1 Examples of Weak Induction
|
|
- Emma Grant
- 6 years ago
- Views:
Transcription
1 More About Mathematical Induction Mathematical induction is designed for proving that a statement holds for all nonnegative integers (or integers beyond an initial one). Here are some extra examples of proofs by induction. 1 Examples of Weak Induction [ ] n(n + 1) Theorem 1.1. For every integer n > 0, n 3 =. [ ] 1(1 + 1) Base case: When n = 1, we have 1 3 = 1 and = 1. Inductive step: [ Let ] k 1 be an arbitrary integer, and suppose that k(k + 1) k 3 =. Then [ ] k(k + 1) k 3 + (k + 1) 3 = + (k + 1) 3 (by induction hypothesis) = k (k + 1) (k + 1)(k + 1) + = (k + k + )(k + 1) = (k + [ ] ) (k + 1) (k + 1)(k + ) =. Theorem 1.. For every integer n, n n. Base case: When n =, we have n = 16 = n. 1
2 Inductive step: Let k be an arbitrary integer, and suppose that k k. Then k+1 = k k (by induction hypothesis) = k + k k + k (since k ) = k + k + k k + k + 1 = (k + 1). Theorem 1.3. For every positive integer n, n 1 = 10 n 1. Base case: n = 1. Then 9 10 n 1 = 9 whereas 10 n 1 = 9. Inductive step: Suppose k 1 = 10 k 1 for some positive integer k. Then k k = 10 k k = k 1 1 = 10 k 1. Theorem 1.. For every positive integer n, ( 1 1 ) ( 1 1 ) ( 1 1 ) (1 1 ) 1 8 n + 1. n+1 Base case: n = 1. Then 1 1 n = 1 while n+1 = = 1 1.
3 ( Inductive step: Suppose 1 1 ) ( 1 1 ) (1 1 ) k some positive integer k. Then ( 1 1 ) ( 1 1 ) (1 1 k ) ( 1 1 k for k+1 ) ( ) ( 1 1 ) k+1 k+1 = k+1 1 k+1 k+1 = k+1 1 k+1 k+1 = 1 ( ) 1 k+1 k ( 3 1 ) 1 k+1, where we have used the fact that k 1 implies that k + 1, which in turn implies that k+1 1, which then implies that 1, which then finally k+1 implies that 1 1. So k+1 ( 1 1 as desired. ) ( 1 1 ) (1 1 k ) ( 1 1 k+1 ) 1 ( ) 1 = k+1 1 = 1 k+1 + 1, k+ 1 k+1 Examples of Strong Induction If ordinary induction is not good enough to prove that a statement is true for every positive integer, then, hopefully, strong induction or the well-ordering principle will work instead. Definition.1. The Fibonacci numbers are defined recursively by F 0 = 0; F 1 = 1; 3
4 for every integer n 1, F n+1 = F n + F n 1. Theorem.. For every n 1, F n ϕ n, where ϕ is the larger zero of the polynomial x x 1. (The constant ϕ is called the golden ratio.) Proof. We use strong induction. Base case 1: n = 1. Then F n = F 1 = 1 and ϕ n = ϕ 1 = 1/ϕ. Given the above characterization of ϕ, observe that ϕ ϕ 1 = 0 so that ϕ = ϕ + 1. Dividing both sides of this equation by 1/ϕ yields 1 = 1/ϕ + 1/ϕ 1/ϕ. So F 1 = 1 ϕ 1. Base case : n =. Then F n = F = F 1 + F 0 = = 1 and ϕ n = ϕ 0 = 1. The theorem holds in this case. Inductive step: Let k be an arbitrary integer larger than 1. Assume that for every positive integer j k, we have that F j ϕ j. Then, since k + 1 >, we may use the recursive definition of the Fibonacci numbers to conclude that F k+1 = F k + F k 1 ϕ k + ϕ k 3 = ϕ k 3 (ϕ + 1) = ϕ k 3 ϕ = ϕ k 1. (by induction hypothesis) Note that in the proof of Theorem., we needed two base cases. That is because since the recursive definition of the Fibonacci numbers rely on the previous two numbers of the sequence, we must independently verify the first two cases of the theorem in order to apply the recurrence validly.
