More About Induction Reading: EC 2.4 Peter J. Haas INFO 150 Fall - PowerPoint PPT Presentation

More About Induction Reading: EC 2.4 Peter J. Haas INFO 150 Fall Semester 2019 Lecture 9 1/ 15

More About Induction Overview Sums as Recursive Sequences Proving Properties of a Recursive Sequence Closed-Form Sequences General Inductive Proofs Fundamental Theorem of Arithmetic Lecture 9 2/ 15

Overview Goal I Connect induction on recursive sequences and sums I Illustrate fancier applications of induction I Use induction in a variety of settings Lecture 9 3/ 15

Sums as Recursive Sequences Recall: with a 1 = 1 and a k = a k − 1 + (2 k − 1), prove that a n = n 2 I Inductive step to show that a m = m 2 : a m = a m − 1 + (2 m − 1) = ( m − 1) 2 + (2 m − 1) using a m − 1 = ( m − 1) 2 = m 2 − 2 m + 1 + 2 m − 1 = m 2 Lecture 9 4/ 15

Sums as Recursive Sequences Recall: with a 1 = 1 and a k = a k − 1 + (2 k − 1), prove that a n = n 2 I Inductive step to show that a m = m 2 : a m = a m − 1 + (2 m − 1) = ( m − 1) 2 + (2 m − 1) using a m − 1 = ( m − 1) 2 = m 2 − 2 m + 1 + 2 m − 1 = m 2 Recall: Prove that P n i =1 i = n ( n +1) 2 I Inductive step to show S ( m ): 1 + 2 + · · · + m = � 1 + 2 + · · · + ( m − 1) � + m = ( m − 1) m + m using S ( m − 1) 2 = m 2 − m = m 2 + m + 2 m = m ( m + 1) 2 2 2 2 Lecture 9 4/ 15

Sums as Recursive Sequences Recall: with a 1 = 1 and a k = a k − 1 + (2 k − 1), prove that a n = n 2 I Inductive step to show that a m = m 2 : a m = a m − 1 + (2 m − 1) = ( m − 1) 2 + (2 m − 1) using a m − 1 = ( m − 1) 2 = m 2 − 2 m + 1 + 2 m − 1 = m 2 Recall: Prove that P n i =1 i = n ( n +1) 2 I Inductive step to show S ( m ): 1 + 2 + · · · + m = � 1 + 2 + · · · + ( m − 1) � + m = ( m − 1) m + m using S ( m − 1) 2 = m 2 − m = m 2 + m + 2 m = m ( m + 1) 2 2 2 2 Both proofs have a common structure I First step: Establish relationship between a m and a m − 1 [or S ( m ) and S ( m − 1)] I Second step: use that information Lecture 9 4/ 15

Sums as Recursive Sequences, Continued Key idea: sums are recursive by nature I Sum of 101 numbers = (sum of 100 numbers) + (final number) I Will write s n instead of a n to emphasize that given sequence is defined by a sum Lecture 9 5/ 15

Sums as Recursive Sequences, Continued Key idea: sums are recursive by nature I Sum of 101 numbers = (sum of 100 numbers) + (final number) I Will write s n instead of a n to emphasize that given sequence is defined by a sum Example I Consider s n = P n i =1 (2 i − 1) = 1 + 3 + 5 + · · · + (2 n − 1) I E.g., s 5 = 1 + 3 + 5 + 7 + 9 (five terms) I Find a recursive description of s n I s n = sum of first ( n − 1) terms plus n th term: n X s n = (2 i − 1) i =1 = 1 + 3 + 5 + · · · + (2 n − 1) � � = 1 + 3 + 5 + · · · + (2 n − 3) + (2 n − 1) = s n − 1 + (2 n − 1) I So s n = s n − 1 + (2 n − 1) is recurrence relation (with s 1 = 1) Lecture 9 5/ 15

