Mathematical Induction - Clemson CECAScecas.clemson.edu/.../M453/Lectures/L1-MathInduction.pdfQED is an abbreviation for the Latin Quod Erat Demonstratum or that which was to be shown

MATH 4530: Analysis One

Mathematical Induction

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

January 10, 2019


Outline

1 Introduction to the Class

2 Mathematical Induction

3 Homework


Introduction to the Class

First, a few comments about this class.

This class is about training you to think in a different way.MATH 4530 is about analysis and you will find the way wethink about problems here is different than the way we handleproblems in MATH 4120 which is about abstract algebra.Our job is to get you to think about the underpinnings of thematerial you have already mastered in calculus more abstractlyand train you to really think through the consequences ofassumptions.

The basic outline of this course is

Sequences of numbersLimits of functionsContinuity and Differentiation of FunctionsDeeper ideas such as compactnessExtremal Theory: minimum and maximum for functionsApproximation Ideas using Taylor polynomials



First, a few comments about this class.

This class is about training you to think in a different way.MATH 4530 is about analysis and you will find the way wethink about problems here is different than the way we handleproblems in MATH 4120 which is about abstract algebra.Our job is to get you to think about the underpinnings of thematerial you have already mastered in calculus more abstractlyand train you to really think through the consequences ofassumptions.

The basic outline of this course is

Sequences of numbersLimits of functionsContinuity and Differentiation of FunctionsDeeper ideas such as compactnessExtremal Theory: minimum and maximum for functionsApproximation Ideas using Taylor polynomials



This class is organized like this

3 Exams and One Final which is cumulative and all have totake it. The days for the exams and the final are already setand you can’t change those so make you travel plansaccordingly.Usually HW is assigned every lectureTwo projects which let you look at some topics in detail aswell as train you how to write mathematics.

Details

Handwritten notes by me. You’ll get used to my handwritingas it is the same stuff you see when I write on the boardNo textbookNo blackboard type stuff: I’ll assign HW via email after eachLecture.I take daily attendance and give you bonus points at the end.



This class is organized like this

3 Exams and One Final which is cumulative and all have totake it. The days for the exams and the final are already setand you can’t change those so make you travel plansaccordingly.Usually HW is assigned every lectureTwo projects which let you look at some topics in detail aswell as train you how to write mathematics.

Details

Handwritten notes by me. You’ll get used to my handwritingas it is the same stuff you see when I write on the boardNo textbookNo blackboard type stuff: I’ll assign HW via email after eachLecture.I take daily attendance and give you bonus points at the end.



Some words about studying

HW is essential. I give about 2 - 4 exercises per lecture so that isabout 80 - 160 exercises in all. In addition, there are two projects.All are designed to make you learn and to see how to think aboutthings on your own. This takes time so do the work and be patient.

I give 3 sample exams on the web site. You should do this toprepare for an exam:

Study hard and then take sample exam 1 as a timed test for 60minutes. You will know what you didn’t get right and so youcan figure out what to study. I don’t give answers to thesetests as this is stuff you should know.After studying again, take the second sample exam as a 60minute timed exam and see how you did. Then study again tofill in the gaps.Take the third sample exam as a timed 60 minute exam. Youshould be able to do well and you are now prepared for the realexam.



Some words about studying

HW is essential. I give about 2 - 4 exercises per lecture so that isabout 80 - 160 exercises in all. In addition, there are two projects.All are designed to make you learn and to see how to think aboutthings on your own. This takes time so do the work and be patient.

I give 3 sample exams on the web site. You should do this toprepare for an exam:

Study hard and then take sample exam 1 as a timed test for 60minutes. You will know what you didn’t get right and so youcan figure out what to study. I don’t give answers to thesetests as this is stuff you should know.After studying again, take the second sample exam as a 60minute timed exam and see how you did. Then study again tofill in the gaps.Take the third sample exam as a timed 60 minute exam. Youshould be able to do well and you are now prepared for the realexam.




Theorem

The Principle of Mathematical InductionFor each natural number n, let P(n) be a statement or propositionabout the numbers n.

P(1) is true: This is called the BASIS STEP

P(k + 1) is true when P(k) is true:

This is called the INDUCTIVE STEP

then we can conclude P(n) is true for all natural numbers n.



Comment

The natural numbers or counting numbers are usually denoted bythe symbol N. The set of all integers, positive, negative and zero isdenoted by Z and the real numbers is denoted by < or R.

Comment

The phrase for all or for every is usually denoted by the symbol ∀.

Comment

The phrase there is or there exists, although we haven’t gottento using that phrase yet, is denoted by ∃.



There are many alternative versions of this.

Theorem

The Principle of Mathematical InductionFor each natural number n, let P(n) be a statement or propositionabout the numbers n. If

If there is a number n0 so that P(n0) is true: BASIS STEP

P(k + 1) is true when P(k) is true for all k ≥ n0:

INDUCTIVE STEP

then we can conclude P(n) is true for all natural numbers n ≥ n0.



