Mathematical Induction James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University January 12, 2017 Outline Introduction to the Class Mathematical Induction
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 we think about problems here is different than the way we handle problems in MATH 4120 which is about abstract algebra. Our job is to get you to think about the underpinnings of the material you have already mastered in calculus more abstractly and train you to really think through the consequences of assumptions. The basic outline of this course is Sequences of numbers Limits of functions Continuity and Differentiation of Functions Deeper ideas such as compactness Extremal Theory: minimum and maximum for functions Approximation Ideas using Taylor polynomials This class is organized like this 3 Exams and One Final which is cumulative and all have to take it. The days for the exams and the final are already set and you can t change those so make you travel plans accordingly. Usually HW is assigned every lecture Two projects which let you look at some topics in detail as well as train you how to write mathematics. Details Handwritten notes by me. You ll get used to my handwriting as it is the same stuff you see when I write on the board No textbook No blackboard type stuff: I ll assign HW via email after each Lecture. 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 is about 80-160 exercises in all. In addition, there are two projects. All are designed to make you learn and to see how to think about things 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 to prepare for an exam: Study hard and then take sample exam 1 as a timed test for 60 minutes. You will know what you didn t get right and so you can figure out what to study. I don t give answers to these tests as this is stuff you should know. After studying again, take the second sample exam as a 60 minute timed exam and see how you did. Then study again to fill in the gaps. Take the third sample exam as a timed 60 minute exam. You should be able to do well and you are now prepared for the real exam. Mathematical Induction The Principle of Mathematical Induction For each natural number n, let P(n) be a statement or proposition about 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.
The natural numbers or counting numbers are usually denoted by the symbol N. The set of all integers, positive, negative and zero is denoted by Z and the real numbers is denoted by R or R. The phrase for all or for every is usually denoted by the symbol. The phrase there is or there exists, although we haven t gotten to using that phrase yet, is denoted by. There are many alternative versions of this. The Principle of Mathematical Induction For each natural number n, let P(n) be a statement or proposition about 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 : BASIS Verify P(1) is true INDUCTIVE Assume P(k) is true for arbitrary k and use that information to prove P(k + 1) is true. We have verified the inductive step. Hence, by the POMI, P(n) holds for all n. You must include this finishing statement as part of your proof and show the QED as above. QED QED is an abbreviation for the Latin Quod Erat Demonstratum or that which was to be shown. We often use the symbol instead of QED. The POMI Template So in outline, we have Proposition Statement Body : BASIS Verification INDUCTIVE Body of Argument Finishing Phrase QED or Note we try to use pleasing indentation strategies and white space to improve readability and understanding.
n! 2 n 1 n 1 BASIS : P(1) is the statement 1! 2 1 1 = 2 0 = 1 which is true. So the basis step is verified. INDUCTIVE : We assume P(k) is true for an arbitrary k > 1. We use k > 1 because in the basis step we found out the proposition holds for k = 1. Hence, we know Now look at P(k + 1). We note k! 2 k 1. (k + 1)! = (k + 1) k! But, by the induction assumption, we know k! 2 k 1. Plugging this fact in, we have (k + 1)! = (k + 1) k! (k + 1) 2 k 1. To finish, we note since k > 1, k + 1 > 2. Thus, we have the final step (k + 1)! = (k + 1) k! (k + 1) 2 k 1 2 2 k 1 = 2 k. This is precisely the statement P(k + 1). Thus P(k + 1) is true and we have verifed the inductive step. Hence, by the POMI, P(n) holds for all n.
Let s change it a bit. n! 3 n 1 n 5 BASIS : P(5) is the statement 5! = 120 3 5 1 = 3 4 = 81 which is true. So the basis step is verified. INDUCTIVE : We assume P(k) is true for an arbitrary k > 5. We use k > 5 because in the basis step we found out the proposition holds for k = 5. Hence, we know Now look at P(k + 1). We note k! 3 k 1. (k + 1)! = (k + 1) k! But, by the induction assumption, we know k! 3 k 1. Plugging this fact in, we have (k + 1)! = (k + 1) k! (k + 1) 3 k 1. To finish, we note since k > 5, k + 1 > 6 > 3. Thus, we have the final step (k + 1)! = (k + 1) k! (k + 1) 3 k 1 3 3 k 1 = 3 k. This is precisely the statement P(k + 1). Thus P(k + 1) is true and we have verifed the inductive step. Hence, by the POMI, P(n) holds for all n 5. Note we use a different form of POMI here.
This one is a bit harder. 1 2 2 2 + 3 2 + ( 1) n+1 n 2 = 1 2 ( 1)n+1 n (n + 1) BASIS : P(1) is what we get when we plug in n = 1 here. This gives 1 2 on the left hand side and 1 2 ( 1)2 1(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 1 2 2 2 + 3 2 + ( 1) k+1 k 2 = 1 2 ( 1)k+1 k (k + 1) Now look at the left hand side of P(k + 1). We note at k + 1, we have the part that stops at k + 1 and a new term that stops at k + 2. We can write this as ( 1 2 2 2 + 3 2 + ( 1) k+1 k 2) + ( 1) k+2 (k + 1) 2. But the first part of this sum corresponds to the induction assumption or hypothesis. 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 ( 1) k+1 (k + 1) k (k + 1) = 1 2 2 ( 1)k+1 (k + 1) {k 2k 2}
The term k 2k 2 = k 2 and so bringing out another 1 and putting it into the ( 1) k+1, we have { } 1 ( 1) k+1 (k + 1) k (k + 1) = 1 2 2 ( 1)k+2 (k + 1)(k + 2) This is precisely the statement P(k + 1). Thus P(k + 1) is true and we have verifed the inductive step. Hence, by the POMI, P(n) holds for all n 5. This argument was a lot harder as we had to think hard about how to manipulate the left hand side. So remember, it can be tricky to see how to finish the induction part of the argument! Homework 1 Provide a careful proof of this proposition. There are also more worked out examples of POMI in the notes. So make sure you read through that material. 1.1 n! 4 n n 9