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# Presentation4

Chapter 5- Sequences and Mathematical Induction (I)

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### Presentation4

1. 1. Proundly presents Task 4 Alex Soh 032184 Fahmi 032799 Lailatulkadariah 033059 Nazri 032515 Nursyafiqah 032251 Shafiq 033035 Nurul Afiqah 032656 Hafizah 033006 Nurul Huda 032405 Thiba 032669
2. 2. Topics:4.1 Sequences - Definition of sequence - Example problem involving sequence. -Types of sequence -Give 1 example sequences use in computer programming4.2 Mathematical Induction 1 -List down the principle of Mathematical Induction. -Explain the method of Proof by mathematical Induction -Give 2 example problems that use for solving mathematical induction.
3. 3. 4.1 Sequence
4. 4. Definition of sequence• A sequence is a list of numbers or a set of integers.• In technical terms, a sequence is a function whose domain is the set of natural numbers and whose range is a subset of the real numbers.• We use the notation 𝑎 𝑛 to denote the image of the integer n.• We call 𝑎 𝑛 a term of the sequence.EXAMPLEConsider the function 𝑎 𝑛 = 2n + 1 (explicit formulae)The list of the terms of the sequence 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ,𝑎5 …… (list of domain)This function describes the sequence 3,5,7,9,11,...... (list of range)4.1 Sequences
5. 5. Example of problems involving sequenceThe first term of an arithmetic sequence is equal to 6 and the common difference isequal to 3. Find a formula for the n th term and the value of the 50 th termSolution• Use the value of the common difference d = 3 and the first term a1 = 6 in the formula for the n th term given above an = a1 + (n - 1 )d = 6 + 3 (n - 1) =3n+3The 50 th term is found by setting n = 50 in the above formula. a50 = 3 (50) + 3 = 1534.1 Sequences
6. 6. TYPES OF sequence① ARITHMETIC PROGRESSION Arithmetic progression is a sequence of the form a ,a+d ,a+2d,……,a+(n-1)d, a+nd where the initial a and the common difference d are real numbers. A arithmetic progression is a discrete analogue of the linear function f(x)=dx+a.4.1 Sequences
7. 7. Example• The sequences {dn} with dn= −1 + 4n and {tn} with tn= 7 − 3n are both arithmetic progressions with initial terms and common differences equal to −1 and 4, and 7 and −3, respectively, If we start at n = 0. The list of terms d0, d1, d2, d3, . . . begins with −1, 3, 7, 11, . . . , and the list of terms t0, t1, t2, t3,… begins with 7, 4, 1,−2, . . . .4.1 Sequences
8. 8. How to find the terms• A nth term of an arithmetic sequence can be defined using the following formula, an = a +(n-1)d4.1 Sequences
9. 9. ExampleYou are given that the first term of an arithmetic sequence is 1and the 41st term is 381.What is the 43rd term? The difference between ai and aj is d·(j −i).How can we use this to solve the given problem? Well sincewe know a1 = 1 and a41 = 381,we have a41=381= 1+40d. So, d=381−1 380 380 40 and a43 − a41 = 2d = 2( 40 )= 20 = 19Therefore, a 43 = 381 + 19 = 400.4.1 Sequences
10. 10. • Alternatively, we can use the equation a n =a+(n-1)d 380 a 43 = 1+ (43-1) ( ) 40 380 = 1 + (42)( ) 40 = 1+399 = 4004.1 Sequences
11. 11. Sum Of nth Terms 𝑛• If m and n are integers m≤n the symbol 𝑘=𝑚 𝑎 𝑘 , read the summation from k equals m to n of a-sub-k, is the sum of all the terms am, am+1, am+2, …, an. We say that am,+ am+1+ am+2+ …+ an is the expanded form of the sum and we write 𝑛 𝑘=𝑚 𝑎 𝑘= a 𝑚+ a 𝑚 + 1+ a 𝑚 + 2+…+a 𝑛 • k= index of the summation • m= lower limit of the summation • n= upper limit of the summation4.1 Sequences
12. 12. Sum Of nth Terms (cotd’)• If we wanted to find the sum of terms ai + ai+1 + ai+2 + ···+ aj, we need to find the average of the terms multiplied by the number of terms ai+ aj ai + ai+1 + ai+2 + ···+ aj = · (j − i + 1) 2Note that if you every get a fractional sum from an arithmetic sequence of integers, you probablydid something wrong!4.1 Sequences
13. 13. ExampleLet there be a sequence defined by Ai={2, 5, 8,11, 14, 17, 20, 23}, where 0<i<9. Find the sum ofthis sequence.Using formula, (2+23) 25 . 8−1+1 = ·8 2 2 8 𝑖=1 A i = 100.4.1 Sequences
14. 14. ② GEOMETRIC PROGRESSION Geometric progression is a sequence of form a,ar,ar²,… arⁿ-1 where initial term a and the common ratio r are real number, n≥04.1 Sequences
