# Solve the first-order linear recurrence relation: S n+1 = 5 S n + 1, with S 0 =1. You may use the general solution given

Solve the first-order linear recurrence relation: S n+1 = 5 S n + 1, with S 0 =1. You may use the general solution given on P.342. Transcribed Image Text: This can be proved by MI – if for some ke P, S = aki+ ak-1 c +

C +… + ac + c then

Sk+1 = aSk + c

= a [a*I + a*=\c + a*~?c+ … + ac +

= ak+’ I + a*c + ak-‘c + . . + a²c + ac + c.

+ a*-\c + a*-²c + . + ac +

„k-2

´c +

+ c

°c +

Thus, (8.2.2) is correct. Hence, if S is any sequence on N

satisfying (8.2.1), then for vn e P

if a = 1, S, = 1″I + 1″-‘c + 1″-?c + … + lc + c

= I + nc,

// And this formula works for So

%3D

and if a + 1, S, = a”I + a”-‘c + a”-?c + … + ac + c

%3D

// and from Theorem 3.6.9

1- a”

с

3 а”1 + с-

1 – a

C

a”

a”I +

=

1 — а

C

= d” |1- +

/| And this formula works for So

1- a

Therefore, the general solution of the recurrence equation

S n+1

aS n + c

for V n e N

(8.2.1)

is given in two parts:

Pg. 342

if a = 1,

Sn = I + nc

for Vn e N;

if a + 1,

S, = a”A +

for V n e N.

When a = 1, any particular solution is obtained by

determining a specific, numerical value for I. In fact, a particular

solution is determined by a specific, numerical value / for any

(particular) entry, S;. Solving the equation

J = I + jc for I,

I = J – jc.

// since S; = I + jc

// where So = I

we get

// One particular “particular solution” has / = 0.

When a + 1, any particular solution is obtained by

determining a specific, numerical value for A; if the starting value

Iis given, then A = I – . In fact, a particular solution is

%3D

determined by a specific, numerical value / for any (particular)

entry, Sj. Solving the equation

J = Aa’

+

1 —а

for A,

we get

A =

// But what if a =

0?

// One particular “particular solution” has A = o.

%3D

Example 8.2.1:

The recurrence equation for the number of moves in the

Towers of Hanoi problem is a first-order linear recurrence

equation:

The Towers of Hanoi

Tn = 2Tµ-1 + 1.

Here a = 2 and c = 1, so = 2

= -1, and an. sequence T

1-a Transcribed Image Text: 8.2 Solving First-Order Linear Recurrence

Equations

A first-order linear recurrence equation relates consecutive entries

in a sequence by an equation of the form

S n+1 = aS n + c

for V n in the domain of S.

(8.2.1)

But let’s assume that the domain of S is N. Let’s also assume

that a * 0; otherwise, S, = c for Vn > 0, and the solutions to

(8.2.1) are not very interesting.

// are they?

We saw in Chap. 3, that when a = 1, any sequence satisfying

(8.2.1) is an arithmetic sequence, and when S is defined on N and

So is some initial value /,

Sn

= I + nc

for Vn e N.

// Theorem 3.6.4

Also, when c = 0, any sequence satisfying (8.2.1) is a geometric

sequence, and when S is defined on N and So is some initial value

I,

Sn = a” I

for V n e N.

// Theorem 3.6.7

Furthermore, Theorem 3.6.8gives a formula for the sum of

the first (n + 1) terms of a geometric series when a + 1:

a”+1

– 1

a°1 + a’I + a²I + … + a”I = I (1 + a + a² + …

+ a”) = 1-

а — 1

In particular, 1 + ½ + (½)* + (‘2) + … + ‘2)” = [(“2)* -1|/[2-1]

(2)’

(‘2) = (C’2)” – 1|/[‘2 – 1]

\n+1

%D

= 2 – (‘½)” .

Equation 8.2.1 says how the sequence continues, after

starting with some initial value / but doesn’t restrict the value of

I; hence, there is an infinite number of solutions.

A general solution is an algebraic description of all such

solution sequences.

If S is any sequence on N satisfying (8.2.1), then denoting So

by I, we have

S1 = aSo + c

S2 = aS 1 + c = a[al + c] + c

= al + c,

— а?1 + ас + с,

S3 = aS2 + c = a

[a* –

+c]

+c = a³I+ a²c + ac + c.

+ ас

So we might conjecture

S, = a”I + a”¯’c+a”=²c + . .. + ac + c

for V n e P. (8.2.2)

n-

This can be proved by MI – if for some ke P, Sk = akI+ a*-1 c +

ak-2 c+ … + ac + c then

Sk+1 = aSk + c

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