Solving by Iteration

8/13/2019 Solving by Iteration

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Section 2: Solving Recurrence

Relations by Iteration As we noted at the end of the last lecture, when

analyzing recurrence relations, we want to rewrite

the general term as a function of the index andindependent of predecessor terms.

This will allow us to compute any arbitrary term

in the sequence without having to compute all the

previous terms.

In this section, we will look at one method for

solving recurrence relations.

8.2.1


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The Method of Iteration

Let {ai} be the sequence defined by:ak= ak1+ 2 with a0= 1.

Plugging values of kinto the relation, we get:

a1= a0+ 2 = 1 + 2

a2= a1+ 2 = 1 + 2 + 2 = 1 + 2(2)

a3= a2+ 2 = 1 + 2 + 2 + 2 = 1 + 3(2)

a4= a3+ 2 = 1 + 2 + 2 + 2 + 2 = 1 + 4(2)

Continuing in this fashion reinforces the apparentpattern that an= 1 + n(2) = 1 + 2n.

This brute forcetechnique is theMethod of

Iteration.

8.2.2


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The Tower of Hanoi

Recall the Tower of Hanoi relation:mk= 2mk1+ 1 with m1= 1.

Plugging values of kinto the relation, we get:

m2= 2m1+ 1 = 2 + 1

m3= 2m2+ 1 = 2(2 + 1) + 1 = 22+ 2 + 1

m4= 2m3+ 1 = 2(22+ 2 + 1) + 1

= 23+ 22+ 2 + 1

m5= 2m4+ 1 = 2(23

+ 22

+ 2 + 1)= 24+ 23+ 22+ 2 + 1.

Thus, weguess: mn= 2n1+ 2n2+...+ 22+ 2 + 1.

This is a Geometric Series, so mn= 2

n

1.

8.2.3


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Another Example

Let {ai} be the sequence given by:ak= ak1+ k with a0= 0.

Solve this recurrence relation and find a100.

Now, a1= a0+ 1 = 1 + 0a2= a1+ 2 = 2 + 1 + 0

a3= a2+ 3 = 3 + 2 + 1 + 0

a4= a3+ 4 = 4 + 3 + 2 + 1 + 0 Thus an= n+ (n1) + (n2) +...+ 3 + 2 + 1 + 0

so an= n(n+1)/2.

Plugging in n= 100: a100

= 100(101)/2 = 5050.

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A Geometric Sequence

Let aand bbe non-zero constants, and consider:

sk= ask-1 withs0= b.

Thus: s1= as0= ab

s2= as1= a(ab) = a2b

s3= as2= a(a2b) = a3b

s4= as3= a(a3b) = a4b.

From this, we can make the conjecture that:sn= a

nb.

Note: if b= 1, thensn= an.

8.2.5


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A Perturbed Geometric Sequence

Let abe non-zero constants, and consider:

sk= ask-1 + 1 withs0= 1.

Thus: s1= as0 + 1 = a+ 1

s2= as1 + 1 = a(a+ 1) + 1 = a2 + a+ 1

s3= as2 + 1 = a(a2 + a+ 1) + 1

= a3 +a2 + a+ 1

s4= as3 + 1 = a(a3 +a2 + a+ 1)= a4+ a3 +a2 + a+ 1.

It appears thatsn= an + an1 +...+a2 + a+ 1, so

sn= (a

n+1

1) / (a

1) .

8.2.6


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Some Observations

In solving these recurrence relations, we point out

the following observations:

1. Each recurrence relation looks only 1 step back; that is

each relation has been of the formsn= F(sn1);2. We have relied on luckto solve the relation, in that we

have needed to observe a pattern of behavior and

formulated the solution based on the pattern;

3. The initial condition has played a role in making thispattern evident;

4. Generating a formula from the generalization of the

pattern looks back to our study of induction.

8.2.7


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Verifying Solutions

Once we guess the form of the solution for a

recurrence relation, we need to verify it is, in fact,

the solution.

We use Mathematical Induction to do this. For example, in the Tower of Hanoi game, we

conjecture that the solution is mn= 2n1.

Basis Step: m1= 1 (by playing the game), and211 = 2 1 = 1, therefore m1= 2

11.

Inductive Step: Recall the recurrence relation is

mk= 2mk1+ 1. Assume mk= 2k1.

8.2.8


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Verifying Solutions (contd.)

Inductive Step: Show mk+1= 2k+11.

Now, mk+1= 2mk+ 1

= 2(2k1) + 1

= 22k2 + 1= 2k+11.

Therefore mk= 2k1 is the solution to

mk= 2mk1+ 1, when m1= 1. QED A similar verification process will work for all

the other formulas we discovered in this section.

8.2.9

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