5 This is also the reason why we need strong induction. When working out the inductive step, we need to assume that the theorem holds for more than just the previous integer. Theorem.3. (Existence and uniqueness of m-ary representations.) Let m be an integer, m. Every natural number n can be uniquely written in the form n = a 0 + a 1 m + a m + + a j m j, where j is a nonnegative integer, 1 a j m 1, and 0 a i m 1 for i = 0, 1,..., j 1. Proof. We proceed by strong induction on n. Base case: n = 0. Then take a 0 = 0 and k = 0. This case trivially holds. Inductive step: Let k be an arbitrary positive integer, k, and suppose that for every positive integer r < k, r can be uniquely written in the form r = a 0 + a 1 m + + a j m j, where j is a nonnegative integer, 1 a j m 1, and 0 a i m 1 for i = 0, 1,..., j 1. Let j be the power of m such that m j k < m j +1. Then 0 k m j < m j +1 m j = m j (m 1). Furthermore, because 0 k m j < k, we can apply the induction hypothesis on k m j to get a unique m-ary representation k m j = a 0 + a 1 m + a m + + a j m j, with j j. If j < j, then k = a 0 + a 1 m + a m + + a j m j + m j is an m-ary representation of k. If j = j, then the fact that k m j < (m 1)m j tells us that a j m so that a j + 1 m 1 and, thus, k = a 0 + a 1 m + a m + + (a j + 1)m j is an m-ary representation of k. In any event, k has its own m-ary representation. We shall now show that this representation is unique. Let k = b 0 + b 1 m + b m + + b l m l be another m-ary representation of k, where 0 b j m 1 5
6 for all j = 1,,..., l and b l 1. If l j + 1, then k < m j +1 b l m l n, which is impossible. If instead l j 1, then k = b 0 + b 1 m + b m + + b l m l (m 1) + (m 1)m + (m 1)m + + (m 1)m l = m l+1 1 (geometric sum formula from calculus) < m j k, which is again impossible. Therefore j = l. If a j < b j, then k = a 0 + a 1 m + a m + + a j 1m j 1 + a j m j (m 1) + (m 1)m + (m ), + + (m 1)m j 1 + a j m j = (m j 1) + a j m j < (a j + 1)m j b j m j n, which is also impossible. Hence b j a j. By symmetry, we must also have a j b j, and so a j = b j. Then k a j m j = a 0 +a 1 m+a m ++a j 1m j 1 = b 0 +b 1 m+b m ++b j 1m j 1, which is incidentally a nonnegative integer smaller than k. By the inductive hypothesis, a i = b i for i = 0, 1,..., j 1. It follows that the m-ary representation of k is unique. 6
Recurrence Relations
Recurrence Relations Recurrence Relations Reading (Epp s textbook) 5.6 5.8 1 Recurrence Relations A recurrence relation for a sequence aa 0, aa 1, aa 2, ({a n }) is a formula that relates each term a k
More informationINDUCTION AND RECURSION. Lecture 7 - Ch. 4
INDUCTION AND RECURSION Lecture 7 - Ch. 4 4. Introduction Any mathematical statements assert that a property is true for all positive integers Examples: for every positive integer n: n!
More informationFall 2017 Test II review problems
Fall 2017 Test II review problems Dr. Holmes October 18, 2017 This is a quite miscellaneous grab bag of relevant problems from old tests. Some are certainly repeated. 1. Give the complete addition and
More informationMath 324 Summer 2012 Elementary Number Theory Notes on Mathematical Induction
Math 4 Summer 01 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If
More informationMATH 324 Summer 2011 Elementary Number Theory. Notes on Mathematical Induction. Recall the following axiom for the set of integers.
MATH 4 Summer 011 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If
More informationMathematical Induction Assignments
1 Mathematical Induction Assignments Prove the Following using Principle of Mathematical induction 1) Prove that for any positive integer number n, n 3 + 2 n is divisible by 3 2) Prove that 1 3 + 2 3 +
More informationTHE ISLAMIC UNIVERSITY OF GAZA ENGINEERING FACULTY DEPARTMENT OF COMPUTER ENGINEERING DISCRETE MATHMATICS DISCUSSION ECOM Eng. Huda M.
