# If $f(x+1) +f(x-1) =\sqrt{3}\,f(x)$ and $f(2) =2$, what is the value of $f(4)$?

My Attempt

$f(2)=2$. So, $f(1) + f(3)=2\sqrt{3}$ and $f(2) + f(4)=\sqrt{3}\,f(3)$. After solving these equations I got the value of $f(3)=2\sqrt{3}$ and $f(4)=4$. But are there any other methods than this? Any suggestions are welcome.

Update:- @ProfessorVector pointed out that the above solutions are only true if $f(1)=0$. After checking I find that it is true. So, my above attempt is a failure. Is there a way to solve this question?

Update 2:- Is there a way to find the period of this function?

• You could always solve for $f(x)$ directly: en.wikipedia.org/wiki/Characteristic_equation_(calculus) – Simply Beautiful Art Sep 17 '17 at 18:11
• That's only true if $f(1)=0$. From where do you get that? – Professor Vector Sep 17 '17 at 18:19
• @SimplyBeautifulArt But this is a second order linear recurrence, so it needs two initial conditions to solve. – dxiv Sep 17 '17 at 18:28
• @SimplyBeautifulArt Right, but there is no second condition in OP's question, so the answer is more like "$f(4)$ can be whatever you want it to be". – dxiv Sep 17 '17 at 18:36
• @SerialKiller Is there a way to solve this question? $\,f(4) =4 - \sqrt{3} \cdot f(1)\,$, so there is no unique solution unless you know $\,f(1)\,$ or some other value of $\,f(n)\,$. – dxiv Sep 17 '17 at 19:07

I'll be more general.

$f(n+1)+f(n-1) = cf(n)$.

Suppose $f(n) = b^n$.

Then $b^{n+1}+b^{n-1} = cb^n$ or $b^2-cb+1 = 0$.

Then $b =\dfrac{c\pm\sqrt{c^2-4}}{2}$.

If $c^2=4$, $b = c/2 = \pm 1$, so $f(n)=1$ or $(-1)^n$.

If $c^2 > 4$, then $b$ has two possible values, one with $|b|>1$ and one with $|b|<1$.

If $c^2 < 4$, then $b$ has two possible complex values $b_1 =\dfrac{c+\sqrt{c^2-4}}{2} =\dfrac{c+i\sqrt{4-c^2}}{2}$ and $b_2 =\dfrac{c-i\sqrt{4-c^2}}{2}$. Note that $b_1b_2 =\dfrac{c+\sqrt{c^2-4}}{2}\dfrac{c-\sqrt{c^2-4}}{2} =\dfrac{4}{4} =1$ (as can also be deduced from the quadratic specifying $b$) and that $|b_k|^2 =\dfrac{c^2+4-c^2}{4} =1$.

Since $|b| = 1$, $b =e^{it} =\cos(t)+i\sin(t)$ where $\cos(t) =c/2$.

In your case, $c = \sqrt{3}$ so $t =\arccos(\sqrt{3}/2) =\pi/6$.

Therefore the two possible solutions are $f(n) =e^{\pm ni\pi/6}$ and any linear combination of these.

So any solution is of the form $ue^{in\pi/6}+ve^{-in\pi/6} =(u+v)\cos(n\pi/6)+(u-v)\sin(n\pi/6)$.

If the solution is real for an $n$ such that $\sin(n\pi/6) \ne 0$, then $u=v$, so it is $f(n)=2u\cos(n\pi/6)$.

If $f(2) = 2$, then $2=f(2) =2u\cos(\pi/3) =u$ so $u = 2$ and the solution is $f(n) =4\cos(n\pi/6)$.

Putting $n=4$, $f(4) =4\cos(4\pi/6) =-2$.

• If the solution is real for an n ... then u=v That's assuming $u,v$ are real, which they don't have to. All that's needed in this case is $u = \overline v$ for $f(n)$ to be real for $\forall n$. – dxiv Sep 17 '17 at 19:30
• How can you suppose that $f(n)=b^n$? – Rory Daulton Sep 18 '17 at 0:07