15. 15. EXAMPLEThe sequences {bn}with bn = (−1)n, {cn} with cn = 2 ・ 5n, and {dn} with dn =6 ・ (1/3)n are geometric progressions with initial term and common ratioequal to 1 and −1; 2 and 5; and 6and 1/3, respectively.If we start at n = 0. The list of terms b0, b1, b2, b3, b4, . . . begins with 1,−1, 1,−1, 1, . . . ; C0,C1,C2,C3,C4 2,10,50,250,1250 d0,d1,d2,d3,d4 6,2,2 ,2 ,2 ….. 3 9 244.1 Sequences
16. 16. Finding nth term• To find the nth of a GP, we must first need to find the ratio of the GP by using the formula 𝑎𝑟 𝑛 + 1 𝑎𝑟 𝑛 Then, we can use ar 𝑛 to find the nth term of a GP4.1 Sequences
17. 17. EXAMPLE• The first term of a geometric sequence of positive integers is 1 and the 11th term is 243. How can you find the 13th term? g11• We know g1= 1 and g11 = 243 so = 243 = r10. This g1 allows us to say gj j-i g13 12 = r , you get = r gi g1 = (r10)6/5 = 7294.1 Sequences
18. 18. Summation in GP• First formula: if a and r are real number and r≠1, then + 𝑛 𝑟 𝑛 1−1 S n= 𝑗=0 𝑎𝑟 𝑗 𝑎( 𝑟−1 ), r≠1 𝑎• Second formula: |r|<1 = 1−𝑟 + 𝑎−𝑟(𝑎𝑟𝑛) 𝑎𝑟 𝑛 1−𝑎 |r|≥1 = = 1−𝑟 𝑟−14.1 Sequences
19. 19. EXAMPLE• Given the term is {1, 3, 9, 27, 81}. Find the sum of the term given. 1 − 3(81) 𝑆5 = = 121 1−34.1 Sequences
20. 20. ③ HARMONIC SEQUENCE• A harmonic sequence is a sequence h1, h2, . . . , 1 1 1 hk such that , , …, is an arithmetic h1 h2 hk sequence.4.1 Sequences
21. 21. ④ FIBONACCI SEQUENCE• The Fibonacci sequence, f0, f1, f2, . . . , is defined by the initial conditions f0 = 0, f1 = 1, and the recurrence relation fn = fn−1 + fn−2 for n = 2, 3, 4, . . .4.1 Sequences
22. 22. RECURRENCE RELATIONS• A recurrence relation for the sequence { an } is an equation that expresses an in terms of one or more of the previous terms of the sequence, namely, a0 , a1 , . . . , an-1, an ,for all integers n with n ≥ n0 , where n0 is a nonnegative integer.• A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation.4.1 Sequences
23. 23. • Its say that the recurrence relation is solved together with the initial conditions when we find an explicit formula, called a close formula, for the terms of the sequence.• Example: Suppose that {an} is the sequence of integers defined by an= n!, the value of the factorial function at the integer n, where n = 1, 2, 3, . . .. Because n! = n((n − 1)(n − 2) . . . 2 ・ 1)• n(n − 1)! = nan-1 , we see that the sequence of factorials satisfies the recurrence relation• an = nan-1 , together with the initial condition a1 = 1.4.1 Sequences
24. 24. EXAMPLE• Find the Fibonacci numbers f2, f3, f4, f5, and f6.• Solution: The recurrence relation for the Fibonacci sequence tells us that we find successive terms by adding the previous two terms. Because the initial conditions tell us that f0 = 0 and f1 = 1, using the recurrence relation in the definition we find that • f2 = f1 + f0 = 1 + 0 = 1, • f3 = f2 + f1 = 1 + 1 = 2, • f4 = f3 + f2 = 2 + 1 = 3, • f5 = f4 + f3 = 3 + 2 = 5, • f6 = f5 + f4 = 5 + 3 = 8.4.1 Sequences
25. 25. ⑤ Special Integer Sequences• It is used to identify a sequence• Example: Find the formulae for the sequences with the following first 5 terms a) 1, 1/2, 1/4, 1/8, 1/16 b) 1,3,5,7,94.1 Sequences
26. 26. • Solution: a) we recognize that the denominator are powers of 2. the sequence with an = (1/2)n , n=0,1,2,... is a possible match. This proposed sequence is a geometric progression with a=1 and r= 1/2 . b) note that each term is obtained by adding 2 to the previous term. The sequence with an = 2n=1, n=0,1,2,.. is a possible match. This proposed sequence is an arithmetic progression with a=1 and d=2.4.1 Sequences
27. 27. 4.1 Sequences
28. 28. 4.1 Sequences
29. 29. Index shifting• Sometimes we shift the index of summation in a sum. This is often done when two sums need to be added but their indices of summation do not match.• Example: we have 5 𝑗2 but we want the index of summation to 𝑗=1 run between 0 and 4 rather than 1 to 5. Then we let k=j-1=0 and 𝑗2 become (k+1)2 . Hence, 5 𝑗=1 𝑗2= 4 (𝑘 𝑘=0 + 1)2 =1 + 4 + 9 + 16 + 25 = 554.1 Sequences
30. 30. Double summation 4 3 4 𝑖=1 𝑗=1 𝑖𝑗 = 𝑖=1(𝑖 + 2𝑖 + 3𝑖) 4 = 𝑖=1 6𝑖 =6+12+18+24 =604.1 Sequences
31. 31. usage on computer programming• A structured series of shots or scenes with a beginning, middle and end , the term sequence can be applied to video, audio or graphics.• Structured programming provides a number of constructs that are used to define the sequence in which the program statements are to be executed.
