Recurrence equation: $u_n = 4u_{n−1} + 4u_{n−2}$ ; is $4x+4 = 4$ the characteristic equation? Given this recurrence equation:
$u_1 = 0, u_2 = 1$
$u_n = 4u_{n−1} + 4u_{n−2}$
Is the correct characteristic equation:
$4x+4 = 4$
EDIT:
Complete solve:
The characteristic equation is
$x^2-4x-4=0$
We solve the quadratic equation...
$\alpha = 5$ 
$\beta=-1$
So:
$u_n = c_1 \alpha^n + c_2 \beta^n$
We solve the equation...
$c_1 = 1/30$
$c_2 = 1/6$
Finally:
$u_n = \dfrac{5^n}{30} + \dfrac{(-1)^n}{6}$
 A: When I am working with recurrence relations and trying to simplify them, I like to move all of the "recurring" terms to one side. Doing this here will give you a homogeneous difference equation. So here you would get,
$u_{n} - 4u_{n-1} - 4u_{n-2} = 0$.
Now to find the characteristic equation, you may want to shift the subscripts to get,
$u_{n+2} - 4u_{n+1} - 4u_{n} = 0$.
Now, we can see that the characteristic equation will be,
$\lambda^{2} - 4 \lambda - 4 = 0$.
A: Your characteristic equation is incorrect. The way you obtain the characteristic equation is to assume $u_n = m^n$. Plug it in to get a quadratic in $m$. Solve for $m$. Get the two roots say $m_1$ and $m_2$. Now the general solution is given by the linear combination namely $u_n = a_1 m_1^n + a_2 m_2^n$. Solve for $a_1$ and $a_2$ using the initial conditions.
This methodology is analogous to plugging in $y=e^{mx}$ when you want to solve a linear ODE. Here you have a difference equation instead of a differential equation.
A: The characteristic equation is the equation satisfied by $\lambda$ if $u_n = \lambda^n$ be a solution. For example, if the recurrence is $a_n = a_{n-1} + a_{n-2}$ then a solution $\lambda^n$ must satisfy $$\lambda^n = \lambda^{n-1} + \lambda^{n-2}.$$ The point is that all these equations reduce to the one equation $$\lambda^2 = \lambda + 1.$$ Back to your question, when I follow the same steps I get something else.
A: Give Wilf's techniques a try... define $U(z) = \sum_{n \ge 0} u_{n + 1} u^z$,
and write the recurrence as:
$$
u_{n + 2} = 4 u_{n + 1} + 4 u_n \qquad u_1 = 0, u_2 = 1
$$
By properties of generating functions:
$$
\frac{U(z) - u_1 - u_2 z}{z^2} = 4 \cdot \frac{U(z) - u_1}{z} + 4 \cdot U(z)
$$
Maxima solves this as:
$$
U(z) = \frac{z}{1 - 4 z - 4 z^2}
     = \frac{1}{2^{5/2}} \cdot \frac{1}{1 - (2^{3/2} + 2) z}
         - \frac{1}{2^{5/2}} \cdot \frac{1}{1 + (2^{3/2} - 2) z}
$$
Two geometric series:
$$
u_{n + 1} = \frac{1}{2^{5/2}} \cdot
              \left( 
                 (2^{3/2} - 2)^n - (-1)^n (2^{3/2} + 2)^n
              \right)
$$