Sums as Recursive Sequences, Continued Key idea: sums are recursive by nature I Sum of 101 numbers = (sum of 100 numbers) + (final number) I Will write s n instead of a n to emphasize that given sequence is defined by a sum Example I Consider s n = P n i =1 (2 i − 1) = 1 + 3 + 5 + · · · + (2 n − 1) I E.g., s 5 = 1 + 3 + 5 + 7 + 9 (five terms) I Find a recursive description of s n I s n = sum of first ( n − 1) terms plus n th term: n X s n = (2 i − 1) i =1 = 1 + 3 + 5 + · · · + (2 n − 1) � � = 1 + 3 + 5 + · · · + (2 n − 3) + (2 n − 1) = s n − 1 + (2 n − 1) I So s n = s n − 1 + (2 n − 1) is recurrence relation (with s 1 = 1) Itt ) i =1 (2 i + 1)? ( Example: What is the recursive form of s n = P n 921 it - Lecture 9 5/ 15

Example: Finding a Closed Form for a Sum Example: Show that P n 3 i = 1 − 1 2 3 n for every positive integer n i =1 1. Let s n = P n 3 i : will show that s n = 1 − 1 2 3 n i =1 2. Recurrence relation: s n = s n − 1 + 2 3 n with s 1 = 2 3 3. s 1 = 1 − 1 3 1 = 2 3 X 1 4. Fix m ≥ 2 & assume that statements up to s m − 1 = 1 − 3 m − 1 have been checked 5. Consider s m : s m = s m − 1 + 2 from recurrence relation 3 m 1 + 2 1 ⇣ ⌘ = 1 − since s m − 1 = 1 − 3 m − 1 3 m 3 m − 1 = 1 − 3 3 m + 2 3 m = 1 − 1 3 m X Lecture 9 6/ 15

Example: Finding a Closed Form for a Sum Example: Show that P n 3 i = 1 − 1 2 3 n for every positive integer n i =1 1. Let s n = P n 3 i : will show that s n = 1 − 1 2 3 n i =1 2. Recurrence relation: s n = s n − 1 + 2 3 n with s 1 = 2 3 3. s 1 = 1 − 1 3 1 = 2 3 X 1 4. Fix m ≥ 2 & assume that statements up to s m − 1 = 1 − 3 m − 1 have been checked 5. Consider s m : s m = s m − 1 + 2 from recurrence relation 3 m 1 + 2 1 ⇣ ⌘ = 1 − since s m − 1 = 1 − 3 m − 1 3 m 3 m − 1 = 1 − 3 3 m + 2 3 m = 1 − 1 3 m X Moral: Finding nice formulas for sums is a special case of finding closed formulas for recurrence relations Lecture 9 6/ 15

Another Example 4in ) Exercise: Show that P n she i =1 (4 i − 1) = n (2 n + 1) ✓ ) Chitti Si 4.1-1=3 1. I ' Cti Me = . . ; is correct for .gs closed formula . ,sm that the , . , . . Assume 2. sni-sm.tt/m-D=cm-Dkcm-iH-ilt4m-l by induction hyp . 3 . 4th I ) " I It 2cm - the - = - ' )t5m -2 Llm -2mW = 4mt2t5m 2 2Mt - - = m( 1) ✓ 2Mt " tm Lm = = Lecture 9 7/ 15

Proving Properties of a Recursive Sequence Key idea: You can prove properties of a sequence even if you can’t find a closed form Lecture 9 8/ 15

Proving Properties of a Recursive Sequence Key idea: You can prove properties of a sequence even if you can’t find a closed form Example: For a n = 2 a n − 1 + a n − 2 with a 1 = 5 , a 2 = 10, show that a n < 3 n for n ≥ 3 1. Calculate: a 1 = 5, a 2 = 10, a 3 = 25, a 4 = 60 2. We see that a 3 < 3 3 and a 4 < 3 4 3. Fix m ≥ 5 such that inequalities up to a m − 1 < 3 m − 1 have been checked 4. Consider a m and show that a m < 3 m : a m = 2 a m − 1 + a m − 2 by recurrence relation < 2 · 3 m − 1 + a m − 2 since a m − 1 < 3 m − 1 is true < 2 · 3 m − 1 + 3 m − 2 since a m − 2 < 3 m − 2 is true < 2 · 3 m − 1 + 3 m − 1 since 3 m − 2 < 3 m − 1 = 3 · 3 m − 1 = 3 m X Lecture 9 8/ 15

