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Math 3336 Recurrence Relations

Definition: A recurrence relation for the sequence {𝑎𝑎𝑛𝑛} is an equation that expresses 𝑎𝑎𝑛𝑛 in terms of one or more of the previous terms of the sequence, namely, 𝑎𝑎0, 𝑎𝑎1, … ,𝑎𝑎𝑛𝑛−1, for all integers 𝑛𝑛 with 𝑛𝑛 ≥ 𝑛𝑛0, where 𝑛𝑛0 is a nonnegative integer.

• A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation.

• The initial conditions for a sequence specify the terms that precede the first term where the recurrence relation takes effect.

Rabbits and the Fibonacci Numbers

Example: A young pair of rabbits (one of each gender) is placed on an island. A pair of rabbits does not breed until they are 2 months old. After they are 2 months old, each pair of rabbits produces another pair each month. Find a recurrence relation for the number of pairs of rabbits on the island after n months, assuming that rabbits never die.

This is the original problem considered by Leonardo Pisano (Fibonacci) in the thirteenth century.

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Solution:

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The Tower of Hanoi

In the late nineteenth century, the French mathematician EÉ douard Lucas invented a puzzle consisting of three pegs on a board with disks of different sizes. Initially all of the disks are on the first peg in order of size, with the largest on the bottom.

Rules: You are allowed to move the disks one at a time from one peg to another as long as a larger disk is never placed on a smaller. oal: Using allowable moves, end up with all the disks on the second peg in order of size with largest on the bottom. Solution:

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• There was a myth created with the puzzle. Monks in a tower in Hanoi are transferring 64 gold disks from one peg to another following the rules of the puzzle. They move one disk each day. When the puzzle is finished, the world will end.

• Using this formula for the 64 gold disks of the myth,

264 −1 = 18,446, 744,073, 709,551,615 days are needed to solve the puzzle, which is more than 500 billion years.

• Reve’s puzzle (proposed in 1907 by Henry Dudeney) is similar but has 4 pegs. There is a well-known unsettled conjecture for the minimum number of moves needed to solve this puzzle.

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Counting Bit Strings Example: Find a recurrence relation and give initial conditions for the number of bit strings of length n without two consecutive 0s. How many such bit strings are there of length five? Solution:

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Linear Homogeneous Recurrence Relations Definition: A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a

n = c

1a

n−1 + c

2a

n−2 + ….. + c

k a

n−k , where c

1, c

2, ….,c

k are

real numbers, and ck ≠ 0

Examples of Recurrence Relations 1. Pn = (1.11)Pn−1

2. 𝑓𝑓𝑛𝑛 = 𝑓𝑓𝑛𝑛−1 + 𝑓𝑓𝑛𝑛−2

3. 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 𝑎𝑎𝑛𝑛−22

4. 𝐻𝐻𝑛𝑛 = 2𝐻𝐻𝑛𝑛−1 + 1

5. 𝑏𝑏𝑛𝑛 = 𝑛𝑛𝑏𝑏𝑛𝑛−1

6. 𝑐𝑐𝑛𝑛 = 3𝑐𝑐𝑛𝑛−1 + 4𝑐𝑐𝑛𝑛−2 + 5𝑐𝑐𝑛𝑛−3

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Solving Linear Homogeneous Recurrence Relations Solving Linear Homogeneous Recurrence Relations of Degree Two

Theorem 1: Let 𝑐𝑐1 and 𝑐𝑐2 be real numbers. Suppose that 𝑟𝑟2 − 𝑐𝑐1𝑟𝑟 − 𝑐𝑐2 = 0 has two distinct roots 𝑟𝑟1 and 𝑟𝑟2. Then the sequence {𝑎𝑎𝑛𝑛} is a solution to the recurrence relation 𝑎𝑎𝑛𝑛 =𝑐𝑐1𝑎𝑎𝑛𝑛−1 + 𝑐𝑐2𝑎𝑎𝑛𝑛−2 if and only if 𝑎𝑎𝑛𝑛 = 𝛼𝛼1𝑟𝑟1𝑛𝑛 + 𝛼𝛼2𝑟𝑟2𝑛𝑛 for 𝑛𝑛 = 0,1,2, … , where 𝛼𝛼1 and 𝛼𝛼2 are constants. Example: What is the solution of the recurrence relation 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 2𝑎𝑎𝑛𝑛−2? Step 1: Write a characteristic equation of a recurrence relation (CERR). If a𝑛𝑛 = 𝑐𝑐1𝑎𝑎𝑛𝑛−1 + 𝑐𝑐2𝑎𝑎𝑛𝑛−2 + ⋯+ 𝑐𝑐𝑘𝑘𝑎𝑎𝑛𝑛−𝑘𝑘, then

𝑟𝑟𝑘𝑘 − 𝑐𝑐1𝑟𝑟𝑘𝑘−1 − 𝑐𝑐2𝑟𝑟𝑘𝑘−2 − ⋯− 𝑐𝑐𝑘𝑘−1𝑟𝑟 − 𝑐𝑐𝑘𝑘 = 0 is the characteristic equation of 𝑎𝑎𝑛𝑛. Step 2: Solve CERR Step 3: Write solution in terms of 𝛼𝛼s. Step 4: Find 𝛼𝛼1 and 𝛼𝛼2 from initial conditions.

