Recurrence Relations: Selected Exercises. Copyright © Peter Cappello2 Exercise 10 (a) A person deposits $1,000 in an account that yields 9% interest compounded

Recurrence Relations: Selected Exercises

Copyright © Peter Cappello 2

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.


Exercise 10 (a) Solution



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.


Exercise 10 (b)



Find an explicit formula for the amount in the account at

the end of n years.


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).



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.


Exercise 10 (c)



How much money will the account contain after 100

years?


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


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

2. 1,1, 2

3. 1, 2, 1

4. 2, 1, 1

5. 2, 2


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




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.


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.


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


Exercise 30 (b)

What are the initial conditions?



What are the initial conditions?

b0 = b1 = 0.

Why do we need 2 initial conditions?


Exercise 30 (c)

How many ternary strings of length 6 contain 2

consecutive 0s?


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.


Exercise 40

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

length n with an even # of 0s.


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



End



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?


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 2011 22

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.


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.


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


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 .


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


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


50 Solution continued

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

then J(n) = 2k + 1.

Check this for J(17).


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





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.

Documents

Recurrence Relations: Selected Exercises. Copyright © Peter Cappello2 Exercise 10 (a) A person deposits $1,000 in an account that yields 9% interest compounded