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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?

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up vote 6 down vote accepted

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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$$R_n = R_{n-1} + R_{n-2} $$ The difference is because of $$R_n and R_{n-2} $$ which determine the equation to be second-order? – optimus Apr 30 '11 at 9:04
Yes${}{}{}{}{}$ – Gerry Myerson Apr 30 '11 at 9:44
The techniques for solving second order difference equations are analogous to those used in solving second order differential equations (complementary function, 'particular integral'..) – Anon445 Apr 30 '11 at 11:55
@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. – Ross Millikan Apr 30 '11 at 14:48

One explanation is that one solves (see Recurrence relation, Wikipedia, under "Solving") the following homogeneous difference equation (or recurrence relation) with constant coefficients


by means of the second degree characteristic equation


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

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