aak-1 + Bak-2 if and only if t satisfies t2 – At – B = 0 by substituting in t terms for ak as follows: tk = Atk-1 + Btk-2
Since t 0, we can divide both sides through by tk-2
t2 = At + B t2 – At – B = 0
Using the characteristic equation
Consider ak = ak-1 + 2ak-2 What is its characteristic equation?
t2 – t – 2 = 0 What are its roots?
t = 2 and t = -1 What are the sequences defined by each
value of t? 1, 2, 22, 23, …, 2n, … 1, -1, 1, -1, …, (-1)n, …
Do these sequences satisfy the recurrence relation?
Finding other sequences
An infinite number of sequences satisfy the recurrence relation, depending on the initial conditions
If r0, r1, r2, … and s0, s1, s2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D R, the following sequence also satisfies the SOLHRRwCC an = Crn + Dsn, for integers n ≥ 0
Finding a sequence with specific initial conditions
Solve the recurrence relation ak = ak-1 + 2ak-2 where a0 = 1 and a1 = 8
The result from the previous slide says that, for any sequence rk and sk that satisfy ak an = Crn + Dsn, for integers n ≥ 0
We have these sequences that satisfy ak
1, 2, 22, 23, …, 2n, … 1, -1, 1, -1, …, (-1)n, …
Thus, we have a0 = 1 = C20 + D(-1)0 = C + D
a1 = 8 = C21 + D(-1)1 = 2C – D Solving for C and D, we get:
C = 3 D = -2
Thus, our final result is an = 3∙2n – 2(-1)n
Distinct Roots Theorem
We can generalize this result Let sequence a0, a1, a2, … satisfy:
ak = Aak-1 + Bak-2 for integers k ≥ 2 If the characteristic equation t2 – At –
B = 0 has two distinct roots r and s, then the sequence satisfies the explicit formula: an = Crn + Dsn
where C and D are determined by a0 and a1
Now we can solve Fibonacci! Fibonacci is defined as follows:
Fk = Fk-1 + Fk-2, k ≥ 2
F0 = 1
F1 = 1 What is its characteristic equation?
t2 – t – 1 = 0 What are its roots?
Thus,
251
t
nn
n DCF
251
251
We just need C and D
So, we plug in F0 = 1 and F1 = 1
Solving for C and D, we get
Substituting in, this yields
251
251
251
251
112
512
51
11
1
00
0
DCDCF
DCDCF
52
51C
52
)51( D
11
251
51
251
51
251
52
)51(2
51
52
51
nnnn
nF
Single-Root Case
Our previous technique only works when the characteristic equation has distinct roots r and s
Consider sequence ak = Aak-1 + Bak-2
If t2 – At – B = 0 only has a single root r, then the following sequences both satisfy ak
1, r1, r2, r3, … rn, … 0, r, 2r2, 3r3, … nrn, …
Single root equation
Our old rule said that if r0, r1, r2, … and s0, s1, s2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D R, the following sequence also satisfies the SOLHRRwCC an = Crn + Dsn, for integers n ≥ 0
For the single root case, this means that the explicit formula is: an = Crn + Dnrn, for integers n ≥ 0
where C and D are determined by a0 and a1
SOLHRRwCC recap
To solve sequence ak = Aak-1 + Bak-2
Find its characteristic equation t2 – At – B = 0
If the equation has two distinct roots r and s Substitute a0 and a1 into an = Crn + Dsn
to find C and D If the equation has a single root r
Substitute a0 and a1 into an = Crn + Dnrn to find C and D
Defining anything recursively
For example, we can define sets recursively
To do so, we need three things:I. Base: A statement of that certain
things are in the setII. Recursion: A set of rules saying how
new objects can be shown to be in the set based on ones that are already known to be in the set
III. Restriction: A statement that no objects belong to the set other than those coming from I and II
Example
The set P of all strings of legal parenthesizations
I. Base: () is in PII. Recursion:
a. If E is in P, so is (E)b. If E and F are in P, so is E F
III. Restriction: No configurations of parentheses are in P other than those derived from I and II
Ackermann function
Even functions can be defined recursively The Ackermann function is famous because
it grows faster than any algebraic function It is defined for all non-negative integers as
follows: A(0,n) = n + 1 A(m,0) = A(m – 1, 1) A(m,n) = A(m – 1, A(m,n – 1))
Find A(1,2) I won't make you find A(4,4), because it is
roughly
65536222
Sets
A set is a collection of elements The set is defined entirely by its
elements Order doesn't matter Repetitions don't matter (but, in general,
we write each element of set only a single time)
Examples { 1, 4, 9, 25 } { 1 } is not the same as 1 { } is the empty set
Notation
For a finite set, we can write all of the elements inside curly braces
For infinite sets, we don't have that luxury
What's in each of the following sets? A = { x R | -3 < x < 4 } B = { x Z | -3 < x < 4 } C = { x Z+ | -3 < x < 4 } D = { x Q | -3 < x < 4 }
Subsets
Perhaps the simplest relationship between sets is the subset relationship
We say that X is a subset of Y iff every element of X is an element of Y
Notation X Y Examples:
Is { 1, 2, 3 } { 1, 2, 3, 4 } ? Is { 1, 2, 3, 4 } { 1, 2, 3 } ? Is { 1, 2, 3 } { 1, 2, 3 } ? Is { } { 1, 2, 3 } ? Is { {1}, {2}, {3} } { 1, 2, 3 } ?
