summary of solving linear, constant-coefficient recurrence

Summary of Solving Linear,Constant-Coefficient Recurrence Relations

Ioan Despi

[email protected]

University of New England

September 27, 2013

Outline

1 The Technique

2 Homogeneous Case

3 Non-Homogeneous Case

4 Examples

Ioan Despi – AMTH140 2 of 12

The Technique

The technique is based on method of characteristic equations.

It relies on the principle of uniqueness:

I any answer which works is the correct answer since there is only onecorrect answer (this can be proved!).

There are two parts of the total solution:

1 The homogeneous part of the solution depends only on what is on the leftof the recurrence relation.

2 The particular part of the total solution depends on what is in RHS andhas the same form as RHS.

We calculate the two parts separately and add them to form the totalsolution.

The Technique

It relies on the principle of uniqueness:

I any answer which works is the correct answer since there is only onecorrect answer (this can be proved!).

The Technique

It relies on the principle of uniqueness:I any answer which works is the correct answer since there is only one

correct answer (this can be proved!).

The Technique

There are two parts of the total solution:1 The homogeneous part of the solution depends only on what is on the left

of the recurrence relation.

The Technique

of the recurrence relation.2 The particular part of the total solution depends on what is in RHS and

has the same form as RHS.

The Technique

of the recurrence relation.2 The particular part of the total solution depends on what is in RHS and

has the same form as RHS.

The Technique

There are four steps in the process:

(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).

I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose

values cannot be determined until (S4).

(S2.) Find the particular solution by guessing a form similar to RHS.

I This step does not produce any additional undetermined coefficients, nordoes it eliminate those from (S1).

(S3.) Combine (add) the homogeneous and particular solutions.

(S4.) Use initial conditions to eliminate the undetermined constants from (S1).

The Technique

I If RHS is already zero, skip next two steps and go to (S4).

I The solution will contain one or more undetermined coefficients whosevalues cannot be determined until (S4).

The Technique

(S2.) Find the particular solution by guessing a form similar to RHS.I This step does not produce any additional undetermined coefficients, nor

does it eliminate those from (S1).

The Technique

Homogeneous Case

1 First solve the characteristic equation.

(a) Distinct roots

Example 𝜆 = 1, 𝜆 = 2

Solution 𝑎𝑛 = 𝐴(1)𝑛 +𝐵(2)𝑛 = 𝐴+𝐵2𝑛

(b) Repeated roots

Example 𝜆 = 3 with multiplicity 2

Solution 𝑎𝑛 = (𝐴𝑛+𝐵)3𝑛

Solution 𝑎𝑛 = (𝐴𝑛2 +𝐵𝑛+ 𝐶)2𝑛

𝜆 = −1 with multiplicity 2

Solution 𝑎𝑛 = 𝐴2𝑛 + (𝐵𝑛+ 𝐶)(−1)𝑛

2 If there are initial conditions, use them to find the values for theconstants 𝐴,𝐵, etc.

Homogeneous Case

1 First solve the characteristic equation.(a) Distinct roots

(b) Repeated roots

Homogeneous Case

(b) Repeated roots

Homogeneous Case

(b) Repeated roots

Homogeneous Case

(b) Repeated roots

Homogeneous Case

(b) Repeated roots

Non-Homogeneous Case

Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it 𝑢𝑛.

If there are initial conditions leave these for the moment.Next find a particular solution, 𝑣𝑛, to the non-homogeneous case.

(a) Solution to the homogeneous case not like the right hand side of therecurrence relation.

F Try substituting something that looks like the right hand side (RHS).Example RHS = 2𝑛, try 𝑣𝑛 = 𝐶2𝑛

Example RHS = 4𝑛, try 𝑣𝑛 = 𝐴𝑛+𝐵Example RHS = 3𝑛𝑛, try 𝑣𝑛 = 3𝑛(𝐴𝑛+𝐵)Example RHS = 𝑛2, try 𝑣𝑛 = 𝐴𝑛2 +𝐵𝑛+ 𝐶

(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.

(i) Distinct roots: Try a solution with an extra 𝑛.Example 𝜆 = 2 and RHS = 2𝑛, try 𝑣𝑛 = 𝐶𝑛2𝑛.

(ii) Repeat Roots: Try a solution with an extra 𝑛2 for multiplicity 2, 𝑛3 formultiplicity 3 etc.

