# 3. Recurrence Recursive Definitions. To construct a recursively defined function:

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 3. RECURRENCE Recurrence 3.1. Recursive Definitions. To construct a recursively defined function: 1. Initial Condition(s) (or basis): Prescribe initial value(s) of the function.. Recursion: Use a fixed procedure (rule) to compute the value of the function at the integer n + 1 using one or more values of the function for integers n. To construct a recursively defined set: 1. Initial Condition(s) (or basis): Prescribe one or more elements of the set.. Recursion: Give a rule for generating elements of the set in terms of previously prescribed elements. In computer programming evaluating a function or executing a procedure is often accomplished by using a recursion. A recursive process is one in which one or more initial stages of the process are specified, and the nth stage of the process is defined in terms of previous stages, using some fixed procedure. In a computer program this is usually accomplished by using a subroutine, such as a For loop, in which the same procedure is repeated a specified number of times; that is, the procedure calls itself. Example The function f(n) = n, where n is a natural number, can be defined recursively as follows: 1. Initial Condition: f(0) = 1,. Recursion: f(n + 1) = f(n), for n 0. For any particular n, this procedure could be programmed by first initializing F = 1, and then executing a loop For i = 1 to n, F = F. Here is how the definition gives us the first few powers of : 1 = 0+1 = 0 = = 1+1 = 1 = = 4 3 = +1 = = 4 = 8

2 3. RECURRENCE Recursive Definition of the Function f(n) = n!. Example The factorial function f(n) = n! is defined recursively as follows: 1. Initial Condition: f(0) = 1. Recursion: f(n + 1) = (n + 1)f(n) Starting with the initial condition, f(0) = 1, the recurrence relation, f(n + 1) = (n + 1)f(n), tells us how to build new values of f from old values. For example, 1! = f(1) = 1 f(0) = 1,! = f() = f(1) =, 3! = f(3) = 3 f() = 6, etc. When a function f(n), such as the ones in the previous examples, is defined recursively, the equation giving f(n + 1) in terms of previous values of f is called a recurrence relation Recursive Definition of the Natural Numbers. Definition The set of natural numbers may be defined recursively as follows. 1. Initial Condition: 0 N. Recursion: If n N, then n + 1 N. There are a number of ways of defining the set N of natural numbers recursively. The simplest definition is given above. Here is another recursive definition for N. Example Suppose the set S is defined recursively as follows: 1. Initial Condition: 0, 1 S,. Recursion: If x, y S, then x + y S.

3 3. RECURRENCE 1 Then S = N. Notice that, in the recursive step, x and y don t have to represent different numbers. Thus, having x = y = 1 S, we get = S. Then we get 1 + = 3 S. And so on. It should be noted that there is an extremal clause in recursively defined sets. If you cannot build a given element in a finite number of applications of the recursion then it is not in the set built from the recursive definition. To prove an element is in a recursively defined set, you must show that element can be built in a finite number of steps. Example Prove that the set S recursively in Example is equal to the set N of natural numbers. Proof. We will show that S = N by showing separately that S N and N S. 1. First we show N S. Prove by induction that n S for every natural number n 0. (a) Basis Step. 0, 1 S by definition. (b) Induction Step. Suppose n S, for some n 1. Then, by the recursion step, n S and 1 S imply n + 1 S. Thus, by the first principle of mathematical induction, n S for every natural number n 0.. Now we show S N. This time we apply the second principle of mathematical induction on n to show that if s S is produced by applying n steps (1 initial condition and n 1 recursive steps), then s N. (a) Basis Step. After one step the only elements produced are 0 and 1, each of which is in N. (b) Induction Step. Suppose n 1 and assume any element in S produced by applying n or fewer steps is also an element of N. Suppose s S is produced by applying n + 1 steps. Since n + 1, there must be two elements x and y in S, such that s = x+y. Both x and y must have been produced by applying fewer than n+1 steps, since s is produced by applying n+1 steps, and we use one step to obtain s from x and y. By the induction hypothesis both x and y are elements of N. Since the sum of two natural numbers is again a natural number we have s N Proving Assertions About Recursively Defined Objects. Assertions about recursively defined objects are usually proved by mathematical induction. Here are three useful versions of induction. In particular, note the third version which we introduce here.

