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where we used the expression for the sum of the square of integers (see
Appendix).
The general solution is then:

yk ( + 1 ) = (2 k + 1 k ) + C
6 k ( + 1 )
From the initial condition y(1) = 1, we deduce that: C = 1.

In-Class Exercise

Pb. 2.19 Find the general solutions for the following difference equations,
assuming that y(1) = 1.
k
a. y(k + 1) – 3ky(k) = 3 .
b. y(k + 1) – ky(k) = k.

2.7 Nonlinear Difference Equations

In this and the following chapter section, we explore a number of nonlinear
difference equations that exhibit some general features typical of certain
classes of solutions and observe other instances with novel qualitative fea-
tures. Our exploration is purely experimental, in the sense that we restrict
our treatment to guided computer runs. The underlying theories of most of
the models presented are the subject of more advanced courses; however,
many educators, including this author, believe that there is virtue in expos-
ing students qualitatively early on to these fascinating and generally new
developments in mathematics.
2.7.1 Computing Irrational Numbers

In this model, we want to exhibit an example of a nonlinear difference equa-
tion whose solution is a sequence that approaches a specific limit, irrespec-
tive, within reasonable constraints, of the initial condition imposed on it. This
type of difference equation has been used to compute a class of irrational
numbers. For example, a well-defined approximation for computing A is
the feedback process:

yk( + 1 ) = 1   yk( ) + A   (2.38)
2  yk() 

This equation’s main features are explored in the following exercise.

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