Understanding Linear Recursive Sequences and Their General Terms

Linear recursive sequences are fundamental in mathematics, with applications ranging from computer science to finance. In this article, we explore how to derive the general term for a specific linear recursive sequence. By following a step-by-step approach, we will uncover the underlying structure and derive a concise formula for the xn.

Step-by-Step Derivation of the General Term

The given sequence is defined by the recurrence relation: x0 c xn 1 axn b

Step 1: Compute the First Few Terms

Let's compute the first few terms of the sequence to identify any patterns:

x0 c x1 a x0 b ac b x2 a x1 b a(ac b) b a2c ab b x3 a x2 b a(a2c ab b) b a3c a2b ab b

Step 2: Identify the Pattern

From the computed terms, we can start to recognize a pattern. The term involving b seems to form a geometric series:

For x1 ac b For x2 a2c ab b a2c (a 1)b For x3 a3c a2b ab b a3c (a2 a 1)b

This suggests that each term can be expressed as a combination of anc and a series involving powers of a.

Step 3: Generalize the Pattern

The general term of the sequence can be expressed as:

xn anc b sum_{k0}^{n-1} ak

Step 4: Simplify the Summation

The summation sum_{k0}^{n-1} ak is a geometric series. For a ≠ 1, this series can be simplified as:

sum_{k0}^{n-1} ak frac{1 - an}{1 - a}

Step 5: Write the General Term

Substituting the summation back into the expression for xn gives:

xn anc b cdot frac{1 - an}{1 - a}

Final General Term

Thus, the general term for the series is:

xn anc frac{b(1 - an)}{1 - a}quad text{for } a ≠ 1

When a 1, the recurrence simplifies to:

xn c nb

Connection to Linear Equations and Vector Spaces

Consider a linear equation Ax r in some vector space V, whether finite or infinite dimensional. The process involves the following steps:

Solve the homogeneous equation Az 0 to find all z satisfying it. Guess an arbitrary solution y satisfying Ay r. Form the general solution as the sum of the solutions from steps 1 and 2. Choose parameters in step 1 to satisfy the constraints.

In this context, V is the sequence space of real or complex numbers, A is the linear map given by xn → xn 1 - axn, and r b is the constant sequence.

Homogeneous Solution

The homogeneous equation is:

xn 1 axn

Its solution is given by:

xn Can for some C ∈ mathbb{R}.

Arbitrary Solution

Guess an arbitrary solution yn satisfying yn 1 ayn b:

If yn d for all n ∈ mathbb{N}_0, then d ad b. Thus, d frac{b}{1-a} For a 1, a different form of solution is yn bn.

General Solution

The general solution is given by:

xn begin{cases} C an frac{b(1 - an)}{1 - a} quad text{if } a ≠ 1 quad C nb quad text{if } a 1 end{cases}

Choosing Parameters

Select parameters C such that the initial condition is satisfied:

For a ≠ 1, C frac{b}{1-a} c implies C frac{c(1-a)}{b}. For a 1, C c.

Substitute these values back into the general solution to obtain the final result.