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My prof gives an example of trying to solve the equations of motion for a series of gliders each connected by springs, with the same spring constant. From looking at the $j-1$ through $j+1$ glider, he got the following: \begin{equation}mx^{\prime\prime}_j = \omega_0^2 (x_{j+1} - 2x_j + x_{j-1}).\end{equation} This I understand.

Solving $x'' = \omega_0 ^2 (A_{j+1} -2A_j + A_{j-1})$ using $x=A_je^{-\imath \omega t}$ gives the solution \begin{equation}\left[2-\left(\frac{\omega}{\omega_0}\right)^2\right]A_j = A_{j+1} + A_{j-1}.\end{equation} I can follow this step.

Why does this mean it has solutions, $A_j = A_+ e^{\imath\, \theta j}$ and $A_- e^{-\imath\,\theta j}$? This is what my lecture notes say.

I thought $A_+$ might be $A_{j+1}$ divided by $\left[2-\left(\frac{\omega}{\omega_0}\right)^2\right]$, but then why did the exponent on the $e$ change?

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  • $\begingroup$ Basically, we want to find $A_j$ as a function of $j$. It turns out that both $A_j=A_+ e^{i\theta j}$ and $A_j=A_- e^{-i\theta j}$ work, for some $A_+$, $A_-$, and $\theta$ independent of $j$. You should be able to find $\theta$ in terms of $\omega$ and $\omega_0$. $\endgroup$
    – leongz
    May 4, 2015 at 7:25

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Basically this is just the solution to the recurrence formula, look here. They show how to solve recurrence relation using difference equation.

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