linear nonhomogeneous recurrence...

Linear Nonhomogeneous RecurrenceRelations

Connection between Homogeneous and NonhomogeneousProblems

Ioan Despi

[email protected]

University of New England

September 23, 2013

Outline

1 Introduction

2 Main theorem

3 Examples

4 Notes

Ioan Despi – AMTH140 2 of 12

Introduction

𝑐𝑚𝑎𝑛+𝑚 + 𝑐𝑚−1𝑎𝑛+𝑚−1 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 = 𝑔(𝑛), 𝑛 ≥ 0 (*)

𝑚∑︁𝑘=0

𝑐𝑘𝑎𝑛+𝑘 = 𝑔(𝑛), 𝑐0𝑐𝑚 ̸= 0

𝑐𝑚𝑎𝑛+𝑚 + 𝑐𝑚−1𝑎𝑛+𝑚−1 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 = 0, 𝑛 ≥ 0 (**)

𝑚∑︁𝑘=0

𝑐𝑘𝑎𝑛+𝑘 = 0, 𝑐0𝑐𝑚 ̸= 0

The solutions of linear nonhomogeneous recurrence relations are closelyrelated to those of the corresponding homogeneous equations.

First of all, remember Corrolary 3, Section 21:

If 𝑣𝑛 and 𝑤𝑛 are two solutions of the nonhomogeneous equation (*),

then 𝜙𝑛 = 𝑤𝑛 − 𝑣𝑛 , 𝑛 ≥ 0 is a solution of the homogeneous equation (**).

Main theorem

Theorem

Consider the following linear constant coefficient recurrence relation

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 = 𝑔(𝑛), 𝑐0𝑐𝑚 ̸= 0, 𝑛 ≥ 0 (*)

and its corresponding homogeneous form

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 = 0 . (**)

If 𝑢𝑛 is the general solution of the homogeneous equation (**), and 𝑣𝑛 is anyparticular solution of the nonhomogeneous equation (*), then

𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛 , 𝑛 ≥ 0

is the general solution of the nonhomogeneous equation (*).

Main theorem

Proof.

For 𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛, we have

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 =

𝑚∑︁𝑖=0

𝑐𝑖𝑎𝑛+𝑖 =

0‖⏞ ⏟

𝑚∑︁𝑖=0

𝑐𝑖𝑢𝑛+𝑖 +

𝑔(𝑛)‖⏞ ⏟

𝑚∑︁𝑖=0

𝑐𝑖𝑣𝑛+𝑖

= 𝑔(𝑛) ,

i.e., 𝑎𝑛 satisfies the non-homogeneous recurrence relation (*).

Since the general solution 𝑢𝑛 of the homogeneous problem has 𝑚arbitrary constants thus so is 𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛.

Hence 𝑎𝑛 is the general solution of (*).

More precisely, for any solution 𝑤𝑛 of (*), since 𝜙𝑛 = 𝑤𝑛 − 𝑣𝑛 satisfies(**), 𝜙𝑛 will just be a special case of the general solution 𝑢𝑛 of (**).

Hence 𝑤𝑛 = 𝜙𝑛 + 𝑣𝑛 is included in the solution 𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛.

Therefore, 𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛 is the general solution of the nonhomogeneous

problem (*).

Main theorem

Proof.

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 =

𝑚∑︁𝑖=0

0‖⏞ ⏟

𝑚∑︁𝑖=0

𝑔(𝑛)‖⏞ ⏟

𝑚∑︁𝑖=0

= 𝑔(𝑛) ,

problem (*).

Main theorem

Proof.

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 =

𝑚∑︁𝑖=0

0‖⏞ ⏟

𝑚∑︁𝑖=0

𝑔(𝑛)‖⏞ ⏟

𝑚∑︁𝑖=0

= 𝑔(𝑛) ,

problem (*).

Main theorem

Proof.

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 =

𝑚∑︁𝑖=0

0‖⏞ ⏟

𝑚∑︁𝑖=0

𝑔(𝑛)‖⏞ ⏟

𝑚∑︁𝑖=0

= 𝑔(𝑛) ,

problem (*).

Main theorem

Proof.