4 3. RECURRENCE 13 Version 1. Second Principle of Induction a. Basis Step: Prove the assertion for the initial conditions. (The assertion may have to be verified for more than one particular value.) b. Induction Step: Assume the assertion is true for integers n, and use the recurrence relation to prove the assertion for the integer n + 1. Version. a. Basis Step: Prove the assertion for the initial conditions. (The assertion may have to be verified for more than one particular value.) b. Induction Step: Assume the assertion is true when the recursive definition has been applied less than or equal to n times for some integer n 0, and use the recurrence relation to prove the assertion when the recursive definition is applied n + 1 times. Version 3. Generalized or Structural Principle of Induction: Use to prove an assertion about a set S defined recursively by using a set X given in the basis and a set of rules using s 1, s,..., s k S for producing new members in the recursive set. a. Basis Step: Prove the assertion for every s X. b. Induction Step: Let s 1, s,..., s k S be arbitrary and assume the assertion for these elements (this is the induction hypothesis). Prove all elements of S produced using the recursive definition and s 1, s,..., s k satisfies the assertion. Example Let S, a subset of N N, be defined recursively by 1. Initial Condition: (0, 0) S. Recursion: If (m, n) S, then (m +, n + 3) S. Prove that if (m, n) S, then m + n is a multiple of 5. Proof. We use the Generalized Principle of Induction. 1. Basis Step: Show the statement for (0, 0): is a multiple of 5. Since 0 = 0 5 this is clear.. Inductive Step: Let (m, n) S and assume m + n is a multiple of 5. Show the statement is true for (m +, n + 3). In other words, show (m + ) + (n + 3) is a multiple of 5. (m + ) + (n + 3) = (m + n) + 5 We know m + n is a multiple of 5 and clearly 5 is a multiple of 5, so the sum must also be a multiple of 5. This proves the induction step.

5 3. RECURRENCE 14 Therefore, by the first principle of mathematical induction if (m, n) S, then m + n is a multiple of 5. Exercise Suppose the set S is defined recursively as follows: 1. 1 S,. If x S, then x S. Prove that S = { n n 0}. Exercise Suppose the set S is defined recursively as follows: 1. 0, 1 S,. If x, y S, then x y S. What are the elements of S? Prove that your answer is correct. Example Suppose f : N R is define recursively by 1. Initial Condition: f(0) = 0. Recursion: f(n + 1) = f(n) + (n + 1), for n 0. Then f(n) = n(n + 1) for all n 0 Proof. 1. Basis Step (n = 0): f(0) = 0, by definition. On the other hand 0(0 + 1) 0(0 + 1) = 0. Thus, f(0) =. n(n + 1). Inductive Step: Suppose f(n) = for some n 0. We must prove (n + 1)((n + 1) + 1) f(n + 1) =. f(n + 1) = f(n) + (n + 1) (recurrence relation) n(n + 1) = + (n + 1) (by the induction hypothesis) [ ] n = (n + 1) + 1 [ ] n + = (n + 1) = (n + 1)((n + 1) + 1)

6 3. RECURRENCE 15 Therefore, by the first principle of mathematical induction f(n) = all n 0. Exercise Suppose f : N R is define recursively as follows: n(n + 1) for 1. f(0) = 0,. f(n + 1) = f(n) + (n + 1), for n 0. Prove that f(n) = n for all n Definition of f n. Definition Let f : A A be a function. Then we define f n recursively as follows 1. Initial Condition: f 1 = f. Recursion: f n+1 = f f n, for n 1. Example Prove that if f is injective, then f n is injective for n 1. Proof. Assume f is injective. 1. Basis Step: f 1 = f is injective by our assumption.. Inductive Step: Let n 1 and assume f n is injective. Prove f n+1 is injective. Recall that to prove f n+1 is injective we must show a, b A[(f n+1 (a) = f n+1 (b)) (a = b)] Assume a, b A and f n+1 (a) = f n+1 (b). f n+1 (a) = f n+1 (b) (recurrence relation) (f f n )(a) = (f f n )(b) (recurrence relation) f(f n (a)) = f(f n (b)) (by the definition of composition) f n (a) = f n (b) (since f is injective) a = b (by the induction hypothesis, f n is injective) Therefore, by the first principle of mathematical induction f n positive integers. is injective for all

7 3. RECURRENCE 16 Exercise Prove that if f is surjective that f n is surjective Example Example Given a real number a 0, define a n for all natural numbers, n, inductively by 1. Initial Condition: a 0 = 1. Recursion: a (n+1) = a n a Theorem m n[a m a n = a m+n ] where m, n are natural numbers. Proof. Proof that ( m)( n)[a m a n = a m+n ], where m, n are natural numbers. We accomplish this by assuming m is an arbitrary natural number and proving n[a m a n = a m+n ] by induction on n. 1. Basis Step (n = 0): Show a m a 0 = a m+0. This follows directly from the initial condition of the definition: a 0 = 1, therefore a m a 0 = a m (1) = a m = a m+0.. Induction Step: Induction hypothesis: Let n be a natural number and assume a m a n = a n+m. Prove a m a n+1 = a m+(n+1). a m a n+1 = a m a n a by the recursive part of the definition: a n+1 = a n a = a m+n a by the induction hypothesis = a (m+n)+1 = a (m+(n+1)) by the recursive part of the definition By induction, n[a m a n = a m+n ]. Since m was an arbitrary natural number the statement is true for all natural numbers m. Here we see a recursive definition for the function f(n) = a n, where n is a natural number and a is an arbitrary nonzero real number followed by a proof of one of the laws of exponents, a m+n = a m a n. This proof uses both mathematical induction and Universal Generalization. We fix m as some arbitrary natural number and then proceed to use induction on n. We do not need to use induction on both m and n simultaneously. When there are two different variables, this is a standard strategy to try. There are circumstances, however, when this strategy doesn t work so that you

