Recurrence Relations: Selected Exercises

Copyright © Peter Cappello Exercise 10 (a) A person deposits $1,000 in an account that yields 9% interest compounded annually. Set up a recurrence relation for the amount in the account at the end of n years. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 10 (a) Solution A person deposits $1,000 in an account that yields 9% interest compounded annually. Set up a recurrence relation for the amount in the account at the end of n years. Let an represent the amount after n years. an = an-1 + 0.09an-1 = 1.09an-1 a0 = 1000. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 10 (b) A person deposits $1,000 in an account that yields 9% interest compounded annually. Find an explicit formula for the amount in the account at the end of n years. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 10 (b) Solution After 1 year, a1 = 1.09a0 = 1.09x1000 = 1000x1.091 After 2 years, a2 = 1.09a1 = 1.09( 1000x(1.09)1 ) = 1000x(1.09)2 After n years, an = 1000x(1.09)n Since an is recursively defined, we prove the formula, for n ≥ 0, by mathematical induction (The problem does not ask for proof). Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 10 (b) Solution Basis n = 0: a0 = 1000 = 1000x(1.09)0 . The 1st equality is the recurrence relation’s initial condition. Induction hypothesis: an = 1000x1.09n. Induction step: Show: an+1 = 1000x1.09n+1. an+1 = 1.09an = 1.09 ( 1000x1.09n ) = 1000x1.09n+1. The 1st equality: the definition of the recurrence relation. The 2nd equality: the induction hypothesis. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 10 (c) A person deposits $1,000 in an account that yields 9% interest compounded annually. How much money will the account contain after 100 years? Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 10 (c) Solution The account contains a100 dollars after 100 years: a100 = 1000x1.09100 = $5,529,041. That is before taxes. With 30% federal + 10% CA on interest earned, it becomes 1000x1.05100 = $131,500  Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 20 A country uses as currency coins with pesos values of 1, 2, 5, & 10 pesos Find a recurrence relation, an, for the # of payment sequences for n pesos. E.g., a bill of 4 pesos could be paid with any of the following sequences: 1, 1, 1, 1 1,1, 2 1, 2, 1 2, 1, 1 2, 2 Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 20 Solution The sequences that start w/ a 1 peso coin differ from the sequences that don’t. Use the sum rule: Partition the set of sequences, based on which kind of coin starts the sequence. It could be a: 1 peso coin, in which case we have an-1 ways to finish the bill 2 peso coin, in which case we have an-2 ways to finish the bill 5 peso coin, in which case we have an-5 ways to finish the bill 10 peso coin, in which case we have an-10 ways to finish the bill The recurrence relation is an = an-1 + an-2 + an-5 + an-10 with 10 initial conditions, a1 = 1, a2 = 2, a3 = 3, a4 = 5, a5 = 9, a6 = 15, a7 = 26, a8 = 44, a9 = 75, a10 = 125. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 30 (a) A string that contains only 0s, 1s, & 2s is called a ternary string. Find a recurrence relation for the # of ternary strings of length n that do not contain 2 consecutive 0s. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 30 (a) Solution Subtract the # of “bad” strings (contain 2 consecutive 0s), bn, , from the # of ternary strings, 3n. To count bn, , use the sum rule: Partition the set of strings, depending on what digit starts the string: The string starts with a 1: bn-1 ways to finish the string. The string starts with a 2: bn-1 ways to finish the string. The string starts with a 0: The remaining string starts with a 0: 3n-2 ways to finish the string. The remaining string starts with a 1: bn-2 ways to finish the string. The remaining string starts with a 2: bn-2 ways to finish the string. Summing, bn = 2bn-1 + 2bn-2 + 3n-2 Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 30 (b) What are the initial conditions? Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 30 (b) Solution What are the initial conditions? b0 = b1 = 0. Why do we need 2 initial conditions? Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 30 (c) How many ternary strings of length 6 contain 2 consecutive 0s? Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 30 (c) Solution The number of such strings is the value of b6. Using bn = 2bn-1 + 2bn-2 + 3n-2, we compute: b0 = b1 = 0. (Initial conditions) b2 = 2b1 + 2b0 + 30 = 2x0 + 2x0 + 30 = 1 b3 = 2b2 + 2b1 + 31 = 2x1 + 2x0 + 31 = 5 b4 = 2b3 + 2b2 + 32 = 2x5 + 2x1 + 32 = 21 b5 = 2b4 + 2b3 + 33 = 2x21 + 2x5 + 33 = 79 b6 = 2b5 + 2b4 + 34 = 2x79 + 2x21 + 34 = 281. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 40 Find a recurrence relation, en, for the # of bit strings of length n with an even # of 0s. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 40 Solution Strings are sequences: Order matters: There is a 1st bit. Use the sum rule: Partition the desired set of bit strings, based on the string’s 1st bit: Strings with an even # of 0s that begin with 1: en-1 Strings with an even # of 0s that begin with 0: 2n-1 - en-1 Summing, en = en-1 + 2n-1 - en-1 = 2n-1 Copyright © Peter Cappello

