Exam 2 Review 8.2, 8.5, 8.6,
Thm. 1 for 2 roots, Thm. 2 for 1 root Theorem 1: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has two distinct roots r 1 and r 2, Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 iff a n =α 1 r 1 n + α 2 r 2 n for n=0, 1, 2… where α 1 and α 2 are constants Theorem 2: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has only one root r 0, Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 iff a n =α 1 r 0 n + α 2 n r 0 n for n=0, 1, 2… where α 1 and α 2 are constants.
Steps for Solving 2 nd degree LHRR-K For degree 2: the characteristic equation is r 2 -c 1 r –c 2 =0 (roots are used to find explicit formula) Find characteristic equation Find roots Basic Solution: a n =α 1 r 1 n + α 2 r 2 n where r 1 and r 2 are roots of the characteristic equation Solve for α 1, α 2 to find solution Prove this is the solution
8.2 examples Ex with two roots: Let a n =7a n-1 – 10 a n-2 for n≥2; a 0 =2, a 1 =1 Find characteristic equation Find solution Ex with one root: a n =8a n-1 -16a n-2 for n≥2; a 0 =2 and a 1 =20 Find characteristic equation Find solution
Ex: a n =7a n-1 – 10 a n-2 for n≥2; a 0 =2, a 1 =1 Prove the solution you just found is a solution
8.5 - unions |A1 A2 A3| =∑|Ai| - ∑|Ai ∩ Aj| + |A1∩ A2 ∩ A3| |A1 A2 A3 A4| =∑|Ai| - ∑|Ai ∩ Aj| + ∑ |Ai∩ Aj ∩ Ak| - |A1∩ A2 ∩ A3∩ A4|
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8.6- optional Let A i =subset containing elements with property P i N(P 1 P 2 P 3 …P n )=|A 1 ∩A 2 ∩…∩A n | N(P 1 ’ P 2 ‘ P 3 ‘…P n ‘)= number of elements with none of the properties P1, P2, …Pn =N - |A1 A2 … An| =N- (∑|Ai| - ∑|Ai ∩ Aj| + … +(-1) n+1 |A1∩ A2 ∩…∩ An|) = N - ∑ N (Pi) + ∑(PiPj) -∑N(PiPjPk) +… +(-1) n N(P1P2…Pn)
Sample applications Ex 1: How many solutions does x 1 +x 2 +x 3 = 11 have where xi is a nonnegative integer with x 1 ≤ 3, x 2 ≤ 4, x 3 ≤ 6 (note: harder than previous > problems) Ex: 2: How many onto functions are there from a set A of 7 elements to a set B of 3 elements Ex. 3: Sieve- primes Ex. 4: Hatcheck-- The number of derangements of a set with n elements is Dn= n![1 - ] Derangement formula will be given.
9.1- Relations Def. of Function: f:A→B assigns a unique element of B to each element of A Def of Relation?
RSAT A relation R on a set A is called: reflexive if (a,a) R for every a A symmetric if (b,a) R whenever (a,b) R for a,b A antisymmetric : (a,b) R and (b,a) R only if a=b for a,b A transitive if whenever (a,b) R and (b,c) R, then (a,c) R for a,b,c A
RSAT A relation R on a set A is called: reflexive if aRa for every a A symmetric if bRa whenever aRb for every a,b A antisymmetric : aRb and bRa only if a=b for a,b A transitive if whenever aRb and bRc, then aRc for every a, b, c A Do Proofs of these****
Combining relations R∩S R S R – S S – R S ο R = {(a,c)| a A, c C, and there exists b B such that (a,b) R and (b,c) S} R n+1 =R n ⃘ R
Thm 1 on 9.1 Theorem 1: Let R be a transitive relation on a set A. Then R n is a subset of R for n=1,2,3,… Proof 8.2– not much on 8.2--just joins and projections
9.3 Representing relations R on A as both matrices and as digraphs (directed graphs) Zero-one matrix operations: join, meet, Boolean product M R R6 = M R5 v M R6 M R5∩R6 = M R5 ^ M R6 M R6 °R5 = M R5 M R6
9.4 Def: Let R be a relation on a set A that may or may not have some property P. (Ex: Reflexive,…) If there is a relation S with property P containing R such that S is a subset of every relation with property P containing R, then S is called the closure of R with respect to P. Find reflexive and symmetric closures
Transitive closures 9.4: Theorem 1: Let R be a relation on a set A. There is a path of length n from a to b iff (a,b) R n --In examples, find paths of length n that correspond to elements in R n
R* Find R*= Sample mid-level proofs: – R* is transitive
9.5 and 9.6 Equivalence Relations: R, S, T Partial orders: R, A, T (see definitions in other notes)
Definitions to thoroughly know & use a divides b a b mod m Relation Reflexive, symmetric, antisymmetric, transitive (not ones like asymmetric, from hw) Equivalence Relation- RST Partial Order- RAT Comparable Total Order
Definitions to apply You won’t have to state word for word, but may need to fill in gaps or apply definitions: – 8.2: outline for Thm. 1: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has two distinct roots r 1 and r 2, Then the sequence {a n } is a solution of the recurrence relation a n = ____________ iff a n = __________ for n=0, 1, 2… where______ (you fill in the gaps on Thm 1) – Ch. 9: Maximal, minimal, greatest, least element
Thereoms to know well and use 9.1: Theorem 1: Let R be a transitive relation on a set A. Then R n is a subset of R for n=1,2,3,… 9.4: Theorem 1: Let R be a relation on a set A. There is a path of length n from a to b iff (a,b) R n 9.4: Thm. 2: The transitive closure of a relation R is R* =
Mid-level proofs to be able to do Prove that a given relation R, S, A, or T using the definitions – Ex: Show (Z+,|) is antisymmemetric – Ex: Show R={(a,b)|a b mod m} on Z+ is transitive Some basic proofs by induction Let R be a transitive relation on a set A. Then R n is a subset of R for n=1,2,3,… R* is transitive Provide a counterexample to disprove that a relations is R, S, A, or T – Ex: Show R={(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,),(4,4)} on {1,2,3,4} is not transitive
Procedures to do Represent relations as ordered pairs, matrices, or digraphs Find Pxy and Jx and composite keys (sec 9.2) Create relations with designated properties (ex: reflexive, but not symmetric Determine whether a relation has a designated property Find closures (ex: reflexive, transitive) Find paths and circuits of a certain length and apply section 9.4 Thm. 1 Calculate R∩S,R S,R – S,S – R,S ο R,R n+1 =R n ⃘ R Given R, describe an ordered pair in R 3 Given an equivalence R on a set S, find the partition… and vice versa Identify examples and non-examples of eq. relations and of posets Create and work with Hasse diagrams: max, min, lub,…