8 3. RECURRENCE 17 would need to use induction on both of the variables (double induction). We will not encounter these kinds of problems in this course, however Fibonacci Sequence. Definition The Fibonacci sequence may be defined recursively as follows: 1. Initial Conditions: F (0) = 0, F (1) = 1. Recursion: = F (n) + F (n 1) for n 1 The famous Fibonacci sequence is defined here using a recursively defined function. The definition of the Fibonacci sequence requires two initial conditions. There are no limits on the number of initial conditions on a recursively defined object only that there be a fixed finite number in each instance. Example Suppose F (n), n 0, denotes the Fibonacci sequence. Prove that 1 < < for all n 3. F (n) Proof. Let R(n) = F (n) for n 1. We will prove by induction that 1 < R(n) < for all n Basis Step (n = 3): R(3) = F (4) F (3) = 3, and 1 < 3 <.

9 3. RECURRENCE 18. Induction Step: Suppose 1 < R(n) < for some n 3. We need to prove 1 < R(n + 1) <. R(n + 1) = F (n + ) = + F (n) by the recursive definition of F (n) = 1 + F (n) or Thus, R(n + 1) = R(n) By the inductive hypothesis 1 < R(n) < ; and so 1 > 1 R(n) > 1. > R(n) > 3 > 1, or 1 < R(n) <. Substituting from above, we have 1 < R(n + 1) <. By the first principle of mathematical induction, 1 < F (n) Example (calculus required) Prove that lim n F (n) limit exists. < for all n 3. = 1 + 5, if the Proof. Let R(n) = F (n) as in Example Then, from the induction step in Example 3.7.1, we see that R(n + 1) = Assume lim R(n) n and let L = lim n F (n) Therefore, L = lim n ( R(n) = lim R(n). n ) 1 = 1 + L. F (n) exists Notice that lim n R(n) = lim n R(n + 1).

10 Since L is a real number, the equation 3. RECURRENCE 19 is equivalent to By the quadratic formula Since L > 0 and since 5 > 1, L = L L L 1 = 0. L = 1 ± 5. L = Exercise Prove that F (n) > ( 3 )n 1 for n 6. [Hint: Show that the statement is true for n = 6 and 7 (basis step), and then show the induction step works for n 7.] Here is another Fibonacci-like sequence. Example Suppose F (n), n 0, is defined recursively as follows: 1. F (0) = 1 and F (1) =,. = F (n) + F (n 1), for n 1. Prove that F (n) = n for all n 0. Proof. (Using the second principle of mathematical induction) 1. Basis Step (n = 0 and n = 1): F (0) = 1 = 0 and F (1) = = 1.. Induction Step: Let n 1. Suppose F (k) = k for 0 k n. Then = F (n) + F (n 1) = n + n 1 = n + n = n = n+1 Thus, by the second principle of mathematical induction, F (n) = n for all n 0.

11 3. RECURRENCE 130 Remarks (1) Here and in the previous exercise we see the slight variation in the basis step from the ones encountered in Module 3.3 Induction; there may be more than one initial condition to verify before proceeding to the induction step. () Notice that in this example as well as in some of the other examples and exercises, we have been asked to prove that a function defined recursively is also given by a relatively simple formula. The problem of solving recurrence relations for these nice formulas (so-called closed form) is an interesting subject in its own right, but it will not be discussed in this course. Exercise Let f : N Z be the function defined recursively by 1. f(0) = 1 and f(1) = 4,. f(n) = 3f(n 1) + 4f(n ), for n. Prove f(n) = ( 4) n Exercise Let f : N Z be the function defined recursively by 1. f(0) = and f(1) = 7,. f(n) = f(n 1) + f(n ), for n. Prove f(n) = 3 n ( 1) n 3.8. Strings. Definition Given a finite set of symbols, Σ, the set of strings, denoted Σ, over Σ is defined recursively as follows: 1. Initial Condition: The empty string λ is in Σ.. Recursion: If w is a string in Σ and a is a symbol in Σ, then wa is in Σ. The set Σ is usually called an alphabet, and strings in Σ are called words in the alphabet Σ. Strings ( words ) on a finite alphabet, Σ, are defined recursively using right concatenation. In other words, every string of symbols from Σ can be built from a smaller string by applying new letters to the right. Remark When a is a symbol from an alphabet, we use the notation a n, where n N, to represent a concatenated with itself n times. In particular, a 0 is understood to represent the empty string, λ.