Proving Properties of a Recursive Sequence Key idea: You can prove properties of a sequence even if you can’t find a closed form Example: For a n = 2 a n − 1 + a n − 2 with a 1 = 5 , a 2 = 10, show that a n < 3 n for n ≥ 3 1. Calculate: a 1 = 5, a 2 = 10, a 3 = 25, a 4 = 60 2. We see that a 3 < 3 3 and a 4 < 3 4 3. Fix m ≥ 5 such that inequalities up to a m − 1 < 3 m − 1 have been checked 4. Consider a m and show that a m < 3 m : I a m = 2 a m − 1 + a m − 2 by recurrence relation < 2 · 3 m − 1 + a m − 2 since a m − 1 < 3 m − 1 is true < 2 · 3 m − 1 + 3 m − 2 since a m − 2 < 3 m − 2 is true < 2 · 3 m − 1 + 3 m − 1 since 3 m − 2 < 3 m − 1 = 3 · 3 m − 1 = 3 m X =3 n Observations because I We checked initially for n = 3 instead of n = 1 (why?) I We checked both n = 3 and n = 4 before inductive step—what if we only not induction For to This checked n = 3? need ay on as is use ) is false . , C and proven I Moral: It is not always su ffi cient to check only one initial statement Lecture 9 8/ 15

Fibonacci Numbers Recall definition I F n = F n − 1 + F n − 2 with F 1 = 1 , F 2 = 1 I Sequence looks like 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , . . . Prove: P ( n ) : “ F 3 n is always even” (i.e., F k is even if k is divisible by 3) 1. By inspection, P (1) is true, since F 3 = 2 2. Fix m ≥ 2 and assume that P (1) , P (2) , . . . , P ( m − 1) have been checked; i.e., F 3 , F 6 , . . . , F 3( m − 1) are all even 3. Consider P ( m ): F 3 m = F 3 m − 1 + F 3 m − 2 by recurrence relation for F 3 m = ( F 3 m − 2 + F 3 m − 3 ) + F 3 m − 2 by recurrence relation for F 3 m − 1 = F 3( m − 1) + 2 F 3 m − 2 rearranging terms and simplifying 4. that F 3( m − 1) is even has been checked & 2 F 3 m − 2 is clearly even 5. F 3 m = F 3( m − 1) + 2 F 3 m − 2 is the sum of two even numbers, and hence is even X Lecture 9 9/ 15

Another Fibonacci Example i =1 ( F i ) 2 = ( F n )( F n +1 )” Prove: P ( n ): “ P n 1. ( F 1 ) 2 = ( F 1 )( F 2 ) is true, since 1 2 = 1 · 1, so P (1) is true 2. Fix a positive integer m and assume that P (1) , P (2) , . . . , P ( m − 1) have been checked to be true 3. Consider P ( m ): m ( F i ) 2 = ( F 1 ) 2 + ( F 2 ) 2 + · · · + ( F m ) 2 X i =1 = ( F 1 ) 2 + ( F 2 ) 2 + · · · + ( F m − 1 ) 2 + ( F m ) 2 m − 1 ! X ( F i ) 2 + ( F m ) 2 = i =1 = ( F m − 1 )( F m ) + ( F m ) 2 by P ( m − 1) = ( F m )( F m + F m − 1 ) = ( F m )( F m +1 ) X by recurrence relation for F m +1 Lecture 9 10/ 15

Exercise , E- 2,4=3 F- I I IT - n - - . - , , Prove: P ( n ): “ F n < 2 n for all positive integers n ” true Ph ) and PCD are ' I File f so 22 I - I - . , that true put Dare Plm 3 and assume - Pick > m . . . , r . Fm = Fm Fm . it 3 . , i ) and up assumed Plm - ' In Plm are I - t since - true < ' 2 zm " zm - - " +2M since a zm < I m - 2 2 - = ✓ In = Lecture 9 11/ 15