Backtrack to find an explicit formula for the sequence defined by the recurrence

relation b_n = 2b_{n – 1} + 1 with initial condition b₁ = 7.

Solution. We begin by substituting the definition of the previous term in the defining formula.

b_n = 2b_{n – 1} + 1 = 2 (2b_{n – 2} + 1) + 1 = 2[2 (2b_{n – 3} + 1) + 1] + 1 = 23 _b n – 3 + 4 + 2 + 1 = 23 _b n – 3 + 22 + 21 + 1.

A pattern is emerging with these rewritting of b_n.

(Note : There are no set rules for how to rewrite these expressions and a certain amount of experimentation may be necessary.)

The backtracking will end at b_n = 2n – 1_b n – (n – 1)+ 2n – 2 + 2n – 3 + ... + 22 + 21 + 1 = 2n – 1_b 1 + 2n – 1 – 1 = 7 . 2n – 1 _{+ 2}n – 1 _{– 1 (using b} 1 = 7) = 8 . 2n – 1 – 1 or 2n + 2 – 1.

Problem 2.13. Write down the first six terms of the sequence defined by a₁ = 1, a_{k + 1} = 3a_k + 1 for k ≥ 1. Guess a formula for a_n and prove that your formula is correct.

Solution. The first six terms are

a₁ = 1

a₂ = 3a₁ + 1 = 3(1) + 1 = 4 a₃ = 3a₂ + 1 = 3(4) + 1 = 13 a₄ = 40, a₅ = 121, a₆ = 364.

Since there is multiplication by 3 at each step, we might suspect that 3n _{is involved in the answer.}

After trial and error, we guess that a_n = 1

2 (3

n _{– 1) and verify this by mathematical induction.}

When n = 1, the formula gives 1

2 (31 – 1) = 1, which is indeed a1, the first term in the sequence.

Now assume that k ≥ 1 and that ak = 1

2 (3k – 1).

We wish to prove that a_{k + 1} = 1₂ (3k + 1 _{– 1)}

We have a_{k + 1} = 3a_k + 1 = 31

2 (3k – 1) + 1

Using the induction hypothesis.

Hence, a_{k + 1} = 1 2 3 k + 1 _– 3 2 + 1 = 1 2(3 k + 1 _{– 1) as required.} By the principle of mathematical induction, our guess is correct.

Problem 2.14. A sequence is defined recursively by a₀ = 1, a₁ = 4 and a_n = 4a_{n – 1} – 4a_{n – 2} for n ≥ 2. Find the first six terms of this sequence. Guess a formula for a_n and establish the validity of your guess.

Solution. Here there are two initial conditions a₀ = 1, a₁ = 4.

Also, the recurrence relation, a_n = 4a_{n – 1} – 4a_{n – 2}, defines the general term as a function of two previous terms.

The first six terms of the sequence are a₀ = 1 a₁ = 4 a₂ = 4a₁ – 4a₀ = 4(4) – 4(1) = 12 a₃ = 4a₂ – 4a₁ = 4(12) – 4(4) = 32 a₄ = 4a₃– 4a₂ = 4(32) – 4(12) = 80 a₅ = 4a₄ – 4a₃ = 4(80) – 4(32) = 192. Finding a general formula for a_n requires some ingenuity. Let us examine some of the first six terms.

We note that a₃ = 32 = 24 + 8 = 3(8) + 4(8)

and a₄ = 80 = 64 + 16 = 4(16) + 16 = 5(16)

and a₅ = 192 = 6(32).

We guess that a_n = (n + 1) 2n_.

To prove this, we use the strong form of mathematical induction (with n₀ = 0). When n = 0, we have (0 + 1) 20 _{= 1(1) = 1, in agreement with the given value for a}

0.

When n = 1, (1 + 1) 21 _{= 4 = a}

1. Now that the formula has been verified for k = 0 and k = 1.

We may assume that k > 1 and that a_n = (n + 1) 2n _{for all n in the interval 0 ≤ n < k.} We wish to prove the formula is valid for n = k.

