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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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the homework tag is not meant to be used as a standalone tag. Please use it in addition to a tag of the correct subject area. –  Willie Wong Apr 30 '11 at 11:52
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2 Answers 2

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
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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
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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. –  Ross Millikan Apr 30 '11 at 14:48
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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.

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