12 3. RECURRENCE 131 Exercise Suppose the alphabet Σ is the set {a, b, c, d}. Then Σ is the set of all words on Σ. Use right concatenation to build the bit string daabc starting with the empty string, λ (use λa = a for any element, a, in Σ). Exercise Now take the same bit string, daabc, and build it using left concatenation. Notice that your steps are not the same; that is, concatenation is not commutative. Regardless, we arrive at the same set of strings, Σ Bit Strings. Example The set S of all bit strings with no more than a single 1 can be defined recursively as follows: 1. Initial Condition: λ, 1 S. Recursion: If w is a string in S, then so are 0w and w0. In this definition we must start with two objects in the set. Then we can build all the bit strings that have at most one 1. We can define various subsets of bit strings using recursively defined sets. Example This example is an extension of example Let the set S be defined recursively by Basis: λ, 1 S Recursion: If w S then 0w S and w0 S. In creating a set recursively, it can help to use a tree diagram to develop more of an intuitive understanding of how this set is built. The following diagram shows how the applying the above definition 4 times gives the elements in the diagram. Some key ideas to keep in mind is that all strings in the tree and all strings that would be in the tree if you kept going are in the set. When a string is repeated, that means there is more than one way to get that element but there is no need to see what is produced from it more than one time.

13 3. RECURRENCE 13 l Initial Condition 1 Apply recursive step w 0w w w0 w 0w w w0 0l=0 l0=0 equal Apply recursive step Apply recursive step Prove S is the set of all bit strings with no more than one 1. Proof. Let A denote the set of all bit strings with no more than one 1 in the string. Then we need to show A = S. First we will show A S. Let a A. Then a is a bit string with either no 1 s or it is a bit string with exactly one 1. Case 1 Suppose a has no 1 s. Then a = 0 n where n is some natural number. We can build a using the recursive definition by starting with λ and applying the recursive step n times. (If we apply the recursive step 0 times, we get λ). Case Suppose a has exactly on 1. Then a = 0 n 10 m for some n, m N. We can build a by starting with 1, which is given by the initial condition, and applying the recursive step (w S) (0w S) n times and applying (w S) (w0 S) m times. This shows we can apply the recursive definition given for S finitely many times to obtain any element of A. Therefore A S. Now we show S A by general induction.

14 3. RECURRENCE 133 basis: The elements given by the basis (initial condition) of the defintion of S are both in A since λ has no 1 s and 1 has one 1. induction step: Let x S and assume x A. We need to show any elements obtained by appling the recursive step one time will also be in A. Notice we obtain 0x and x0 when we apply the recursive step one time to x. Since x is in A we know the string x has either no ones or a single one. 0x and x0 do not add any more 1 s to the bit string, so they are also in A. Thus by the principle of mathematical induction x S(x A). This completes the proof that S = A. Example Here s an example of proving a recursively defined set of bit strings is the same as set of bit strings defined in a non-recursive manner: Let A be the set of all bit strings of the form 0 n 1 n, where n can be any natural number. Note that this is the same as A = {0 n 1 n n N} = {λ, 01, 0011, , ,...} and is slightly different from the set described in Exercise Now define B by the following recursive definition: Basis: λ B Recursive Step: If w B then 0w1 B Prove that A = B. Proof. First we prove A B. Let a A. Then a = 0 n 1 n for some n N. If we use the recursive definition of B we see λ = by the basis step and we claim that if we apply the recursive step n times to λ we will build to the element 0 n 1 n. This will demonstrate that we can apply the recursive definition to find a using a finite number of steps. Thus a B. To prove the claim we use the first principle of induction on the predicate P (n) = If we apply the recursive step n times to λ we will build to the element 0 n 1 n. Exercise Prove the above claim. Now we need to prove B A. We will do this using generalized induction, which gives us a formal proof of this statement. Basis: Notice the element, λ, created by the initial step (or basis step) in the definition of B is also an element of A (λ = ). Induction Step: Let x B be such that x A as well. Show that any element of B obtained from x by applying the recursive step one time is also an element of A.

