104 CHAPTER 4 A Differential Approach to Uniform Subdivision
without loss of generality, that the leading coefficient β
m
is one, the differential
operator for this equation can be factored into linear terms:
m
i =0
β
i
D [x]
i
==
m
i =1
(D[x] − α
i
), (4.19)
where the α
i
are roots of the polynomial equation
m
i =0
β
i
x
i
== 0. Observe that if
the coefficients
β
i
are real then these roots α
i
are either real or complex conjugates.
We now proceed to our main challenge: constructing the difference mask
d
k
[x]
for the differential operator
!
m
i =1
(D[x] −α
i
) on the grid
1
2
k
Z. As in the previous cases,
our approach is to factor the differential operator and then develop difference
masks for each of these factors. The product of these difference masks provides the
desired discrete operator.
Focusing on a single factor of the form
(D[x] −α), we observe that the differential
equation
(D[x] −α)p[x] == 0 has a solution of the form p[x] = e
αx
. As in the polyno-
mial case, the difference mask associated with
(D[x] − α) should annihilate samples
of this solution taken on the grid
1
2
k
Z. Given that samples of e
αx
taken on the grid
1
2
k
Z form the geometric sequence {..., η[k, α]
−2
, η[k, α]
−1
, 1, η[k, α], η[k, α]
2
, ...} where
η[k, α] = e
2
−k
∗α
,
the difference mask (1 − η[k, α]x) yields the desired cancellation.
All that remains is to choose an appropriate normalization for this difference
mask to account for the grid size. In the polynomial case, this normalization for the
difference mask
(1 − x) was the factor 2
k
. On the grid
1
2
k
Z, discrete approximations
p
k
[x] to the function x have the form
i
i
2
k
x
i
. Now, the product 2
k
(1 − x)p
k
[x] is
the generating function
i
x
i
, which is exactly the discretization for the constant
function
1. Thus, the difference mask 2
k
(1 − x) acts analogously to the differential
operator
D [x]: it annihilates constants and maps the function x to the function 1.
In the exponential case, the correct normalizing constant is
α
η[k,α]−1
. The reader
may verify this choice by observing that if
p[x] is the constant function 1 then
(D[x] −α)p[x] == −α. Likewise, multiplying the difference mask (
α
η[k,α]−1
)(1 −η[k, α]x)
by the generating function
i
x
i
yields the generating function −α
i
x
i
. This nor-
malization is compatible with the polynomial case,
α == 0, because the limit as
α → 0 of the expression
α
η[k,α]−1
is exactly the constant 2
k
of the polynomial case.
If the root
α has multiplicity n, solutions to the differential equation
(D[x] − α)
n
p[x] == 0 also satisfy equation 4.17. For this differential equation, the
reader may verify that the space of solutions is simply the span of the functions
p[x] = x
j
e
α x
where j = 0 ...n−1. Again, the difference mask (1 − η[k, α]x)
n
annihilates