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Linear difference equations with constant coefficients

1. The forward shift operator


Many probability computations can be put in terms of recurrence relations that have to be satisfied by suc-
cessive probabilities. The theory of difference equations is the appropriate tool for solving such problems.
This theory looks a lot like the theory for linear differential equations with constant coefficients.
In order to simplify notation we introduce the forward shift operator E, that takes a term un and shifts the
index one step forward to un+1.We write

The general linear difference equation of order r with constant coefficients is

where Φ(E) is a polynomial of degree r in E and where we may as sume that the coefficient of E' is 1.

2. Homogeneous difference equations
The simplest class of difference equations of the form (1) has f(n) = 0, that is simply

These are called homogeneouse quations.
When Φ(E) =(E-λ1)(E-λ2)...(E-λr) where the λi are constants that are all distinct from each other,
one can prove that the most general solution to the homogeneous equation is

where a1,a2,...,ar are arbitrary constants.
When Φ(E) contains a repeated factor (E-λα)h ,the corresponding part of the general solution becomes

In order to find the n'th term of a linear difference equation of order r, one can of course start by r initial
values, and the solve recursively for any given n. Thus, if we want our solution to satisfy certain initial conditions
 we may first determine the general solution, and then (if possible) make it satisfy the initial conditions. There
can be no more than r such initial conditions, but they need not (as when we compute the solution recursively)
necessarily be conditions on μ0,... ,μr-1,but can be on any press ion.html">set of r values .

Example 1. Solve μn+2n = 0.

The equation can be written in the form

or

The general solution is therefore

where a, b, c are constants.

Example 2. Find the general solution to the equation


and hence obtain the particular solution satisfying the conditions

The equation may be written in the form

The general solution is therefore

where a, b, c, d are constants.
For the particular side conditions we have

whence a = 0, b =1, c = -2, d =1, so the particular solution is

3. Non-homogeneous difference equations


When solving linear differential equations with constant coefficients one first finds the general solution for
the homogeneous equation, and then adds any particular solution to the non-homogeneous one. The same
recipe works in the case of difference equations, i.e. first find the general solution to

and a particular solution to

and add the two together for the general solution to the latter equation. Thus to solve these more general
equations, the only new problem is to identify some particular solutions. We will only give a few examples
here, not attempting to treat this problem in anygenerality.

(i) f(n) = kμn , μ ≠λi, i =1, 2,... ,r
In this case one can show that


is a particular solution to Let namely Then

Example 3. The general solution of

is where a and b are arbitrary constants.
a non-repeated factor of Φ(E)
In this case a particular solution is given by

where Φ'(μ) denotes
Example 4. The general solution of

is

where a, b are arbitrary constants.
a repeated factor of Φ(E)
Suppose now that (E-λi) is repeated h times in Φ(E). Then


where is a particular solution of the equation Φ(E)μn=kμn
 

Example 5. The general solution of the equation

is

with a, b, c, d are arbitrary constants
(iv) f(n) is a polynomial in n

First write ƒ as a polynomial in the factorial powers n (k) ,so

Now define the difference operator , by Using the symbolic relationship
E=1+Δ we can re-express Φ(E) as Ψ(Δ) Still arguing symbolically,aparticular solution is obtained by

provided that we can make any sense out of 1/Ψ(Δ) The way this will be done is by expanding 1/Ψ(Δ)
in powers of Δ and using long division . The fol lowing rules are needed:
 Δn(r)=rn(r-1)


and

Example 6. Find a particular solution of the equation

First write Thus we get

Example 7. Find a particular solution of the equation

The required solution is

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