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 =

𝑚∑︁𝑖=0

0‖⏞ ⏟

𝑚∑︁𝑖=0

𝑔(𝑛)‖⏞ ⏟

𝑚∑︁𝑖=0

= 𝑔(𝑛) ,

problem (*).

Main theorem

Proof.

𝑐𝑚𝑎𝑛+𝑚 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 =

𝑚∑︁𝑖=0

0‖⏞ ⏟

𝑚∑︁𝑖=0

𝑔(𝑛)‖⏞ ⏟

𝑚∑︁𝑖=0

= 𝑔(𝑛) ,

problem (*).Ioan Despi – AMTH140 5 of 12

Example 1

Example

Find a particular solution of 𝑎𝑛+2 − 5𝑎𝑛 = 2 × 3𝑛 for 𝑛 ≥ 0.

Solution.

As the r.h.s. is 2 × 3𝑛, we try the special solution in the form of𝑎𝑛 = 𝐶3𝑛, with the constant 𝐶 to be determined.

The substitution of 𝑎𝑛 = 𝐶3𝑛 into the recurrence relation thus gives

𝐶 · 3𝑛+2⏟ ⏞ ‖

𝑎𝑛+2

− 5 · 𝐶 · 3𝑛⏟ ⏞ ‖𝑎𝑛

= 2 × 3𝑛 ,

i.e., 4𝐶 = 2 or 𝐶 =1

2. Hence 𝑎𝑛 =

2× 3𝑛 for 𝑛 ≥ 0 is a particular

solution.

Example 1

Example

Solution.

𝐶 · 3𝑛+2⏟ ⏞ ‖

𝑎𝑛+2

− 5 · 𝐶 · 3𝑛⏟ ⏞ ‖𝑎𝑛

= 2 × 3𝑛 ,

i.e., 4𝐶 = 2 or 𝐶 =1

2. Hence 𝑎𝑛 =

solution.

Example 1

Example

Solution.

𝐶 · 3𝑛+2⏟ ⏞ ‖

𝑎𝑛+2

− 5 · 𝐶 · 3𝑛⏟ ⏞ ‖𝑎𝑛

= 2 × 3𝑛 ,

i.e., 4𝐶 = 2 or 𝐶 =1

2. Hence 𝑎𝑛 =

solution.

Example 2

Example

Find a particular solution of 𝑓𝑛+1 − 2𝑓𝑛 + 3𝑓𝑛−4 = 6𝑛, 𝑛 ≥ 4.

Solution.

As the r.h.s. is 6𝑛, we try the similar form

𝑓𝑛 = 𝐴𝑛 + 𝐵 ,

with constants 𝐴 and 𝐵 to be determined.Hence 𝑓𝑛 be a solution requires

6𝑛 = 𝑓𝑛+1 − 2𝑓𝑛 + 3𝑓𝑛−4

=(︀𝐴(𝑛 + 1) + 𝐵

)︀− 2(𝐴𝑛 + 𝐵) + 3

(︀𝐴(𝑛− 4) + 𝐵

)︀= 2𝐴𝑛 + (2𝐵 − 11𝐴)

i.e.2𝐴 = 6

2𝐵 − 11𝐴 = 0⇔ 𝐴 = 3

𝐵 = 33/2

Therefore our particular solution is 𝑓𝑛 = 3𝑛 +33

Example 2

Example

Solution.

𝑓𝑛 = 𝐴𝑛 + 𝐵 ,

with constants 𝐴 and 𝐵 to be determined.

Hence 𝑓𝑛 be a solution requires

6𝑛 = 𝑓𝑛+1 − 2𝑓𝑛 + 3𝑓𝑛−4

=(︀𝐴(𝑛 + 1) + 𝐵

)︀− 2(𝐴𝑛 + 𝐵) + 3

(︀𝐴(𝑛− 4) + 𝐵

)︀= 2𝐴𝑛 + (2𝐵 − 11𝐴)

i.e.2𝐴 = 6

2𝐵 − 11𝐴 = 0⇔ 𝐴 = 3

𝐵 = 33/2

Example 2

Example

Solution.