15 3. RECURRENCE 134 If we apply the recursive step to x one time the only element we get 0x1. Since x is an element of A we know x = 0 n 1 n for some n N. So then 0x1 = 0(0 n 1 n )1 = 0 n+1 1 n+1 which we see is also an element of A. Thus by the principle of generalized induction x B(x A). This completes that proof that A = B. Exercise What kinds of bit strings would we have if the initial condition in Example is changed to 1 S only? So the definition would be 1. Initial Condition: 1 S,. Recursion: If w is a string in S, then so are 0w and w0. Exercise What kinds of strings do we get from the following recursive definition? 1. Initial Conditions: λ, 1 S,. Recursion: If w is a string in S, then so is 0w. Exercise Find a recursive definition for the set of bit strings T = {0 r 1 s r, s N}. Exercise Prove your answer for Exercise is correct. Exercise Find a recursive definition for the set of all bit strings containing no adjacent 1 s. (For example, is allowed, but is not.)

### 3. Mathematical Induction

3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

### Appendix F: Mathematical Induction

Appendix F: Mathematical Induction Introduction In this appendix, you will study a form of mathematical proof called mathematical induction. To see the logical need for mathematical induction, take another

### Section 6-2 Mathematical Induction

6- Mathematical Induction 457 In calculus, it can be shown that e x k0 x k k! x x x3!! 3!... xn n! where the larger n is, the better the approximation. Problems 6 and 6 refer to this series. Note that

### Introduction. Appendix D Mathematical Induction D1

Appendix D Mathematical Induction D D Mathematical Induction Use mathematical induction to prove a formula. Find a sum of powers of integers. Find a formula for a finite sum. Use finite differences to

### 2. Methods of Proof Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try.

2. METHODS OF PROOF 69 2. Methods of Proof 2.1. Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try. Trivial Proof: If we know q is true then

### Mathematical Induction. Lecture 10-11

Mathematical Induction Lecture 10-11 Menu Mathematical Induction Strong Induction Recursive Definitions Structural Induction Climbing an Infinite Ladder Suppose we have an infinite ladder: 1. We can reach

### Sample Induction Proofs

Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Sample Induction Proofs Below are model solutions to some of the practice problems on the induction worksheets. The solutions given

### Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 5, due Wednesday, February 22 5.1.4 Let P (n) be the statement that 1 3 + 2 3 + + n 3 = (n(n + 1)/2) 2 for the positive integer n. a) What

### SECTION 10-2 Mathematical Induction

73 0 Sequences and Series 6. Approximate e 0. using the first five terms of the series. Compare this approximation with your calculator evaluation of e 0.. 6. Approximate e 0.5 using the first five terms

### Applications of Methods of Proof

CHAPTER 4 Applications of Methods of Proof 1. Set Operations 1.1. Set Operations. The set-theoretic operations, intersection, union, and complementation, defined in Chapter 1.1 Introduction to Sets are

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

### Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition

Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing

### Formal Languages and Automata Theory - Regular Expressions and Finite Automata -

Formal Languages and Automata Theory - Regular Expressions and Finite Automata - Samarjit Chakraborty Computer Engineering and Networks Laboratory Swiss Federal Institute of Technology (ETH) Zürich March

### Recursive Algorithms. Recursion. Motivating Example Factorial Recall the factorial function. { 1 if n = 1 n! = n (n 1)! if n > 1

Recursion Slides by Christopher M Bourke Instructor: Berthe Y Choueiry Fall 007 Computer Science & Engineering 35 Introduction to Discrete Mathematics Sections 71-7 of Rosen cse35@cseunledu Recursive Algorithms

### Cartesian Products and Relations

Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special

### Worksheet on induction Calculus I Fall 2006 First, let us explain the use of for summation. The notation

Worksheet on induction MA113 Calculus I Fall 2006 First, let us explain the use of for summation. The notation f(k) means to evaluate the function f(k) at k = 1, 2,..., n and add up the results. In other

### Basic Proof Techniques

Basic Proof Techniques David Ferry dsf43@truman.edu September 13, 010 1 Four Fundamental Proof Techniques When one wishes to prove the statement P Q there are four fundamental approaches. This document

### Cardinality. The set of all finite strings over the alphabet of lowercase letters is countable. The set of real numbers R is an uncountable set.

Section 2.5 Cardinality (another) Definition: The cardinality of a set A is equal to the cardinality of a set B, denoted A = B, if and only if there is a bijection from A to B. If there is an injection

### Reading 13 : Finite State Automata and Regular Expressions

CS/Math 24: Introduction to Discrete Mathematics Fall 25 Reading 3 : Finite State Automata and Regular Expressions Instructors: Beck Hasti, Gautam Prakriya In this reading we study a mathematical model

### Collatz Sequence. Fibbonacci Sequence. n is even; Recurrence Relation: a n+1 = a n + a n 1.

Fibonacci Roulette In this game you will be constructing a recurrence relation, that is, a sequence of numbers where you find the next number by looking at the previous numbers in the sequence. Your job

### Pythagorean Triples. Chapter 2. a 2 + b 2 = c 2

Chapter Pythagorean Triples The Pythagorean Theorem, that beloved formula of all high school geometry students, says that the sum of the squares of the sides of a right triangle equals the square of the

### Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Spring 2012 Homework # 9, due Wednesday, April 11 8.1.5 How many ways are there to pay a bill of 17 pesos using a currency with coins of values of 1 peso, 2 pesos,

### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion

CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a

### CONTENTS 1. Peter Kahn. Spring 2007

CONTENTS 1 MATH 304: CONSTRUCTING THE REAL NUMBERS Peter Kahn Spring 2007 Contents 2 The Integers 1 2.1 The basic construction.......................... 1 2.2 Adding integers..............................

