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?

  • $\begingroup$ You could always solve for $f(x)$ directly: en.wikipedia.org/wiki/Characteristic_equation_(calculus) $\endgroup$ – Simply Beautiful Art Sep 17 '17 at 18:11
  • 7
    $\begingroup$ That's only true if $f(1)=0$. From where do you get that? $\endgroup$ – Professor Vector Sep 17 '17 at 18:19
  • 3
    $\begingroup$ @SimplyBeautifulArt But this is a second order linear recurrence, so it needs two initial conditions to solve. $\endgroup$ – dxiv Sep 17 '17 at 18:28
  • 1
    $\begingroup$ @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". $\endgroup$ – dxiv Sep 17 '17 at 18:36
  • 1
    $\begingroup$ @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 $\,f(n)\,$. $\endgroup$ – 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 $.

  • 1
    $\begingroup$ 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 $\forall n$. $\endgroup$ – dxiv Sep 17 '17 at 19:30
  • 1
    $\begingroup$ How can you suppose that $f(n)=b^n$? $\endgroup$ – Rory Daulton Sep 18 '17 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.