Questions and Answers #5 Sequences

Questions and Answers Sheet 5 Sequences Question #1 Determine the z-transform for each of the following sequences and indicate the corresponding region of convergence for each of them. 1 𝑛 𝑥[𝑛] = (− ) 𝑢[−𝑛 − 2] 3 Answer: We have 1 𝑛 𝑥[𝑛] = (− ) 𝑢[−𝑛 − 2] 3 Apply Z-transform on given sequence, we get ∞ 1 𝑛 𝑋(𝑧) = ∑ (− ) 𝑢[−𝑛 − 2]𝑧 −𝑛 3 𝑛=−∞ 𝑛 −2 1 = ∑ (− ) 𝑧 −𝑛 3 −∞ ∞ 1 −𝑛 = ∑ (− ) 𝑧 𝑛 3 𝑛=2 put 𝑛1 = 𝑛 − 2 ∞ 1 −𝑛−2 𝑛 +2 𝑋(𝑧) = ∑ (− ) 𝑧 1 3 = 𝑛1 =0 1 −𝑛1 2 ∑∞ 9𝑧 𝑛1=0(−3) 𝑧 𝑛1 = 9𝑧 1 −1 2 ∑∞ 𝑛1 =0((−3) 𝑧) = 9𝑧 2 ( 𝑛1 1 ) 1 −1 1 − (− ) 𝑧 3 3𝑧 = 1 1 + 𝑧 −1 3 1 1 Hence, the sequence converges if |𝑧| < so that 𝑅𝑂𝐶 = |𝑧| < . ROC is not including the unit |z|=1 3 circle, so that it doesn't have Fourier transform. Question #2 Find the limit of the sequence 𝑑𝑛 = 3𝑛 2𝑛+3 4 −𝑛 𝑛! Answer: We have 3𝑛 2𝑛+3 𝑑𝑛 = −𝑛 4 𝑛! Simplify (3 × 2 × 4)𝑛 𝑑𝑛 = 23 × 𝑛! 3 8 × (24)𝑛 𝑛! Some of terms of series 1 24 242 243 8× +8× +8× +8× +⋯ 1 11 21 31 2 3 𝑥 𝑥 𝑥 𝑒𝑥 = 1 + + + +⋯ 11 21 31 𝑥 = 24 Given series limit 8 × 𝑒 24 𝑑𝑛 = Question #3 Which of the sequences are bounded? 2𝑛 7𝑛 3(−1)𝑛 1 9𝑛 Answer: a) 2n it is a montonic increasing for every n it is not bounded 𝑛 = 1,2,3, .. 2𝑛 = 2,4,6, … Thus, option a isn't bounded b) 7𝑛 It is also not bounded c) 3(−1)𝑛 This sequence contain only 2 elements for n=odd it gives -3 for n=even it gives 3 It is bounded above and below both so, it is bounded sequence. 1 d) 𝑛 9 It is montonic decreasing sequence and it converges at 𝑛 → ∞ 1 lim ( 𝑛 ) = 0 𝑛→∞ 9 Thus, it's also a bounded sequence. Question #4 Find the Z transform of the sequence 𝑠𝑒𝑛 π𝑘 4 Answer: π𝑘 π𝑘 Let 𝑢𝑛 = 𝑠𝑒𝑛 = sin 4 4 [∵ 𝑠𝑒𝑛 = 𝑠𝑖𝑛]n some language Let 𝑢𝑛 = sin π𝑘 4 for my convenience. Therefore Z transform is given by π 𝑍 sin π𝑘 4 𝑍 [sin ]= π+1 4 𝑍 2 − 2𝑍 cos 4 1 𝑍× √2 = 2− √2(𝑍 √2𝑍 + 1) 𝑍 = √2(𝑍 2 − √2𝑍 + 1) ∴ 𝑍 [sin ∴ 𝑍 [𝑠𝑒𝑛 π𝑘 4 π𝑘 4 ]= ]= 𝑍 √2(𝑍 2 −√2𝑍+1) 𝑍 √2(𝑍 2 −√2𝑍+1) is the Z transfered of the sequence [sin Find the limit of the sequence 2 [𝑎𝑛 ] = [ ] 2 √𝑛 + 7𝑛 − √𝑛2 + 2𝑛 Answer: = 2 √𝑛2 + 7𝑛 − √𝑛2 + 2𝑛 2(√𝑛2 + 7𝑛 + √𝑛2 + 2𝑛) (√𝑛2 + 7𝑛 − √𝑛2 + 2𝑛)(√𝑛2 + 7𝑛 + √𝑛2 + 2𝑛) 2(√𝑛2 + 7𝑛 + √𝑛2 + 2𝑛) 𝑛2 + 7𝑛 − 𝑛2 − 2𝑛 2(√𝑛2 + 7𝑛 + √𝑛2 + 2𝑛) = 5𝑛 = 2 7 2 = (√1 + + √1 + ) 5 𝑛 𝑛 Now, lim 𝑎𝑛 = lim 𝑛→∞ 𝑛→∞ 2 7 2 (√1 + + √1 + ) 5 𝑛 𝑛 2 = (1 + 1) 5 4 = 5 Thus, limit of 𝑎𝑛 is 4 5 Question #6 Determine the convergence or divergence of the sequence. ∑∞ 𝑛=10 [ 1 1 4𝑛2 −16𝑛+15 𝑒 𝑛−1 + ( ) 2 π ] Answer: ∞ 1 1 𝑒 𝑛−1 ∑ [ 2 + ( ) ] 4𝑛 − 16𝑛 + 15 2 π 𝑛=10 ∞ ∞ = ∑ 𝑢𝑛 + ∑ 𝑣𝑛 𝑛=10 𝑛=10 4 ] is the required Z transform of the sequence [𝑠𝑒𝑛 Question #5 𝑎𝑛 = π𝑘 π𝑘 4 ] Here, 1 1 𝑒 𝑛−1 , 𝑣 = ( ) 𝑛 4𝑛2 − 16𝑛 + 15 2 π Now, take Raabe's test 𝑢𝑛 lim 𝑛 ( − 1) 𝑛→∞ 𝑢𝑛 + 1 4(𝑛 + 1)2 − 16(𝑛 + 1) + 15 = lim 𝑛 ( − 1) 𝑛→∞ 4𝑛2 − 16𝑛 + 15 4𝑛2 + 4 + 8𝑛 − 16𝑛 − 16 + 15 = lim 𝑛 ( − 1) 𝑛→∞ 4𝑛2 − 16𝑛 + 15 (4𝑛2 − 8𝑛 + 3)(4𝑛2 − 16𝑛 + 15) = lim 𝑛 ( ) 𝑛→∞ 4𝑛2 − 16𝑛 + 15 8𝑛 − 12 = lim 𝑛 ( 2 ) 𝑛→∞ 4𝑛 − 16𝑛 + 15 12 8− 𝑛 ) = lim ( 16 15 𝑛→∞ 4− + 2 𝑛 𝑛 =2>1 So, we have ∑∞ 𝑛=10 𝑢𝑛 is convergent 𝑢𝑛 = ∞ ∑ 𝑛=10 1 𝑒 𝑛−1 ( ) 2 π By Ratio test lim | n → vn + 1 | vn 1 e n +1−1 ( ) 2 = lim |  | n → 1 e n −1 ( ) 2  e = lim | | 1 n →  By ration test this sequence is convergent. If both series are convergent, their sum is also convergent ∞ ∞ ∑ 𝑢𝑛 + ∑ 𝑣𝑛 𝑛=10 ∞ 𝑛=10 = ∑ (𝑢𝑛 + 𝑣𝑛 ) 𝑛=10 = ∑∞ 𝑛=10 [ 1 4𝑛2 −16𝑛+15 1 + ( 𝑒 𝑛−1 2 π )] is convergent series. Question #7 Find the limits of the sequences: 1) 𝑏𝑛 = ( 3+𝑛 𝑛 𝑛 ) 5 2) 𝑎𝑛 = 𝑛𝑛 Answer: 1) 𝑏𝑛 = ( 3+𝑛 𝑛 𝑛 ) lim 𝑏𝑛 = lim ( 𝑛→∞ 𝑛→∞ 3+𝑛 𝑛 3 𝑛 ) = lim (1 + ) 𝑛→∞ 𝑛 𝑛 We know that 𝑎 𝑛 lim (1 + ) = 𝑒 𝑎 𝑛→∞ 𝑛 = 3𝑒 3 = 20.08 3+𝑛 𝑛 lim ( ) = 𝑒 3 = 20.08 𝑛→∞ 𝑛 5 2) 𝑎𝑛 = 𝑛𝑛 1 5 5 lim 𝑎𝑛 = lim 𝑛𝑛 = lim (𝑛𝑛 ) 𝑛→∞ 5 𝑛→∞ 𝑛→∞ =1 =1 Question #8 Find the Z transform of the sequence 3𝑘 + (−3)𝑘 + 3𝑘 Answer: 3𝑘 + (−3)𝑘 + 3𝑘 𝑍 𝑍[𝑎𝑛 ] = 𝑍−𝑎 𝑍 𝑍[𝑛] = (𝑍 − 1)2 𝑘 𝑍[3 + (−3)𝑘 + 3𝑘] = 𝑍[3𝑘 ] + 𝑍[(−3)𝑘 ] + 3𝑍[𝑘] 𝑍 𝑍 𝑍 = + +3 (𝑍 − 1)2 𝑍−3 𝑍+3 𝑍 𝑍 3𝑍 = + + 𝑍 − 3 𝑍 + 3 (𝑍 − 1)2 Question #9 Show that the following sequences converge 𝑎𝑛 = 5 + 1 𝑛 Answer: 1 𝑛 Calculate the limit 𝑎𝑛 = 5 + lim 𝑎𝑛 = lim 5 + 𝑛→∞ 𝑛→∞ 1 𝑛 ⇒ lim 𝑎𝑛 = 5 + lim 𝑛→∞ 1 𝑛→∞ 𝑛 1 =5+0 =5 ∞ Hence, the sequence is decreasing and it converges to 5. lim 𝑎𝑛 = 5 + 𝑛→∞ Question #10 Suppose f : (𝑀, 𝑑) → (𝑁, 𝑝) is continuous. Prove