Solve the initial value problem for the sequence $\left \{ u_{n}| n \in \mathbb{N} \right \}$ satisfying the recurrence relation:
$u_n − 5u_{n-1} + 6u_{n−2} = 0 $ with $u_0 = 1$ and $u_1 = 1$.
Ive gotten the general solution to be $u_n = A(2)^n + B(3)^n$.
Once I sub the initial values:
$u_0 = 2A + 3B = 1$
$u_1 = 2A + 3B = 1$
And Im unsure on how to solve this system. Any help appreciated, thanks.
Best Answer
These problems of generalized Fibonacci sequences seem to show up here weekly. Here, then, is a generalized solution for solving all such problems with a recursion formula of the type $f_n=af_{n-1}+bf_{n-2}$ and arbitrary initial conditions.
There have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\in\mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” $Applied \ Probability \ Trust$; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)
We have extended Maynard's analysis to include arbitrary $f_0,f_1\in\mathbb{R}$. It is relatively straightforward to show that
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta}= \left(f_1-\frac{af_0}{2}\right)F_n+\frac{af_0}{2}L_n$$
where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$, $F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$, and $L_n=\frac{\alpha^n+\beta^n}{\alpha+\beta}$.
The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1, f_1=1, \text{ and } f_0=0 \text{ or }2$.
This can also be expressed as
$$f_n=f_1F_n+bf_0F_{n-1}$$