Section 5 1 Sequences A sequence of numbers

Section 5. 1 Sequences A sequence of numbers is an ordered list of numbers. In this section we mostly consider infinite sequences of the form a 1, a 2, a 3, …, an, … An explicit formula or general formula for a sequence is a rule that shows how the value of ak depend on k. Example 1. if ak = k/(k + 1), then the sequence is 2. if bk = (- ½)k, then the sequence is 1 -1 2, 2 1 4 3, 3 4, 4 5, 5 6, , -1 8 , 1 16 , … …

Finding an Explicit formula to Fit Given Initial Terms Examples: 1. 4, 7, 10, 13, 16, … 2. 3, 4, 6, 9, 13, 18, … 3. 1, 0, -1, 0, 1, … 4. 1, 1, 2, 3, 5, 8, 13, 21, …

An explicit formula for the sequence 1, 0, -1, 0, … without using trigonometric functions where n = 1, 2, 3, 4, … Explanation: The given sequence can be viewed as the term wise product of two sequences;

The sequence has an explicit formula The sequence also has an explicit formula, but it is more complicated where

Fibonacci Sequence for n = 0, 1, 2, 3, 4, …

The Summation Notation

The Product Notation Factorial notation n! = n×(n – 1)×(n – 2)×···× 2× 1 for n > 0 0! is defined to be 1. F. Y. I A generalization of the factorial function is the Gamma function

The Product Notation Factorial notation n! = n×(n – 1)×(n – 2)×···× 2× 1 for n > 0 0! is defined to be 1. F. Y. I A generalization of the factorial function is the Gamma function and it is not hard to see that Γ(n) = (n – 1)!

Properties of Summation and Products Theorem If am, am+1, am+2, … and bm, bm+1, bm+2, … are sequences of real numbers and c is a constant, then the following equations are true

Two major types of arguments Deductive Reasoning • applies general principles to reach specific conclusions. • logically progresses from general evidence to a particular truth or conclusion. Inductive Reasoning • exams a systematic list of examples to impute a general principle • starts with a particular observation that is believed to be a demonstrative model for a truth or principle that is assumed to apply generally.

Examples Deductive Reasoning • In the 19 th century, Adams and Le. Verrier applied Newton's theory (general principle) to deduce the existence, mass, position, and orbit of Neptune from perturbations in the observed orbit of Uranus. Inductive Reasoning • By thinking about phenomena such as how apples fall and how the planets move, Isaac Newton induced his theory of gravity

Claim Any whole number of cents of at least 8¢ can be obtained using only 3¢ and 5¢. Number of cents How to obtain it 8¢ 3¢ + 5¢ 9¢ 3¢ + 3¢ 10¢ 5¢+ 5¢ 11¢ 3¢ + 5¢ 12¢ 3¢ + 3¢ 13¢ 3¢ + 5¢ 14¢ 3¢ + 5¢

Observe another pattern 1=1 1+3= 4 1+3+5=9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 A pattern: The sum of the first n positive odd numbers is n 2 Mathematical formulation P(n) : 1 + 3 + 5 + … + (2 n-1) = n 2 We would like to show that P(n) is true for all positive whole number n.

Observe another pattern 13 = 1 13 + 23 = 9 13 + 23 + 33 = 36 13 + 23 + 33 + 43 = 100 13 + 23 + 33 + 43 + 53 = 225 1=1 1+2=3 1+2+3=6 1 + 2 + 3 + 4 = 10 1 + 2 + 3 + 4 + 5 = 15 A pattern: the sum of the first n cubes is equal to the square of the sum of the first n natural numbers. Mathematical formulation P(n) : 13 + 23 + … + n 3 = (1 + … + n)2 We would like to show that P(n) is true for all positive whole number n.

Mathematical Induction The method of induction is like building an infinite chain of dominoes where one is positioned behind another in such a way that if any given domino falls backwards, it makes the one behind it fall backwards too. Now imagine that we push the first domino backwards, … Human Domino

Well-ordering principle of whole numbers For any non-empty (finite or infinite) set of whole numbers, there must be a smallest element in that set. Note: this principle is not true for real numbers or even fractions. Example: Consider the set of fractions {1/q : q is a positive integer} = This set does not have a smallest element.

Well-ordering principle of whole numbers An application: If a and b are positive integers, then there are integers s and t such that as + bt = GCD(a, b) To prove this result, we have to construct a nonempty subset S of the positive integers. Then according to the well-ordering principle, S must have a smallest element k. We finally prove that this smallest non-zero k is the GCD of a and b.

We would like to prove that P(n) is true for all whole number n. By contradiction: Suppose to the contrary that P(n) is not true for all n, then we have a non-empty subset A of whole numbers, A = {n : P(n) is false} By the well-ordering principle, A must have a smallest element. Let this smallest element be k. Since we know that P(1) is true, k must be >1. k 0 1 2 . . . k -1

k 0 1 2 . . . k -1 The smallest “bad” number. The red dots are “bad” numbers These are all “good” numbers If we can prove that P(k-1) P(k), then we have a contradiction and the proof is complete. In other words, our assumption that “P(n) is not true for all n” is false.

