Recurrence Relations: Selected Exercises

1. Recurrence Relations: Selected Exercises

2. 10 (a) A person deposits $1,000 in an account that yields 9% interest compounded annually. a) Set up a recurrence relation for the amount in the account at the end ofn years.

3. 10 (a) Solution Let an represent the amount after n years. an = an-1 + 0.09an-1 = 1.09an-1 a0 = 1000.

4. 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 ofn years.

5. 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 anis recursively defined, we prove the formula, for n ≥ 0, by mathematical induction (The problem does not ask for proof).

6. 10 (b) Solution Basis n = 0: a0 = 1000 = 1000x(1.09)0 . The 1st equality is the recurrence relation’s initial condition. Show: an = 1000x1.09nan+1 = 1000x1.09n+1. • Assumean = 1000x1.09n . • an+1= 1.09an= 1.09 (1000x1.09n) = 1000x1.09n+1. The 1st equality is from the definition of the recurrence relation. The 2nd equality is from the induction hypothesis.

7. 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. 10 (c) Solution The account will contain 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.

9. 20 A country uses as currency: • coins with pesos values of 1, 2, 5, & 10 pesos • bills with pesos values of 5, 10, 20, 50, & 100. 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

10. 20 Solution For n pesos, our 1st (order matters) currency object can be a coin or a bill. Sequences that start w/ a 1 peso coin are different from other sequences: Use the sum principle to decompose this problem into disjoint sub-problems, based on which kind of currency object starts the sequence. If the 1st currency object is a coin, 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 If there were only coins, the recurrence relation would be 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

11. 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 So, 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.

12. 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.

13. 30 (a) Solution We subtract the # of “bad” strings, bn, , from the # of ternary strings, 3n. We use the sum principle to decompose the problem into disjoint sub-problems, depending on what digit starts the string: Case the string starts with a 1: bn-1 ways to finish the string. Case the string starts with a 2: bn-1 ways to finish the string. Case the string starts with a 0: Case the remaining string starts with a 0: 3n-2 ways to finish the string. Case the remaining string starts with a 1: bn-2 ways to finish the string. Case the remaining string starts with a 2: bn-2 ways to finish the string. Summing, bn = 2bn-1 + 2bn-2 +3n-2

14. 30 (b) b) What are the initial conditions?

15. 30 (b) Solution b0 =b1 = 0. Why do we need 2 initial conditions?

16. 30 (c) How many ternary strings of length 6 contain 2 consecutive 0s?

17. 30 (c) Solution The number of such strings is b6. Using bn = 2bn-1 + 2bn-2 +3n-2, we compute: b0 =b1 = 0. (Initial conditions) b2 =2b1 +2b0 +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.

18. 40 Find a recurrence relation, en, for the # of bit strings of length n with an even # of 0s.

19. 40 Solution Strings are sequences: Order matters: There is a 1st bit. Use the sum principle to decompose the problem into disjoint sub-problems, based on their 1st bit: The strings with an even # of 0s that begin with 1: en-1 The 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? Remember this question when we study binomial coefficients.

21. 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.

22. 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 1 2 3 4 5 12 3 4 5 12345 12345 So, J(5) = 3. Continuing, for each value of n, results in the following table.

23. 49 Solution

24. 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.

25. 50 Solution

26. 50 Solution continued

27. 50 Solution continued So, if n = 2m + k, where m, k N & k<2m, then J(n) = 2k + 1. Check this for J(17).