Questions and Answers #6 Sequences

Questions and Answers Sheet 6 Sequences Question #1 Give an example of such f(x) and prove, why it works 1 A non constant function 𝑓: [0,1] → 𝑅 such that 𝑓(𝑥) ≥ 0 for any 𝑥 ∈ [0,1], but ∫0 𝑓(𝑥)𝑑𝑥 = 0 Answer: 1 ∫ 𝑓(𝑥)𝑑𝑥 = 0 0 𝑏 𝑏 𝑏 And we know that ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 + ∫𝑎 𝑓(𝑥)𝑑𝑥 , 𝑎 ≤ 𝑐 ≤ 𝑏 0,0  x  1 ( f ( x) =  ) 1, x = 1 Now, let Thus, f(x) is non-constant and 𝑓(𝑥) ≥ 0∀𝑥 ∈ [0,1] 1 1−ξ Now, ∫0 𝑓(𝑥)𝑑𝑥 = ∫0 1 1−ξ i.e. ∫0 𝑓(𝑥)𝑑𝑥 = lim (∫0 ξ→0 1−ξ = lim (∫ ξ→0 1 𝑓(𝑥)𝑑𝑥 + ∫1−ξ 𝑓(𝑥)𝑑𝑥 , where ξ → 0 1 𝑓(𝑥)𝑑𝑥 + ∫1−ξ 𝑓(𝑥)𝑑𝑥 ) 1 0𝑑𝑥 + ∫ 1𝑑𝑥 ) 0 1−ξ 1 = lim(0 + 𝑥1−ξ ) = lim ξ = 0 ξ→0 So, ξ→0 1 ∫0 𝑓(𝑥)𝑑𝑥 =0 Question #2 Find the limits of the following sequences 𝑎𝑛 = 𝑛5/𝑛 Answer: lim 𝑎𝑛 𝑛→∞ = lim 𝑛5/𝑛 𝑛→∞ 5 = lim 𝑒 log 𝑛𝑛 𝑛→∞ 5 = lim 𝑒 𝑛 log 𝑛 𝑛→∞ Apply L'Hospital Rule 1 5𝑛 = lim 𝑒 1 𝑛→∞ 5 𝑒∞ = = 𝑒0 = 1 ∴ lim 𝑎𝑛 = 1 𝑛→∞ Question #3 Use the approach in Gauss's Problem to find the following sums of arithmetic sequences. 1+2+3+4+...+1001 Answer: Consider the series 1+2+3+4+...+1001 . For the series 1+2+3+4+...+1001, it is clear that there are 1001 terms, which is odd. So, add the series to itself in reverse order. Note that each pair has a sum of 1002 and there are 1001 pairs which gives: 1002(1001)=1003002 Since this is the sum of the series added to itself, the sum of the original series is half this sum. 1 The sum of 1+2+3+4+...+1001 is then (1003002) = 501501 2 This is the answer. Question #4 Find the missing numbers in the following sequences: 7, 9, 12, 16, 21, 34, 42, 59 Answer: A sequence is the arrangement of a set of numbers in a particular order. We have the sequennce 7,9,12,16,21,(a),34,42,(b),59 As this sequence follows a certain pattern 7+2=9 9+3=12 12+4=16 16+5=21 And so on (after (b) the numbers are decreasing) a=21+6 or 34-7 a=27 b=42+9 b=51 Thus, we have the answer 27 and 51 Question #5 Prove that lim 𝑒 𝑛+1 𝑛 𝑛→∞ ( 𝑛+1 𝑛 =𝑒 ) is the exponential of e Answer: lim 𝑒 𝑛+1 𝑛 𝑛→∞ =𝑒 lim 𝑎𝑛 = 𝑎 convergent if ∀> 0 𝑛→∞ there exist 𝑎𝑛 𝑁0 ∈ 𝑁 such that |𝑎𝑛 − 𝑎| < ϵ, when 𝑛 > 𝑁0 Choose ϵ > 0 such that 𝑛+1 ) 𝑛 |𝑒 ( ⇒ |𝑒 − 𝑒| < ϵ 𝑛+1 −1 𝑛 − 1| < ϵ/𝑒 |𝑒 1/𝑛 − 1| < ϵ/𝑒 𝑒 1/𝑛 − 1 < |𝑒 1/𝑛 − 1| < ϵ/𝑒 𝑒 1/𝑛 < ϵ/𝑒 + 1 1 ϵ < ln ( + 1) 𝑛 𝑒 1 𝑛> ϵ ln ( + 1) 𝑒 𝑁0 > 1 ϵ 𝑒 ln( +1) So, ∀ϵ > 0 we can find and 𝑁0 ∈ 𝑁 such that |𝑒 Hence, lim 𝑒 𝑛+1 𝑛 𝑛→∞ 𝑛+1 𝑛 − 𝑒| < ϵ when 𝑛 > 𝑁0 =𝑒 Question #6 Prove, using the definitio of a convergent sequence lim ln ( 𝑛→∞ 𝑛 )=0 𝑛+1 Answer: lim ln ( 𝑛→∞ 𝑛 )=0 𝑛+1 Let | ln( |ln( e n ) | n +1 n )| n +1 e 𝑛 ⇒ < 𝑒ϵ 𝑛+1 1 ⇒1− < 𝑒ϵ 𝑛+1 1 ⇒ 1 − 𝑒ϵ < 𝑛+1 1 ⇒ > 𝑛+1 1 − 𝑒ϵ 1 ⇒ −1 >𝑛 1 − 𝑒ϵ ϵ 𝑒 ⇒ ϵ > 𝑛 1−𝑒 Given, ϵ > 0 , choose 𝑛0 = | 𝑒ϵ 1−𝑒 ϵ | Therefore, given any ϵ > 0 , we can find a positive integer 𝑛0 | ln( n ) | n +1 Question #7 Calculate the Z transform of the sequences 1 𝑘 1) (− ) 5 2) cos 𝑘 π 3) 3k Answer: We have the sequence 1 𝑘 1) (− ) 5 1 𝑘 𝑥[𝑘] = (− ) 5 Formula for z-transform is ∞ 𝑍(𝑥[𝑘]) = ∑(𝑥[𝑘])𝑧 −𝑘 𝑘=0 Thus, ∞ 1 𝑘 𝑍(𝑥[𝑘]) = ∑ (− ) 𝑧 −𝑘 5 𝑘=0 1 𝑘 1 5𝑧 5𝑧 𝑍(𝑥[𝑘]) = ∑∞ 𝑘=0 (− ) = 1 − 1 + (5𝑧)2 − ⋯ Above series is geometric series 1 Thus, 𝑍(𝑥[𝑘]) = 1 1−(− ) 5𝑧 5𝑧 ⇒ 𝑍(𝑥[𝑘]) = 5𝑧 + 1 Question #8 Given the first 5 terms, write down the general term (a) 80, 74, 68, 62, 56... 1 1 (b) 3, 1, , , 1 3 9 27 (c) 4 5 ,− 8 , 16 11 17 ,… ,− 32 64 , 23 29 ,… Answer: We are given the first terms and we have to find the general form 1) 80, 74, 68, 62, 56... 𝑎2 − 𝑎 = 𝑎3 − 𝑎2 = 𝑎4 − 𝑎3 = 𝑎5 − 𝑎4 74 − 80 = 687 − 74 = 62 − 68 = 56 − 62 −6 = −6 = −6 = −6 Hence, we have 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 80 + (𝑛 + 1)(−6) 80 − 6𝑛 + 6 = 86 − 6𝑛 - general term, where n is positive integers Question #9 What is the next three terms of the following sequences? a. 2,4, 8, 16, 32,... b. 3,5, 7,9, 11, 13,... c. 1,4,9,16,... d. 1,4, 11, 29, 76, 199,... Answer: a. 2,4, 8, 16, 32,... Here 2=2 22 = 4 23 = 8 24 = 16 25 = 32 Thus, the next three terms are 26 = 64, 27 = 128, 28 = 256 b. 3,5, 7,9, 11, 13,... Here, 3+2=5 5+2=7 7+2=9 9 + 2 = 11 Therefore, the next three terms are 11 + 2 = 13 , 13 + 2 = 15 , 15 + 2 = 17 Question #10 Find 𝑎5 and 𝑎𝑛 for the geometric sequences. 𝑎4 = 1080, 𝑟 = −6 Answer: Given: 𝑎4 = 1080, 𝑟 = −6 Formula: a_n=ar^{n-1} 𝑎4 = 𝑎𝑟 3 ⇒ 1080 = 𝑎(−6)3 ⇒ 𝑎 = − 1080 216 𝑎=5 To find 𝑎5 and 𝑎𝑛 𝑎5 = 𝑎𝑟 4 = −5(−6)4 = −5(1296) 𝑎5 = −6480 𝑎𝑛 = 𝑎𝑟 𝑛−1 = −5(−6)𝑛−1 𝑎𝑛 = −5(−6)𝑛−1 Question #11 Find 𝑎5 and 𝑎𝑛 for the geometric sequence. 