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Let $f(x)$ be a function such that $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$. Then for what value of $k$ is the relation $f(x + k) = f(x)$ necessarily true for every real $x$?

The answer/solution suggested in my module is like this: "this is a bit involved but can be proved that $k=6$". Could anybody explain me this?

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Just add two consecutive relations: $$ f(x+2)+f(x)=f(x+1) $$ $$ f(x+3)+f(x+1)=f(x+2) $$ Then you'll get $$ f(x+3)+f(x)=0$$ vor every real $x$. You have also $f(x+6)+f(x+3)=0$ for every real $x$, and therefore $f(x)=f(x+6)$ for every real $x$.


Another way. Consider the recurrence relation $x_{n+1}+x_{n-1}=x_n$, $x_0, x_1 \in \Bbb{R}$. Then the general term of the sequence is $$ x_n=a \varepsilon^n +b \varepsilon^n$$ where $a,b$ are real numbers determined by $x_0,x_1$ and $\varepsilon $ verifies the characteristic equation $x^2-x+1=0$. This implies that $\varepsilon^3=-1$ and $\varepsilon^6=1$. I think this approach may prove that we can construct some function $f$ for which $k=6$ is the least $k$ for which $f(x)=f(x+6)$ for every $x \in \Bbb{R}$.

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    $\begingroup$ @Ganit Each of the first two relations added were a result of rather than applying the function to $x$, instead applying the function to $x+1$. After all, if $f(x+1)+f(x-1)=f(x)$ is true for all $x$, then it should also be true that $f(25+1)+f(25-1)=f(25)$ and $f(y+1)+f(y-1)=f(y)$ and in particular $f((x+1)+1)+f((x+1)-1)=f((x+1))$. Similarly for $x+2$. Now... from here, rearrange these so everything is on the left. $f(x+2)-f(x+1)+f(x)=0$ and $f(x+3)-f(x+2)+f(x+1)=0$. Add these two together and note the cancellations that occur. This is why we have $f(x+3)+f(x)=0$. $\endgroup$ – JMoravitz Mar 26 at 15:38
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    $\begingroup$ @Ganit go through the same process to note that since $f(x+3)+f(x)=0$ and $f(x+6)+f(x+3)=0$, subtracting the one from the other gives $f(x+6)-f(x)=0$ and so $f(x+6)=f(x)$. $\endgroup$ – JMoravitz Mar 26 at 15:39
  • $\begingroup$ The approach below the horizontal rule certainly suggests defining $f(x) = e^{\pi i x}/3$ or $f(x) = e^{-\pi i x}/3$ as a counterexample for any $k$ but a multiple of 6. Or, in terms of real-valued functions, use a linear combination such as $f(x) = \cos(\pi x/3)$. $\endgroup$ – Daniel Schepler Mar 26 at 15:41
  • $\begingroup$ @JMoravitz : Thanks for such detailed clarification ! $\endgroup$ – Ganit Mar 26 at 16:36
  • $\begingroup$ @JMoravitz : Thanks for such detailed clarification ! $\endgroup$ – Ganit Mar 26 at 16:36
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There are many ways to get the desired result. Beni’s first method is about as simple as they come, but a slightly different one occurred to me first:

$$\begin{align*} 2f(x)&=f(x-1)+f(x)+f(x+1)\\ &=f(x-2)+f(x-1)+2f(x)+f(x+1)+f(x+2)\\ &=f(x-2)+3f(x)+f(x+2)\;, \end{align*}$$

so $$f(x-2)+f(x)+f(x+2)=0\;.\tag{1}$$ Replacing $x$ by $x+2$, we have $$f(x)+f(x+2)+f(x+4)=0\;.\tag{2}$$ Now just subtract $(1)$ from $(2)$ to get $$f(x+4)-f(x-2)=0\;,$$ or $f(x+4)=f(x-2)$ for all $x\in\Bbb R$, which is of course equivalent to $f(x+6)=f(x)$ for all $x\in\Bbb R$.

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