Example 𝜆 = 3 with multiplicity 2 and RHS = 3𝑛, try

𝑣𝑛 = 𝐶𝑛23𝑛

Example 𝜆 = 1 with multiplicity 3 and RHS = 4 = 4× 1𝑛, try

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it 𝑢𝑛.If there are initial conditions leave these for the moment.

Next find a particular solution, 𝑣𝑛, to the non-homogeneous case.

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it 𝑢𝑛.If there are initial conditions leave these for the moment.Next find a particular solution, 𝑣𝑛, to the non-homogeneous case.

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it 𝑢𝑛.If there are initial conditions leave these for the moment.Next find a particular solution, 𝑣𝑛, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the

recurrence relation.

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

recurrence relation.F Try substituting something that looks like the right hand side (RHS).

Example RHS = 2𝑛, try 𝑣𝑛 = 𝐶2𝑛

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

Example RHS = 4𝑛, try 𝑣𝑛 = 𝐴𝑛+𝐵

Example RHS = 3𝑛𝑛, try 𝑣𝑛 = 3𝑛(𝐴𝑛+𝐵)Example RHS = 𝑛2, try 𝑣𝑛 = 𝐴𝑛2 +𝐵𝑛+ 𝐶

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

Example RHS = 4𝑛, try 𝑣𝑛 = 𝐴𝑛+𝐵Example RHS = 3𝑛𝑛, try 𝑣𝑛 = 3𝑛(𝐴𝑛+𝐵)

Example RHS = 𝑛2, try 𝑣𝑛 = 𝐴𝑛2 +𝐵𝑛+ 𝐶

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.(i) Distinct roots: Try a solution with an extra 𝑛.

Example 𝜆 = 2 and RHS = 2𝑛, try 𝑣𝑛 = 𝐶𝑛2𝑛.

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

Example 𝜆 = 2 and RHS = 2𝑛, try 𝑣𝑛 = 𝐶𝑛2𝑛.(ii) Repeat Roots: Try a solution with an extra 𝑛2 for multiplicity 2, 𝑛3 for

multiplicity 3 etc.

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

multiplicity 3 etc.

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

multiplicity 3 etc.

𝑣𝑛 = 𝐶𝑛3 · 13 = 𝐶𝑛3

Once you have decided what a possible solution to the nonhomogeneouscase could be, substitute it into the original recurrence relation.

Then find values for the constants and put these values back into theexpressions for 𝑣𝑛.

Now add 𝑢𝑛 and 𝑣𝑛 to obtain the general solution. That is,

𝑎𝑛 = 𝑢𝑛 + 𝑣𝑛.

If there are initial conditions, substitute them in now to find the values ofthe constants that were part of 𝑣𝑛.

Finally, check your answer by substituting it into the original recurrencerelation.

Examples

Example

Find the general solution of the recurrence relation

𝑎𝑛+2 − 5𝑎𝑛+1 + 6𝑎𝑛 = 0, 𝑛 ≥ 0 .

Solution.

The associated characteristic equation

𝜆2 − 5𝜆 + 6 = 0

has two (distinct) roots 𝜆1 = 2 and 𝜆2 = 3.

Hence the general solution of the recurrence relation is

𝑎𝑛 = 𝐴2𝑛 + 𝐵3𝑛, 𝑛 ≥ 0 ,

where 𝐴 and 𝐵 are arbitrary constants.

Examples

Example

𝑎𝑛+2 − 5𝑎𝑛+1 + 6𝑎𝑛 = 0, 𝑛 ≥ 0 .

Solution.

𝜆2 − 5𝜆 + 6 = 0

𝑎𝑛 = 𝐴2𝑛 + 𝐵3𝑛, 𝑛 ≥ 0 ,

Examples

Example

𝑎𝑛+2 − 5𝑎𝑛+1 + 6𝑎𝑛 = 0, 𝑛 ≥ 0 .

Solution.

𝜆2 − 5𝜆 + 6 = 0

𝑎𝑛 = 𝐴2𝑛 + 𝐵3𝑛, 𝑛 ≥ 0 ,

Examples

Example

Find a particular solution of the previous recurrence relation

𝑎𝑛+2 − 5𝑎𝑛+1 + 6𝑎𝑛 = 0, 𝑛 ≥ 0

such that it satisfies the initial conditions 𝑎0 = 2 and 𝑎1 = 3.

