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I was given an example

$R_n = R_{n-1} + R_{n-2} $

This equation is given as an second-order equation.

Why is it so?

2 Answers 2

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The fact that it is second-order refers to the fact that the largest difference in indices is $2$. For example,

$ R_{n+4}=3R_{n+1}^2+R_n $

is a fourth-order difference equation and

$ R_{n+3}=2R_{n+2}\cdot R_{n+1} $

is a second order difference equation.

If you're familiar with ODEs, the terminology is analogous.

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    @liangteh: the reason for interest in the order is that it tells you how many old terms can influence a new one. For a second order equation you need two initial conditions. For the first of GleasSpty's examples you would need 4.2011-04-30
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One explanation is that one solves (see Recurrence relation, Wikipedia, under "Solving") the following homogeneous difference equation (or recurrence relation) with constant coefficients

$a_{n}+Aa_{n-1}+Ba_{n-2}=0,$

by means of the second degree characteristic equation

$r^2+Ar+B=0,$

pretty much as one woud solve a homogeneous second-order linear ordinary differential equation with constant coefficients.