or disprove that f is uniformly continuous ⇔ f maps a Cauchy seq. to Cauchy seq. Answer: We have 𝑓 (𝑀,𝑑) → (𝑁, ρ) Let ξ > 0 and m,n be any two points. Since f is uniformly continuous, there exists δ > 0 , s.t. for every 𝑚, 𝑛 ∈ 𝑀 , distance between (𝑚, 𝑛) < δ dis 𝑡𝑎𝑛 ce between [𝑓(𝑚), 𝑓(𝑛)] < ξ. Since x_m is a cauchy sequence in M for every 𝑚 ∈ 𝑀 s.t. 𝑚, 𝑛 ≥ 𝑛0 ⇒ ρ[𝑓(𝑚), 𝑓(𝑛)] < δ ⇒ ρ[𝑓(𝑥𝑚 ), 𝑓(𝑥𝑛 )] < ξ Hence, image of a cauchy seq. under a uniform continuous function is again a cauchy sequence. (This property is not true foe arbitrary continuous function) Question #11 Find the first four terms of each of the recursively defined sequences 𝑑𝑘 = 𝑘(𝑑𝑘−1 )2 for all integers 𝑘 ≥ 1 𝑑0 = 3 Answer: The recursive definition of the given sequence is 𝑑𝑘 = 𝑘(𝑑𝑘−1 )2 for all integers 𝑘 ≥ 1 The first term of the sequence is 𝑑0 = 3 Evaluate the second term 𝑑1 using 𝑑𝑘 = 𝑘(𝑑𝑘−1 )2 and 𝑑0 = 3 𝑑1 = (1)(𝑑1−1 )2 = (𝑑0 )2 = 32 =9 Now, evaluate the third term 𝑑2 using 𝑑𝑘 = 𝑘(𝑑𝑘−1 )2 and 𝑑1 = 9 𝑑2 = (2)(𝑑2−1 )2 = 2(𝑑1 )2 = 2(9)2 = 162 Now, evaluate the fourth term 𝑑3 using 𝑑𝑘 = 𝑘(𝑑𝑘−1 )2 and 𝑑2 = 162 𝑑3 = (3)(𝑑3−1 )2 = 3(𝑑2 )2 = 3(162)2 = 78732 Thus, the first four terms are: 3,9,162,78732 Question #12 Consider the sequences a. 𝑎𝑛 = −2,5,12,19, … b. 𝑏𝑛 = 3,6,12,24,48, … (a) Find the next two terms of the sequence. (b) Find a recurrence relation that generates the sequence. (c) Find an explicit formula for the nth term of the sequence. Answer: 1)Given sequence is an arythmetic sequence First term 𝑎1 = −2 Common difference d=5-(-2)=5+2=7 Next two terms are 𝑎5 = 𝑎1 + (5 − 1)𝑑 = −2 + (4)7 = −2 + 28 𝑎5 = 26 𝑎6 = 𝑎1 + (6 − 1)𝑑 = −2 + (5)7 = −2 + 35 𝑎5 = 33 Recurrence formula: 𝑎1 = −2 𝑎𝑛 = 𝑎𝑛−1 + 7 Explicit formula 𝑎5 = 𝑎1 + (𝑛 − 1)𝑑 2)Given sequence is a multiple of 2 First term 𝑎1 = 3 Next two terms: 𝑎6 = 2𝑎5 = 2(48) = 96 𝑎7 = 2𝑎6 = 2(96) = 192 Recurrence formula 𝑎1 = 3 𝑎𝑛 = 2𝑎𝑛−1 Explicit formula 𝑎𝑛 = 3(2)𝑛−1 Question #13 Determine whether the given sequence is convergent or divergent 𝑎𝑛 = ln(𝑛 + 1) √𝑛 Answer: 𝑎𝑛 = ln(𝑛+1) √𝑛 Now, lim 𝑎𝑛 = lim 𝑛→∞ 𝑛→∞ ln(𝑛 + 1) ∞ , [ ] 𝑓𝑜𝑟𝑚 ∞ √𝑛 1 (𝑛 + 1) = lim = 1 𝑛→∞ 2√ 𝑛 2√ 𝑛 = lim 𝑛→∞ 𝑛 + 1 2 √𝑛 = lim 1 𝑛→∞ 𝑛 (1 + ) 𝑛 2 √𝑛 = lim 𝑛→∞ √𝑛√𝑛 (1 + 1) 𝑛 2 = =0 1 ∞ (1 + ) 𝑛 ln(𝑛+1) Thus, lim =0 𝑛→∞ √𝑛 Hence, 𝑎𝑛 converhes to 0 Question #14 Give the first six terms of the following sequences. 