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Step 5: Write the solution. Example: What is the solution of the recurrence relation 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 + 6𝑎𝑎𝑛𝑛−2? Step 1: Write a characteristic equation of a recurrence relation (CERR). If a𝑛𝑛 = 𝑐𝑐1𝑎𝑎𝑛𝑛−1 + 𝑐𝑐2𝑎𝑎𝑛𝑛−2 + ⋯+ 𝑐𝑐𝑘𝑘𝑎𝑎𝑛𝑛−𝑘𝑘, then

𝑟𝑟𝑘𝑘 − 𝑐𝑐1𝑟𝑟𝑘𝑘−1 − 𝑐𝑐2𝑟𝑟𝑘𝑘−2 − ⋯− 𝑐𝑐𝑘𝑘−1𝑟𝑟 − 𝑐𝑐𝑘𝑘 = 0 is the characteristic equation of 𝑎𝑎𝑛𝑛. Step 2: Solve CERR Step 3: Write solution in terms of 𝛼𝛼s. Step 4: Find 𝛼𝛼1 and 𝛼𝛼2 from initial conditions.

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Step 5: Write the solution Example: Find an explicit formula for the Fibonacci numbers. Step 1: Write a characteristic equation of a recurrence relation (CERR). If a𝑛𝑛 = 𝑐𝑐1𝑎𝑎𝑛𝑛−1 + 𝑐𝑐2𝑎𝑎𝑛𝑛−2 + ⋯+ 𝑐𝑐𝑘𝑘𝑎𝑎𝑛𝑛−𝑘𝑘, then

𝑟𝑟𝑘𝑘 − 𝑐𝑐1𝑟𝑟𝑘𝑘−1 − 𝑐𝑐2𝑟𝑟𝑘𝑘−2 − ⋯− 𝑐𝑐𝑘𝑘−1𝑟𝑟 − 𝑐𝑐𝑘𝑘 = 0 is the characteristic equation of 𝑎𝑎𝑛𝑛. Step 2: Solve CERR Step 3: Write solution in terms of 𝛼𝛼s. Step 4: Find 𝛼𝛼1 and 𝛼𝛼2 from initial conditions.

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Step 5: Write the solution

Theorem 2: Let 𝑐𝑐1 and 𝑐𝑐2 be real numbers with 𝑐𝑐2 ≠ 0 . Suppose that 𝑟𝑟2 − 𝑐𝑐1𝑟𝑟 − 𝑐𝑐2 = 0 has only one root 𝑟𝑟0. A sequence {𝑎𝑎𝑛𝑛} is a solution of the recurrence relation 𝑎𝑎𝑛𝑛 = 𝑐𝑐1𝑎𝑎𝑛𝑛−1 +𝑐𝑐2𝑎𝑎𝑛𝑛−2 if and only if 𝑎𝑎𝑛𝑛 = 𝛼𝛼1𝑟𝑟0𝑛𝑛 + 𝛼𝛼2𝑛𝑛𝑟𝑟0𝑛𝑛 for 𝑛𝑛 = 0,1,2, … , where 𝛼𝛼1 and 𝛼𝛼2 are constants.

Example: What is the solution of the recurrence relation 𝑎𝑎𝑛𝑛 = 6𝑎𝑎𝑛𝑛−1 − 9𝑎𝑎𝑛𝑛−2 with initial conditions 𝑎𝑎0 = 1 and 𝑎𝑎1 = 6? Step 1: Write a characteristic equation of a recurrence relation (CERR).

𝑟𝑟𝑘𝑘 − 𝑐𝑐1𝑟𝑟𝑘𝑘−1 − 𝑐𝑐2𝑟𝑟𝑘𝑘−2 − ⋯− 𝑐𝑐𝑘𝑘−1𝑟𝑟 − 𝑐𝑐𝑘𝑘 = 0 Step 2: Solve CERR Step 3: Write solution in terms of 𝛼𝛼s. Step 4: Find 𝛼𝛼1 and 𝛼𝛼2 from initial conditions.

