# $g(x)$ is a function such that $g(x+1)+g(x-1)=g(x)$, $x \in \mathbb{R}$. For what value of $p$, $g(x+p)=g(x)$.

$g(x)$ is a function such that $g(x+1)+g(x-1)=g(x)$, $x \in \mathbb{R}$.For what value of $p$, $g(x+p)=g(x)$.

$g(x+2)+g(x)=g(x+1)$

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Is $p \ne 0$? if that's the case how do you know that i exists? – Mercy King Aug 21 '12 at 7:54

This is really a question about sequences hence assume that $x_{n+2}-x_{n+1}+x_n=0$ for every $n$, for some sequence $(x_n)_{n\in\mathbb Z}$. The characteristic equation of this recursion is $r^2-r+1=0$, with roots $r=\mathrm e^{\pm\mathrm i\pi/3}$, hence $x_n=A\mathrm e^{n\mathrm i\pi/3}+B\mathrm e^{-n\mathrm i\pi/3}$ for some $A$ and $B$. Now, $\mathrm e^{2\mathrm i\pi}=1$ hence, for every $(A,B)$, $x_{n+6}=x_n$. No period smaller than $6$ is valid for every such sequence $(x_n)_{n\in\mathbb Z}$, as the example of $x_n=\cos(n\pi/3)$ shows.

Edit: The link with the original question is as follows. For every $0\leqslant x\lt1$, $x_n=g(x+n)$ defines a sequence $(x_n)_{n\in\mathbb Z}$ as above, hence there exists two parameters $A(x)$ and $B(x)$ such that $$g(x+n)=A(x)\mathrm e^{n\mathrm i\pi/3}+B(x)\mathrm e^{-n\mathrm i\pi/3},$$ for every integer $n$. There is no relation whatsoever between the families $(g(x+n))_{n\in\mathbb Z}$ and $(g(x'+n))_{n\in\mathbb Z}$ for $x\ne x'$ in $[0,1)$. Hence, each solution $g:\mathbb R\to\mathbb C$ is described uniquely by two functions $A:[0,1)\to\mathbb C$ and $B:[0,1)\to\mathbb C$ and any two such functions $(A,B)$ define uniquely a solution $g_{A,B}$.

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The OP talked about functions $g:\ {\mathbb R}\to{\mathbb R}$ (or $\to{\mathbb C}$). This aspect is insufficiently addressed in the answers so far. E.g.: Which functions of period $6$ satisfy the given functional equation? – Christian Blatter Aug 21 '12 at 11:00
@ChristianBlatter See Edit. – Did Aug 21 '12 at 12:25

Now that we know, by @did's answer, that the answer is $6$, we ought to be able to prove it directly. Here is a solution: Note

$$g(x+3) = g(x+2) - g(x+1) = (g(x+1) - g(x)) - g(x+1) = - g(x).$$

Therefore

$$g(x+6) = -g(x+3) = g(x).$$

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After @did did it already I'd like to add a matrix-view into it.
In the same way we define a sequence $g_0,g_1,g_2,...$ , assuming we had already found some $g_0$ and $g_1$. Then $g_2 = g_1 - g_0$, in matrix-notation
$$\begin{matrix} & & \begin{bmatrix}-1\\ 1\end{bmatrix} \\ & . \\ \begin{bmatrix}g_0&g_1\end{bmatrix} & = & \begin{bmatrix} g_2 \end{bmatrix} \end{matrix}$$
To make it iterable, we prefix a column at the coefficients-matrix:
$$\begin{matrix} & & \begin{bmatrix}0&-1\\1&1\end{bmatrix} \\ &. \\ \begin{bmatrix}g_0&g_1\end{bmatrix} & = & \begin{bmatrix}g_1 & g_2\end{bmatrix} \end{matrix}$$
Let's call the resp. matrices $G_0,G_1$ and the coefficients-matrix $C$ so we have $$G_k \cdot C = G_{k+1}$$

We see, that if we iterate this we can generate a sequence of values $g_k$ for the function $g(x)$ which have the required relation between that elements - after we have assumed some initial values $g_0,g_1$
To find, whether this is periodic we must ask, whether for some p the formula comes out to be $$G_0 \cdot C^p = G_0$$ or said differently, whether for some p we'll have $C^p = I$ where $I$ is the identity matrix.