### C H A P T E R Regular Expressions regular expression

7 CHAPTER Regular Expressions Most programmers and other power-users of computer systems have used tools that match text patterns. You may have used a Web search engine with a pattern like travel cancun

### Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

### CHAPTER 3. Methods of Proofs. 1. Logical Arguments and Formal Proofs

CHAPTER 3 Methods of Proofs 1. Logical Arguments and Formal Proofs 1.1. Basic Terminology. An axiom is a statement that is given to be true. A rule of inference is a logical rule that is used to deduce

### Mathematical induction. Niloufar Shafiei

Mathematical induction Niloufar Shafiei Mathematical induction Mathematical induction is an extremely important proof technique. Mathematical induction can be used to prove results about complexity of

### Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.

Advice for homework: Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Even if a problem is a simple exercise that doesn t

### The Language of Mathematics

CHPTER 2 The Language of Mathematics 2.1. Set Theory 2.1.1. Sets. set is a collection of objects, called elements of the set. set can be represented by listing its elements between braces: = {1, 2, 3,

### Math212a1010 Lebesgue measure.

Math212a1010 Lebesgue measure. October 19, 2010 Today s lecture will be devoted to Lebesgue measure, a creation of Henri Lebesgue, in his thesis, one of the most famous theses in the history of mathematics.

### Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note 11

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note Conditional Probability A pharmaceutical company is marketing a new test for a certain medical condition. According

### Mathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson

Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement

### Iteration, Induction, and Recursion

CHAPTER 2 Iteration, Induction, and Recursion The power of computers comes from their ability to execute the same task, or different versions of the same task, repeatedly. In computing, the theme of iteration

### HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!

Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following

### Theory of Computation Prof. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras

Theory of Computation Prof. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Lecture No. # 31 Recursive Sets, Recursively Innumerable Sets, Encoding

### LEARNING OBJECTIVES FOR THIS CHAPTER

CHAPTER 2 American mathematician Paul Halmos (1916 2006), who in 1942 published the first modern linear algebra book. The title of Halmos s book was the same as the title of this chapter. Finite-Dimensional

### 1.4 Factors and Prime Factorization

1.4 Factors and Prime Factorization Recall from Section 1.2 that the word factor refers to a number which divides into another number. For example, 3 and 6 are factors of 18 since 3 6 = 18. Note also that

### Notes on Determinant

ENGG2012B Advanced Engineering Mathematics Notes on Determinant Lecturer: Kenneth Shum Lecture 9-18/02/2013 The determinant of a system of linear equations determines whether the solution is unique, without

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### You know from calculus that functions play a fundamental role in mathematics.

CHPTER 12 Functions You know from calculus that functions play a fundamental role in mathematics. You likely view a function as a kind of formula that describes a relationship between two (or more) quantities.

### WRITING PROOFS. Christopher Heil Georgia Institute of Technology

WRITING PROOFS Christopher Heil Georgia Institute of Technology A theorem is just a statement of fact A proof of the theorem is a logical explanation of why the theorem is true Many theorems have this

### Student Outcomes. Lesson Notes. Classwork. Discussion (10 minutes)

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8 Student Outcomes Students know the definition of a number raised to a negative exponent. Students simplify and write equivalent expressions that contain

### Handout #1: Mathematical Reasoning

Math 101 Rumbos Spring 2010 1 Handout #1: Mathematical Reasoning 1 Propositional Logic A proposition is a mathematical statement that it is either true or false; that is, a statement whose certainty or

### 8 Primes and Modular Arithmetic

8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.

### Mathematical Induction. Mary Barnes Sue Gordon

Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by

### INCIDENCE-BETWEENNESS GEOMETRY

INCIDENCE-BETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The full

### P (A) = lim P (A) = N(A)/N,

1.1 Probability, Relative Frequency and Classical Definition. Probability is the study of random or non-deterministic experiments. Suppose an experiment can be repeated any number of times, so that we

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### So let us begin our quest to find the holy grail of real analysis.

