Mathematical Induction Reading: EC 2.3 Peter J. Haas INFO 150 - PowerPoint PPT Presentation

Mathematical Induction Reading: EC 2.3 Peter J. Haas INFO 150 Fall Semester 2019 Lecture 8 1/ 16

Mathematical Induction Overview First Examples Sentences About the Positive Integers Formal Mathematical Induction Examples Strong versus Weak Induction Lecture 8 2/ 16

Overview What is induction? I Ordinary usage: infer the future based on the past (”the sun will rise tomorrow”) I Mathematical induction: prove the truth of the next integer, from the past I Exploit fact that recurrence relations are often relatively simple I The perfect tool for attacking complexity Lecture 8 3/ 16

Overview What is induction? I Ordinary usage: infer the future based on the past (”the sun will rise tomorrow”) I Mathematical induction: prove the truth of the next integer, from the past I Exploit fact that recurrence relations are often relatively simple I The perfect tool for attacking complexity Example I 1 + 2 + · · · + 100 = 5 , 050 I 1 + 2 + · · · + 100 + 101 =? I Since first answer is correct: 1 + 2 + · · · + 100 + 101 = (1 + 2 + · · · + 100) + 101 = (5 , 050) + 101 = 5 , 151 Lecture 8 3/ 16

Induction as a Game Example I a k = a k − 1 + (2 k − 1) with a 1 = 1 I Try some values: a 1 = 1, a 2 = 4, a 3 = 9, a 4 = 16, a 5 = 25 I It looks like a n = n 2 , but can you prove this? I Try a Big Honking Table: n 2 Is a n = n 2 ? n Formula for a n Value of a n 1 a 1 = 1 1 1 yes 2 a 2 = a 1 + (2 · 2 − 1) 1 + 3 = 4 4 yes . . . . . . . . . . . . . . . 49 a 49 = a 48 + (2 · 49 − 1) 2304 + 97 = 2401 2401 yes 50 a 50 = a 49 + (2 · 50 − 1) 2401 + 99 = 2500 2500 yes Look at the last row: a 50 = a 49 + (2 · 50 − 1) = 49 2 + (2 · 50 − 1) Generalize: a m = ( m − 1) 2 + (2 · m − 1) Lecture 8 4/ 16

A First Inductive Proof Proposition The sequence defined recursively by a 1 = 1 and a k = a k − 1 + (2 k − 1) has closed form a n = n 2 . Proof: 1. Check every row in the table up through m − 1 2. For the m th row, a m = a m − 1 + (2 m − 1) 3. Since we have checked row m − 1, we know that a m − 1 = ( m − 1) 2 4. Substitute and solve: " " a m = a m − 1 + (2 m − 1) = ( m − 1) 2 + (2 m − 1) = m 2 − 2 m + 1 + 2 m − 1 = m 2 , and so row m of the table checks out. 5. We therefore know that, repeating this procedure, every row will check out Lecture 8 5/ 16

Example, Continued A tabular view of the general case n 2 Is a n = n 2 ? Formula for a n Value of a n n 1 a 1 = 1 1 1 yes 2 a 2 = a 1 + (2 · 2 − 1) 1 + 3 = 4 4 yes 3 a 3 = a 2 + (2 · 3 − 1) 4 + 5 = 9 9 yes . . . . . . . . . . . . . . . ( m − 1) 2 ( m − 1) 2 m − 1 a m − 1 = a m − 2 + � 2 · ( m − 1) − 1 � yes r � � mr m a m = a m − 1 + 2 · m − 1 ??? mr yes A from slide previous Example: given row n = 29 , check row n = 30 n 2 Is a n = n 2 ? n Formula for a n Value of a n � � 29 2 29 2 29 a 29 = a 28 + 2 · 29 − 1 yes a 30 = a 29 + (2 · 30 − 1) = 29 2 + (2 · 30 − 1) = 841 + 59 = 900 = 30 2 X Lecture 8 6/ 16