Solution.

From the general solution obtained in the previous example, the initialconditions give rise to the following two equations

𝑎0 = 𝐴 + 𝐵 = 2 , 𝑎1 = 𝐴× 2 + 𝐵 × 3 = 3

Solving the above 2 equations we obtain 𝐴 = 3 and 𝐵 = −1.Hence the particular solution satisfying the initial conditions is

𝑎𝑛 = 3 × 2𝑛 − 3𝑛, 𝑛 ≥ 0

Examples

Example

𝑎𝑛+2 − 5𝑎𝑛+1 + 6𝑎𝑛 = 0, 𝑛 ≥ 0

Solution.

𝑎0 = 𝐴 + 𝐵 = 2 , 𝑎1 = 𝐴× 2 + 𝐵 × 3 = 3

Solving the above 2 equations we obtain 𝐴 = 3 and 𝐵 = −1.

Hence the particular solution satisfying the initial conditions is

𝑎𝑛 = 3 × 2𝑛 − 3𝑛, 𝑛 ≥ 0

Examples

Example

𝑎𝑛+2 − 5𝑎𝑛+1 + 6𝑎𝑛 = 0, 𝑛 ≥ 0

Solution.

𝑎0 = 𝐴 + 𝐵 = 2 , 𝑎1 = 𝐴× 2 + 𝐵 × 3 = 3

Solving the above 2 equations we obtain 𝐴 = 3 and 𝐵 = −1.Hence the particular solution satisfying the initial conditions is

𝑎𝑛 = 3 × 2𝑛 − 3𝑛, 𝑛 ≥ 0

Examples

Example

Find the general solution for the recurrence relation

𝑤𝑛+2 − 4𝑤𝑛+1 + 4𝑤𝑛 = 2𝑛+2 , 𝑛 ≥ 0

Solution.

The associated characteristic equation 𝜆2 − 4𝜆 + 4 = (𝜆− 2)2 = 0 has adouble root 𝜆1 = 2.Hence the general solution of the corresponding homogeneous problem is𝑢𝑛 = 2𝑛(𝐴 + 𝐵𝑛) for 𝑛 ≥ 0, where 𝐴 and 𝐵 are arbitrary constants.Since the nonhomogeneous term is 2𝑛+2 = 4 × 2𝑛, a particular solution 𝑣𝑛will take the form 𝑣𝑛 = 𝐶𝑛22𝑛 because 2 is a double root of theassociated characteristic equation.Hence, substituting 𝑣𝑛 into the nonhomogeneous recurrence relation𝑣𝑛+2 − 4𝑣𝑛+1 + 4𝑣𝑛 = 2𝑛+2, we obtain

𝐶[︁4(𝑛 + 2)2 − 4(𝑛 + 1)2 × 2 + 4𝑛2

]︁= 4

which then simplifies to just 2𝐶 = 1.

Examples

Example

𝑤𝑛+2 − 4𝑤𝑛+1 + 4𝑤𝑛 = 2𝑛+2 , 𝑛 ≥ 0

Solution.

The associated characteristic equation 𝜆2 − 4𝜆 + 4 = (𝜆− 2)2 = 0 has adouble root 𝜆1 = 2.

Hence the general solution of the corresponding homogeneous problem is𝑢𝑛 = 2𝑛(𝐴 + 𝐵𝑛) for 𝑛 ≥ 0, where 𝐴 and 𝐵 are arbitrary constants.Since the nonhomogeneous term is 2𝑛+2 = 4 × 2𝑛, a particular solution 𝑣𝑛will take the form 𝑣𝑛 = 𝐶𝑛22𝑛 because 2 is a double root of theassociated characteristic equation.Hence, substituting 𝑣𝑛 into the nonhomogeneous recurrence relation𝑣𝑛+2 − 4𝑣𝑛+1 + 4𝑣𝑛 = 2𝑛+2, we obtain

𝐶[︁4(𝑛 + 2)2 − 4(𝑛 + 1)2 × 2 + 4𝑛2

]︁= 4

Examples

Example

𝑤𝑛+2 − 4𝑤𝑛+1 + 4𝑤𝑛 = 2𝑛+2 , 𝑛 ≥ 0

Solution.