𝑓𝑛 = 𝐴𝑛 + 𝐵 ,

6𝑛 = 𝑓𝑛+1 − 2𝑓𝑛 + 3𝑓𝑛−4

=(︀𝐴(𝑛 + 1) + 𝐵

)︀− 2(𝐴𝑛 + 𝐵) + 3

(︀𝐴(𝑛− 4) + 𝐵

)︀= 2𝐴𝑛 + (2𝐵 − 11𝐴)

i.e.2𝐴 = 6

2𝐵 − 11𝐴 = 0⇔ 𝐴 = 3

𝐵 = 33/2

Example 2

Example

Solution.

𝑓𝑛 = 𝐴𝑛 + 𝐵 ,

6𝑛 = 𝑓𝑛+1 − 2𝑓𝑛 + 3𝑓𝑛−4

=(︀𝐴(𝑛 + 1) + 𝐵

)︀− 2(𝐴𝑛 + 𝐵) + 3

(︀𝐴(𝑛− 4) + 𝐵

)︀= 2𝐴𝑛 + (2𝐵 − 11𝐴)

i.e.2𝐴 = 6

2𝐵 − 11𝐴 = 0⇔ 𝐴 = 3

𝐵 = 33/2

Example 3

Example

Find the particular solution of 𝑎𝑛+3 − 7𝑎𝑛+2 + 16𝑎𝑛+1 − 12𝑎𝑛 = 4𝑛𝑛 with

𝑎0 = −2 , 𝑎1 = 0 , 𝑎2 = 5 .

Solution.

We first find the general solution 𝑢𝑛 for the corresponding homogeneousproblem.

Then we look for a particular solution 𝑣𝑛 for the nonhomogeneousproblem without concerning ourselves with the initial conditions.

Once these two are done, we obtain the general solution 𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛 forthe nonhomogeneous recurrence relation, and we just need to use theinitial conditions to determine the arbitrary constants in the generalsolution 𝑎𝑛 so as to derive the final particular solution.

Example 3

Example

𝑎0 = −2 , 𝑎1 = 0 , 𝑎2 = 5 .

Solution.

Example 3

Example

𝑎0 = −2 , 𝑎1 = 0 , 𝑎2 = 5 .

Solution.

Example 3

Example

𝑎0 = −2 , 𝑎1 = 0 , 𝑎2 = 5 .

Solution.

Example 3

(a) The associated characteristic equation 𝜆3 − 7𝜆2 + 16𝜆− 12 = 0 can beshown to admit the following roots 𝜆1 = 3, 𝑚1 = 1, (simple root),𝜆2 = 2, 𝑚2 = 2, (double root):

𝜆3 − 7𝜆2 + 16𝜆− 12 = 𝜆3 − 3𝜆2 − 4𝜆2 + 16𝜆− 12 =

𝜆2(𝜆− 3) − 4(𝜆2 − 4𝜆 + 3) = 𝜆2(𝜆− 3) − 4(𝜆− 3)(𝜆− 1) =

(𝜆− 3)(𝜆2 − 4𝜆 + 4) = (𝜆− 3)(𝜆− 2)2

I The general solutions for the corresponding homogeneous problem thusreads

𝑢𝑛 = 𝐴3𝑛 + (𝐵 + 𝐶𝑛)2𝑛, 𝑛 ≥ 0 .

I That is, 𝑢𝑛 solves 𝑎𝑛+3 − 7𝑎𝑛+2 + 16𝑎𝑛+1 − 12𝑎𝑛 = 0.

Example 3

𝜆3 − 7𝜆2 + 16𝜆− 12 = 𝜆3 − 3𝜆2 − 4𝜆2 + 16𝜆− 12 =

𝜆2(𝜆− 3) − 4(𝜆2 − 4𝜆 + 3) = 𝜆2(𝜆− 3) − 4(𝜆− 3)(𝜆− 1) =

(𝜆− 3)(𝜆2 − 4𝜆 + 4) = (𝜆− 3)(𝜆− 2)2

𝑢𝑛 = 𝐴3𝑛 + (𝐵 + 𝐶𝑛)2𝑛, 𝑛 ≥ 0 .

