1. Week 6 - Friday

2.  What did we talk about last time?  Solving recurrence relations

4.  This is a donkey made of toothpicks  Move exactly one toothpick to make an identical donkey oriented in another direction

5. Student Lecture

7.  Second-order linear homogeneous relations with constant coefficients are recurrence relations of the following form:  a k = Aa k-1 + Ba k-2 where A, B  R and B  0  These relations are:  Second order because they depend on a k-1 and a k-2  Linear because a k-1 and a k-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

8.  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 reasons 1. We can solve it 2. It lets us get an explicit formula for Fibonacci!

9.  Which of the following are second-order linear homogeneous recurrence relations with constant coefficients? a) a k = 3a k-1 + 2a k-2 b) b k = b k-1 + b k-2 + b k-3 c) c k = (1/2)c k-1 – (3/7)c k-2 d) d k = d 2 k-1 + d k-1 d k-2 e) e k = 2e k-2 f) f k = 2f k-1 + 1 g) g k = g k-1 + g k-2 h) h k = (-1)h k-1 + (k-1)h k-2

10.  We will use a tool to solve a SOLHRRwCC called its characteristic equation  The characteristic equation of a k = Aa k-1 + Ba k-2 is:  t 2 – At – B = 0 where t  0  Note that the sequence 1, t, t 2, t 3, …, t n satisfies a k = Aa k-1 + Ba k-2 if and only if t satisfies t 2 – At – B = 0

11.  We can see that 1, t, t 2, t 3, …, t n satisfies a k = Aa k-1 + Ba k-2 if and only if t satisfies t 2 – At – B = 0 by substituting in t terms for a k as follows:  t k = At k-1 + Bt k-2  Since t  0, we can divide both sides through by t k-2  t 2 = At + B  t 2 – At – B = 0

12.  Consider a k = a k-1 + 2a k-2  What is its characteristic equation?  t 2 – t – 2 = 0  What are its roots?  t = 2 and t = -1  What are the sequences defined by each value of t?  1, 2, 2 2, 2 3, …, 2 n, …  1, -1, 1, -1, …, (-1) n, …  Do these sequences satisfy the recurrence relation?

13.  An infinite number of sequences satisfy the recurrence relation, depending on the initial conditions  If r 0, r 1, r 2, … and s 0, s 1, s 2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D  R, the following sequence also satisfies the SOLHRRwCC  a n = Cr n + Ds n, for integers n ≥ 0

14.  Solve the recurrence relation a k = a k-1 + 2a k-2 where a 0 = 1 and a 1 = 8  The result from the previous slide says that, for any sequence r k and s k that satisfy a k  a n = Cr n + Ds n, for integers n ≥ 0  We have these sequences that satisfy a k  1, 2, 2 2, 2 3, …, 2 n, …  1, -1, 1, -1, …, (-1) n, …  Thus, we have  a 0 = 1 = C2 0 + D(-1) 0 = C + D  a 1 = 8 = C2 1 + D(-1) 1 = 2C – D  Solving for C and D, we get:  C = 3  D = -2  Thus, our final result is an = 3∙2 n – 2(-1) n

15.  We can generalize this result  Let sequence a 0, a 1, a 2, … satisfy:  a k = Aa k-1 + Ba k-2 for integers k ≥ 2  If the characteristic equation t 2 – At – B = 0 has two distinct roots r and s, then the sequence satisfies the explicit formula:  a n = Cr n + Ds n  where C and D are determined by a 0 and a 1

16.  Fibonacci is defined as follows:  F k = F k-1 + F k-2, k ≥ 2  F 0 = 1  F 1 = 1  What is its characteristic equation?  t 2 – t – 1 = 0  What are its roots?  Thus,

17.  So, we plug in F 0 = 1 and F 1 = 1  Solving for C and D, we get  Substituting in, this yields

18.  Our previous technique only works when the characteristic equation has distinct roots r and s  Consider sequence a k = Aa k-1 + Ba k-2  If t 2 – At – B = 0 only has a single root r, then the following sequences both satisfy a k  1, r 1, r 2, r 3, … r n, …  0, r, 2r 2, 3r 3, … nr n, …

19.  Our old rule said that if r 0, r 1, r 2, … and s 0, s 1, s 2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D  R, the following sequence also satisfies the SOLHRRwCC  a n = Cr n + Ds n, for integers n ≥ 0  For the single root case, this means that the explicit formula is:  a n = Cr n + Dnr n, for integers n ≥ 0  where C and D are determined by a 0 and a 1

20.  To solve sequence a k = Aa k-1 + Ba k-2  Find its characteristic equation t 2 – At – B = 0  If the equation has two distinct roots r and s  Substitute a 0 and a 1 into a n = Cr n + Ds n to find C and D  If the equation has a single root r  Substitute a 0 and a 1 into a n = Cr n + Dnr n to find C and D

22.  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 set II. 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

23.  The set P of all strings of legal parenthesizations I. Base: () is in P II. 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

24.  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

26.  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

27.  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 }

28.  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 } ?

29.  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 } ?

30.  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 X Y Y X X Y Y X X Y Y

31.  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}}

32.  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?

33.  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 A c is the set of all elements of U that are not in A

34.  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 A  BA  B  B – A AcAc

35.  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)

36.  Two sets A and B are considered disjoint if A  B =   Sets A 1, A 2, … A n are mutually disjoint (or nonoverlapping) if A i  A j =  for all i  j  A collection of nonempty sets {A 1, A 2, … A n } is a partition of set A iff: 1. A = A 1  A 2  …  A n 2. A 1, A 2, … A n are mutually disjoint

37.  Given a set A, the power set of A, written P (A) or 2 A 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 2 n

38.  An ordered n-tuple (x 1, x 2, … x n ) 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 R 2 )

40.  Inclusion of Intersection:  For all sets A and B A  B  AA  B  A A  B  B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

41.  The basic way to prove that X is a subset of Y 1. Suppose that x is a particular but arbitrarily chosen element of X 2. 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

42.  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  Y 2. x  X  Y  x  X  x  Y 3. x  X – Y  x  X  x  Y 4. x  X c  x  X 5. (x, y)  X  Y  x  X  y  Y

43. 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  AA  B  A QED  Premise  Definition of intersection  Specialization  By generalization  Definition of subset

44. NameLawDual Commutative A  B = B  AA  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   = AA  U = A Complement A  A c = UA  A c =  Double Complement(A c ) c = A Idempotent A  A = AA  A = A Universal Bound A  U = UA   =  De Morgan’s (A  B) c = A c  B c (A  B) c = A c  B c Absorption A  (A  B) = AA  (A  B) = A Complements of U and  U c =  c = U Set Difference A – B = A  B c

45.  To prove that X = Y  Prove that X  Y and  Prove that Y  X

46. Theorem: 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)

47.  Let x be some element in (A  B)  (A  C) x  (A  B)  x  (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  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

48.  Prove that, for any set A, A   =   Hint: Use a proof by contradiction

50.  More on set theory  Russell’s paradox

51.  Homework 4 is due Monday  Keep reading Chapter 6