1) 𝑔1 = 2 and 𝑔2 = 1 2) 𝑐1 = 4, 𝑐2 = 5 and 𝑐𝑛 = 𝑐𝑛−1 × 𝑐𝑛−2 for 𝑛 ≥ 3 3) 𝑏1 = 1, 𝑏2 = 3 and 𝑏𝑛 = 𝑏𝑛−1 − 𝑏𝑛−2 for 𝑛 ≥ 3 4) 𝑑1 = 1, 𝑑2 = 1 and 𝑑𝑛 = (𝑑𝑛−1 )2 + 𝑑𝑛−2 for 𝑛 ≥ 3 Answer: Given 1) 𝑔1 = 2 and 𝑔2 = 1, 𝑔𝑛 = 𝑛𝑔𝑛−1 + 9𝑛−2 𝑔3 = 3𝑔2 + 𝑔1 = 3 × 1 + 2 = 5 𝑔4 = 4𝑔3 + 𝑔2 = 4 × 5 + 1 = 21 𝑔5 = 5𝑔4 + 𝑔3 = 5 × 21 + 5 = 110 𝑔6 = 6𝑔5 + 𝑔4 = 6 × 110 + 21 = 660 + 21 = 681 so first six terms are: 2,1,5,21,110,681 2) 𝑐1 = 4, 𝑐2 = 5 and 𝑐𝑛 = 𝑐𝑛−1 × 𝑐𝑛−2 for 𝑛 ≥ 3 𝑐3 = 𝑐2 × 𝑐1 = 5 × 4 = 20 𝑐4 = 𝑐3 × 𝑐2 = 20 × 5 = 100 𝑐5 = 𝑐4 × 𝑐3 = 100 × 20 = 2000 𝑐6 = 𝑐5 × 𝑐4 = 2000 × 100 = 200000 Thus, the first six terms are: 4,5,20,100,2000,200000 Question #15 Show that the following sequence converges 1 1 𝑎𝑛 = (1 − 𝑛 ) 3 3 Answer: 1 1 3 3𝑛 𝑎𝑛 = (1 − ) Itis known that a monotonic decreasing sequence bounded is convergent and it converges to the greatest lower bound and a monotonic increasing sequence bounded above converges to the least upper bound. For the given sequence 1 1 1 1 𝑎𝑛+1 − 𝑎𝑛 = (1 − 𝑛+1 ) − (1 − 𝑛 ) 3 3 3 3 1 1 = 𝑛 − 𝑛+1 3 3 1 1 = 𝑛 (1 − ) 3 3 2 1 = × 𝑛>0 3 3 Since, 𝑎𝑛+1 > 𝑎𝑛 , therefore the sequence is monotonic increasing. Now, For all 𝑛 ≥ 1 1 (1 − 𝑛 ) < 1 3 1 1 1 (1 − 𝑛 ) < 3 3 3 1 𝑎𝑛 < 3 So, the sequence is bounded above with the upper bound 1 3 Hence, being monotonic increasing sequence bounded above, 1 3 being upper bound, the given sequence 𝑎𝑛 is convergent Question #16 You have the sequence 𝑠𝑛 = 3𝑛−4𝑛3 5𝑛2 +2 1 and 𝑡𝑛 = cos ( ) . What conclusion can you make about each of the 𝑛 sequences? Answer: 3 3𝑛 − 4𝑛3 𝑛 − 4𝑛 𝑠𝑛 = = 2 5𝑛2 + 2 5+ 2 𝑛 3 − 4𝑛 𝑛 lim 𝑠𝑛 = lim 2 𝑛→∞ 𝑛→∞ 5+ 2 𝑛 0−∞ = = −∞ 5+0 lim 𝑠𝑛 does not exists definitely so. That 𝑠𝑛 is a divergent sequence. 