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Linear Recurrence Relations

Part I

Jorge Cobb

The University of Texas at Dallas

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Solving recurrence relations Setting up a recurrence relation is important –

it corresponds to modeling a problem Solving it (providing a sequence that satisfies

the recurrence) is also needed Sometimes, this can be done through

iteration/induction For certain types of recurrence relations,

there are systematic methods for solving them

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Linear recurrence relations of degree k

an = c1an-1 + c2an-2 + ... + ckan-k with c1,c2,...,ck real numbers and ck≠0

Linear: The right-hand side is a sum of weighted previous

terms of the sequence – the weights do not depend on the sequence (but not necessarily constant, may be a function of n)

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an = c1an-1 + c2an-2 + ... + ckan-k with c1,c2,...,ck real numbers and ck≠0

Homogeneous: No terms appear on the right hand side that are not

multiples of a previous term

Of degree k: The recurrence goes back k terms, i.e., the earliest

previous term on the right hand side is an-k

Constant coefficients: The multipliers of the previous terms are all constants, not

functions that depend on n

Linear homogeneous recurrence relations of degree k with constant coefficients

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Classifying recurrences

an=2an-1 + an-22

an=an-1an-2

an=an-1+an-2

an=1.05an-1

an=nan-1

an=2an-1+1

an=an-1+an-4

Classification doesn’t depend on initial values

Not linear

Non-homogeneous

Not linearYes, degree 2

Coefficients are not constant

Yes, degree 1

Yes, degree 4

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First degree linear homogeneous recurrence relations with const. coef.

an = c1an-1

Recall the compound interest example Through iterative expansion, an = c1an-1 = c1(c1an-2) = c1

2an-2 = c12(c1an-3) =

c13an-3 = ... = c1

na0

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Approach for a general solution

an = c1an-1 + c2an-2 + ... + ckan-k

Pretend that there is a solution of the form an=rn for some constant r

rn = c1rn-1 + c2rn-2 + ... + ckrn-k

rk – c1rk-1 – c2rk-2 - ... – ck-1r – ck = 0 This is the characteristic equation of the

recurrence relation; the numbers r that satisfy it are the characteristic roots of the recurrence relation

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Second degree linear homogeneous recurrence relations

Assumptions

Recurrence relation: an = c1an-1 + c2an-2 Characteristic equation has two distinct roots r1, r2.

Theorem 1

{an} is a solution of the recurrence

if and only if

{an} is of the form an = b1r1n + b2r2

n for all n≥0, and for some constants b1, b2

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Proving Theorem 1 We need to prove both directions:

If {an} is a solution, then it must be of the form an= b1r1

n + b2r2n

for some appropriately chosen constants b1

and b2

If {an} is of the form an= b1r1

n + b2r2n

for all n≥0, then it must be a solution.

We prove the second one first (harder )

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Proving Theorem 1, first part Show an = b1r1

n + b2r2n, n≥2, is a solution

Because r2 – c1r – c2 = 0 and r1 and r2 are roots,

Therefore,

)()(

)()(

2212

222112

11

222

2112

122

1111

2211

crcrbcrcrb

rbrbcrbrbc

acaca

nn

nnnn

nnn

22221

21211 , rcrcrcrc

nn

nnn

rbrb

rrbrrba

2211

22

222

21

211

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Proving Theorem 1, first part

What about n=1 and n=0?

These are just the initial conditions of our solution

210

220

110

2211122

1111

bbrbrba

rbrbrbrba

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Proving Theorem 1, second part

Show that if {an} is a solution it must be of the form an = b1r1

n + b2r2n (we must find b1 and b2)

Note: a second-degree linear homogeneous recurrence has a unique solution {an} if a0 and a1 are specified

Because there is a single way to generate the terms an from the two initial values

Therefore, if we show that there is a solution of the form an = b1r1

n + b2r2n, n ≥0, that satisfies the initial

values a0 and a1, this must be the solution that was given to us, and we are done.

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Proving Theorem 1, second part

We first show that there are b1 and b2 such that b1r1

0 + b2r20 = a0 and b1r1

1 + b2r21 = a1

These simplify to b1 + b2 = a0

b1r1 + b2r2 = a1, (two linear equations, two unknowns)

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1012101212122120 )()(

rr

raabraabrrarbrba

12

120

12

1011020

12

1010201 rr

ara

rr

raarara

rr

raaabab

201 bab

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Consider the sequence {a’n}, where a’n = b1r1

n + b2r2n, and a’0 = a0, a’1 = a1

From the first part of the theorem, we have shown that a’n = b1r1

n + b2r2n is a solution for any b1

and b2

We have shown (previous slide) that b1 and b2 can be chosen so that the satisfy a’0 = a0, a’1 = a1

Thus, {a’n} is a solution, just like {an}, and with the same initial conditions.

However, the initial conditions determine the rest of the sequence.

Hence, {a’n} = {an} and {an} is in the desired form.

Proving Theorem 1, second part

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Notes about the proof

Our proof of the second part also serves as the means to find the solution

It depended on r1 ≠ r2

The characteristic roots r1 and r2 may be complex numbers (the proof and the solution are still valid)

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Example solution

an = an-1 + 2an-2, a0=2 and a1=7 Characteristic equation

r2 – r – 2 = 0 The quadratic formula for ax2+bx+c = 0,

gives roots r1=(1-(3))/2=-1 and r2=(1+(3))/2 = 2

a

acbbx

2

42

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Example solution

Therefore,

33

9

)1(2

)1(27

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1012

rr

raab

13

3

)1(2

722

12

1201

rr

arab

nnnnnnn rbrba )1(2323)1(12211