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• Linear Nonhomogeneous Recurrence Relations

Connection between Homogeneous and Nonhomogeneous Problems

Ioan Despi

despi@turing.une.edu.au

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 closely related 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 (**).

Ioan Despi – AMTH140 3 of 12

• 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 any particular solution of the nonhomogeneous equation (*), then

𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛 , 𝑛 ≥ 0

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

Ioan Despi – AMTH140 4 of 12

• 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 𝑚 arbitrary 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 (*).

Ioan Despi – AMTH140 5 of 12

• 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 𝑚 arbitrary 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 (*).

Ioan Despi – AMTH140 5 of 12

• 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 𝑚 arbitrary 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 (*).

Ioan Despi – AMTH140 5 of 12

• 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 𝑚 arbitrary 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 (*).

Ioan Despi – AMTH140 5 of 12

• 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 𝑚 arbitrary 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 (*).

Ioan Despi – AMTH140 5 of 12

• 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 𝑚 arbitrary 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 (*).

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 of 𝑎𝑛 = 𝐶3

𝑛, with the constant 𝐶 to be determined.

The substitution of 𝑎𝑛 = 𝐶3 𝑛 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 =

1

2 × 3𝑛 for 𝑛 ≥ 0 is a particular

solution.

Ioan Despi – AMTH140 6 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 of 𝑎𝑛 = 𝐶3

𝑛, with the constant 𝐶 to be determined.

The substitution of 𝑎𝑛 = 𝐶3 𝑛 into the recurrence relation thus gives

𝐶 · 3𝑛+2⏟ ⏞ ‖

𝑎𝑛+2

− 5 · 𝐶 · 3𝑛⏟ ⏞ ‖ 𝑎

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