Example 3

𝜆3 − 7𝜆2 + 16𝜆− 12 = 𝜆3 − 3𝜆2 − 4𝜆2 + 16𝜆− 12 =

𝜆2(𝜆− 3) − 4(𝜆2 − 4𝜆 + 3) = 𝜆2(𝜆− 3) − 4(𝜆− 3)(𝜆− 1) =

(𝜆− 3)(𝜆2 − 4𝜆 + 4) = (𝜆− 3)(𝜆− 2)2

𝑢𝑛 = 𝐴3𝑛 + (𝐵 + 𝐶𝑛)2𝑛, 𝑛 ≥ 0 .

Example 3

(b) Since the r.h.s. of the nonhomogeneous recurrence relation is 4𝑛 · 𝑛 ,

which fits into the description of 4𝑛× (first order polynomial in 𝑛) ,

we’ll try a particular solution in a similar form, i.e.,

𝑣𝑛 = 4𝑛(𝐷𝑛 + 𝐸).

The substitution of 𝑣𝑛 into the original recurrence relation then gives

4𝑛 · 𝑛 = 𝑣𝑛+3 − 7𝑣𝑛+2 + 16𝑣𝑛+1 − 12𝑣𝑛

= 4𝑛+3(𝐷(𝑛 + 3) + 𝐸) − 7 × 4𝑛+2(𝐷(𝑛 + 2) + 𝐸)

+ 16 × 4𝑛+1(𝐷(𝑛 + 1) + 𝐸) − 12 × 4𝑛(𝐷𝑛 + 𝐸), i.e.,

𝑛 = 64(𝐷𝑛+ 3𝐷 + 𝐸)− 112(𝐷𝑛+ 2𝐷 + 𝐸) + 64(𝐷𝑛+𝐷 + 𝐸)− 12(𝐷𝑛+ 𝐸)

= 4𝐷𝑛+ 4𝐸 + 32𝐷 .

Hence we have

4𝐷 = 1 , 4𝐸 + 32𝐷 = 0 ⇔ 𝐷 =1

4, 𝐸 = −2

and consequently 𝑣𝑛 = 4𝑛(𝑛4 − 2) .

Example 3

(c) The general solution for the nonhomogeneous problem is then given by𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛, i.e.

𝑎𝑛 = 4𝑛(︁𝑛

4− 2

)︁+ 𝐴3𝑛 + (𝐵 + 𝐶𝑛)2𝑛 , 𝑛 ≥ 0 .

(d) We now determine 𝐴, 𝐵, 𝐶 by the initial conditions and the use ofthe solution expression in (c)

Initial Conditions Induced Equations Solutions

𝑎0 = −2𝑎1 = 0𝑎2 = 5

𝐴 + 𝐵 − 2 = −23𝐴 + 2𝐵 + 2𝐶 − 7 = 0

9𝐴 + 4𝐵 + 8𝐶 = 29

𝐴 = 1𝐵 = −1𝐶 = 3

Finally the particular solution satisfying both the nonhomogeneousrecurrence relations and the initial conditions is given by

𝑎𝑛 = 4𝑛(︀𝑛4 − 2

)︀+ 3𝑛 + (3𝑛− 1)2𝑛 , 𝑛 ≥ 0 .

Example 3

(c) The general solution for the nonhomogeneous problem is then given by𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛, i.e.

𝑎𝑛 = 4𝑛(︁𝑛

4− 2

)︁+ 𝐴3𝑛 + (𝐵 + 𝐶𝑛)2𝑛 , 𝑛 ≥ 0 .

(d) We now determine 𝐴, 𝐵, 𝐶 by the initial conditions and the use ofthe solution expression in (c)

Initial Conditions Induced Equations Solutions

𝑎0 = −2𝑎1 = 0𝑎2 = 5

𝐴 + 𝐵 − 2 = −23𝐴 + 2𝐵 + 2𝐶 − 7 = 0

9𝐴 + 4𝐵 + 8𝐶 = 29

𝐴 = 1𝐵 = −1𝐶 = 3

Finally the particular solution satisfying both the nonhomogeneousrecurrence relations and the initial conditions is given by

𝑎𝑛 = 4𝑛(︀𝑛4 − 2

)︀+ 3𝑛 + (3𝑛− 1)2𝑛 , 𝑛 ≥ 0 .