32. 32. Usage on computer programming FLOWCHART 1 2 34.1 Sequences
33. 33. Pseudocode Statement-1 Statement-2 Statement-3 Example: input a b= 5 + 2 * a print b4.1 Sequences
34. 34. 4.2 Mathematical Induction
35. 35. Principle of Mathematical InductionTo prove that P(n) is true for all positive intergers n, where P(n) is a propositionalfunction by 2 steps: BASIS: we verify that P(1) is true @ show that a initial value is true for all Z+ of the propositional function (inductive hypothesis ) INDUCTIVE: we show that the conditional statement ∀k (P(k) → P(k+1)) is true for all Z+ of k.4.2 Mathematical Induction 1
36. 36. • Similarly, we can say that mathematical induction is a method for proving a property defined that the property for integer n is true for all values of n that are greater than or equal to some initial interger. P(1)^∀k(P(k) → P(k+1))) → ∀nP(n)4.2 Mathematical Induction 1
37. 37. Method of proof• The proofs of the basis and inductive steps shown in the example illustrate 2 different ways to show an equation is true Transforming LHS and RHS independently until they seem to be equal. Transforming one side of equation until it is seen to be the same as the other side of the equation. 4.2 Mathematical Induction 1
38. 38. Problem Solving • Example 1 Using Mathematical Induction, prove that 𝑛(𝑛+1) 1+2+…+n = 2 , for all integers n≥14.2 Mathematical Induction 1
39. 39. Solution: we know that the property P(n) is the equation shown at the previouspage. Hence, we need to prove it using BASIS and INDUCTION steps. BASIS: Show that P(1) is true for the LHS and RHS of the equation. 𝑛(𝑛+1) LHS= 1+2+…+n RHS= , n≥1 =1 2 1(1+1) = 2 2 = 2 =1 ∴Since LHS = RHS, we have proven P(1) is true.
40. 40. INDUCTIVE: we need to show that the equation can take any value k and it’s successivevalue (k+1) by defining P(k) [the inductive hypothesis] and P(k+1) for k≥1. If P(k) is true then P(k+1) is true.Inductive Hypothesis P(k+1)= 1+2+…+k+(k+1) P(k+1)= 1+2+…+k+(k+1)P(k)= 1+2+…+k (𝑘+1)(𝑘+1+1) = P(k)+ (k+1) = 𝑘(𝑘+1) 2 𝑘2+𝑘 = 2 (𝑘+1)(𝑘+2) = 2 + (k+1) = 𝑘2+𝑘 2 2 𝑘2+𝑘 +2(𝑘+1) = 2 𝑘 +3𝑘+2 = = 2 -Eq.1 2 2(assume that it is true) 𝑘 +𝑘+2𝑘+2 = 2 𝑘2+3𝑘+2 = 2 -Eq.2 ∴Since Eq.1 and Eq.2 is identical for both LHS and RHS, therefore P(k+1) is true.
41. 41. • Example 2 Show that the sum of the first n odd integers is n2 Example: If n = 5, 1+3+5+7+9 = 25 = 52 Here, we know that d= n2-n1 =2 and a=1 Hence, an = 1+(n-1)2 𝑛 Sn = 𝑖=0(2𝑛 − 1) = 1+ (n)2 𝑛 P(n)= 𝑖=0(2𝑛 − 1) = 1+2n-2 𝑛 n2 = 𝑖=0(2𝑛 − 1) =2n-1
42. 42. Solution: Again, we have defined the general formula for the sum of the nth term ofthe odd positive integers as P(n) BASIS: Show that P(1) is true for the LHS and RHS of the equation. LHS RHS 𝑛 n 2= 1 2 P(n) = 𝑖=1(2𝑛 − 1) , n≥1 =1 P(1) = 1 (2(1) − 1) 𝑖=1 =1 ∴Since LHS = RHS, we have proven P(1) is true.
43. 43. INDUCTIVE: Again we need to show that if P(k) is true then P(k+1) is true. P(k+1)= 1+3+…+(2k) 𝑘+1 = 𝑖=0 (2𝑖 − 1)Inductive Hypothesis = 𝑘+1 2 (LHS)P(k)= 1+3+…+(2k-1) 𝑘 𝑘+1 = 𝑖=1(2𝑖 − 1) 𝑖=1 (2𝑖 − 1)= 𝑘 + 1 2 = 𝑘2 2(k+1)-1+ 𝑘 𝑖=1(2𝑖 − 1) = 𝑘 + 1 2(assume that it is true) 2k+1+k2= 𝑘 + 1 2 k2+2k+1= k2 +2k+1 (RHS) ∴Since LHS = RHS, therefore P(k+1) is true.