Download
1 / 51
510 likes | 710 Vues
Week 6 - Friday. CS322. Last time. What did we talk about last time? Solving recurrence relations. Questions?. Logical warmup. This is a donkey made of toothpicks Move exactly one toothpick to make an identical donkey oriented in another direction.
E N D
Week 6 - Friday CS322
Last time • What did we talk about last time? • Solving recurrence relations
Logical warmup • This is a donkey made of toothpicks • Move exactly one toothpick to make an identical donkey oriented in another direction
Solving Second-Order Linear Homogeneous Relations with Constant Coefficients Student Lecture
Solving Second-Order Linear Homogeneous Relations with Constant Coefficients
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 reasons • We can solve it • 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? • ak = 3ak-1 + 2ak-2 • bk = bk-1 + bk-2+ bk-3 • ck = (1/2)ck-1 – (3/7)ck-2 • dk = d2k-1 + dk-1dk-2 • ek = 2ek-2 • fk = 2fk-1 + 1 • gk = gk-1 + gk-2 • 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,
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
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: • Base: A statement of that certain things are in the set • 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 • 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 • Base: () is in P • Recursion: • If E is in P, so is (E) • If E and F are in P, so is E F • 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
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 Yiff 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 Yiff 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 XY, there are 3 possibilities: X X Y X Y 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 • xX 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 = YiffX 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 AorB • The intersection of A and B, written A B is the set of all elements of U that are in AandB • 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 Aiff: • A = A1 A2 … An • A1, A2, … An are mutually disjoint
Power set • Given a set A, the power set of A, writtenP(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 Y • Suppose that x is a particular but arbitrarily chosen element of X • 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: • x X Y x X x Y • x X Y x X x Y • x X – Y x X x Y • x Xc x X • (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 A QED • Premise • Definition of intersection • Specialization • By generalization • Definition of subset
Proving set equivalence • To prove that X = Y • Prove that X Y and • Prove that Y X
Example proof of equivalence 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)
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
Proof example • Prove that, for any set A, A = • Hint: Use a proof by contradiction
Next time… • More on set theory • Russell’s paradox