𝑎3 = 2, 𝑟 = 1/3 Answer: Since the n-th term in GP is given by 𝑎𝑛 = 𝑎1 𝑟 𝑛−1 13−1 𝑎3 = 𝑎1 3 1 2 = 𝑎1 9 𝑎1 = 18 1 5−1 𝑎5 = 18 ( ) 3 18 = 81 2 = 9 Question #12 Find the limits of the following sequences ∞ π (cos ((2𝑛 + 1) )) 2 𝑛=1 π𝑛 ∞ ( 2𝑛+1 ) 𝑒 𝑛=1 ∞ ( 𝑛2 + 3 ) 𝑛3 + 𝑛2 − 1 𝑛=1 (𝑛 sin π ∞ ) 𝑛 𝑛=1 Answer: ∞ π 1) (cos ((2𝑛 + 1) )) 2 𝑛=1 π lim (cos(2𝑛 + 1) ) 𝑛→∞ 2 π = cos(∞) 2 = cos ∞ = 𝑂𝑛𝑒(∵ cos ∞ = 𝑛𝑜𝑡 𝑢𝑛𝑖𝑞𝑢𝑒) π = lim [cos(2𝑛 + 1) ] = 𝑜𝑛𝑒 𝑛→∞ π𝑛 2) ( ∞ 2 ) 𝑒 2𝑛+1 𝑛=1 𝑛 lim π 𝑛→∞ 𝑒 2𝑛+1 𝑛 π lim 𝑛→∞ (𝑒 2 )𝑛 ×𝑒 π𝑛 𝑛 1 lim ( ) × 𝑛→∞ 𝑒 2 𝑒 π ∞ 1 = ( 2) × = 0 𝑒 𝑒 π𝑛 = lim 2𝑛+1 = 0 𝑛→∞ 𝑒 Question #13 Calculate the Z transform of the sequences 1 𝑘 1) (− ) 5 2) (cos 𝑘 π) 3) 3k Answer: 1 𝑘 1) (− ) 5 1 𝑘 𝑓(𝑘) = (− ) 5 Therefore, ∞ 1 𝑘 1 𝑘 𝑍𝑓(𝑘) = 𝑍 (− ) = ∑ (− ) 𝑧 −𝑘 5 5 𝑘=0 1 1 2 + (− ) − ⋯ 5𝑧 5𝑧 1 𝑘 1 ⇒ 𝑍 (− ) = 1 5 1 − (− ) 5𝑧 5𝑧 1 = , |𝑧| > 5𝑧 + 1 5 5𝑧 1 Thus, the answer is and the region of convergence is |𝑧| ≻ =1− 5𝑧+1 5 2) (cos 𝑘 π) ∞ 𝑍𝑓(𝑘) = 𝑍 cos 𝑘 π = ∑ cos(𝑘π) 𝑧 −𝑘 𝑘=0 ∞ 𝑒 𝑖π𝑘 + 𝑒 −𝑖π𝑘 −𝑘 ⇒ cos(𝑘π) = ∑ [ ]𝑧 2 𝑘=0 ∞ = ∞ 1 1 ∑ 𝑒 𝑖π𝑘 𝑧 −𝑘 + ∑ 𝑒 −𝑖π𝑘 𝑧 −𝑘 2 2 𝑘=0 ∞ 𝑘=0 ∞ 𝑘 1 𝑒 𝑖π 1 𝑒 −𝑖π = ∑( ) + ∑( ) 2 𝑧 2 𝑧 1 = 2 𝑘=0 𝑘 𝑘=0 1 1 1 + 𝑖π 𝑒 𝑒 −𝑖π 2 1− 1− 𝑧 𝑧 1 1 1 1 = + (−1) 2 (−1) 2 1− 1− 𝑧 𝑧 1 𝑧 1 𝑧 = + 2𝑧 +1 2𝑧 + 1 𝑧 = , |𝑧| ≻ 1 𝑧+1 𝑧 Thus, the answer is , and the region of convergence is |z|>-1 𝑧+1 Question #14 Write the first 5 terms of the sequence with general terms 𝑎𝑛 and indicate whether they are convergent or divergent 1) 𝑎𝑛 = 3𝑛−5 𝑛+8 2) 𝑎𝑛 = (2𝑛 + 1) × 3𝑛 3) 𝑎𝑛 = 1 2𝑛 + 𝑛 = 𝑛 + 2−𝑛 Answer: 3𝑛−5 1) 𝑎𝑛 = 𝑛+8 3 − 5 −2 𝑎1 = = 1+8 9 6−5 1 𝑎2 = = 2 + 8 10 9−5 4 𝑎3 = = 3 + 8 11 12 − 5 7 𝑎4 = = 4+8 12 15 − 5 10 𝑎5 = = 5+8 13 Test convergence: lim 𝑎𝑛 𝑛→∞ 3𝑛 − 5 𝑛+8 5 3− 𝑛 = lim 8 𝑛→∞ 1+ 𝑛 =3≠0 Therefore, the series diverges 2) 𝑎𝑛 = (2𝑛 + 1) × 3𝑛 𝑎1 = 3 × 3 = 9 𝑎2 = 5 × 32 = 45 𝑎3 = 7 × 33 = 189 𝑎4 = 9 × 34 = 729 𝑎5 = 11 × 35 = 2673 𝑎𝑛+1 𝐿 = lim 𝑛→∞ 𝑎 + 𝑛 (2𝑛 + 3) × 3𝑛+1 = lim 𝑛→∞ (2𝑛 + 1) × 3𝑛 2𝑛 + 3 = 3 lim 𝑛→∞ 2𝑛 + 1 =3>0 Thus, this series diverges = lim 𝑛→∞ Question #15 Write recurrence relations that describe the following sequences. 1, 4, 13, 40, 121,... 