Apply the method of Exercise 9.1.21 to solve the following...

Apply the method of Exercise 9.1.21 to solve the following iterative equations: (a) $u^{(k+2)}=-u^{(k+1)}+2 u^{(k)}, \quad u^{(0)}=1, \quad u^{(1)}=2$. (b) $12 u^{(k+2)}=u^{(k+1)}+u^{(k)}, \quad u^{(0)}=-1, \quad u^{(1)}=2$. (c) $u^{(k+2)}=4 u^{(k+1)}+u^{(k)}, \quad u^{(0)}=1, \quad u^{(1)}=-1$. (d) $u^{(k+2)}=2 u^{(k+1)}-2 u^{(k)}, \quad u^{(0)}=1, \quad u^{(1)}=3$. (e) $u^{(k+3)}=2 u^{(k+2)}+u^{(k+1)}-2 u^{(k)}, \quad u^{(0)}=0, \quad u^{(1)}=2, \quad u^{(2)}=3$. (f) $u^{(k+3)}=u^{(k+2)}+2 u^{(k+1)}-2 u^{(k)}, \quad u^{(0)}=0, \quad u^{(1)}=1, \quad u^{(2)}=1$.

Apply the method of Exercise 9.1.21 to solve the following iterative equations:
(a) $u^{(k+2)}=-u^{(k+1)}+2 u^{(k)}, \quad u^{(0)}=1, \quad u^{(1)}=2$.
(b) $12 u^{(k+2)}=u^{(k+1)}+u^{(k)}, \quad u^{(0)}=-1, \quad u^{(1)}=2$.
(c) $u^{(k+2)}=4 u^{(k+1)}+u^{(k)}, \quad u^{(0)}=1, \quad u^{(1)}=-1$.
(d) $u^{(k+2)}=2 u^{(k+1)}-2 u^{(k)}, \quad u^{(0)}=1, \quad u^{(1)}=3$.
(e) $u^{(k+3)}=2 u^{(k+2)}+u^{(k+1)}-2 u^{(k)}, \quad u^{(0)}=0, \quad u^{(1)}=2, \quad u^{(2)}=3$.
(f) $u^{(k+3)}=u^{(k+2)}+2 u^{(k+1)}-2 u^{(k)}, \quad u^{(0)}=0, \quad u^{(1)}=1, \quad u^{(2)}=1$.

Key Concepts

Iterative Equations

Iterative equations are relations that define a sequence recursively, where each term is defined as a function of one or more of its previous terms. These equations model many dynamic processes and are crucial in areas such as numerical analysis and discrete dynamical systems.

Linear Recurrence Relations

A linear recurrence relation is an equation that recursively defines a sequence where each term is a linear combination of previous terms plus a potential constant term. They are a fundamental tool in discrete mathematics, often requiring the determination of the sequence’s behavior from given initial conditions.

Characteristic Equation

The characteristic equation is a polynomial derived from a linear recurrence relation by assuming solutions of an exponential form. Finding the roots of this polynomial is a key step in solving the recurrence, as these roots help construct the general solution of the sequence.

Method of Characteristic Roots

This method involves solving the characteristic equation to obtain its roots, which are then used to write the general solution of a homogeneous linear recurrence relation. Depending on the multiplicity and nature (real or complex) of the roots, the structure of the solution can vary significantly.

Initial Conditions

Initial conditions provide the starting values for the recursive sequence, and are essential for determining the specific constants in the general solution of a recurrence relation. They ensure that the solution is uniquely tailored to the given problem setup.

Repeated and Complex Roots

When the characteristic equation yields repeated or complex roots, special techniques are employed to construct the general solution. Repeated roots involve multiplying by powers of the index, while complex roots lead to solutions expressed in terms of trigonometric functions, reflecting oscillatory behavior in the sequence.

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