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Linear difference equations with constant coeffici
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 successive
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 u_{n} and shifts the
index one step forward to u_{n+1}. We write
Eu_{n} = u_{n+1}
E^{2}u_{n} = EEu_{n} = Eu_{n+1} = Eu_{n+2}
...
E^{r}u_{n}= u_{n+r}
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 assume that the coefficient of E^{r} is 1.
2. Homogeneous difference equations
The simplest class of difference equations of the form (1) has f (n) = 0, that
is simply
Φ((E)u_{n} = 0.
These are called homogeneous equations.
When Φ(E) = (E−λ_{1})(E−λ_{2})...(E−λ_{r}) where the λ_{i}
are constants that are all distinct from each other,
one can prove that the most general solution to the homogeneous equation is
u_{n} =a_{1}λ_{1}^{n}+ a_{2}λ_{2}^{n}+...+a_{r}λ_{r}^{n}
where a_{1}, a_{2},..., a_{r} are arbitrary constants.
WhenΦ(E) contains a repeated factor (E−λ_{a} )^{h}, the
corresponding part of the general solution becomes
where n^{(k)} = n(n−1)(n−2)...(n−k +1) = n!/(n−k)!.
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 giv en 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 u_{0},... , u_{r−1},
but can be on any set of r values.
Example 1. Solve u_{n+2}−u_{n} = 0.
The equation can be written in the form
(E^{2}−1)u_{n} = 0
or
(E−1)(E +1)u_{n} = 0
The general solution is therefore
u_{n} = a(−1)^{n} + b1^{n
}where a, b, c are constants.
Example 2. Find the general solution to the equation
u_{n+4} − 9u_{n+3} + 30u_{n+2}
− 44u_{n+1} + 24u_{n }= 0
and hence obtain the particular solution satisfying the conditions
u_{0} = 1, u_{1} = 5, u_{2} = 1, u_{3} = − 45.
The equation may be written in the form
(E^{4} − 9E^{3} + 30E^{2} − 44E + 24)u_{n} = 0
(E− 2)^{3}(E − 3)u_{n} = 0.
The general solution is therefore
u_{n} = 2^{n}(a + bn + cn(n− 1)) + d3^{n}
where a, b, c, d are constants.
For the particular side conditions we have
u_{0} = a + d = 1,
u_{1} = 2a + 2b + 3d = 5
u_{2} = 4a + 8b + 8c + 9d = 1
u_{3} = 8a + 24b + 48c + 27d =− 45
whence a = 0, b = 1, c = − 2, d = 1, so the particular solution is
u_{n} = 2n^{n}(3 − 2n) + 3^{n}.
3. Nonhomogeneous 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
nonhomogeneous one. The same
recipe works in the case of difference equations, i.e. first find the general
solution to
Φ(E)u_{n} = 0
and a particular solution to
Φ(E) = f (n)
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 any generality.
(i) f (n) = kμ ^{n}, μ ≠ λ_{i}, i =
1, 2,... , r
In this case one can show that
is a particular solution to Φ(E)u_{n} = kμ ^{n}. Let namely Φ(E)
=Σa_{i}E^{i}. Then
Example 3. The general solution of
u_{n+2}− 5u_{n+1} + 6u_{n} = 3( 4^{n})
is u_{n} = a2^{n} + b3^{n} +
4^{n} where a and b are arbitrary
constants.
(ii) f (n) = kμ^{ n}, μ =λ_{ i}, λ_{
i}a nonrepeated factor of Φ(E)
In this case a particular solution is given by
where Φ'(μ) denotes
Example 4. The general solution of
u_{n+2} − 5u_{n+1 }+ 6u_{n} = 3( 2^{n})
is
where a, b are arbitrary constants.
(iii) f (n) = kμ^{ n}, μ = λ _{i},λ
_{i} a repeated factor of Φ(E)
Suppose now that (E − λ_{ i} is repeated h times in
Φ(E). Then
where n^{(k)} = n(n − 1)...(n −
k +1), is a particular solution of the equation Φ(E) u_{n} = kμ
^{n}.
Example 5. The general solution of the equation
(E−2)^{3}(E−3)u_{n} = 5( 2^{n})
is
with a, b, c, d are arbitrary constants
(iv) f (n) is a polynomial in n
First write f as a polynomial in the factorial powers n ^{(k)}, so
f (n) = a_{0 }+ a_{1}^{n} + a_{2}^{n(2 )
}+ ...
Now define the difference operator , by Δu_{n} = u_{n+1}−
u_{n} = (E−1) u_{n}. Using the symbolic relationship
E = 1 + Δ, we can reexpress Φ(E)as Ψ(Δ).
Still arguing symbolically , a particular solution is obtained by
provided that we can make any sense out of.The
way this will be done is by expanding in
powers
of , and using long division . The following rules are needed :
n^{(r)} = rn^{(r}^{−1)
}Δ^{2}n^{(r)} = r^{(2 )}n^{(r2)}
...
Δ^{k} n^{(r)} = r^{(k)}n^{(rk)}) for k≤r,
= 0 for k > r
and
Example 6. Find a particular solution of the
equation
u_{n+2}7u_{n}+1 +12u_{n}= 3n^{2 }+ 2n + 2.
First write 3n^{2} + 2n + 2 = 3n^{(2 )} + 5n^{(1 )} + 2
and E^{2}  7E +12 = (2Δ )( 3Δ ). Thus we get
Example 7. Find a particular solution of the
equation
u_{n+3}  5u_{n+2} + 7u_{n+1}  3u_{n} = n^{2}
+ 4n +1.
The required solution is
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