1. Recurrence Relations: Selected Exercises

2. Copyright © Peter Cappello2 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.

3. Copyright © Peter Cappello3 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 a n represent the amount after n years. a n = a n-1 + 0.09a n-1 = 1.09a n-1 a 0 = 1000.

4. Copyright © Peter Cappello4 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.

5. Copyright © Peter Cappello5 Exercise 10 (b) Solution After 1 year, a 1 = 1.09a 0 = 1.09x1000 = 1000x1.09 1 After 2 years, a 2 = 1.09a 1 = 1.09( 1000x(1.09) 1 ) = 1000x(1.09) 2 After n years, a n = 1000x(1.09) n Since a n is recursively defined, we prove the formula, for n ≥ 0, by mathematical induction (The problem does not ask for proof).

6. Copyright © Peter Cappello6 Exercise 10 (b) Solution Basis n = 0: a 0 = 1000 = 1000x(1.09) 0. The 1 st equality is the recurrence relation’s initial condition. Induction hypothesis: a n = 1000x1.09 n. Induction step: Show: a n+1 = 1000x1.09 n+1. a n+1 = 1.09a n = 1.09 ( 1000x1.09 n ) = 1000x1.09 n+1. The 1 st equality: the definition of the recurrence relation. The 2 nd equality: the induction hypothesis.

7. Copyright © Peter Cappello7 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?

8. Copyright © Peter Cappello8 Exercise 10 (c) Solution The account contains a 100 dollars after 100 years: a 100 = 1000x1.09 100 = $5,529,041. That is before taxes. With 30% federal + 10% CA on interest earned, it becomes 1000x1.05 100 = $131,500 

9. Copyright © Peter Cappello9 Exercise 20 A country uses as currency coins with pesos values of 1, 2, 5, & 10 pesos Find a recurrence relation, a n, 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 2.1,1, 2 3.1, 2, 1 4.2, 1, 1 5.2, 2

10. Copyright © Peter Cappello10 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 a n-1 ways to finish the bill 2 peso coin, in which case we have a n-2 ways to finish the bill 5 peso coin, in which case we have a n-5 ways to finish the bill 10 peso coin, in which case we have a n-10 ways to finish the bill The recurrence relation is a n = a n-1 + a n-2 + a n-5 + a n-10 with 10 initial conditions, a 1 = 1, a 2 = 2, a 3 = 3, a 4 = 5, a 5 = 9, a 6 = 15, a 7 = 26, a 8 = 44, a 9 = 75, a 10 = 125.

11. Copyright © Peter Cappello11 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.

12. Copyright © Peter Cappello12 Exercise 30 (a) Solution Subtract the # of “bad” strings (contain 2 consecutive 0s), b n,, from the # of ternary strings, 3 n. To count b n,, use the sum rule: Partition the set of strings, depending on what digit starts the string: 1.The string starts with a 1: b n-1 ways to finish the string. 2.The string starts with a 2: b n-1 ways to finish the string. 3.The string starts with a 0: 1.The remaining string starts with a 0: 3 n-2 ways to finish the string. 2.The remaining string starts with a 1: b n-2 ways to finish the string. 3.The remaining string starts with a 2: b n-2 ways to finish the string. Summing, b n = 2b n-1 + 2b n-2 + 3 n-2

13. Copyright © Peter Cappello13 Exercise 30 (b) What are the initial conditions?

14. Copyright © Peter Cappello14 Exercise 30 (b) Solution What are the initial conditions? b 0 = b 1 = 0. Why do we need 2 initial conditions?

15. Copyright © Peter Cappello15 Exercise 30 (c) How many ternary strings of length 6 contain 2 consecutive 0s?

16. Copyright © Peter Cappello16 Exercise 30 (c) Solution The number of such strings is the value of b 6. Using b n = 2b n-1 + 2b n-2 + 3 n-2, we compute: b 0 = b 1 = 0. (Initial conditions) b 2 = 2b 1 + 2b 0 + 3 0 = 2x0 + 2x0 + 3 0 = 1 b 3 = 2b 2 + 2b 1 + 3 1 = 2x1 + 2x0 + 3 1 = 5 b 4 = 2b 3 + 2b 2 + 3 2 = 2x5 + 2x1 + 3 2 = 21 b 5 = 2b 4 + 2b 3 + 3 3 = 2x21 + 2x5 + 3 3 = 79 b 6 = 2b 5 + 2b 4 + 3 4 = 2x79 + 2x21 + 3 4 = 281.

17. Copyright © Peter Cappello17 Exercise 40 Find a recurrence relation, e n, for the # of bit strings of length n with an even # of 0s.

