Linear Nonhomogeneous Recurrence amth140/Lectures/Lecture_26/ آ  The solutions of linear nonhomogeneous

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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