Another Example Recursive sequence: a n = a n − 1 + 2 · n with a 1 = 2 Proposed closed form: a n = n ( n + 1) Tabular setup Formula for a n Value of a n n ( n + 1) Is a n = n ( n + 1)? n 1 a 1 = 2 2 1 · 2 yes 2 a 2 = a 1 + 2 · 2 2 + 4 = 6 2 · 3 yes . . . . . . . . . . . . . . . m − 1 a m − 1 = a m − 2 + 2 · ( m − 1) ( m − 1) m ( m − 1) m yes Yes m a m = a m − 1 + 2 · m ? m ( m + 1) ??? ✓ th ) Check row m : , ) dm - it many Lm m = - 1) m th m = am 2M = - it am= Lecture 8 7/ 16

Closed Formulas for Sums i =1 i = 1 + 2 + · · · + n = n ( n +1) Claim: P n 2 P n n ( n +1) sum = n ( n +1) n i =1 i or 1 + 2 + · · · + n Simplified sum ? 2 2 P 1 1 · 2 1 i =1 i = 1 1 2 = 1 Yes P 2 2 · 3 2 i =1 i = 1 + 2 1 + 2 = 3 2 = 3 Yes P 3 3 · 4 3 i =1 i = (1 + 2) + 3 3 + 4 = 6 2 = 6 Yes P 4 4 · 5 4 i =1 i = (1 + 2 + 3) + 4 6 + 4 = 10 2 = 10 Yes . . . . . . . . . . . . . . . P 34 34 · 35 34 i =1 i = (1 + 2 + · · · + 33) + 34 561 + 34 = 595 = 595 Yes 2 343+35595433=630 I ( it Yes Ei " 3521=630 t Lt - 35 . - i - I . . . . . - . . . . . . . . . . P m − 1 ( m − 1) m ( m − 1) m m − 1 i =1 i = 1 + 2 = · · · + ( m − 1) Yes 2 2 P m m ( m +1) m i =1 i = 1 + 2 = · · · + m ??? ??? 2 ! Lm - 1) -21Mt ,¥ tm DMTLM C Hht M t C m - m - - - I - - Check row m : = - th ) 2 m ( ( m - 1) MLMTD ✓ = = Check row 146: - - I 2 2 this Do exercise as an Lecture 8 8/ 16

Sentences About the Positive Integers Definition A statement about the positive integers is a predicate P ( n ) with the set of positive integers as its domain. Example: Which are statements about the positive integers? x I I n 2 + n is even I 100 − n I 100 − n > 83 ✓ I John has fewer than n apples in his refrigerator Example: For each predicate, write the sentence when n = 2 and n = 30 and determine whether it is true or false T - T EGO ) EH - I E ( n ) is the statement “ n 2 + n is even” - - , f 6130 ) I GCL ) T = I G ( n ) is the statement “100 − n > 83” ) ✓ , 5121--11-2=2 I S ( n ) is the statement “1 + 2 + · · · + n = n ( n +1) ” 2 ✓ t 30=465 =3 5130 ) It = - - - Lecture 8 9/ 16

Sentences, Continued Example I P ( n ) is “If there are n students in the class, the room will be too small” the 35 students the class there - D= Ii it If I What is P (35)? in are , be small in the class will too room students , I What is P ( m − 1)? there be too will If , are small ma room the Example I S ( n ) is “1 + 4 + 9 + · · · + n 2 = n ( n +1)(2 n +1) ” 6 sense E. I Rewrite using Σ notation ie . I Write S (1), S (2), and S (3) and determine if true or false D= if , T= 2¥ it # I ✓ SC It ✓ Scs ) = ' ¥t¥ = I 245=14 3641 - ✓ e SC 3) 2 it i = , I Write S ( m − 1) and simplify it - is Hit tem son . 6 . . . ul ) = 6 Lecture 8 10/ 16