18. Copyright © Peter Cappello18 Exercise 40 Solution Strings are sequences: Order matters: There is a 1 st bit. Use the sum rule: Partition the desired set of bit strings, based on the string’s 1 st bit: Strings with an even # of 0s that begin with 1: e n-1 Strings with an even # of 0s that begin with 0: 2 n-1 - e n-1 Summing, e n = e n-1 + 2 n-1 - e n-1 = 2 n-1 Does this answer suggest an alternate explanation? Does this question relate to our study binomial coefficients?

19. Copyright © Peter Cappello19 Use the Binomial Theorem ( x + y ) n = Σ j=0 to n C( n, j )x n-j y j = C( n, 0 )x n y 0 + C( n,1 )x n-1 y 1 + … + C( n, j )x n-j y j + … + C( n, n )x 0 y n. Evaluate at x = 1, y = -1: (1 – 1) n = Σ j=0 to n C( n, j )1 n-j ( -1 ) j = C( n, 0 ) - C( n,1 ) + C( n,2 ) - C( n,3 ) +-... (-1) n C( n, n ). C( n,1 ) + C( n,3 ) + C( n,5 )... = C( n, 0 ) + C( n,2 ) + C( n,4 )... = 2 n-1 Example: C(4,0) + C(4,2) + C(4,4) = C(4,1) + C(4,3) = 2 3 The # of bit strings of length 4 that have an even number of 0s is 2 3.

20. Copyright © Peter Cappello20 End

21. Exercise 40 Solution Strings are sequences: Order matters: There is a 1 st bit. Use the sum rule: Partition the desired set of bit strings, based on the string’s 1 st bit: Strings with an even # of 0s that begin with 1: e n-1 Strings with an even # of 0s that begin with 0: 2 n-1 - e n-1 Summing, e n = e n-1 + 2 n-1 - e n-1 = 2 n-1 Copyright © Peter Cappello21

22. Copyright © Peter Cappello22 49 The variation we consider begins with people numbered 1, …, n, standing around a circle. In each stage, every 2 nd 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.

23. Copyright © Peter Cappello23 49 Solution Put 5 people, named 1, 2, 3, 4, & 5, in a circle. Starting with 1, kill every 2 nd 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.

24. Copyright © Peter Cappello24 49 Solution nJ(n)n 1193 21105 33117 41129 531311 651413 7715 81161

25. Copyright © Peter Cappello25 50 Use the values you found in Exercise 49 to conjecture a formula for J(n). Hint: Write n = 2 m + k, where m, k  N & k < 2 m.

26. Copyright © Peter Cappello 201126 50 Solution nJ(n)n 1 = 2 0 + 019 = 2 3 + 13 2 = 2 1 + 0110 = 2 3 + 25 3 = 2 1 + 1311 = 2 3 + 37 4 = 2 2 + 0112 = 2 3 + 49 5 = 2 2 + 1313 = 2 3 + 511 6 = 2 2 + 2514 = 2 3 + 613 7 = 2 2 + 3715 = 2 3 + 715 8 = 2 3 + 0116 = 2 4 + 01

27. Copyright © Peter Cappello 201127 50 Solution continued nJ(n)n 1 = 2 0 + 01 = 2*0 + 19 = 2 3 + 13 = 2*1 + 1 2 = 2 1 + 01 = 2*0 + 110 = 2 3 + 25 = 2*2 + 1 3 = 2 1 + 13 = 2*1 + 111 = 2 3 + 37 = 2*3 + 1 4 = 2 2 + 01 = 2*0 + 112 = 2 3 + 49 = 2*4 + 1 5 = 2 2 + 13 = 2*1 + 113 = 2 3 + 511 = 2*5 + 1 6 = 2 2 + 25 = 2*2 + 114 = 2 3 + 613 = 2*6 + 1 7 = 2 2 + 37 = 2*3 + 115 = 2 3 + 715 = 2*7 + 1 8 = 2 3 + 01 = 2*0 + 116 = 2 4 + 01 = 2*0 + 1

28. Copyright © Peter Cappello28 50 Solution continued So, if n = 2 m + k, where m, k  N & k < 2 m, then J(n) = 2k + 1. Check this for J(17).

29. Copyright © Peter Cappello29 Exercise 20 Solution continued But, we also can use bills. If the 1 st currency object is a bill, it could be a 1.5 peso, in which case we have a n-5 ways to finish the bill 2.10 peso, in which case we have a n-10 ways to finish the bill 3.20 peso, in which case we have a n-20 ways to finish the bill 4.50 peso, in which case we have a n-50 ways to finish the bill 5.100 peso, in which case we have a n-100 ways to finish the bill Using both coins & bills, we have a n = a n-1 + a n-2 + a n-5 + a n-10 + a n-5 + a n-10 + a n-20 + a n-50 + a n-100 = a n-1 + a n-2 + 2a n-5 + 2a n-10 + a n-20 + a n-50 + a n-100, with 100 initial conditions, which I will not produce.