How to make this difference equation continuous? $$ F_n=F_{n-1}-F_{n-2} $$
How can I convert this oscillating sequence into a continuous function? IE get it in terms of n.
 A: This sequence has exactly 6 values:
($a$,$\ $ $b$, $\ $ $b-a$, $\ $ $-a$, $\ $ $-b$, $\ $ $a-b$) in cycle.
So, there is a reason to build trigonometric function with period of $6$:
$$
F(n) = A \sin \frac{\pi n}{3} + B \cos\frac{ \pi n}{3},
$$

If $F_0 = a$, $F_1 = b$, then
$$
\left\{
\begin{array}{c}
A\cdot 0 + B\cdot 1 = a;\\
A\cdot \dfrac{\sqrt{3}}{2} + B \cdot \dfrac{1}{2} = b; \\
A\cdot \dfrac{\sqrt{3}}{2} - B \cdot \dfrac{1}{2} = b-a; \\
\end{array}
\right.
$$ 
$B=a$, $A=\dfrac{2b-a}{\sqrt{3}}$.
So,
$$
F(n) = \dfrac{2b-a}{\sqrt{3}} \cdot \sin \frac{\pi n}{3} \;+ \; a \cdot \cos\frac{\pi n}{3},
$$
where $a=F_0$, $b=F_1$.
A: Using the solution process of Recurrence relation,
we have $r^2-r+1=0\implies r=\frac{1\pm\sqrt3i}2$
So, if $\lambda_1=\frac{1+\sqrt3i}2, \lambda_2=\frac{1-\sqrt3i}2$
$$F_n=A\cdot \lambda_1^n+B\cdot \lambda_2^n$$ where $A,B$ are arbitrary constants
Let $\frac12=r\cos\theta,\frac{\sqrt3}2=r\sin\theta$ where $r>0$ so that 
Squaring & adding we get $r^2=1\implies r=1$ as $r>0$
$\implies \cos\theta=\frac12>0$ and $\sin\theta=\frac{\sqrt3}2>0$
On division $\tan\theta=\sqrt3\implies \theta=\frac\pi3$
So, $\lambda_1=r(\cos\theta+i\sin\theta)=\cos\frac\pi3+i\sin\frac\pi3$
and $\lambda_2=r(\cos\theta-i\sin\theta)=\cos\frac\pi3-i\sin\frac\pi3=\cos(-\frac\pi3)+i\sin(-\frac\pi3)$ as $\cos(-x)=\cos x,\sin(-x)=-\sin x$
$\implies \lambda_1^n=(\cos\frac\pi3+i\sin\frac\pi3)^n=\cos \frac{n\pi}3+i\sin\frac{n\pi}3$ using De Moivre's formula
Similarly, $\lambda_2^n=\{\cos(-\frac\pi3)+i\sin(-\frac\pi3)\}^n=\cos(-\frac{n\pi}3)+i\sin(-\frac{n\pi}3)=\cos \frac{n\pi}3-i\sin\frac{n\pi}3$
$$\implies F_n=A\left(\cos \frac{n\pi}3+i\sin\frac{n\pi}3\right)+B\left(\cos \frac{n\pi}3-i\sin\frac{n\pi}3\right)$$
$$=(A+B)\cos \frac{n\pi}3+i(A-B)\sin\frac{n\pi}3$$