We can diagonalize $C$ and find the eigenvalues having the complex values $$\lambda_0 = {1- \sqrt{3}î\over 2} \\ \lambda_1 = {1+ \sqrt{3}î\over 2}$$ and these are just $6'th$ roots of the complex unit, so we simply check what $C^6$ is and find that indeed $$C^6 = I\\p=6$$ which is the solution. (Clearly, we can look at the characteristic polynomial and find $\mathcal P(C):= x^2-x+1$ which was mentioned in @did's answer and which we have to solve for its roots.)

This is the same result as in the previous answer, but it shows, how one could proceed, if the problem-parameters were different - and were possibly even more complicated (linear) compositions of the function, for instance $g(x)= 1 \cdot g(x-1) - 2\cdot g(x-2)+3\cdot g(x-3)$ and similarly.

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Unfortunately, $g(x)= 1 \cdot g(x-1) - 2\cdot g(x-2)+3\cdot g(x-3)$ has no periodic solution except $g(x)=0$ for every $x$. – Did Aug 21 '12 at 9:29
@did: upps... well some common integer power of the eigenvalues should simultaneously arrive at 1, which is clearly some very special requirement... (besides that their absolute value should be the unit) thanks for the reminding! – Gottfried Helms Aug 21 '12 at 9:45

As Sean Eberhard has shown such a function necessarily has period $6$. It remains to investigate which functions of period 6 actually satisfy the given functional equation.

Every reasonable function $g$ of period $6$ can be developed into a Fourier series $$g(x)=\sum_{k\in{\mathbb Z}} c_k e^{k\pi ix/3}$$ with complex coefficients $c_k$. Put $e^{\pi i/3}=:\omega$. Then $$g(x+1)-g(x)+g(x-1)=\sum_k c_k(\omega^k -1 +\omega^{-k})e^{k\pi ix/3}\ .$$ The RHS is $\equiv 0$ iff for all $k\in{\mathbb Z}$ at least one of $c_k$ and $\omega^k -1 +\omega^{-k}=2\cos{k\pi\over3}-1$ is zero. The latter is the case iff $k=\pm1$ modulo $6$. Therefore the function $g$ has to be of the form $$g(x)=\sum_{l\in{\mathbb Z}}c_{6l+1}e^{(6l+1)\pi i x/3}+\sum_{l\in{\mathbb Z}}c_{6l-1}e^{(6l-1)\pi i x/3}=C(x)e^{\pi i x/3}+D(x)e^{-\pi i x/3}\ ,$$ where now $C(x)$ and $D(x)$ are arbitrary complex functions of period $1$. The most general real function satisfying our functional equation is therefore given by $$g(x)=A(x)\cos{\pi x\over3}+B(x)\sin{\pi x\over3}\ ,$$ where $A(x)$ and $B(x)$ are arbitrary real functions of period $1$.

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$g(x+1)+g(x-1)=g(x)$

$g(x+1)=g(x)-g(x-1)$ -------(1)

$g(x+2)=g(x+1)-g(x)$ -------(2)

$g(x+3)=g(x+2)-g(x+1)$ -------(3)

from (1) and (2) we have,

$g(x+2)=-g(x-1)$ -------(2b)

from (2) and (3) we have,

$g(x+3)=-g(x)$ -------(3b)

from (2b) or (3b) we have,

$g(x+6)=-g(x+3)$ -------(3c)

from, (3b) and (3c) we have,

$g(x+6)=g(x)$

we can now conclude that $p=6$

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