Sentences, Continued 15,93=19 Example 9=11 a , I Define sequence by a 1 = 11 and a k = a k − 1 + 4 I R ( n ) is “ a n = 4 n + 7” I Write R (1), R (2), and R (3) and determine if true or false ✓ 4.247=15 91=4 V AE 7=11 Rh ) Ru ) It : : - ✓ 93=4.31-7--19 ) Rls : I Write R ( m − 1) and simplify it 3 4Mt - Dtt = 4. cm = - D am Acm : , Lecture 8 11/ 16

The Principle of Mathematical Induction The Principle of Mathematical Induction Let P ( n ) be a statement about the positive integers. If one can prove that " Bus ? i. P (1) is true; and → case ii. for every integer m ≥ 2, whenever P (1) , P (2) , . . . , P ( m − 1) are all true, it follows that P ( m ) is true, then P ( n ) is true for every positive integer n . i =1 i = n ( n +1) Example: Prove that P n for every positive integer n 2 i =1 i = n ( n +1) 1. Let P ( n ) = “ P n ” 2 2. P (1) states that 1 = 1(1+1) , which is clearly true 2 3. Let m ≥ 2 and suppose that P (1) , P (2) , . . . , P ( m − 1) are true 4. Consider P ( m ): m X i = 1 + 2 + · · · + m = (1 + 2 + · · · + ( m − 1)) + m i =1 = ( m − 1) m + m since S ( m − 1) is true 2 � ( m − 1) + 2 � = ( m − 1) m + 2 m = m = m ( m + 1) . 2 2 2 5. Thus P ( m ) is true, completing the induction Lecture 8 12/ 16

Examples i =1 2 i − 1 = 2 n − 1” Example 1: Prove P ( n ) = “ P n ✓ : ! 's j ' i. PhD ? , I - all true = - 1) are that . , Plm Pll ) and suppose . Let . 32 , - ' - ! m - ' I z poms : Iii ' ' In I hi 20=1--2 . - ' in ' " it 2. I - . a t = - V - 3. = am I - result induction follows by so Example 2: Given a 1 = 1 and a k = a k − 1 + (2 k − 1) for k ≥ 2, prove that a n = n 2 for p C n ) all n ≥ 1 " ' ' ' an : ' n are all true ✓ get - D Pll ) : and suppose that put , . , Plm i. . . Let - 1) is miss PL 'm a - - d.my#2m-l=Lm-D2f2m since . I - true assumed Plm ) am : 3. T K Hmt 4M - I so that am . ; - ) t mh - = m2 ✓ s Lecture 8 13/ 16

Exercise Given a 1 = 2 and a k = a k − 1 + 2 k for k ≥ 2, prove that a n = n ( n + 1) for all n ≥ 1 Curti ) P Ch ) i n ane ✓ - fit D= 2 I P LD go : I - D true . Pcm PCI ) are and . Let assume me 2 . . . , , 2 Ams want to show what mcm xD ← we Plm ) am : = 3 - ' I . Plm = ( - 1) km since ) t am Dti t Lm m am . , ÷÷÷÷ : Lecture 8 14/ 16

All Puppies are the Same Color Puppy Theorem Let P ( n ) be the predicate that any set of n puppies are all the same color. Then P ( n ) holds for n ≥ 1. Inductive Proof: 1. Base case: P (1) holds trivially (a puppy is the same color as itself) Line , for some 2. Assume that P (1) , P (2) , . . . , P ( m − 1) all hold for some m > 1 32 ) 3. Inductive argument for P ( m ), i.e. for a set of m puppies: M argument 3.1 Remove one puppy (say, Whiskers), leaving m − 1 puppies hold 3.2 By induction, the m − 1 puppies have the same color (say, white), since must P ( m − 1) holds for all 3.3 Pick one of those puppies (say, Bella), who must be white 3.4 Remove Bella and add back Whiskers, leaving m − 1 puppies 32 m , 3.5 Since P ( m − 1) is assumed to hold, all of the puppies are the same color, f- so Whiskers is white also but 3.6 Hence all of the puppies are white, and hence are the same color fails Pls ) , i.e for argument for man . , Lecture 8 15/ 16