Copyright © Peter Cappello End Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 40 Solution Strings are sequences: Order matters: There is a 1st bit. Use the sum rule: Partition the desired set of bit strings, based on the string’s 1st bit: Strings with an even # of 0s that begin with 1: en-1 Strings with an even # of 0s that begin with 0: 2n-1 - en-1 Summing, en = en-1 + 2n-1 - en-1 = 2n-1 Does this answer suggest an alternate explanation? Does this question relate to our study binomial coefficients? Copyright © Peter Cappello

Use the Binomial Theorem ( x + y )n = Σj=0 to n C( n, j )xn-jyj = C( n, 0 )xny0 + C( n,1 )xn-1y1 + … + C( n, j )xn-jyj + … + C( n, n )x0yn. Evaluate at x = 1, y = -1: (1 – 1)n = Σj=0 to n C( n, j )1n-j( -1 )j = C( n, 0 ) - C( n,1 ) + C( n,2 ) - C( n,3 ) +- . . . (-1)nC( n, n ). C( n,1 ) + C( n,3 ) + C( n,5 ) . . . = C( n, 0 ) + C( n,2 ) + C( n,4 ) . . . = 2n-1 Example: C(4,0) + C(4,2) + C(4,4) = C(4,1) + C(4,3) = 23 The # of bit strings of length 4 that have an even number of 0s is 23. Copyright © Peter Cappello

Copyright © Peter Cappello 2011 49 The variation we consider begins with people numbered 1, …, n, standing around a circle. In each stage, every 2nd person still alive is killed until only 1 survives. We denote the number of the survivor by J(n). Determine the value of J(n) for 1  n  16. Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 49 Solution Put 5 people, named 1, 2, 3, 4, & 5, in a circle. Starting with 1, kill every 2nd person until only 1 person is left. The sequence of killings is: 1 2 3 4 5 So, J(5) = 3. Continuing, for each value of n, results in the following table. Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 49 Solution n J(n) 1 9 3 2 10 5 11 7 4 12 13 6 14 15 8 16 Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 50 Use the values you found in Exercise 49 to conjecture a formula for J(n). Hint: Write n = 2m + k, where m, k N & k < 2m . Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 50 Solution n J(n) 1 = 20 + 0 1 9 = 23 + 1 3 2 = 21 + 0 10 = 23 + 2 5 3 = 21 + 1 11 = 23 + 3 7 4 = 22 + 0 12 = 23 + 4 9 5 = 22 + 1 13 = 23 + 5 11 6 = 22 + 2 14 = 23 + 6 13 7 = 22 + 3 15 = 23 + 7 15 8 = 23 + 0 16 = 24 + 0 Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 50 Solution continued n J(n) 1 = 20 + 0 1 = 2*0 + 1 9 = 23 + 1 3 = 2*1 + 1 2 = 21 + 0 10 = 23 + 2 5 = 2*2 + 1 3 = 21 + 1 11 = 23 + 3 7 = 2*3 + 1 4 = 22 + 0 12 = 23 + 4 9 = 2*4 + 1 5 = 22 + 1 13 = 23 + 5 11 = 2*5 + 1 6 = 22 + 2 14 = 23 + 6 13 = 2*6 + 1 7 = 22 + 3 15 = 23 + 7 15 = 2*7 + 1 8 = 23 + 0 16 = 24 + 0 Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 50 Solution continued So, if n = 2m + k, where m, k N & k < 2m , then J(n) = 2k + 1. Check this for J(17). Copyright © Peter Cappello 2011

Exercise 20 Solution continued But, we also can use bills. If the 1st currency object is a bill, it could be a 5 peso, in which case we have an-5 ways to finish the bill 10 peso, in which case we have an-10 ways to finish the bill 20 peso, in which case we have an-20 ways to finish the bill 50 peso, in which case we have an-50 ways to finish the bill 100 peso, in which case we have an-100 ways to finish the bill Using both coins & bills, we have an = an-1 + an-2 + an-5 + an-10 + an-5 + an-10 + an-20 + an-50 + an-100 = an-1 + an-2 + 2an-5 + 2an-10 + an-20 + an-50 + an-100 , with 100 initial conditions, which I will not produce. Copyright © Peter Cappello