# Recurrence relation

Recurrent redirects here; for the meaning of "recurrent" in contemporary hit radio, see Recurrent rotation.

In mathematics, a recurrence relation, also known as a difference equation, is an equation which defines a sequence recursively: each term of the sequence is defined as a function of the preceding terms.

For example (the logistic map):

[itex]x_{n+1} = r x_n (1 - x_n) \,[itex]

Some simply defined recurrence relations can have very complex (chaotic) behaviours and are sometimes studied by physicists and mathematicians in a field of mathematics known as nonlinear analysis.

Solving a recurrence relation means obtaining a non-recursive function of n.

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## Linear homogeneous recurrence relations with constant coefficients

The term linear means that each term of the sequence is defined as a linear function of the preceding terms. The coefficients and the constants may depend on n, even non-linearly.

A special case is when the coefficients do not depend on n.

Homogeneous means that the constant term of the relation is zero.

In order to obtain a unique solution to the linear recurrence there must be some initial conditions, as the first number in the sequence can not depend on other numbers in the sequence and must be set to some value.

## Solving linear recurrence relations

Solutions to recurrence relations are found by systematic means, often by using generating functions (formal power series) or by noticing the fact that rn is a solution for particular values of r.

For second-order recurrence relations in the form:

[itex]a_{n}=Aa_{n-1}+Ba_{n-2} \,[itex]

we have the solution rn:

[itex]r^{n}=Ar^{n-1}+Br^{n-2} \,[itex]

Dividing through by [itex]r^{n-2}[itex] we get:

[itex]r^2=Ar+B \,[itex]
[itex]r^2-Ar-B=0 \,[itex]

This is known as the characteristic equation of the recurrence relation. Solve for r to obtain the two roots [itex]\lambda_1, \lambda_2 [itex], and if these roots are distinct, we have the solution

[itex]a_n = C\lambda_1^n+D\lambda_2^n \,[itex]

while if they are identical (when A2+4B=0), we have

[itex]a_n = C\lambda^n+Dn\lambda^n \,[itex]

where C and D are constants.

Additionally, if the equation is of the form [itex]a_{n}=Aa_{n-1}+B[itex] you can substitute 2 for n and get [itex]r^2=Ar+B[itex] as above. The constants C and D can be found from the "side conditions" that are often given as [itex]a_0=a[itex], [itex]a_1=b[itex].

Different solutions are obtained depending on the nature of the roots of the characteristic equation.

If the recurrence is inhomogeneous, a particular solution can be found by the method of undetermined coefficients and the solution is the sum of the solution of the homogeneous and the particular solutions.

Interestingly, the method for solving linear differential equations is similar to the method above — the "intelligent guess" for linear differential equations is ex.

This is not a coincidence. If you consider the Taylor series of the solution to a linear differential equation:

[itex]

\sum_{n=0}^{\infin} \frac{f^{(n)}(a)}{n!} (x-a)^{n} [itex]

you see that the coefficients of the series are given by the n-th derivative of f(x) evaluated at the point a. The differential equation provides a linear difference equation relating these coefficients.

This equivalence can be used to quickly solve for the recurrence relationship for the coefficients in the power series solution of a linear differential equation.

## Example: Fibonacci numbers

The Fibonacci numbers are defined using a linear recurrence relation:

[itex]F_{n} = F_{n-1}+F_{n-2} \,[itex]
[itex]F_{0} = 0 \,[itex]
[itex]F_{1} = 1 \,[itex]

and has solution (letting [itex]\Phi = {1+\sqrt{5} \over 2}[itex] be the golden ratio)

[itex]F_n = {\Phi^n \over \sqrt{5}} - {(1-\Phi)^n \over \sqrt{5}}[itex]

The initial conditions are:

[itex]F_{0} = 0 \,[itex]
[itex]F_{1} = 1 \,[itex]

Therefore, the sequence of Fibonacci numbers is:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ...

## Relationship to differential equations

When solving an ordinary differential equation numerically, one typically encounters a recurrence relation. For example, when solving the initial value problem

with Euler's method and a step size h, one calculates the values [itex]y_0=y(t_0)[itex], [itex]y_1=y(t_0+h),[itex] [itex]y_2=y(t_0+2h),...[itex] by the recurrence

[itex] y_{n+1} = y_n + hf(t_n,y_n). [itex]

Systems of linear first order differential equations can be discretized exactly analytically using the methods shown in the discretization article.

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