THE ISLAMIC UNIVERSITY OF GAZA ENGINEERING FACULTY DEPARTMENT OF COMPUTER ENGINEERING DISCRETE MATHMATICS DISCUSSION ECOM 2011 Eng. Huda M. Dawoud December, 2015 Section 1: Mathematical Induction 3. Let
More informationDefinition: A sequence is a function from a subset of the integers (usually either the set
Math 3336 Section 2.4 Sequences and Summations Sequences Geometric Progression Arithmetic Progression Recurrence Relation Fibonacci Sequence Summations Definition: A sequence is a function from a subset
More informationHomework #2 Solutions Due: September 5, for all n N n 3 = n2 (n + 1) 2 4
Do the following exercises from the text: Chapter (Section 3):, 1, 17(a)-(b), 3 Prove that 1 3 + 3 + + n 3 n (n + 1) for all n N Proof The proof is by induction on n For n N, let S(n) be the statement
More informationMathematical Fundamentals
Mathematical Fundamentals Sets Factorials, Logarithms Recursion Summations, Recurrences Proof Techniques: By Contradiction, Induction Estimation Techniques Data Structures 1 Mathematical Fundamentals Sets
More informationMathematical Induction
Mathematical Induction MAT30 Discrete Mathematics Fall 018 MAT30 (Discrete Math) Mathematical Induction Fall 018 1 / 19 Outline 1 Mathematical Induction Strong Mathematical Induction MAT30 (Discrete Math)
More informationLecture Overview. 2 Weak Induction
COMPSCI 30: Discrete Mathematics for Computer Science February 18, 019 Lecturer: Debmalya Panigrahi Lecture 11 Scribe: Kevin Sun 1 Overview In this lecture, we study mathematical induction, which we often
More informationMEETING 9 - INDUCTION AND RECURSION
MEETING 9 - INDUCTION AND RECURSION We do some initial Peer Instruction... Predicates Before we get into mathematical induction we will repeat the concept of a predicate. A predicate is a mathematical
More informationCSCE 222 Discrete Structures for Computing. Dr. Hyunyoung Lee
CSCE 222 Discrete Structures for Computing Sequences and Summations Dr. Hyunyoung Lee Based on slides by Andreas Klappenecker 1 Sequences 2 Sequences A sequence is a function from a subset of the set of
More informationChapter 5.1: Induction
Chapter.1: Induction Monday, July 1 Fermat s Little Theorem Evaluate the following: 1. 1 (mod ) 1 ( ) 1 1 (mod ). (mod 7) ( ) 8 ) 1 8 1 (mod ). 77 (mod 19). 18 (mod 1) 77 ( 18 ) 1 1 (mod 19) 18 1 (mod
More informationDiscrete Mathematics. Spring 2017
Discrete Mathematics Spring 2017 Previous Lecture Principle of Mathematical Induction Mathematical Induction: rule of inference Mathematical Induction: Conjecturing and Proving Climbing an Infinite Ladder
More informationAlgorithm Analysis Recurrence Relation. Chung-Ang University, Jaesung Lee
Algorithm Analysis Recurrence Relation Chung-Ang University, Jaesung Lee Recursion 2 Recursion 3 Recursion in Real-world Fibonacci sequence = + Initial conditions: = 0 and = 1. = + = + = + 0, 1, 1, 2,
More informationSection Summary. Sequences. Recurrence Relations. Summations. Examples: Geometric Progression, Arithmetic Progression. Example: Fibonacci Sequence
Section 2.4 Section Summary Sequences. Examples: Geometric Progression, Arithmetic Progression Recurrence Relations Example: Fibonacci Sequence Summations Introduction Sequences are ordered lists of elements.
More informationMathematical Induction
Mathematical Induction MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Mathematical Induction Fall 2014 1 / 21 Outline 1 Mathematical Induction 2 Strong Mathematical
More informationProblem Set 5 Solutions
Problem Set 5 Solutions Section 4.. Use mathematical induction to prove each of the following: a) For each natural number n with n, n > + n. Let P n) be the statement n > + n. The base case, P ), is true
More informationInduction. Induction. Induction. Induction. Induction. Induction 2/22/2018
The principle of mathematical induction is a useful tool for proving that a certain predicate is true for all natural numbers. It cannot be used to discover theorems, but only to prove them. If we have
More informationChapter 11 - Sequences and Series
Calculus and Analytic Geometry II Chapter - Sequences and Series. Sequences Definition. A sequence is a list of numbers written in a definite order, We call a n the general term of the sequence. {a, a
More informationUSA Mathematical Talent Search Round 4 Solutions Year 20 Academic Year
1/4/20. Consider a sequence {a n } with a 1 = 2 and a n = a2 n 1 a n 2 for all n 3. If we know that a 2 and a 5 are positive integers and a 5 2009, then what are the possible values of a 5? Since a 1 and
More informationPart III, Sequences and Series CS131 Mathematics for Computer Scientists II Note 16 RECURRENCES
CS131 Part III, Sequences and Series CS131 Mathematics for Computer Scientists II Note 16 RECURRENCES A recurrence is a rule which defines each term of a sequence using the preceding terms. The Fibonacci
More informationNote that r = 0 gives the simple principle of induction. Also it can be shown that the principle of strong induction follows from simple induction.