1 Section 5.2 The Complete Ordered Field: Purpose of Section We present an axiomatic description of the real numbers as a complete ordered field. The axioms which describe the arithmetic of the real numbers

### x if x 0, x if x < 0.

Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the

### APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

### CHAPTER 3 Numbers and Numeral Systems

CHAPTER 3 Numbers and Numeral Systems Numbers play an important role in almost all areas of mathematics, not least in calculus. Virtually all calculus books contain a thorough description of the natural,

### Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2 Proofs Intuitively, the concept of proof should already be familiar We all like to assert things, and few of us

### Open and Closed Sets

Open and Closed Sets Definition: A subset S of a metric space (X, d) is open if it contains an open ball about each of its points i.e., if x S : ɛ > 0 : B(x, ɛ) S. (1) Theorem: (O1) and X are open sets.

### CONTINUED FRACTIONS AND FACTORING. Niels Lauritzen

CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents

### Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012

Some series converge, some diverge. Series Convergence Tests Math 22 Calculus III D Joyce, Fall 202 Geometric series. We ve already looked at these. We know when a geometric series converges and what it

### Section IV.1: Recursive Algorithms and Recursion Trees

Section IV.1: Recursive Algorithms and Recursion Trees Definition IV.1.1: A recursive algorithm is an algorithm that solves a problem by (1) reducing it to an instance of the same problem with smaller

### Solutions for Practice problems on proofs

Solutions for Practice problems on proofs Definition: (even) An integer n Z is even if and only if n = 2m for some number m Z. Definition: (odd) An integer n Z is odd if and only if n = 2m + 1 for some

### CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

### Computability Theory

CSC 438F/2404F Notes (S. Cook and T. Pitassi) Fall, 2014 Computability Theory This section is partly inspired by the material in A Course in Mathematical Logic by Bell and Machover, Chap 6, sections 1-10.

### Math 223 Abstract Algebra Lecture Notes

Math 223 Abstract Algebra Lecture Notes Steven Tschantz Spring 2001 (Apr. 23 version) Preamble These notes are intended to supplement the lectures and make up for the lack of a textbook for the course

### Computational Geometry Lab: FEM BASIS FUNCTIONS FOR A TETRAHEDRON

Computational Geometry Lab: FEM BASIS FUNCTIONS FOR A TETRAHEDRON John Burkardt Information Technology Department Virginia Tech http://people.sc.fsu.edu/ jburkardt/presentations/cg lab fem basis tetrahedron.pdf

### 26 Integers: Multiplication, Division, and Order

26 Integers: Multiplication, Division, and Order Integer multiplication and division are extensions of whole number multiplication and division. In multiplying and dividing integers, the one new issue

### Fractions and Decimals

Fractions and Decimals Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles December 1, 2005 1 Introduction If you divide 1 by 81, you will find that 1/81 =.012345679012345679... The first

### MATH10040 Chapter 2: Prime and relatively prime numbers

MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive

### Exercises with solutions (1)

Exercises with solutions (). Investigate the relationship between independence and correlation. (a) Two random variables X and Y are said to be correlated if and only if their covariance C XY is not equal

### 1 Introduction to Counting

1 Introduction to Counting 1.1 Introduction In this chapter you will learn the fundamentals of enumerative combinatorics, the branch of mathematics concerned with counting. While enumeration problems can

### Automata and Formal Languages

Automata and Formal Languages Winter 2009-2010 Yacov Hel-Or 1 What this course is all about This course is about mathematical models of computation We ll study different machine models (finite automata,

### 4 Domain Relational Calculus

4 Domain Relational Calculus We now present two relational calculi that we will compare to RA. First, what is the difference between an algebra and a calculus? The usual story is that the algebra RA is

### 6.2 Permutations continued

6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of

### Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

### Mathematical Induction

Mathematical Induction (Handout March 8, 01) The Principle of Mathematical Induction provides a means to prove infinitely many statements all at once The principle is logical rather than strictly mathematical,

### Math 115 Spring 2011 Written Homework 5 Solutions

. Evaluate each series. a) 4 7 0... 55 Math 5 Spring 0 Written Homework 5 Solutions Solution: We note that the associated sequence, 4, 7, 0,..., 55 appears to be an arithmetic sequence. If the sequence

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### Polynomial Invariants

Polynomial Invariants Dylan Wilson October 9, 2014 (1) Today we will be interested in the following Question 1.1. What are all the possible polynomials in two variables f(x, y) such that f(x, y) = f(y,

### MATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.

MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0-534-40596-7. Systems of Linear Equations Definition. An n-dimensional vector is a row or a column

### This unit will lay the groundwork for later units where the students will extend this knowledge to quadratic and exponential functions.