The POMI Template

State the Proposition Here

Proof:

BASISVerify P(1) is true

INDUCTIVEAssume P(k) is true for arbitrary k and use thatinformation to prove P(k + 1) is true.

We have verified the inductive step. Hence, bythe POMI, P(n) holds for all n.

QEDYou must include this finishing statement as partof your proof and show the QED as above.

Comment

QED is an abbreviation for the Latin Quod Erat Demonstratum or thatwhich was to be shown. We often use the symbol instead of QED.



The POMI Template

So in outline, we haveProposition Statement Body

Proof:

BASIS Verification

INDUCTIVE Body of Argument

Finishing Phrase

QED or

Comment

Note we try to use pleasing indentation strategies and white space toimprove readability and understanding.



Theorem

n! ≥ 2n−1 ∀n ≥ 1

Proof

BASIS : P(1) is the statement 1! ≥ 21−1 = 20 = 1 which is true. Sothe basis step is verified.

INDUCTIVE : We assume P(k) is true for an arbitrary k > 1. We usek > 1 because in the basis step we found out the proposition holds fork = 1. Hence, we know

k! ≥ 2k−1.

Now look at P(k + 1). We note

(k + 1)! = (k + 1) k!



Proof

But, by the induction assumption, we know k! ≥ 2k−1. Plugging this factin, we have

(k + 1)! = (k + 1) k! ≥ (k + 1) 2k−1.

To finish, we note since k > 1, k + 1 > 2. Thus, we have the final step

(k + 1)! = (k + 1) k! ≥ (k + 1) 2k−1 ≥ 2× 2k−1 = 2k .

This is precisely the statement P(k + 1). Thus P(k + 1) is true and wehave verifed the inductive step. Hence, by the POMI, P(n) holds for alln.



Let’s change it a bit.

Theorem

n! ≥ 3n−1 ∀n ≥ 5

Proof

BASIS : P(5) is the statement 5! = 120 ≥ 35−1 = 34 = 81 which istrue. So the basis step is verified.

INDUCTIVE : We assume P(k) is true for an arbitrary k > 5. We usek > 5 because in the basis step we found out the proposition holds fork = 5. Hence, we know

k! ≥ 3k−1.

Now look at P(k + 1). We note

(k + 1)! = (k + 1) k!



Proof

But, by the induction assumption, we know k! ≥ 3k−1. Plugging this factin, we have

(k + 1)! = (k + 1) k! ≥ (k + 1) 3k−1.

To finish, we note since k > 5, k + 1 > 6 > 3. Thus, we have the finalstep

(k + 1)! = (k + 1) k! ≥ (k + 1) 3k−1 ≥ 3× 3k−1 = 3k .

This is precisely the statement P(k + 1). Thus P(k + 1) is true and wehave verifed the inductive step. Hence, by the POMI, P(n) holds for alln ≥ 5.

Comment

Note we use a different form of POMI here.



This one is a bit harder.

Theorem

12 − 22 + 32 − · · ·+ (−1)n+1n2 =1

2(−1)n+1 n (n + 1) ∀ n ≥ 1

Proof

BASIS : P(1) is what we get when we plug in n = 1 here. This gives 12

on the left hand side and 12 (−1)21(2) on the right hand side. We can see

1 = 1 here and so P(1) is true.

INDUCTIVE : We assume P(k) is true for an arbitrary k > 1. Hence,we know

12 − 22 + 32 − · · ·+ (−1)k+1k2 =1

2(−1)k+1 k (k + 1)



Proof

Now look at the left hand side of P(k + 1). We note at k + 1, we havethe part that stops at k + 1 and a new term that stops at k + 2. We canwrite this as(

12 − 22 + 32 − · · ·+ (−1)k+1k2)

+ (−1)k+2(k + 1)2.

But the first part of this sum corresponds to the induction assumption orhypothesis. We plug this into the left hand side of P(k + 1) to get

1

2(−1)k+1 k (k + 1) + (−1)k+2(k + 1)2.

Now factor out the common (−1)k+1(k + 1) to get

(−1)k+1(k + 1)

{1

2k − (k + 1)

}=

1

2(−1)k+1(k + 1) {k − 2k − 2}



Proof

The term k − 2k − 2 = −k − 2 and so bringing out another −1 andputting it into the (−1)k+1, we have

(−1)k+1(k + 1)

{1

2k − (k + 1)

}=

1

2(−1)k+2(k + 1)(k + 2)

This is precisely the statement P(k + 1). Thus P(k + 1) is true and wehave verifed the inductive step. Hence, by the POMI, P(n) holds for alln ≥ 1.

Comment

This argument was a lot harder as we had to think hard about how tomanipulate the left hand side. So remember, it can be tricky to see howto finish the induction part of the argument!


Homework

Homework 1

Provide a careful proof of this proposition. There are also moreworked out examples of POMI in the notes. So make sure you readthrough that material.

1.1

Theorem

n! ≥ 4n ∀n ≥ 9

Documents

Mathematical Induction - Clemson CECAScecas.clemson.edu/.../M453/Lectures/L1-MathInduction.pdfQED is an abbreviation for the Latin Quod Erat Demonstratum or that which was to be shown