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Step 5: Write the solution Theorem 3: Let 𝑐𝑐1, 𝑐𝑐2, … , 𝑐𝑐𝑘𝑘 be real numbers. Suppose that the characteristic equation

𝑟𝑟𝑘𝑘 − 𝑐𝑐1𝑟𝑟𝑘𝑘−1 − 𝑐𝑐2𝑟𝑟𝑘𝑘−2 − ⋯− 𝑐𝑐𝑘𝑘−1𝑟𝑟 − 𝑐𝑐𝑘𝑘 = 0

Has k distinct roots 𝑟𝑟1, 𝑟𝑟2, … , 𝑟𝑟𝑘𝑘. Then a sequence {𝑎𝑎𝑛𝑛} is a solution of the recurrence relation 𝑎𝑎𝑛𝑛 = 𝑐𝑐1𝑎𝑎𝑛𝑛−1 + 𝑐𝑐2𝑎𝑎𝑛𝑛−2 + ⋯𝑐𝑐𝑘𝑘𝑎𝑎𝑛𝑛−𝑘𝑘 if and only if 𝑎𝑎𝑛𝑛 = 𝛼𝛼1𝑟𝑟0𝑛𝑛 + 𝛼𝛼2𝑟𝑟0𝑛𝑛 + ⋯+ 𝑎𝑎𝑘𝑘𝑟𝑟𝑘𝑘𝑛𝑛 for 𝑛𝑛 = 0,1,2, … , where 𝛼𝛼1,𝛼𝛼2, … ,𝛼𝛼𝑘𝑘 are constants.

Example: Find the solution to the recurrence relation 𝑎𝑎𝑛𝑛 = 6𝑎𝑎𝑛𝑛−1 − 11𝑎𝑎𝑛𝑛−2 + 6𝑎𝑎𝑛𝑛−3. Step 1: Write a characteristic equation of a recurrence relation (CERR).

𝑟𝑟𝑘𝑘 − 𝑐𝑐1𝑟𝑟𝑘𝑘−1 − 𝑐𝑐2𝑟𝑟𝑘𝑘−2 − ⋯− 𝑐𝑐𝑘𝑘−1𝑟𝑟 − 𝑐𝑐𝑘𝑘 = 0 Step 2: Solve CERR Step 3: Write solution in terms of 𝛼𝛼s. Step 4: Find 𝛼𝛼1 and 𝛼𝛼2 from initial conditions.

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Step 5: Write the solution Theorem 4: Let c1, c2 ,…, ck be real numbers. Suppose that the characteristic equation

rk – c1r

k−1 –⋯ – ck = 0

has t distinct roots r1, r2, …, rt with multiplicities m1, m2, …, mt, respectively so that mi ≥ 1 for i = 0, 1, 2, …,t and m1 + m2 + … + mt = k. Then a sequence {an} is a solution of the recurrence relation

an = c

1a

n−1 + c

2a

n−2 + ….. + c

k a

n−k

if and only if

for n = 0, 1, 2, …, where α

i,j are constants for 1≤ i ≤ t and 0≤ j ≤ m

i−1.

Example: Suppose that the roots of the characteristic equation of a linear homogeneous recurrence relation are 2, 2, 2, 5, 5, and 9. What is the form of the general solution?

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Example: Find the solution to the recurrence relation 𝑎𝑎𝑛𝑛 = −3𝑎𝑎𝑛𝑛−1 − 3𝑎𝑎𝑛𝑛−2 − 𝑎𝑎𝑛𝑛−3 with initial conditions 𝑎𝑎0 = 1, 𝑎𝑎1 = −2, and 𝑎𝑎2 = −1. Step 1: Write a characteristic equation of a recurrence relation (CERR).

𝑟𝑟𝑘𝑘 − 𝑐𝑐1𝑟𝑟𝑘𝑘−1 − 𝑐𝑐2𝑟𝑟𝑘𝑘−2 − ⋯− 𝑐𝑐𝑘𝑘−1𝑟𝑟 − 𝑐𝑐𝑘𝑘 = 0 Step 2: Solve CERR Step 3: Write solution in terms of 𝛼𝛼s. Step 4: Find 𝛼𝛼𝑠𝑠 from initial conditions.

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Step 5: Write the solution

Linear Nonhomogeneous Recurrence Relations with Constant Coefficients. Definition: A linear nonhomogeneous recurrence relation with constant coefficients is a

recurrence relation of the form: a

n = c

1a

n−1 + c

2a

n−2 + ….. + c

k a

n−k + F(n)

,

where c1, c

2, ….,c

k are real numbers, and F(n) is a function not identically zero depending

only on n. The recurrence relation a

n = c

1a

n−1 + c

2a

n−2 + ….. + c

k a

n−k ,

is called the associated homogeneous recurrence relation.

Example: The following are linear nonhomogeneous recurrence relations with constant coefficients:

an = a

n−1 + 2

n

,

an = a

n−1 + a

n−2 + n

2 + n + 1,

an = 3a

n−1 + n3

n ,

an = a

n−1 + a

n−2 + a

n−3 + n!

Theorem 5: If {an

(p)} is a particular solution of the nonhomogeneous linear recurrence

relation with constant coefficients

an = c

1a

n−1 + c

2a

n−2 + ⋯ + c

k a

n−k + F(n)

,

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then every solution is of the form {an

(p) + a

n

(h)}, where {a

n

(h)} is a solution of the associated

homogeneous recurrence relation

an = c

1a

n−1 + c

2a

n−2 + ⋯ + c

k a

n−k .

Example: Find all solutions of the recurrence relation an = 3an−1 + 2n.

What is the solution with a1 = 3?