Proof by mathematical induction using a strong hypothesis Occasionally a proof by mathematical induction is made easier by using a strong hypothesis: To show P(n) [a statement form that depends on variable
More informationCMSC250 Homework 9 Due: Wednesday, December 3, Question: Total Points: Score:
Name & UID: Circle Your Section! 0101 (10am: 3120, Ladan) 0102 (11am: 3120, Ladan) 0103 (Noon: 3120, Peter) 0201 (2pm: 3120, Yi) 0202 (10am: 1121, Vikas) 0203 (11am: 1121, Vikas) 0204 (9am: 2117, Karthik)
More informationWorksheet 1. Difference
Worksheet Differences Remember: The difference of a sequence u n is: u n u n+ u n The falling factorial (power), n to the k falling, when k > 0, is: n k n(n )... (n k + ), Please complete the following
More informationInfinite Continued Fractions
Infinite Continued Fractions 8-5-200 The value of an infinite continued fraction [a 0 ; a, a 2, ] is lim c k, where c k is the k-th convergent k If [a 0 ; a, a 2, ] is an infinite continued fraction with
More informationCSC 344 Algorithms and Complexity. Proof by Mathematical Induction
CSC 344 Algorithms and Complexity Lecture #1 Review of Mathematical Induction Proof by Mathematical Induction Many results in mathematics are claimed true for every positive integer. Any of these results
More informationSequences of Real Numbers
Chapter 8 Sequences of Real Numbers In this chapter, we assume the existence of the ordered field of real numbers, though we do not yet discuss or use the completeness of the real numbers. In the next
More informationNotes on Continued Fractions for Math 4400
. Continued fractions. Notes on Continued Fractions for Math 4400 The continued fraction expansion converts a positive real number α into a sequence of natural numbers. Conversely, a sequence of natural
More informationEECS 1028 M: Discrete Mathematics for Engineers
EECS 1028 M: Discrete Mathematics for Engineers Suprakash Datta Office: LAS 3043 Course page: http://www.eecs.yorku.ca/course/1028 Also on Moodle S. Datta (York Univ.) EECS 1028 W 18 1 / 16 Sequences and
More informationCOM S 330 Lecture Notes Week of Feb 9 13
Monday, February 9. Rosen.4 Sequences Reading: Rosen.4. LLM 4.. Ducks 8., 8., Def: A sequence is a function from a (usually infinite) subset of the integers (usually N = {0,,, 3,... } or Z + = {,, 3, 4,...
More informationBinomial Coefficient Identities/Complements
Binomial Coefficient Identities/Complements CSE21 Fall 2017, Day 4 Oct 6, 2017 https://sites.google.com/a/eng.ucsd.edu/cse21-fall-2017-miles-jones/ permutation P(n,r) = n(n-1) (n-2) (n-r+1) = Terminology
More informationX. Numerical Methods
X. Numerical Methods. Taylor Approximation Suppose that f is a function defined in a neighborhood of a point c, and suppose that f has derivatives of all orders near c. In section 5 of chapter 9 we introduced
More informationLegendre s Equation. PHYS Southern Illinois University. October 18, 2016
Legendre s Equation PHYS 500 - Southern Illinois University October 18, 2016 PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 1 / 11 Legendre s Equation Recall We are trying
More informationCOL106: Data Structures and Algorithms (IIT Delhi, Semester-II )
1 Solve the following recurrence relations giving a Θ bound for each of the cases 1 : (a) T (n) = 2T (n/3) + 1; T (1) = 1 (Assume n is a power of 3) (b) T (n) = 5T (n/4) + n; T (1) = 1 (Assume n is a power
More informationON QUADRAPELL NUMBERS AND QUADRAPELL POLYNOMIALS
Hacettepe Journal of Mathematics and Statistics Volume 8() (009), 65 75 ON QUADRAPELL NUMBERS AND QUADRAPELL POLYNOMIALS Dursun Tascı Received 09:0 :009 : Accepted 04 :05 :009 Abstract In this paper we
More informationQuiz 3 Reminder and Midterm Results
Quiz 3 Reminder and Midterm Results Reminder: Quiz 3 will be in the first 15 minutes of Monday s class. You can use any resources you have during the quiz. It covers all four sections of Unit 3. It has
More informationF 2k 1 = F 2n. for all positive integers n.
