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$g(x)$ 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?2012-08-21

5 Answers 5

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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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    @ChristianBlatter See Edit.2012-08-21
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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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    @did: upps... well some common integer power of the eigenvalues should simultaneously arrive at 1, which is clearly some very special requirement... ($b$esi$d$es that their absolute value should be the unit) thanks for the reminding!2012-08-21
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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 $p=6$