# Recurrence Relations Selected Exercises 10 a A person

- Slides: 27

Recurrence Relations: Selected Exercises

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 of n years. 2

10 (a) Solution Let an represent the amount after n years. an = an-1 + 0. 09 an-1 = 1. 09 an-1 a 0 = 1000. 3

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. 4

10 (b) Solution After 1 year, a 1 = 1. 09 a 0 = 1. 09 x 1000 = 1000 x 1. 091 After 2 years, a 2 = 1. 09 a 1 = 1. 09(1000 x(1. 09)1) = 1000 x(1. 09)2 After n years, an = 1000 x(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). 5

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

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? 7

10 (c) Solution The account will contain a 100 dollars after 100 years: a 100 = 1000 x 1. 09100 = $5, 529, 041. That is before taxes . With 30% federal + 10% CA on interest earned, it becomes 1000 x 1. 05100 = $131, 500. 8

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 2. 1, 1, 2 3. 1, 2, 1 4. 2, 1, 1 5. 2, 2 9

20 Solution For n pesos, our 1 st (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 subproblems, based on which kind of currency object starts the sequence. If the 1 st 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, 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 10

20 Solution continued But, we also can use bills. If the 1 st currency object is a bill, it could be a 1. 2. 3. 4. 5. 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-20 + an-50 + an-100 = an-1 + an-2 + 2 an-5 + 2 an-10 + an-20 + an-50 + an-100 , with 100 initial conditions, which I will not produce. 11

30 (a) A string that contains only 0 s, 1 s, & 2 s is called a ternary string. Find a recurrence relation for the # of ternary strings of length n that do not contain 2 consecutive 0 s. 12

30 (a) Solution We subtract the # of “bad” strings, bn, , from the # of ternary strings, 3 n. We use the sum principle to decompose the problem into disjoint subproblems, 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: 3 n-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 = 2 bn-1 + 2 bn-2 + 3 n-2 13

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

30 (b) Solution b 0 = b 1 = 0. Why do we need 2 initial conditions? 15

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

30 (c) Solution The number of such strings is b 6. Using bn = 2 bn-1 + 2 bn-2 + 3 n-2, we compute: b 0 = b 1 = 0. (Initial conditions) b 2 = 2 b 1 + 2 b 0 + 30 = 1 b 3 = 2 b 2 + 2 b 1 + 31 = 2 x 1 + 2 x 0 + 31 = 5 b 4 = 2 b 3 + 2 b 2 + 32 = 2 x 5 + 2 x 1 + 32 = 21 b 5 = 2 b 4 + 2 b 3 + 33 = 2 x 21 + 2 x 5 + 33 = 79 b 6 = 2 b 5 + 2 b 4 + 34 = 2 x 79 + 2 x 21 + 34 = 281. 17

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

40 Solution Strings are sequences: Order matters: There is a 1 st bit. Use the sum principle to decompose the problem into disjoint subproblems, based on their 1 st bit: The strings with an even # of 0 s that begin with 1: en-1 The strings with an even # of 0 s that begin with 0: 2 n-1 - en-1 Summing, en = en-1 + 2 n-1 - en-1 = 2 n-1 Does this answer suggest an alternate explanation? Remember this question when we study binomial coefficients. 19

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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. 21

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

49 Solution n 1 2 3 4 5 6 7 8 J(n) 1 1 3 5 7 1 n 9 10 11 12 13 14 15 16 J(n) 3 5 7 9 11 13 15 1 23

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. 24

50 Solution n 1 = 20 + 0 2 = 21 + 0 3 = 21 + 1 4 = 22 + 0 5 = 22 + 1 6 = 22 + 2 7 = 22 + 3 8 = 23 + 0 J(n) 1 1 3 5 7 1 n 9 = 23 + 1 10 = 23 + 2 11 = 23 + 3 12 = 23 + 4 13 = 23 + 5 14 = 23 + 6 15 = 23 + 7 16 = 24 + 0 J(n) 3 5 7 9 11 13 15 1 25

50 Solution continued n 1 = 20 + 0 2 = 21 + 0 3 = 21 + 1 4 = 22 + 0 5 = 22 + 1 6 = 22 + 2 7 = 22 + 3 8 = 23 + 0 J(n) 1 = 2*0 + 1 3 = 2*1 + 1 5 = 2*2 + 1 7 = 2*3 + 1 1 = 2*0 + 1 n 9 = 23 + 1 10 = 23 + 2 11 = 23 + 3 12 = 23 + 4 13 = 23 + 5 14 = 23 + 6 15 = 23 + 7 16 = 24 + 0 J(n) 3 = 2*1 + 1 5 = 2*2 + 1 7 = 2*3 + 1 9 = 2*4 + 1 11 = 2*5 + 1 13 = 2*6 + 1 15 = 2*7 + 1 1 = 2*0 + 1 26

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

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