Question 1 (Fibonacci Identity, 15 points). Recall that the Fibonacci numbers are defined by F 1 = F 2 = 1 and F n+2 = F n+1 + F n for all n 0. Prove that for all positive integers n. n F 2k 1 = F 2n We
More informationA NICE DIOPHANTINE EQUATION. 1. Introduction
A NICE DIOPHANTINE EQUATION MARIA CHIARA BRAMBILLA Introduction One of the most interesting and fascinating subjects in Number Theory is the study of diophantine equations A diophantine equation is a polynomial
More informationLogic and Discrete Mathematics. Section 6.7 Recurrence Relations and Their Solution
Logic and Discrete Mathematics Section 6.7 Recurrence Relations and Their Solution Slides version: January 2015 Definition A recurrence relation for a sequence a 0, a 1, a 2,... is a formula giving a n
More informationConsider an infinite row of dominoes, labeled by 1, 2, 3,, where each domino is standing up. What should one do to knock over all dominoes?
1 Section 4.1 Mathematical Induction Consider an infinite row of dominoes, labeled by 1,, 3,, where each domino is standing up. What should one do to knock over all dominoes? Principle of Mathematical
More information(Infinite) Series Series a n = a 1 + a 2 + a a n +...
(Infinite) Series Series a n = a 1 + a 2 + a 3 +... + a n +... What does it mean to add infinitely many terms? The sequence of partial sums S 1, S 2, S 3, S 4,...,S n,...,where nx S n = a i = a 1 + a 2
More informationRecurrence Relations and Recursion: MATH 180
Recurrence Relations and Recursion: MATH 180 1: Recursively Defined Sequences Example 1: The sequence a 1,a 2,a 3,... can be defined recursively as follows: (1) For all integers k 2, a k = a k 1 + 1 (2)
More information3.3 Accumulation Sequences
3.3. ACCUMULATION SEQUENCES 25 3.3 Accumulation Sequences Overview. One of the most important mathematical ideas in calculus is that of an accumulation of change for physical quantities. As we have been
More informationMath 3000 Section 003 Intro to Abstract Math Homework 6
Math 000 Section 00 Intro to Abstract Math Homework 6 Department of Mathematical and Statistical Sciences University of Colorado Denver, Spring 01 Solutions April, 01 Please note that these solutions are
More informationCSE 311: Foundations of Computing. Lecture 14: Induction
CSE 311: Foundations of Computing Lecture 14: Induction Mathematical Induction Method for proving statements about all natural numbers A new logical inference rule! It only applies over the natural numbers
More informationSection 4.1: Sequences and Series
Section 4.1: Sequences and Series In this section, we shall introduce the idea of sequences and series as a necessary tool to develop the proof technique called mathematical induction. Most of the material
More informationCarmen s Core Concepts (Math 135)
Carmen s Core Concepts (Math 135) Carmen Bruni University of Waterloo Week 4 1 Principle of Mathematical Induction 2 Example 3 Base Case 4 Inductive Hypothesis 5 Inductive Step When Induction Isn t Enough
More informationASSIGNMENT 12 PROBLEM 4
ASSIGNMENT PROBLEM 4 Generate a Fibonnaci sequence in the first column using f 0 =, f 0 =, = f n f n a. Construct the ratio of each pair of adjacent terms in the Fibonnaci sequence. What happens as n increases?