That is, we wish to prove that a_k = (k + 1) 2k_. Since k ≥ 2, we know that ak = 4ak – 1 – 4ak – 2.

Applying the induction hypothesis to k – 1 and to k – 2 (each of which is in the range 0 ≤ n < k). We have a_{k – 1} = k 2k – 1 _{and a}

k – 2 = (k – 1) 2k – 2. Thus, a_k = 4(k 2k – 1_{) – 4(k – 1) 2}k – 2

= 2k 2k _{– k 2}k _{+ 2}k

= k 2k _{+ 2}k _{= (k + 1) 2}k _{as required.}

By principle of mathematical induction, the formula is valid for all n ≥ 0.

Problem 2.15. Solve the recurrence relation a_n = a_{n – 1} + 2, n ≥ 2 subject to initial condition a₁ = 3.

Solution. We backtrack the value of a_n by substituting the expression of a_{n – 1}, a_{n – 2} and so on, until a pattern is clear.

Given a_n = a_{n – 1} + 2 ...(1) Replacing n by n – 1 in (1), we obtain a_{n – 1} = a_{n – 2} + 2 From (1), a_n = a_{n – 1} + 2 = (a_{n – 2} + 2) + 2 = a_{n – 2} + 2.2 ...(2) Replacing n by n – 2 in (1), we obtain a_{n – 2} = a_{n – 3} + 2 So, from (2), a_n = (a_{n – 3} + 2) + 2.2 = a_{n – 3} + 3.2 In general a_n = a_{n – k} + k . 2 For k = n – 1, a_n = a_{n – (n – 1)} + (n – 1) . 2 = a₁ + (n – 1) . 2 = 3 + (n – 1) . 2 which is an explicit formula.

2.2. THE SECOND-ORDER LINEAR Homogeneous Recurrence Relation with constant coefficients

Let k ∈ Z+ _{and C}

n (≠ 0), Cn – 1, Cn – 2, ... Cn – k (≠ 0) be real numbers. If an, for n ≥ 0, is a discrete function, then

is a linear recurrence relation (with constant coefficients) of order k. When f(n) = 0, for all n ≥ 0, the relation is called homogeneous ; other wise, it is non-homogeneous.

The homogeneous relation of order two :

C_na_n + C_{n – 1}a_{n – 1} + C_{n – 2}a_{n – 2} = 0, n ≥ 2.

A solution of the form a_n = Crn_{, where C ≠ 0 and r ≠ 0 substituting a}

n = Crn into C_na_n + C_{n – 1}a_{n – 1} + C_{n – 2}a_{n – 2} = 0.

We obtain C_nCrn _{+ C}

n – 1Crn – 1+ Cn – 2Crn – 2 = 0, with C, r _{≠ 0, this becomes}

C_nr2 _{+ C}

n – 1r + Cn – 2 = 0,

a quadratic equation which is called the characteristic equation.

2.3. THE NON HOMOGENEOUS RECURRENCE RELATIONS

The recurrence relations

a_n + C_{n – 1}a_{n – 1} = f(n), n _{≥ 1} ...(1)

a_n + C_{n – 1}a_{n – 1} + C_{n – 2}a_{n – 2} = f(n), n ≥ 2 ...(2) where C_{n – 1} and C_{n – 2} are constants, C_{n – 1≠ 0, in (1), Cn – 2≠ 0, and f(n) is not identically 0.}

Although there is no general method for solving all non homogeneous relations, for certain func- tions f(n) we shall find a successful technique.