Algebra I Overview View unit yearlong overview here Many of the concepts presented in Algebra I are progressions of concepts that were introduced in grades 6 through 8. The content presented in this course

### MITES Physics III Summer Introduction 1. 3 Π = Product 2. 4 Proofs by Induction 3. 5 Problems 5

MITES Physics III Summer 010 Sums Products and Proofs Contents 1 Introduction 1 Sum 1 3 Π Product 4 Proofs by Induction 3 5 Problems 5 1 Introduction These notes will introduce two topics: A notation which

### 1.2. Successive Differences

1. An Application of Inductive Reasoning: Number Patterns In the previous section we introduced inductive reasoning, and we showed how it can be applied in predicting what comes next in a list of numbers

### Overview. Essential Questions. Precalculus, Quarter 4, Unit 4.5 Build Arithmetic and Geometric Sequences and Series

Sequences and Series Overview Number of instruction days: 4 6 (1 day = 53 minutes) Content to Be Learned Write arithmetic and geometric sequences both recursively and with an explicit formula, use them

### 3 0 + 4 + 3 1 + 1 + 3 9 + 6 + 3 0 + 1 + 3 0 + 1 + 3 2 mod 10 = 4 + 3 + 1 + 27 + 6 + 1 + 1 + 6 mod 10 = 49 mod 10 = 9.

SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012) (1) (Exercise 11, Page 107) Which of the following is the correct UPC for Progresso minestrone soup? Show why the other numbers are not valid

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### Properties of Stabilizing Computations

Theory and Applications of Mathematics & Computer Science 5 (1) (2015) 71 93 Properties of Stabilizing Computations Mark Burgin a a University of California, Los Angeles 405 Hilgard Ave. Los Angeles, CA

### Indiana State Core Curriculum Standards updated 2009 Algebra I

Indiana State Core Curriculum Standards updated 2009 Algebra I Strand Description Boardworks High School Algebra presentations Operations With Real Numbers Linear Equations and A1.1 Students simplify and

### We give a basic overview of the mathematical background required for this course.

1 Background We give a basic overview of the mathematical background required for this course. 1.1 Set Theory We introduce some concepts from naive set theory (as opposed to axiomatic set theory). The

### Deterministic Finite Automata

1 Deterministic Finite Automata Definition: A deterministic finite automaton (DFA) consists of 1. a finite set of states (often denoted Q) 2. a finite set Σ of symbols (alphabet) 3. a transition function

### Pick s Theorem. Tom Davis Oct 27, 2003

Part I Examples Pick s Theorem Tom Davis tomrdavis@earthlink.net Oct 27, 2003 Pick s Theorem provides a method to calculate the area of simple polygons whose vertices lie on lattice points points with

### Lecture 3: Finding integer solutions to systems of linear equations

Lecture 3: Finding integer solutions to systems of linear equations Algorithmic Number Theory (Fall 2014) Rutgers University Swastik Kopparty Scribe: Abhishek Bhrushundi 1 Overview The goal of this lecture

### Discrete Mathematics. Hans Cuypers. October 11, 2007

Hans Cuypers October 11, 2007 1 Contents 1. Relations 4 1.1. Binary relations................................ 4 1.2. Equivalence relations............................. 6 1.3. Relations and Directed Graphs.......................

### This asserts two sets are equal iff they have the same elements, that is, a set is determined by its elements.

3. Axioms of Set theory Before presenting the axioms of set theory, we first make a few basic comments about the relevant first order logic. We will give a somewhat more detailed discussion later, but

### Lesson 6: Proofs of Laws of Exponents

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 Student Outcomes Students extend the previous laws of exponents to include all integer exponents. Students base symbolic proofs on concrete examples to

### ARITHMETICAL FUNCTIONS II: CONVOLUTION AND INVERSION

ARITHMETICAL FUNCTIONS II: CONVOLUTION AND INVERSION PETE L. CLARK 1. Sums over divisors, convolution and Möbius Inversion The proof of the multiplicativity of the functions σ k, easy though it was, actually

### Proof of Infinite Number of Fibonacci Primes. Stephen Marshall. 22 May Abstract

Proof of Infinite Number of Fibonacci Primes Stephen Marshall 22 May 2014 Abstract This paper presents a complete and exhaustive proof of that an infinite number of Fibonacci Primes exist. The approach

### This chapter is all about cardinality of sets. At first this looks like a

CHAPTER Cardinality of Sets This chapter is all about cardinality of sets At first this looks like a very simple concept To find the cardinality of a set, just count its elements If A = { a, b, c, d },

### 9.2 Summation Notation

9. Summation Notation 66 9. Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a

### Induction Problems. Tom Davis November 7, 2005

Induction Problems Tom Davis tomrdavis@earthlin.net http://www.geometer.org/mathcircles November 7, 2005 All of the following problems should be proved by mathematical induction. The problems are not necessarily