Proper subsets
We say that X is a proper subset of Y iff every element of X is an element of Y, but there is some element of Y that is not an element of X
Notation X Y Examples:
Is { 1, 2, 3 } { 1, 2, 3, 4 } ? Is { 1, 2, 3, 4 } { 1, 2, 3 } ? Is { 1, 2, 3 } { 1, 2, 3 } ? Is { } { 1, 2, 3 } ? Is { {1}, {2}, {3} } { 1, 2, 3 } ?
Venn diagrams
Venn diagrams show relationships between sets They can help build intuition about relationships,
but they do not prove anything Also, their usefulness is limited to 2 or 3 sets Beyond that, they become difficult to interpret Example: For X Y, there are 3 possibilities:
X Y X Y
X
Y
vs.
The idea of being an element of a set is strongly tied to the idea of one set being the subset of another
These two relationships are different, however x X means that x is an element of X Y X means that Y is a subset of X Which of the following are true?
2 {1, 2, 3} {2} {1, 2, 3} 2 {1, 2, 3} {2} {1, 2, 3} {2} {{1}, {2}} {2} {{1}, {2}}
Set equality
We say that two sets are equal if they contain exactly the same elements
Another way of saying this is that X = Y iff X Y and Y X
Examples: A = { n Z | n = 2p, p Z } B = the set of all even integers C = { m Z | m = 2q – 2, q Z } D = { k Z | k = 3r + 1, r Z }
Does A = B? Does A = C? Does A = D?
Set operations
We usually discuss sets within some superset U called the universe of discourse
Assume that A and B are subsets of U
The union of A and B, written A B is the set of all elements of U that are in either A or B
The intersection of A and B, written A B is the set of all elements of U that are in A and B
The difference of B minus A, written B – A, is the set of all elements of U that are in B and not in A
The complement of A, written Ac is the set of all elements of U that are not in A
Examples
Let U = {a, b, c, d, e, f, g} Let A = {a, c, e, g} Let B = {d, e, f, g} What are:
A B A B B – A Ac
The empty set
There is a set with no elements in it called the empty set
We can write the empty set { } or It comes up very often For example, {1, 3, 5} {2, 4, 6} = The empty set is a subset of every
other set (including the empty set)
Disjoint sets and partitions
Two sets A and B are considered disjoint if A B =
Sets A1, A2, … An are mutually disjoint (or nonoverlapping) if Ai Aj = for all i j
A collection of nonempty sets {A1, A2, … An} is a partition of set A iff:1. A = A1 A2 … An
2. A1, A2, … An are mutually disjoint
Power set
Given a set A, the power set of A, written P(A) or 2A is the set of all subsets of A
Example: B = {1, 3, 6}P(B) = {, {1}, {3}, {6}, {1,3},
{1,6}, {3,6}, {1,3,6}} Let n be the number of elements in
A, called the cardinality of A Then, the cardinality of P(A) is 2n
Cartesian product
An ordered n-tuple (x1, x2, … xn) is an ordered sequence of n elements, not necessarily from the same set
The Cartesian product of sets A and B, written A x B is the set of all ordered 2-tuples of the form (a, b), a A, b B
Thus, (x, y) points are elements of the Cartesian product R x R (sometimes written R2)
Basic subset relations
Inclusion of Intersection: For all sets A and B A B A A B B
Inclusion in Union: For all sets A and B A A B B A B
Transitive Property of Subsets: If A B and B C, then A C
Element argument
The basic way to prove that X is a subset of Y1. Suppose that x is a particular but
arbitrarily chosen element of X2. Show that x is an element of Y
If every element in X must be in Y, by definition, X is a subset of Y
Procedural versions
We want to leverage the techniques we've already used in logic and proofs
The following definitions help with this goal:1. x X Y x X x Y2. x X Y x X x Y3. x X – Y x X x Y4. x Xc x X5. (x, y) X Y x X y Y
Example proof
Theorem: For all sets A and B, A B A
Proof: Let x be some
element in A B x A x B x A Thus, all elements in
A B are in A A B AQED
Premise
Definition of intersection
Specialization By generalization
Definition of subset
Laying down the law (again)
Name Law Dual
Commutative A B = B A A B = B A
Associative (A B) C = A (B C) (A B) C = A (B C)
Distributive A (B C) = (A B) (A C) A (B C) = (A B) (A C)
Identity A = A A U = A
Complement A Ac = U A Ac =
Double Complement (Ac)c = A
Idempotent A A = A A A = A
Universal Bound A U = U A =
De Morgan’s (A B)c = Ac Bc (A B)c = Ac Bc
Absorption A (A B) = A A (A B) = A
Complements of U and
Uc = c = U
Set Difference A – B = A Bc
Example proof of equivalenceTheorem: For all sets A,B, and C, A (B C) = (A B) (A C)Proof: Let x be some element in A (B C) x A x (B C) Case 1: Let x A
x A x B x A B x A x C x A C x A B x A C x (A B) (A C)
Case 2: Let x B C x B x C x B x A x B x A B x C x A x C x A C x A B x A C x (A B) (A C)
In all possible cases, x (A B) (A C), thus A (B C) (A B) (A C)
Proof of equivalence continued Let x be some element in (A B) (A C) x (A B) x (A C) Case 1: Let x A
x A x B C x A (B C)
Case 2: Let x A x A B x A x B x B x A C x A x C x C x B x C x B C x A x B C x A (B C)
In all possible cases, x A (B C), thus (A B) (A C) A (B C) Since both A (B C) (A B) (A C) and (A B) (A C) A (B C), A (B
C) = (A B) (A C)QED