1, -3, 5, -11, 21, -43... Answer: 1) 1, 4, 13, 40, 121, … 𝑎1 = 1, 𝑎2 = 4, 𝑎3 = 13 𝑎𝑛+1 = 3𝑎𝑛 + 1 2) 1, −3, 5, −11, 21, −43 … 𝑎1 = 1, 𝑎2 = −3, 𝑎3 = 5 𝑎𝑛+1 = (−1)𝑛+1 2𝑎𝑛 + (−1)𝑛 Question #16 Write a formula or rule that generates the terms of each of the following sequences that begin with the given list. Begin indexing at n = 1 . a) 𝑎𝑛 = −5,12, −19,26, −33, … b) 𝑏𝑛 = 3,9,27,81,243, … c) 𝑐𝑛 = −5 12 −19 26 −33 3 , 9 , 27 , , 81 243 ,… Answer: (a) 𝑎𝑛 = −5,12, −19,26, −33, … Let 𝑎1 = −5 𝑎𝑛 | a + 7, if n is even | =  n −1 −(| an −1 | +7), if n is odd (b) 𝑏𝑛 = 3,9,27,81,243, … 𝑏𝑛 = 3𝑛 Question #17 Which term in the sequences will be the first greater than 250 a) 9, 13, 17, 21,.... b) 2, 10, 18, 26 ,.... c) 1, 7, 13, 19,.... Answer: a) 9, 13, 17, 21, … . 𝑎 =9, 𝑑=4 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑎𝑛 = 𝑎 + (𝑛 − 1)4 = 𝑎 + 4𝑛 − 4 = 4𝑛 + 5 Now, put 𝑎𝑛 = 250 4𝑛 + 5 = 250 4𝑛 = 245 𝑛 = 61.25 Put n=62 in the equation 𝑎62 = 4 × 62 + 5 𝑎62 = 253, it is greater that 250 b) 2, 10, 18, 26 , … . 𝑎 =2, 𝑑=8 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑎𝑛 = 2 + (𝑛 − 1)8 = 2 + 8𝑛 − 8 = 8𝑛 − 6 Put 𝑎𝑛 = 250 8𝑛 − 6 = 250 8𝑛 = 256 𝑛 = 32 Now, put n=32 in the equation 𝑎32 = 8 × 32 − 6 = 250 it is equal to 250, thus, try to put n=33 𝑎33 = 8 × 33 − 6 = 258 it is greater than 250 Question #18 Which of these sequences satisfies a second order linear homogeneous recurrence relation? 3𝑛+1 + 𝑛3𝑛−1 2𝑛 + 𝑛3𝑛 1 2𝑛 ( ) 𝑛+1 𝑛 𝑛2 + 2𝑛 Answer: 1) The second term is of second degree 2) The second term is of second degree 3) The sequence is not linear 4) The last one is of degree Thus, none of these sequences are linear. Question #19 Which of them are bounded? a) 𝑛 14𝑛+1 b) 13n c) 10𝑛 d) (−11)𝑛 Answer: a) = 𝑛 14𝑛+1 1 1 𝑛 14+ lim 𝑥𝑛 = lim 𝑛→∞ 1 1 𝑛→∞ 14+𝑛 = 1 14 It is convergent, so it is bounded b) 13n 𝑥𝑛 → It is not bounded c) 10𝑛 𝑥𝑛 → ∞ It is not bounded d) (−11)𝑛 𝑥2𝑛 → ∞ 𝑎𝑠 𝑛 → ∞ 𝑥2𝑛+1 → −∞ 𝑎𝑠 𝑛 → ∞ Thus, it is not bounded. Question #20 Which of the sequences are bounded? 6𝑛 (−7)𝑛 5𝑛 1 8𝑛 Answer: Every convergent sequence is bounded or monotonic. Since 6n isn't convergent, it isn't bounded (−7)𝑛 is not monotonic, thus, it's not bounded 5𝑛 isn't convergent, and it's not bounded 1 1 𝑛 ( 𝑛) = ( ) 8 8 is also not bounded.