More informationSEQUENCES AND SERIES
A sequence is an ordered list of numbers. SEQUENCES AND SERIES Note, in this context, ordered does not mean that the numbers in the list are increasing or decreasing. Instead it means that there is a first
More informationq-counting hypercubes in Lucas cubes
Turkish Journal of Mathematics http:// journals. tubitak. gov. tr/ math/ Research Article Turk J Math (2018) 42: 190 203 c TÜBİTAK doi:10.3906/mat-1605-2 q-counting hypercubes in Lucas cubes Elif SAYGI
More informationMath Assignment 11
Math 2280 - Assignment 11 Dylan Zwick Fall 2013 Section 8.1-2, 8, 13, 21, 25 Section 8.2-1, 7, 14, 17, 32 Section 8.3-1, 8, 15, 18, 24 1 Section 8.1 - Introduction and Review of Power Series 8.1.2 - Find
More informationIB Mathematics HL Year 2 Unit 11: Completion of Algebra (Core Topic 1)
IB Mathematics HL Year Unit : Completion of Algebra (Core Topic ) Homewor for Unit Ex C:, 3, 4, 7; Ex D: 5, 8, 4; Ex E.: 4, 5, 9, 0, Ex E.3: (a), (b), 3, 7. Now consider these: Lesson 73 Sequences and
More informationRENEWAL THEORY STEVEN P. LALLEY UNIVERSITY OF CHICAGO. X i
RENEWAL THEORY STEVEN P. LALLEY UNIVERSITY OF CHICAGO 1. RENEWAL PROCESSES A renewal process is the increasing sequence of random nonnegative numbers S 0,S 1,S 2,... gotten by adding i.i.d. positive random
More informationMath 192r, Problem Set #3: Solutions
Math 192r Problem Set #3: Solutions 1. Let F n be the nth Fibonacci number as Wilf indexes them (with F 0 F 1 1 F 2 2 etc.). Give a simple homogeneous linear recurrence relation satisfied by the sequence
More informationCHAPTER 8 Advanced Counting Techniques
96 Chapter 8 Advanced Counting Techniques CHAPTER 8 Advanced Counting Techniques SECTION 8. Applications of Recurrence Relations 2. a) A permutation of a set with n elements consists of a choice of a first
More informationDirect Proof Universal Statements
Direct Proof Universal Statements Lecture 13 Section 4.1 Robb T. Koether Hampden-Sydney College Wed, Feb 6, 2013 Robb T. Koether (Hampden-Sydney College) Direct Proof Universal Statements Wed, Feb 6, 2013
More informationSum of Squares. Defining Functions. Closed-Form Expression for SQ(n)
CS/ENGRD 2110 Object-Oriented Programming and Data Structures Spring 2012 Thorsten Joachims Lecture 22: Induction Overview Recursion A programming strategy that solves a problem by reducing it to simpler
More informationCS 220: Discrete Structures and their Applications. Mathematical Induction in zybooks
CS 220: Discrete Structures and their Applications Mathematical Induction 6.4 6.6 in zybooks Why induction? Prove algorithm correctness (CS320 is full of it) The inductive proof will sometimes point out
More informationMathematical Induction
Mathematical Induction Let s motivate our discussion by considering an example first. What happens when we add the first n positive odd integers? The table below shows what results for the first few values
More informationHomework 4, 5, 6 Solutions. > 0, and so a n 0 = n + 1 n = ( n+1 n)( n+1+ n) 1 if n is odd 1/n if n is even diverges.
2..2(a) lim a n = 0. Homework 4, 5, 6 Solutions Proof. Let ɛ > 0. Then for n n = 2+ 2ɛ we have 2n 3 4+ ɛ 3 > ɛ > 0, so 0 < 2n 3 < ɛ, and thus a n 0 = 2n 3 < ɛ. 2..2(g) lim ( n + n) = 0. Proof. Let ɛ >
More informationMCS 256 Discrete Calculus and Probability Exam 5: Final Examination 22 May 2007
MCS 256 Discrete Calculus and Probability SOLUTIONS Exam 5: Final Examination 22 May 2007 Instructions: This is a closed-book examination. You may, however, use one 8.5 -by-11 page of notes, your note
More informationSection 4.2: Mathematical Induction 1
Section 4.: Mathematical Induction 1 Over the next couple of sections, we shall consider a method of proof called mathematical induction. Induction is fairly complicated, but a very useful proof technique,
More informationChapter Summary. Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms
1 Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms 2 Section 5.1 3 Section Summary Mathematical Induction Examples of
More informationFinal Exam Review. 2. Let A = {, { }}. What is the cardinality of A? Is
1. Describe the elements of the set (Z Q) R N. Is this set countable or uncountable? Solution: The set is equal to {(x, y) x Z, y N} = Z N. Since the Cartesian product of two denumerable sets is denumerable,
More informationCHAPTER 4 SOME METHODS OF PROOF
CHAPTER 4 SOME METHODS OF PROOF In all sciences, general theories usually arise from a number of observations. In the experimental sciences, the validity of the theories can only be tested by carefully
More information1 Recursive Algorithms
400 lecture note #8 [ 5.6-5.8] Recurrence Relations 1 Recursive Algorithms A recursive algorithm is an algorithm which invokes itself. A recursive algorithm looks at a problem backward -- the solution
More informationName CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page (16 points) Circle T if the corresponding statement is True or F if it is False.
Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page ( points) Circle T if the corresponding statement is True or F if it is False T F GCD(,0) = 0 T F For every recursive algorithm, there is
More informationSOLUTION SETS OF RECURRENCE RELATIONS
SOLUTION SETS OF RECURRENCE RELATIONS SEBASTIAN BOZLEE UNIVERSITY OF COLORADO AT BOULDER The first section of these notes describes general solutions to linear, constant-coefficient, homogeneous recurrence
More informationSolution Set 2. Problem 1. [a] + [b] = [a + b] = [b + a] = [b] + [a] ([a] + [b]) + [c] = [a + b] + [c] = [a + b + c] = [a] + [b + c] = [a] + ([b + c])
Solution Set Problem 1 (1) Z/nZ is the set of equivalence classes of Z mod n. Equivalence is determined by the following rule: [a] = [b] if and only if b a = k n for some k Z. The operations + and are
More information21 Induction. Tom Lewis. Fall Term Tom Lewis () 21 Induction Fall Term / 14
21 Induction Tom Lewis Fall Term 2010 Tom Lewis () 21 Induction Fall Term 2010 1 / 14 Outline 1 The method of induction 2 Strong mathematical induction Tom Lewis () 21 Induction Fall Term 2010 2 / 14 Pessimists
More informationON THE POSSIBLE QUANTITIES OF FIBONACCI NUMBERS THAT OCCUR IN SOME TYPES OF INTERVALS
Acta Math. Univ. Comenianae Vol. LXXXVII, 2 (2018), pp. 291 299 291 ON THE POSSIBLE QUANTITIES OF FIBONACCI NUMBERS THAT OCCUR IN SOME TYPES OF INTERVALS B. FARHI Abstract. In this paper, we show that
More informationMath 0230 Calculus 2 Lectures
Math 00 Calculus Lectures Chapter 8 Series Numeration of sections corresponds to the text James Stewart, Essential Calculus, Early Transcendentals, Second edition. Section 8. Sequences A sequence is a
More informationSolutions to Exercises 4
Discrete Mathematics Lent 29 MA2 Solutions to Eercises 4 () Define sequence (b n ) n by b n ( n n..., where we use n ) for > n. Verify that b, b 2 2, and that, for every n 3, we have b n b n b n 2. Solution.
More informationDiscrete Math, Spring Solutions to Problems V
Discrete Math, Spring 202 - Solutions to Problems V Suppose we have statements P, P 2, P 3,, one for each natural number In other words, we have the collection or set of statements {P n n N} a Suppose
More informationSec$on Summary. Sequences. Recurrence Relations. Summations. Ex: Geometric Progression, Arithmetic Progression. Ex: Fibonacci Sequence
Section 2.4 Sec$on Summary Sequences Ex: Geometric Progression, Arithmetic Progression Recurrence Relations Ex: Fibonacci Sequence Summations 2 Introduc$on Sequences are ordered lists of elements. 1, 2,
More information1 Proof by Contradiction
In these notes, which will serve as a record of what we covered in weeks 3-4, we discuss various indirect methods of proof. While direct proof is often preferred it is sometimes either not possible or
More informationMATH 55 - HOMEWORK 6 SOLUTIONS. 1. Section = 1 = (n + 1) 3 = 2. + (n + 1) 3. + (n + 1) 3 = n2 (n + 1) 2.
MATH 55 - HOMEWORK 6 SOLUTIONS Exercise Section 5 Proof (a) P () is the statement ( ) 3 (b) P () is true since ( ) 3 (c) The inductive hypothesis is P (n): ( ) n(n + ) 3 + 3 + + n 3 (d) Assuming the inductive
More informationGenerating Functions
Generating Functions Prachi Pendse January 9, 03 Motivation for Generating Functions Question How many ways can you select 6 cards from a set of 0? ( + ) 0 = ( + )( + )( + )... = 0 + 0 9 + 4 8 + 0 ( )
More informationRevision Problems for Examination 1 in Algebra 1
Centre for Mathematical Sciences Mathematics, Faculty of Science Revision Problems for Examination 1 in Algebra 1 Arithmetics 1 Determine a greatest common divisor to the integers a) 5431 and 1345, b)
More information3 Finite continued fractions
MTH628 Number Theory Notes 3 Spring 209 3 Finite continued fractions 3. Introduction Let us return to the calculation of gcd(225, 57) from the preceding chapter. 225 = 57 + 68 57 = 68 2 + 2 68 = 2 3 +
More informationChapter 8: Taylor s theorem and L Hospital s rule
Chapter 8: Taylor s theorem and L Hospital s rule Theorem: [Inverse Mapping Theorem] Suppose that a < b and f : [a, b] R. Given that f (x) > 0 for all x (a, b) then f 1 is differentiable on (f(a), f(b))
More informationCHAPTER III THE PROOF OF INEQUALITIES
CHAPTER III THE PROOF OF INEQUALITIES In this Chapter, the main purpose is to prove four theorems about Hardy-Littlewood- Pólya Inequality and then gives some examples of their application. We will begin
More informationPUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime.