Notes1 In all the examples in this lecture, it is easy to verify that the 𝑔(𝑛)

function in (*) is in the form of

𝑔(𝑛) = 𝜇𝑛(𝛼𝑘𝑛𝑘 + · · · + 𝛼1𝑛 + 𝛼0) ,

where 𝜇 is not a root of the associated characteristic equation.

I If this were not the case, we would have to use different forms to try forthe particular solutions.

I These will be the topics of the next lecture.2 If 𝑔(𝑛) = 𝜇𝑛

1𝑛 + 𝜇𝑛2 (3𝑛2 + 1), for instance, with 𝜇1 and 𝜇2 neither being a

root of the characteristic equation, then the particular solution

should be tried in the form

𝑣𝑛 = 𝜇𝑛1 (𝐴1𝑛 + 𝐴0) + 𝜇𝑛

2 (𝐵2𝑛2 + 𝐵1𝑛 + 𝐵0) .

3 If 𝑔(𝑛) = cos(𝛼𝑛) · 𝑛, for another instance, then we can treat it as

𝑔(𝑛) =(𝑒𝑖𝛼𝑛 + 𝑒−𝑖𝛼𝑛)

2𝑛 =

2× 𝜇𝑛

1 +𝑛

2× 𝜇𝑛

in which 𝜇1 = 𝑒𝑖𝛼 and 𝜇2 = 𝑒−𝑖𝛼.Alternatively, we could try the particular solution in the form

𝑣𝑛 = sin(𝛼𝑛)(𝐴1𝑛 + 𝐴0) + cos(𝛼𝑛)(𝐵1𝑛 + 𝐵0) .

𝑔(𝑛) = 𝜇𝑛(𝛼𝑘𝑛𝑘 + · · · + 𝛼1𝑛 + 𝛼0) ,

where 𝜇 is not a root of the associated characteristic equation.I If this were not the case, we would have to use different forms to try for

the particular solutions.

I These will be the topics of the next lecture.2 If 𝑔(𝑛) = 𝜇𝑛

1𝑛 + 𝜇𝑛2 (3𝑛2 + 1), for instance, with 𝜇1 and 𝜇2 neither being a

root of the characteristic equation, then the particular solution

𝑣𝑛 = 𝜇𝑛1 (𝐴1𝑛 + 𝐴0) + 𝜇𝑛

2 (𝐵2𝑛2 + 𝐵1𝑛 + 𝐵0) .

2𝑛 =

2× 𝜇𝑛

1 +𝑛

2× 𝜇𝑛

𝑔(𝑛) = 𝜇𝑛(𝛼𝑘𝑛𝑘 + · · · + 𝛼1𝑛 + 𝛼0) ,

the particular solutions.I These will be the topics of the next lecture.

2 If 𝑔(𝑛) = 𝜇𝑛1𝑛 + 𝜇𝑛

2 (3𝑛2 + 1), for instance, with 𝜇1 and 𝜇2 neither being aroot of the characteristic equation, then the particular solution

𝑣𝑛 = 𝜇𝑛1 (𝐴1𝑛 + 𝐴0) + 𝜇𝑛

2 (𝐵2𝑛2 + 𝐵1𝑛 + 𝐵0) .

2𝑛 =

2× 𝜇𝑛

1 +𝑛

2× 𝜇𝑛

𝑔(𝑛) = 𝜇𝑛(𝛼𝑘𝑛𝑘 + · · · + 𝛼1𝑛 + 𝛼0) ,

𝑣𝑛 = 𝜇𝑛1 (𝐴1𝑛 + 𝐴0) + 𝜇𝑛

2 (𝐵2𝑛2 + 𝐵1𝑛 + 𝐵0) .

2𝑛 =

2× 𝜇𝑛

1 +𝑛

2× 𝜇𝑛

𝑔(𝑛) = 𝜇𝑛(𝛼𝑘𝑛𝑘 + · · · + 𝛼1𝑛 + 𝛼0) ,

𝑣𝑛 = 𝜇𝑛1 (𝐴1𝑛 + 𝐴0) + 𝜇𝑛

2 (𝐵2𝑛2 + 𝐵1𝑛 + 𝐵0) .

2𝑛 =

2× 𝜇𝑛

1 +𝑛

2× 𝜇𝑛

linear nonhomogeneous recurrence...

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