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Chapter 5- Sequences and Mathematical Induction (I)

Chapter 5- Sequences and Mathematical Induction (I)

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  • 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. 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. 4.1 Sequence
  • 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. 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. 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. 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. 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. 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. • 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. • 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. 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. 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. • 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. 4.1 Sequences
  • 28. 4.1 Sequences
  • 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. 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. 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. Usage on computer programming FLOWCHART 1 2 34.1 Sequences
  • 33. Pseudocode Statement-1 Statement-2 Statement-3 Example: input a b= 5 + 2 * a print b4.1 Sequences
  • 34. 4.2 Mathematical Induction
  • 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. • 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. 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. 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. 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. 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. • 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. 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. 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.

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