Let P(n) be a predicate in terms of the variable n. To show that P(n) is true for all (positive integer) n, • check that P(1) is true. • prove that P(k-1) P(k) where k is treated as a constant > 1. Remarks: • some books use P(k) P(k+1), which is equivalent, and it is a matter of taste on which one to use. • the step “assume P(k-1) is true” is just for your own reference. • sometimes we have to start from an integer greater than 1.

Examples: 1. Show that 12 + 22 + 32 + ∙∙∙ + n 2 = n(n + 1)(2 n + 1)/6 2. Show that 13 + 23 + 33 + ∙∙∙ + n 3 = [n(n + 1)/2]2 3. Show that 14 + 24 + 34 + ∙∙∙ + n 4 = n(n + 1)(2 n + 1)(3 n 2 + 3 n – 1)/30 4. Show that 15 + 25 + 35 + ∙∙∙ + n 5 = n 2(n + 1)2(2 n 2 + 2 n – 1)/12 5. Show that 6. Show that every number n > 11 can be written as the from 4 a + 5 b where a, b are non-negative integers.

7. Show that n 3 + 2 n is divisible by 3 for any whole number n. 8. Show that 7 n – 2 n is divisible by 5 for any whole number n. 9. Show that (n + 1)! > 2 n for all whole number n > 1. 10. Show that whenever x is not divisible by π. 11. Show that for all whole number n ≥ 1 Hint:

An induction paradox A teacher told her students that there would be a pop quiz next week (Mon – Fri), and nobody could predict in advance which day it would be on. A bright student started to analyze this way. “If no one can predict in advance, then the quiz cannot be on Friday. And if we know right now that the quiz cannot be on Friday, then we also know right now that it cannot be on Thursday either. By induction, the quiz cannot be on any day of the week at all. ” He then discussed his logic with everyone in class, and they all believed that he was right, hence no one prepared for the quiz. Nevertheless, the pop quiz came on Tuesday – unexpected to everybody. Therefore the teacher was also right. How could this happen?

A classical mistake in Mathematical Induction “Theorem: For any positive integer n, any group of n people must be of the same gender. ” Incorrect proof (by induction on n): When n = 1 this statement is clearly true. Assume that this statement is true for n = k 1. Suppose that we are given k+1 people {s 1, s 2, …, sk+1}, then by induction assumption {s 1, s 2, …, sk} must be of the same gender, say female. In particular, s 2 is a female. Also by induction assumption {s 2, …, sk+1} must also be of the same gender. Since s 2 is a female, {s 2, …, sk+1} must all be female. Hence {s 1, s 2, …, sk+1} are of the same gender. Q. E. D. Where is the mistake?

Strong induction. These inductions are called strong because • we need to check a few more initial cases • we need to make a stronger assumption • we will obtain a stronger result. Example: We construct a sequence {an} by the following rule • a 1 = 5, • a 2 = 10, • an = a(n-1) + a(n-2) when n 3 Our predicate P(n) is: “an is divisible by 5. ” Our prediction is: P(n) is true for all n.

Suppose to the contrary that P(n) is not true for all n, then the set A = { n : P(n) is false } is not empty. By the well-ordering principle, A must have a smallest element, say k. And k must be 3. The smallest “bad” number. These are all “good” numbers

We may try to prove: P(k-1) P(k) But this won’t work because ak depends on a(k-1) and a(k-2) Hence we must use a stronger condition P(k-1) P(k-2) P(k) where k is a constant 3. Another example: If an = a(n-1) + a(n-2) + a(n-3) , then we need to • check that P(1), P(2), P(3) are all true, • prove that P(k-1) P(k-2) P(k-3) P(k) whenever k is a constant 4

Example: Every whole number 2 is either a prime or a product of primes. P(n) is the predicate: n is a prime or a product of primes. • check that P(2) is true, • prove that P(2) P(3) … P(k-1) P(k) The strongest type of induction: • check that P(a) is true • prove that P(a) P(a+1) P(a+2) … P(k-1) P(k) whenever k is a constant > a. • conclude that P(n) is true for all n a

This type of induction applies when P(n) depends on some previous terms P(k) with k < n but not a fixed number of previous terms. Example: Prove that every whole number greater than 1 is either prime or is a product of primes. Proof: Since 2 is clearly a prime number, the initial case is true. Assume that this statement is true for all positive integers greater than 1 and less than some n (n >2). To show that this is also true for n, we consider two cases: (1) n is prime there is nothing to prove in this case.

Prove that every whole number greater than 1 is either prime or is a product of primes. Proof: Since 2 is clearly a prime number, the initial case is true. Assume that this statement is true for all positive integers greater than 1 and less than some n (n>2). To show that this is also true for n, we consider two cases: (1) n is prime there is nothing to prove. (2) n is not prime then n is composite by definition and there must be whole numbers a and b such that n = a×b and n > a, b > 1 By our strong induction assumption, a and b are both products of prime numbers, so n is also a product of prime numbers. Q. E. D.