When C_{n – 1} = – 1, (1) gives, for the non homogeneous relation a_n – a_{n – 1} = f(n), we have a₁ = a₀ + f(1) a₂ = a₁ + f(2) = a₀ + f(1) + f(2) a₃ = a₂ + f(3) = a₀ + f(1) + f(2) + f(3) --- a_n = a₀ + f(1) + ... + f(n) = a₀ + 1 ( ) n i f i =

∑

We can solve this type of relation in terms of n, if we find a suitable summation formula for

1 ( ) n i f i =

∑

(a) The non homogeneous first-order relation a_n + C_{n – 1}a_{n – 1} = krn

where k is a constant and n _{∈ Z}+_.

(b) If rn _{is not a solution of the associated homogeneous relation a}

n + Cn – 1an – 1 = 0, then an(P) = Arn_{, where A is a constant. When r}n _{is a solution of the associated homogeneous relation, then}

a_n(P) = Bn rn, for B a constant.

(c) The non-homogeneous second order relation a_n + C_{n – 1}a_{n – 1} + C_{n – 2}a_{n – 2} = k rn_, where k is a constant. Here,

(i) a_n(p) _{= A r}n_{, for A a constant, if r}n _{is not a solution of the associated homogeneous relation,} (ii) a_n(p) _{= Bn r}n_{, where B is a constant, if a}

n(h) = C1rn + C2r1n where r1≠ r,

(iii) a_n(p) _{= Cn}2_rn_{, for C a constant, when a}

n(h) = (C1 + C2n) rn. 2.3.1. Characteristic Equation Method :

This method can be used to solve any constant order linear recurrence equation with constant coefficient. This recurrence relation may be homogeneous or non-homogeneous. Before attempting to solve any such problem, let us first, understand what is characteristic equation for a given recurrence equation and how to find it.

A recurrence equation of the mentioned type can be arrranged in standard form as :

A_n + C₁A_{n – 1} + C₂A_{n – 2} + C₃ A_{n – 3} = R.H.S. ...(1) Where C₁, C₂, C₃ are constant coefficients and R.H.S. has one of the following forms :

Form Examples

Homogeneous 0

A constant to the nth power 2n_{, π}– s_{, 2}– n_,

2n A polynomial in n 3, n2_{, n}2 _{– n, n}3 _{+ 2n – 1}

A product of a constant to the 2n _(n2 _{+ 2n – 1), (n – 1) n}6_{, n 6}n nth power and a polynomial in n

A linear combination of any of the above (2n _{+ 3}n/2_{) (n}2 _{+ 2n – 1) + 5}

In the recurrence equation (1), assigning R.H.S. = 0, we get

A_n + C₁A_{n – 1} + C₂A_{n – 2} + C₃A_{n – 3} = 0 ...(2)

This equation (2) gives the homogeneous part of the given recurrence equation. Every recurrence equation has a homogeneous part. If the recurrence relation is homogeneous then it has only homogene- ous part and solving such equation is one step process. On the other hand, if the given recurrence equation is non-homogeneous then its homogeneous part is obtained by assigning R.H.S. equal to zero.

A characteristic equation corresponds to homogeneous part of the given recurrence relation.

The characteristic equation of (2) is given as :

x3 _{+ C}

1x2 + C2x + C3 = 0 ...(3)

This has been obtained by the following procedure : (i) Find the order of the recurrence equation here it is 3. (ii) Take any variable (say x) and substitute

A_n, A_{n – 1}, A_{n – 2}, by x3_{, x}2_{, x respectively in the homogeneous part of the recurrence equation.}

Equation so obtained is called characteristic equation of the given recurrence equation.

Example (1) : The characteristic equation of the recurrence equation C_n = 3C_{n – 1} – 2C_{n – 2} is given by

x2 _{– 3x + 2 = 0.}

Example (2) : The characteristic equation of the recurrence equation f_n = f_{n – 1} + f_{n – 2} is given by x2 _{– x – 1 = 0.}

Example (3) : The characteristic equation of the recurrence equation

A_n – 5A_{n – 1} + 6A_{n – 2} = 2n _{+ n} is given by x2 _{– 5x + 6 = 0.}