PUTNAM TRAINING NUMBER THEORY (Last updated: December 11, 2017) Remark. This is a list of exercises on Number Theory. Miguel A. Lerma Exercises 1. Show that the sum of two consecutive primes is never twice
More informationReading 5 : Induction
CS/Math 40: Introduction to Discrete Mathematics Fall 015 Instructors: Beck Hasti and Gautam Prakriya Reading 5 : Induction In the last reading we began discussing proofs. We mentioned some proof paradigms
More information8. Sequences, Series, and Probability 8.1. SEQUENCES AND SERIES
8. Sequences, Series, and Probability 8.1. SEQUENCES AND SERIES What You Should Learn Use sequence notation to write the terms of sequences. Use factorial notation. Use summation notation to write sums.
More informationSubtraction games. Chapter The Bachet game
Chapter 7 Subtraction games 7.1 The Bachet game Beginning with a positive integer, two players alternately subtract a positive integer < d. The player who gets down to 0 is the winner. There is a set of
More informationMathematical Structures Combinations and Permutations
Definitions: Suppose S is a (finite) set and n, k 0 are integers The set C(S, k) of k - combinations consists of all subsets of S that have exactly k elements The set P (S, k) of k - permutations consists
More informationLecture 6: Recurrent Algorithms:
Lecture 6: Recurrent Algorithms: Divide-and-Conquer Principle Georgy Gimel farb COMPSCI 220 Algorithms and Data Structures 1 / 20 1 Divide-and-conquer 2 Math induction 3 Telescoping 4 Examples 5 Pros and
More informationAlgorithms CMSC The Method of Reverse Inequalities: Evaluation of Recurrent Inequalities
Algorithms CMSC 37000 The Method of Reverse Inequalities: Evaluation of Recurrent Inequalities László Babai Updated 1-19-2014 In this handout, we discuss a typical situation in the analysis of algorithms:
More informationRecurrences COMP 215
Recurrences COMP 215 Analysis of Iterative Algorithms //return the location of the item matching x, or 0 if //no such item is found. index SequentialSearch(keytype[] S, in, keytype x) { index location
More informationFour Basic Sets. Divisors
Four Basic Sets Z = the integers Q = the rationals R = the real numbers C = the complex numbers Divisors Definition. Suppose a 0 and b = ax, where a, b, and x are integers. Then we say a divides b (or
More informationn n P} is a bounded subset Proof. Let A be a nonempty subset of Z, bounded above. Define the set
1 Mathematical Induction We assume that the set Z of integers are well defined, and we are familiar with the addition, subtraction, multiplication, and division. In particular, we assume the following
More informationn f(k) k=1 means to evaluate the function f(k) at k = 1, 2,..., n and add up the results. In other words: n f(k) = f(1) + f(2) f(n). 1 = 2n 2.
Handout on induction and written assignment 1. MA113 Calculus I Spring 2007 Why study mathematical induction? For many students, mathematical induction is an unfamiliar topic. Nonetheless, this is an important
More informationCS173 Lecture B, September 10, 2015
CS173 Lecture B, September 10, 2015 Tandy Warnow September 11, 2015 CS 173, Lecture B September 10, 2015 Tandy Warnow Examlet Today Four problems: One induction proof One problem on simplifying a logical
More informationGeneral Comments on Proofs by Mathematical Induction
Fall 2015: CMSC250 Notes on Mathematical Induction Proofs General Comments on Proofs by Mathematical Induction All proofs by Mathematical Induction should be in one of the styles below. If not there should
More informationWhat is Zeckendorf s Theorem?
What is Zeckendorf s Theorem? Nik Henderson July 23, 2016 Abstract While Fibonacci numbers can quite easily be classified as a complete sequence, they have the unusual property that a particular explicitly
More information12 Sequences and Recurrences
12 Sequences and Recurrences A sequence is just what you think it is. It is often given by a formula known as a recurrence equation. 12.1 Arithmetic and Geometric Progressions An arithmetic progression
More information