If $f(x+1)=\frac12(f(x)+f(x-2))$ for every $x$ then $f(x)$ converges when $x\to\infty$? And what would be the limit?

Question

I was bored in class one day and something popped into my head. It was a problem that I could not solve, so I am turning to the arbitrarily large yet finite wisdom of the internet for help. Here it is: $$f(1)=a, f(2)=b, f(3)= c,$$ $$f(x+1)=\frac{f(x)+f(x-2)} 2$$ I wanted to find a way to tell $$1) \text{Whether or not $\lim_{x \to \infty}f(x)$ converges for arbitrary $a, b, \text{and } c,$ and,}$$ $$2) \text{If it does, what it converges to.}$$ I started by guessing that there might be an exponential that follows this pattern. I said that $a^{x+1} = \frac{a^x + a^{x-2}} 2$. If you reduce this, you get a cubic equation, the answers to which are 1 and two complex numbers. Obviously none of these help my current quandry. Any help would be much appreciated.

Answer 1

you get a cubic equation, the answers to which are 1 and two complex numbers. Obviously none of these help my current quandry.

Actually they do, obviously, since the characteristic equation is $2r^3=r^2+1$, which is equivalent to $$(r-1)(2r^2+r+1)=0$$ whose roots are $r=1$, $r=z$ and $r=\bar z$, with $$z=\frac{-1+i\sqrt 7}4$$ Thus, for some suitable $(A,B,C)$ fully determined by the initial conditions $(f(1),f(2),f(3))$,

$$f(n)=A+B\,z^n+C\,\bar z^n$$

Furthermore, $$|z|=|\bar z|=\frac{\sqrt2}2<1$$ hence, for every initial conditions, $$\lim_{n\to\infty}f(n)=A$$ If $(a,b,c)$ are all real valued, a further simplification is that $C=\bar B$ hence $f(n)=A+2\Re(Bz^n)$, or, equivalently,

$$f(n)=A+(-1)^n\varrho^n(B_1\cos(n\vartheta)+B_2\sin(n\vartheta))$$

with $(A,B_1,B_2)$ real valued and fully determined by the initial conditions $(f(1),f(2),f(3))$, and $$\varrho=\frac{\sqrt2}2\qquad\vartheta=\arctan\sqrt 7$$