Recurrence Relations: Selected Exercises

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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.

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10 (a) Solution

Let an represent the amount after n years.

an = an-1 + 0.09an-1 = 1.09an-1

a0 = 1000.

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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.

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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).

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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

.

1. Assume an = 1000x1.09n .

2. 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.

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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?

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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.

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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

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

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20 Solution continued

But, we also can use bills.

If the 1st currency object is a bill, it could be a1. 5 peso, in which case we have an-5 ways to finish the bill

2. 10 peso, in which case we have an-10 ways to finish the bill

3. 20 peso, in which case we have an-20 ways to finish the bill

4. 50 peso, in which case we have an-50 ways to finish the bill

5. 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.

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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.

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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

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30 (b)

b) What are the initial conditions?

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30 (b) Solution

b0 = b1 = 0.

Why do we need 2 initial conditions?

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30 (c)

How many ternary strings of length 6 contain 2

consecutive 0s?

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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.

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40

Find a recurrence relation, en, for the # of bit strings of

length n with an even # of 0s.

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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.

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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.

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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

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

So, J(5) = 3.

Continuing, for each value of n, results in the following table.

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49 Solution

n J(n) n J(n)

1 1 9 3

2 1 10 5

3 3 11 7

4 1 12 9

5 3 13 11

6 5 14 13

7 7 15 15

8 1 16 1

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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 .

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50 Solution

n J(n) n J(n)

1 = 20 + 0 1 9 = 23 + 1 3

2 = 21 + 0 1 10 = 23 + 2 5

3 = 21 + 1 3 11 = 23 + 3 7

4 = 22 + 0 1 12 = 23 + 4 9

5 = 22 + 1 3 13 = 23 + 5 11

6 = 22 + 2 5 14 = 23 + 6 13

7 = 22 + 3 7 15 = 23 + 7 15

8 = 23 + 0 1 16 = 24 + 0 1

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50 Solution continued

n J(n) n J(n)

1 = 20 + 0 1 = 2*0 + 1 9 = 23 + 1 3 = 2*1 + 1

2 = 21 + 0 1 = 2*0 + 1 10 = 23 + 2 5 = 2*2 + 1

3 = 21 + 1 3 = 2*1 + 1 11 = 23 + 3 7 = 2*3 + 1

4 = 22 + 0 1 = 2*0 + 1 12 = 23 + 4 9 = 2*4 + 1

5 = 22 + 1 3 = 2*1 + 1 13 = 23 + 5 11 = 2*5 + 1

6 = 22 + 2 5 = 2*2 + 1 14 = 23 + 6 13 = 2*6 + 1

7 = 22 + 3 7 = 2*3 + 1 15 = 23 + 7 15 = 2*7 + 1

8 = 23 + 0 1 = 2*0 + 1 16 = 24 + 0 1 = 2*0 + 1

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50 Solution continued

So, if n = 2m + k, where m, k N & k < 2m ,

then J(n) = 2k + 1.

Check this for J(17).