The associated characteristic equation 𝜆2 − 4𝜆 + 4 = (𝜆− 2)2 = 0 has adouble root 𝜆1 = 2.Hence the general solution of the corresponding homogeneous problem is𝑢𝑛 = 2𝑛(𝐴 + 𝐵𝑛) for 𝑛 ≥ 0, where 𝐴 and 𝐵 are arbitrary constants.

Since the nonhomogeneous term is 2𝑛+2 = 4 × 2𝑛, a particular solution 𝑣𝑛will take the form 𝑣𝑛 = 𝐶𝑛22𝑛 because 2 is a double root of theassociated characteristic equation.Hence, substituting 𝑣𝑛 into the nonhomogeneous recurrence relation𝑣𝑛+2 − 4𝑣𝑛+1 + 4𝑣𝑛 = 2𝑛+2, we obtain

𝐶[︁4(𝑛 + 2)2 − 4(𝑛 + 1)2 × 2 + 4𝑛2

]︁= 4

Examples

Example

𝑤𝑛+2 − 4𝑤𝑛+1 + 4𝑤𝑛 = 2𝑛+2 , 𝑛 ≥ 0

Solution.

The associated characteristic equation 𝜆2 − 4𝜆 + 4 = (𝜆− 2)2 = 0 has adouble root 𝜆1 = 2.Hence the general solution of the corresponding homogeneous problem is𝑢𝑛 = 2𝑛(𝐴 + 𝐵𝑛) for 𝑛 ≥ 0, where 𝐴 and 𝐵 are arbitrary constants.Since the nonhomogeneous term is 2𝑛+2 = 4 × 2𝑛, a particular solution 𝑣𝑛will take the form 𝑣𝑛 = 𝐶𝑛22𝑛 because 2 is a double root of theassociated characteristic equation.

Hence, substituting 𝑣𝑛 into the nonhomogeneous recurrence relation𝑣𝑛+2 − 4𝑣𝑛+1 + 4𝑣𝑛 = 2𝑛+2, we obtain

𝐶[︁4(𝑛 + 2)2 − 4(𝑛 + 1)2 × 2 + 4𝑛2

]︁= 4

Examples

Example

𝑤𝑛+2 − 4𝑤𝑛+1 + 4𝑤𝑛 = 2𝑛+2 , 𝑛 ≥ 0

Solution.

The associated characteristic equation 𝜆2 − 4𝜆 + 4 = (𝜆− 2)2 = 0 has adouble root 𝜆1 = 2.Hence the general solution of the corresponding homogeneous problem is𝑢𝑛 = 2𝑛(𝐴 + 𝐵𝑛) for 𝑛 ≥ 0, where 𝐴 and 𝐵 are arbitrary constants.Since the nonhomogeneous term is 2𝑛+2 = 4 × 2𝑛, a particular solution 𝑣𝑛will take the form 𝑣𝑛 = 𝐶𝑛22𝑛 because 2 is a double root of theassociated characteristic equation.Hence, substituting 𝑣𝑛 into the nonhomogeneous recurrence relation𝑣𝑛+2 − 4𝑣𝑛+1 + 4𝑣𝑛 = 2𝑛+2, we obtain

𝐶[︁4(𝑛 + 2)2 − 4(𝑛 + 1)2 × 2 + 4𝑛2

]︁= 4

Examples

Hence we have 𝐶 = 12 and 𝑣𝑛 = 𝑛22𝑛−1.

The general solution 𝑤𝑛 of the nonhomogeneous recurrence relation isthus 𝑤𝑛 = 𝑢𝑛 + 𝑣𝑛, and hence takes the form

𝑤𝑛 = 2𝑛(𝐴 + 𝐵𝑛) + 𝑛2 · 2𝑛−1

= 2𝑛[︁𝐴 + 𝐵𝑛 +

]︁, 𝑛 ≥ 0 .

Examples

Hence we have 𝐶 = 12 and 𝑣𝑛 = 𝑛22𝑛−1.

The general solution 𝑤𝑛 of the nonhomogeneous recurrence relation isthus 𝑤𝑛 = 𝑢𝑛 + 𝑣𝑛, and hence takes the form

𝑤𝑛 = 2𝑛(𝐴 + 𝐵𝑛) + 𝑛2 · 2𝑛−1

= 2𝑛[︁𝐴 + 𝐵𝑛 +

]︁, 𝑛 ≥ 0 .

summary of solving linear, constant-coefficient recurrence

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