𝑛→∞ 1 𝑡𝑛 = cos ( ) 𝑛 1 1 lim 𝑡𝑛 = lim cos ( ) = cos ( ) = cos(0) = 1 𝑛→∞ 𝑛→∞ 𝑛 ∞ So that 𝑡𝑛 converges to 1 Question #17 The infinite sequence will be defined as recurrent relation: 1,10,3,20,9,30,27,40, ... The sequence is a composition of some arithmetic and/or geometric sequences. How many elements will become non-recurrent part? Answer: We have the sequence 1,10,3,20,9,30,27,40, ... 𝑎1 = 1 𝑎2 = 10 𝑎3 = 3 = 3𝑎1 = 3𝑎3−2 𝑎4 = 20 = 𝑎2 + 10 = 𝑎4−2 + 10 𝑎5 = 9 = 3𝑎3 = 3𝑎5−2 𝑎6 = 30 = 𝑎4 + 10 = 𝑎6−2 + 10 𝑎7 = 27 = 3𝑎5 = 3𝑎7−2 𝑎8 = 40 = 𝑎6 + 10 = 𝑎8 − 2 + 10 thus, the relation of the given sequence is 𝑎1 = 1 𝑎2 = 10 and 3a , if n is odd (an =  n − 2 ) for all 𝑛 ≥ 3 an − 2 + 10, if n is even Hence, we can say that 2 elements will become non-recurrence part. Answer is 2 Question #18 For each of the sequences, write a formula for the general term. Use the formula to find 𝑎2 0 1) 2,7,12,17,... 2) 6,1,-1,-9,... 3) 𝑎0 = 4, 𝑑 = −0.3 3 1 4 4 4) 𝑎0 = , 𝑑 = − Answer: a) 2,7,12,17,... Here 𝑎0 = 2, 𝑎1 = 7, 𝑎2 = 12, 𝑎3 = 17, … Notice that 𝑎1 = 𝑎0 + 5, 𝑎2 = 𝑎1 + 5, 𝑎3 = 𝑎2 + 5 and so on Thus, here d=5 𝑎𝑛 = 𝑎0 + 𝑑(𝑛 − 1) So, 𝑎20 = 2 + 5(20 − 1) = 97 b) 6,1,-1,-9,... We have the formula 𝑛−1 𝑎𝑛 = 𝑎𝑛−1 − (5 + 3 (∑(−1)𝑛 𝑛)) 𝑛=0 By this formula the value of 𝑎2 0 can be found easily c) 𝑎0 = 4, 𝑑 = −0.3 𝑎2 0 = 4 + (0.3)(20 − 1) = −1.7 3 d) 𝑎0 = , 𝑑 = − 4 3 1 4 4 1 4 𝑎2 0 = + (− ) (20 − 1) = −4 Question #19 Prove that lim 1 𝑛→∞ 𝑛2 =0 Answer: Use the definition of convergent 1 lim 2 = 0 𝑛→∞ 𝑛 Let ϵ be any positive number We have to show that | 1 − 0 | n2 1 Now, take 𝑁 ∈ 𝑁, < 𝑁, then ϵ 1 1 ∀𝑛 > 𝑁 (𝑖𝑒. < ) 𝑛 𝑁 and | ( 1 𝑛2 1 1 1 1 − 0 |=| 2 |= 2   2 n n n N is always positive) Hence, | lim 1 𝑛→∞ 𝑛2 1 − 0 | n2 =0 Question #20 We have the sequence 1,2,3,5,8 What is the next number in the sequence? Answer: From the information, that we have, we can say, that it's the Fibonacci sequence The number that needs to be found in the sequence is obtained by adding the two numbers before to it. Th objective of the problem is obtained by using the concept of Fibonacci sequence which is obtained by adding the two numbers before the number to it. Obtain the number in the sequence Given following sequence 1,2,3,5,8 The required value is 1+2=3 2+3=5 3+5=8 5 + 8 = 13 Thus, 13 is the answer.