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Let f be defined from real to real

$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$

Now how to find the period of this function f(x)? Can someone provide me a purely algebraic method to solve this problem please?

Update:My method

An elementary algebraic approach to the problem :

$f(x-1)+f(x+1)=\sqrt { 3 } f(x)$

Replace $x$ with $x+1$ and $x-1$ respectively.

We get $f(x)+f(x+2)=\sqrt { 3 } f(x+1)$ and $f(x-2)+f(x)=\sqrt { 3 } f(x-1)$

From these three equations we get $f(x-2)+f(x+2)=0$

Putting $x=x+2$ and adding with last equation we get $f(x-2)+f(x+4)=0$....(1)

Similarly $f(x-4)+f(x+2)=0$.....(2)

Put $x=x-6$ in (1)

We get $f(x-8)+f(x-2)=0$.....(3)

From (1) and (3) we get $f(x-8)=f(x+4)$

So the period of $f(x)$ is 12

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    $\begingroup$ What is the period of $f(x)\equiv 0$? $\endgroup$
    – vadim123
    Commented May 9, 2015 at 22:42
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    $\begingroup$ Isnt that a constant function ? 0 is the period.Why? $\endgroup$
    – user220382
    Commented May 9, 2015 at 22:44
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    $\begingroup$ Because this satisfies your condition. $\endgroup$
    – vadim123
    Commented May 9, 2015 at 22:44
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    $\begingroup$ vadim123 is not mistaken, there is a trivial constant solution to your equation. $0+0=\sqrt{3} \cdot 0$. It is not necessary the only solution. $\endgroup$
    – Ian
    Commented May 9, 2015 at 22:46
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    $\begingroup$ Nice trick. I don't think I could have come up with it from scratch. $\endgroup$
    – Ian
    Commented May 10, 2015 at 3:07

2 Answers 2

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For each fixed $y \in [0,1)$, your equation is a second order linear recurrence relation with constant coefficients. This suggests that the first step should be to solve the characteristic equation $\lambda^2-\sqrt{3} \lambda + 1 = 0$. You find $\lambda_1,\lambda_2 = \frac{\sqrt{3} \pm i}{2}=e^{\pm i \pi/6}$. This means that, for $y \in [0,1)$ and $n \in \mathbb{Z}$:

$$f(y+n)=c_1 e^{i n \pi/6} + c_2 e^{-i n \pi/6}$$

where $c_1,c_2$ are specified by the choice of $f(y)$ and $f(y-1)$. This tells you that the period is no larger than $\frac{2 \pi}{\pi/6} = 12$. It can be smaller: for instance your equation has a trivial constant solution $f \equiv 0$.

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  • $\begingroup$ Presently I don't know recurrence.I would be happier with an algebraic solution.Let me know if you know any algebraic solution which can be obtained by replacing x with x+a or x-a in the equation.Anyway thanks. $\endgroup$
    – user220382
    Commented May 9, 2015 at 23:19
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    $\begingroup$ @SanchayanDutta One way or another you're going to have to find the form that I wrote, as far as I can tell. You might be able to do it by simply assuming a sinusoid form, but that's just doing what I wrote without as much motivation. The method I described comes about by finding exponential solutions, which in your case would satisfy $f(x+1)=cf(x)$ for some number $c$. You might perhaps rewrite the equation as $\cos(\pi/6) f(x) = \frac{f(x-1)+f(x+1)}{2}$ and get something out of that. $\endgroup$
    – Ian
    Commented May 9, 2015 at 23:20
  • $\begingroup$ Actually I want to learn your technique.Can you provide me some links so that I can learn recurrence from scratch?Btw I'm just in high school.Not even through with Precalculus. $\endgroup$
    – user220382
    Commented May 10, 2015 at 2:31
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    $\begingroup$ @SanchayanDutta This seems to be a pretty good treatment: www3.cs.stonybrook.edu/~rezaul/Fall-2012/CSE548/… $\endgroup$
    – Ian
    Commented May 10, 2015 at 2:37
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Algebraic (non-coanstant)? NO.
But, for example $$ f(x) = \sin\frac{\pi x}{6} $$ is a solution.

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  • $\begingroup$ Sorry for the confusion I might have caused by "algebraic".I mean an algebraic technique rather than using recurrence. $\endgroup$
    – user220382
    Commented May 10, 2015 at 2:16

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